Neighboring nodes in the networkTest if directed graph is connectedWhy is NeighborhoodGraph so slow?How to...

Brothers & sisters

Is "remove commented out code" correct English?

Could gravitational lensing be used to protect a spaceship from a laser?

Reserved de-dupe rules

Doing something right before you need it - expression for this?

Should I tell management that I intend to leave due to bad software development practices?

Withdrawals from HSA

What's the point of deactivating Num Lock on login screens?

Watching something be written to a file live with tail

AES: Why is it a good practice to use only the first 16bytes of a hash for encryption?

What's the difference between 'rename' and 'mv'?

What is going on with Captain Marvel's blood colour?

Is the Joker left-handed?

Etiquette around loan refinance - decision is going to cost first broker a lot of money

Forgetting the musical notes while performing in concert

Today is the Center

90's TV series where a boy goes to another dimension through portal near power lines

What mechanic is there to disable a threat instead of killing it?

Why is the 'in' operator throwing an error with a string literal instead of logging false?

What is the intuition behind short exact sequences of groups; in particular, what is the intuition behind group extensions?

How can I make my BBEG immortal short of making them a Lich or Vampire?

Twin primes whose sum is a cube

Python: return float 1.0 as int 1 but float 1.5 as float 1.5

Can I use a neutral wire from another outlet to repair a broken neutral?



Neighboring nodes in the network


Test if directed graph is connectedWhy is NeighborhoodGraph so slow?How to add new nodes to an existing graph with fixed (coordinates) nodes?How do I upload a graph as an adjacency list and find the betweenness centrality?Arranging “ranked” nodes of a graph symmetricallyVertexLabels with Graph PropertiesNetwork with Radial Gradient Fill NodesHow to format vertices and control placement in a directed graphHow to label a large number of vertices using a list of namesColor the nodes according to certain valuesHighlight all paths in a graph below some threshold lengthSandbox algorithm for multifractal analysis of complex networks













5












$begingroup$


Consider the graph:



graph = {1 <-> 2, 1 <-> 4, 1 <-> 5, 1 <-> 8, 1 <-> 10, 1 <-> 26, 1 <-> 37, 1 <-> 42, 1 <-> 62, 1 <-> 86, 1 <-> 93, 1 <-> 100, 2 <-> 3, 2 <-> 7, 2 <-> 9, 2 <-> 12, 2 <-> 14, 2 <-> 17, 2 <-> 18, 2 <-> 25, 2 <-> 36, 2 <-> 41, 2 <-> 46, 2 <-> 50, 2 <-> 55, 2 <-> 72, 2 <-> 75, 3 <-> 6, 3 <-> 28, 3 <-> 34, 3 <-> 63, 4 <-> 13, 4 <-> 21, 5 <-> 20, 5 <-> 35, 5 <-> 40, 5 <-> 45, 5 <-> 48, 5 <-> 74, 6 <-> 31, 6 <-> 70, 9 <-> 11, 9 <-> 54, 9 <-> 67, 11 <-> 16, 11 <-> 24, 11 <-> 58, 11 <-> 60, 11 <-> 61, 11 <-> 65, 11 <-> 69, 12 <-> 27, 13 <-> 15, 13 <-> 33, 13 <-> 76, 14 <-> 30, 15 <-> 19, 15 <-> 96, 15 <-> 98, 16 <-> 57, 16 <-> 90, 19 <-> 22, 19 <-> 23, 19 <-> 39, 19 <-> 80, 19 <-> 83, 21 <-> 38, 22 <-> 59, 22 <-> 82, 25 <-> 29, 25 <-> 56, 25 <-> 94, 26 <-> 32, 26 <-> 43, 26 <-> 71, 27 <-> 47, 30 <-> 77, 30 <-> 78, 33 <-> 79, 33 <-> 97, 39 <-> 49, 39 <-> 51, 40 <-> 44, 40 <-> 73, 42 <-> 68, 48 <-> 52, 48 <-> 81, 50 <-> 53, 50 <-> 64, 50 <-> 89, 56 <-> 66, 56 <-> 92, 59 <-> 91, 62 <-> 88, 67 <-> 87, 74 <-> 95, 82 <-> 84, 82 <-> 85, 82 <-> 99};

net = Graph[graph, VertexShapeFunction -> "Name"]


Let's choose any node 'g' in the graph:



g=19;


Let 'r' denote the distance (counted in the number of nodes) from the node 'g':



d = GraphDiameter[net]
r = Range[1, d]


How to count all neighboring nodes within radius 'r' from the node 'g' ?



For example for node g=19 we have 6 nodes for r=1 (nodes: 80,83,22,39,23,15). For r=2 we have 7 nodes: 59,82,49,51,98,96,13.










share|improve this question











$endgroup$

















    5












    $begingroup$


    Consider the graph:



    graph = {1 <-> 2, 1 <-> 4, 1 <-> 5, 1 <-> 8, 1 <-> 10, 1 <-> 26, 1 <-> 37, 1 <-> 42, 1 <-> 62, 1 <-> 86, 1 <-> 93, 1 <-> 100, 2 <-> 3, 2 <-> 7, 2 <-> 9, 2 <-> 12, 2 <-> 14, 2 <-> 17, 2 <-> 18, 2 <-> 25, 2 <-> 36, 2 <-> 41, 2 <-> 46, 2 <-> 50, 2 <-> 55, 2 <-> 72, 2 <-> 75, 3 <-> 6, 3 <-> 28, 3 <-> 34, 3 <-> 63, 4 <-> 13, 4 <-> 21, 5 <-> 20, 5 <-> 35, 5 <-> 40, 5 <-> 45, 5 <-> 48, 5 <-> 74, 6 <-> 31, 6 <-> 70, 9 <-> 11, 9 <-> 54, 9 <-> 67, 11 <-> 16, 11 <-> 24, 11 <-> 58, 11 <-> 60, 11 <-> 61, 11 <-> 65, 11 <-> 69, 12 <-> 27, 13 <-> 15, 13 <-> 33, 13 <-> 76, 14 <-> 30, 15 <-> 19, 15 <-> 96, 15 <-> 98, 16 <-> 57, 16 <-> 90, 19 <-> 22, 19 <-> 23, 19 <-> 39, 19 <-> 80, 19 <-> 83, 21 <-> 38, 22 <-> 59, 22 <-> 82, 25 <-> 29, 25 <-> 56, 25 <-> 94, 26 <-> 32, 26 <-> 43, 26 <-> 71, 27 <-> 47, 30 <-> 77, 30 <-> 78, 33 <-> 79, 33 <-> 97, 39 <-> 49, 39 <-> 51, 40 <-> 44, 40 <-> 73, 42 <-> 68, 48 <-> 52, 48 <-> 81, 50 <-> 53, 50 <-> 64, 50 <-> 89, 56 <-> 66, 56 <-> 92, 59 <-> 91, 62 <-> 88, 67 <-> 87, 74 <-> 95, 82 <-> 84, 82 <-> 85, 82 <-> 99};

    net = Graph[graph, VertexShapeFunction -> "Name"]


    Let's choose any node 'g' in the graph:



    g=19;


    Let 'r' denote the distance (counted in the number of nodes) from the node 'g':



    d = GraphDiameter[net]
    r = Range[1, d]


    How to count all neighboring nodes within radius 'r' from the node 'g' ?



    For example for node g=19 we have 6 nodes for r=1 (nodes: 80,83,22,39,23,15). For r=2 we have 7 nodes: 59,82,49,51,98,96,13.










    share|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$


      Consider the graph:



      graph = {1 <-> 2, 1 <-> 4, 1 <-> 5, 1 <-> 8, 1 <-> 10, 1 <-> 26, 1 <-> 37, 1 <-> 42, 1 <-> 62, 1 <-> 86, 1 <-> 93, 1 <-> 100, 2 <-> 3, 2 <-> 7, 2 <-> 9, 2 <-> 12, 2 <-> 14, 2 <-> 17, 2 <-> 18, 2 <-> 25, 2 <-> 36, 2 <-> 41, 2 <-> 46, 2 <-> 50, 2 <-> 55, 2 <-> 72, 2 <-> 75, 3 <-> 6, 3 <-> 28, 3 <-> 34, 3 <-> 63, 4 <-> 13, 4 <-> 21, 5 <-> 20, 5 <-> 35, 5 <-> 40, 5 <-> 45, 5 <-> 48, 5 <-> 74, 6 <-> 31, 6 <-> 70, 9 <-> 11, 9 <-> 54, 9 <-> 67, 11 <-> 16, 11 <-> 24, 11 <-> 58, 11 <-> 60, 11 <-> 61, 11 <-> 65, 11 <-> 69, 12 <-> 27, 13 <-> 15, 13 <-> 33, 13 <-> 76, 14 <-> 30, 15 <-> 19, 15 <-> 96, 15 <-> 98, 16 <-> 57, 16 <-> 90, 19 <-> 22, 19 <-> 23, 19 <-> 39, 19 <-> 80, 19 <-> 83, 21 <-> 38, 22 <-> 59, 22 <-> 82, 25 <-> 29, 25 <-> 56, 25 <-> 94, 26 <-> 32, 26 <-> 43, 26 <-> 71, 27 <-> 47, 30 <-> 77, 30 <-> 78, 33 <-> 79, 33 <-> 97, 39 <-> 49, 39 <-> 51, 40 <-> 44, 40 <-> 73, 42 <-> 68, 48 <-> 52, 48 <-> 81, 50 <-> 53, 50 <-> 64, 50 <-> 89, 56 <-> 66, 56 <-> 92, 59 <-> 91, 62 <-> 88, 67 <-> 87, 74 <-> 95, 82 <-> 84, 82 <-> 85, 82 <-> 99};

      net = Graph[graph, VertexShapeFunction -> "Name"]


      Let's choose any node 'g' in the graph:



      g=19;


      Let 'r' denote the distance (counted in the number of nodes) from the node 'g':



      d = GraphDiameter[net]
      r = Range[1, d]


      How to count all neighboring nodes within radius 'r' from the node 'g' ?



      For example for node g=19 we have 6 nodes for r=1 (nodes: 80,83,22,39,23,15). For r=2 we have 7 nodes: 59,82,49,51,98,96,13.










      share|improve this question











      $endgroup$




      Consider the graph:



      graph = {1 <-> 2, 1 <-> 4, 1 <-> 5, 1 <-> 8, 1 <-> 10, 1 <-> 26, 1 <-> 37, 1 <-> 42, 1 <-> 62, 1 <-> 86, 1 <-> 93, 1 <-> 100, 2 <-> 3, 2 <-> 7, 2 <-> 9, 2 <-> 12, 2 <-> 14, 2 <-> 17, 2 <-> 18, 2 <-> 25, 2 <-> 36, 2 <-> 41, 2 <-> 46, 2 <-> 50, 2 <-> 55, 2 <-> 72, 2 <-> 75, 3 <-> 6, 3 <-> 28, 3 <-> 34, 3 <-> 63, 4 <-> 13, 4 <-> 21, 5 <-> 20, 5 <-> 35, 5 <-> 40, 5 <-> 45, 5 <-> 48, 5 <-> 74, 6 <-> 31, 6 <-> 70, 9 <-> 11, 9 <-> 54, 9 <-> 67, 11 <-> 16, 11 <-> 24, 11 <-> 58, 11 <-> 60, 11 <-> 61, 11 <-> 65, 11 <-> 69, 12 <-> 27, 13 <-> 15, 13 <-> 33, 13 <-> 76, 14 <-> 30, 15 <-> 19, 15 <-> 96, 15 <-> 98, 16 <-> 57, 16 <-> 90, 19 <-> 22, 19 <-> 23, 19 <-> 39, 19 <-> 80, 19 <-> 83, 21 <-> 38, 22 <-> 59, 22 <-> 82, 25 <-> 29, 25 <-> 56, 25 <-> 94, 26 <-> 32, 26 <-> 43, 26 <-> 71, 27 <-> 47, 30 <-> 77, 30 <-> 78, 33 <-> 79, 33 <-> 97, 39 <-> 49, 39 <-> 51, 40 <-> 44, 40 <-> 73, 42 <-> 68, 48 <-> 52, 48 <-> 81, 50 <-> 53, 50 <-> 64, 50 <-> 89, 56 <-> 66, 56 <-> 92, 59 <-> 91, 62 <-> 88, 67 <-> 87, 74 <-> 95, 82 <-> 84, 82 <-> 85, 82 <-> 99};

      net = Graph[graph, VertexShapeFunction -> "Name"]


      Let's choose any node 'g' in the graph:



      g=19;


      Let 'r' denote the distance (counted in the number of nodes) from the node 'g':



      d = GraphDiameter[net]
      r = Range[1, d]


      How to count all neighboring nodes within radius 'r' from the node 'g' ?



      For example for node g=19 we have 6 nodes for r=1 (nodes: 80,83,22,39,23,15). For r=2 we have 7 nodes: 59,82,49,51,98,96,13.







      graphs-and-networks






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 14 hours ago









      J. M. is away

      98.9k10311467




      98.9k10311467










      asked 14 hours ago









      ralphralph

      1707




      1707






















          5 Answers
          5






          active

          oldest

          votes


















          4












          $begingroup$

          I will choose a bit better GraphLayout for a tree:



          net = Graph[graph, VertexLabels -> "Name", GraphLayout -> "RadialEmbedding"];


          I suggest don't just count directly - get an object - a subgraph - of your query, so you can then run various computations on it and don't need count all over again based on different criteria w/ a different code.



          nei[v_, d_] := NeighborhoodGraph[net, v, d]


          Take distance 1:



          nei[19, 1]


          enter image description here



          and see it is right:



          HighlightGraph[net, nei[19, 1]]


          enter image description here



          Now you can compute whatever you need:



          VertexList[nei[19, 1]]
          Length[%] - 1



          {19, 15, 22, 23, 39, 80, 83}



          6




          For the distance 2:



          VertexList[nei[19, 1]]
          VertexList[nei[19, 2]]
          Complement[%, %%]
          Length[%]



          {19, 15, 22, 23, 39, 80, 83}



          {19, 13, 15, 22, 23, 39, 49, 51, 59, 80, 82, 83, 96, 98}



          {13, 49, 51, 59, 82, 96, 98}



          7




          Timings for large graphs



          net = RandomGraph[BarabasiAlbertGraphDistribution[20000, 1]];

          nei[v_, d_] := NeighborhoodGraph[net, v, d]

          dist15:=Length[Complement[VertexList[nei[#,15]],VertexList[nei[#,14]]]&@RandomInteger[1000]]

          Table[AbsoluteTiming[dist15;][[1]], 5]



          {0.097359, 0.094737, 0.092589, 0.08872, 0.087478}







          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15}).
            $endgroup$
            – ralph
            10 hours ago










          • $begingroup$
            @ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
            $endgroup$
            – Vitaliy Kaurov
            10 hours ago










          • $begingroup$
            Please forgive me. I meant about 200,000 no 20,000 nodes.
            $endgroup$
            – ralph
            10 hours ago










          • $begingroup$
            @Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
            $endgroup$
            – Vitaliy Kaurov
            9 hours ago










          • $begingroup$
            @VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
            $endgroup$
            – Szabolcs
            7 hours ago





















          3












          $begingroup$

          You could build it using BreadthFirstScan:



          net = RandomGraph[BarabasiAlbertGraphDistribution[200000, 1]];

          distance =
          GroupBy[Reap[
          BreadthFirstScan[net,
          19, {"DiscoverVertex" -> (Sow[#3 -> #1] &)}]][[2, 1]],
          First -> Last, Association[{"length" -> Length[#], "set" -> #}] &];


          Get length:



          distance[3, "length"]



          1194




          distance[[All, "length"]]



          <|0 -> 1, 1 -> 214, 2 -> 1194, 3 -> 3058, 4 -> 5826, 5 -> 10069, 6
          -> 15110, 7 -> 19992, 8 -> 23821, 9 -> 24910, 10 -> 24767, 11 -> 21459, 12 -> 17869, 13 -> 13525, 14 -> 9119, 15 -> 5146, 16 -> 2406,
          17 -> 1025, 18 -> 337, 19 -> 106, 20 -> 34, 21 -> 11, 22 -> 1|>




          and set
          distance[21, "set"]




          {182224, 145742, 171910, 124658, 125540, 128520, 196392, 166986,
          159530, 196846, 144772}







          share|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Amazingly fast, halmir! Any idea why this solution is so much faster than GraphDistance, which I would have thought works by BreadthFirstScan internally?
            $endgroup$
            – Roman
            7 hours ago






          • 1




            $begingroup$
            @Roman I had the conviction that GraphDistance compute the entire GraphDistanceMatrix even if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.
            $endgroup$
            – Szabolcs
            6 hours ago










          • $begingroup$
            @Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity of GraphDistance is quadratic in the graph size even when given just one vertex. That should not be so.
            $endgroup$
            – Szabolcs
            6 hours ago












          • $begingroup$
            @halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
            $endgroup$
            – Szabolcs
            6 hours ago












          • $begingroup$
            @Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
            $endgroup$
            – Szabolcs
            6 hours ago



















          2












          $begingroup$

          To count how many nodes there are at every distance (unsorted Association): use this if you want to Lookup a particular distance:



          Counts@GraphDistance[net, g]



          <|4 -> 4, 5 -> 12, 3 -> 7, 6 -> 26, 7 -> 20, 2 -> 7, 8 -> 15, 1 -> 6, 0 -> 1, 9 -> 2|>




          Look them all up in order:



          BinCounts[GraphDistance[net, g], {0, d, 1}]



          {1, 6, 7, 7, 4, 12, 26, 20, 15, 2, 0, 0}







          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15})
            $endgroup$
            – ralph
            10 hours ago










          • $begingroup$
            Yes if you want only short distances then @szabolcs has better tools available. This GraphDistance solution is only good if you want the distances to all nodes in the graph.
            $endgroup$
            – Roman
            9 hours ago



















          2












          $begingroup$


          How to count all neighboring nodes within radius 'r' from the node 'g' ?




          Use IGraph/M.



          IGNeighborhoodSize does precisely this and is probably your fastest bet, but I do not have time to benchmark it against other solutions right now.



          If you want to do it for multiple distances in one go, use IGDistanceCounts,



          IGDistanceCounts[graph, {vertex}]


          This gives you the counts of other vertices found at all (unweighted) distances. You can then simply Accumulate that list to get the result for all r at the same time.



          For weighted distances, use IGDistanceHistogram.






          share|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. And how to count the same as the 'IGDistanceCounts[graph, {vertex}]' formula but for weighted networks?
            $endgroup$
            – ralph
            8 hours ago












          • $begingroup$
            @ralph As I said above, use IGDistanceHistogram
            $endgroup$
            – Szabolcs
            7 hours ago



















          2












          $begingroup$

          For weighted network:



          g1 = {4798 <-> 2641, 4798 <-> 2310, 4798 <-> 4721, 2310 <-> 1942,2310 <-> 961, 4721 <-> 4507, 4721 <-> 4779, 4779 <-> 4336, 4779 <-> 3238, 4336 <-> 3277, 4336 <-> 3514, 3277 <-> 2923, 2923 <-> 2772, 2923 <-> 2401, 2772 <-> 2, 2772 <-> 2771, 3514 <-> 3042, 3514 <-> 2739, 3042 <-> 3007, 3042 <-> 1655, 2739 <-> 2277, 2739 <-> 1895, 2 <-> 5, 2 <-> 3, 3277 <-> 100, 5 <-> 6, 5 <-> 7, 5 <-> 8, 5 <-> 9};

          w1 = {10, 20, 20, 4, 35, 3, 4, 6, 17, 7, 13, 2, 2, 7, 2, 1, 3, 5, 3, 6,4, 6, 2, 1, 1, 1, 1, 1, 1};

          w2=Table[1, 29];

          net1 = Graph[g1, EdgeWeight -> w1, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]

          net2 = Graph[g1, EdgeWeight -> w2, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]

          s = RandomSample[VertexList[net1], 15];

          Mr = Table[IGDistanceCounts[net1, {s[[i]]}], {i, 1, Length[s]}] (*for non weighted*)

          Mr2 = IGDistanceHistogram[net1, 9] (*for weighted graph ?*)

          Mr3 = IGDistanceHistogram[net2, 9] (*for non weighted graph ? Mr3==Mr *)





          share|improve this answer









          $endgroup$














            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "387"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194581%2fneighboring-nodes-in-the-network%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            I will choose a bit better GraphLayout for a tree:



            net = Graph[graph, VertexLabels -> "Name", GraphLayout -> "RadialEmbedding"];


            I suggest don't just count directly - get an object - a subgraph - of your query, so you can then run various computations on it and don't need count all over again based on different criteria w/ a different code.



            nei[v_, d_] := NeighborhoodGraph[net, v, d]


            Take distance 1:



            nei[19, 1]


            enter image description here



            and see it is right:



            HighlightGraph[net, nei[19, 1]]


            enter image description here



            Now you can compute whatever you need:



            VertexList[nei[19, 1]]
            Length[%] - 1



            {19, 15, 22, 23, 39, 80, 83}



            6




            For the distance 2:



            VertexList[nei[19, 1]]
            VertexList[nei[19, 2]]
            Complement[%, %%]
            Length[%]



            {19, 15, 22, 23, 39, 80, 83}



            {19, 13, 15, 22, 23, 39, 49, 51, 59, 80, 82, 83, 96, 98}



            {13, 49, 51, 59, 82, 96, 98}



            7




            Timings for large graphs



            net = RandomGraph[BarabasiAlbertGraphDistribution[20000, 1]];

            nei[v_, d_] := NeighborhoodGraph[net, v, d]

            dist15:=Length[Complement[VertexList[nei[#,15]],VertexList[nei[#,14]]]&@RandomInteger[1000]]

            Table[AbsoluteTiming[dist15;][[1]], 5]



            {0.097359, 0.094737, 0.092589, 0.08872, 0.087478}







            share|improve this answer











            $endgroup$













            • $begingroup$
              Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15}).
              $endgroup$
              – ralph
              10 hours ago










            • $begingroup$
              @ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
              $endgroup$
              – Vitaliy Kaurov
              10 hours ago










            • $begingroup$
              Please forgive me. I meant about 200,000 no 20,000 nodes.
              $endgroup$
              – ralph
              10 hours ago










            • $begingroup$
              @Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
              $endgroup$
              – Vitaliy Kaurov
              9 hours ago










            • $begingroup$
              @VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
              $endgroup$
              – Szabolcs
              7 hours ago


















            4












            $begingroup$

            I will choose a bit better GraphLayout for a tree:



            net = Graph[graph, VertexLabels -> "Name", GraphLayout -> "RadialEmbedding"];


            I suggest don't just count directly - get an object - a subgraph - of your query, so you can then run various computations on it and don't need count all over again based on different criteria w/ a different code.



            nei[v_, d_] := NeighborhoodGraph[net, v, d]


            Take distance 1:



            nei[19, 1]


            enter image description here



            and see it is right:



            HighlightGraph[net, nei[19, 1]]


            enter image description here



            Now you can compute whatever you need:



            VertexList[nei[19, 1]]
            Length[%] - 1



            {19, 15, 22, 23, 39, 80, 83}



            6




            For the distance 2:



            VertexList[nei[19, 1]]
            VertexList[nei[19, 2]]
            Complement[%, %%]
            Length[%]



            {19, 15, 22, 23, 39, 80, 83}



            {19, 13, 15, 22, 23, 39, 49, 51, 59, 80, 82, 83, 96, 98}



            {13, 49, 51, 59, 82, 96, 98}



            7




            Timings for large graphs



            net = RandomGraph[BarabasiAlbertGraphDistribution[20000, 1]];

            nei[v_, d_] := NeighborhoodGraph[net, v, d]

            dist15:=Length[Complement[VertexList[nei[#,15]],VertexList[nei[#,14]]]&@RandomInteger[1000]]

            Table[AbsoluteTiming[dist15;][[1]], 5]



            {0.097359, 0.094737, 0.092589, 0.08872, 0.087478}







            share|improve this answer











            $endgroup$













            • $begingroup$
              Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15}).
              $endgroup$
              – ralph
              10 hours ago










            • $begingroup$
              @ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
              $endgroup$
              – Vitaliy Kaurov
              10 hours ago










            • $begingroup$
              Please forgive me. I meant about 200,000 no 20,000 nodes.
              $endgroup$
              – ralph
              10 hours ago










            • $begingroup$
              @Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
              $endgroup$
              – Vitaliy Kaurov
              9 hours ago










            • $begingroup$
              @VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
              $endgroup$
              – Szabolcs
              7 hours ago
















            4












            4








            4





            $begingroup$

            I will choose a bit better GraphLayout for a tree:



            net = Graph[graph, VertexLabels -> "Name", GraphLayout -> "RadialEmbedding"];


            I suggest don't just count directly - get an object - a subgraph - of your query, so you can then run various computations on it and don't need count all over again based on different criteria w/ a different code.



            nei[v_, d_] := NeighborhoodGraph[net, v, d]


            Take distance 1:



            nei[19, 1]


            enter image description here



            and see it is right:



            HighlightGraph[net, nei[19, 1]]


            enter image description here



            Now you can compute whatever you need:



            VertexList[nei[19, 1]]
            Length[%] - 1



            {19, 15, 22, 23, 39, 80, 83}



            6




            For the distance 2:



            VertexList[nei[19, 1]]
            VertexList[nei[19, 2]]
            Complement[%, %%]
            Length[%]



            {19, 15, 22, 23, 39, 80, 83}



            {19, 13, 15, 22, 23, 39, 49, 51, 59, 80, 82, 83, 96, 98}



            {13, 49, 51, 59, 82, 96, 98}



            7




            Timings for large graphs



            net = RandomGraph[BarabasiAlbertGraphDistribution[20000, 1]];

            nei[v_, d_] := NeighborhoodGraph[net, v, d]

            dist15:=Length[Complement[VertexList[nei[#,15]],VertexList[nei[#,14]]]&@RandomInteger[1000]]

            Table[AbsoluteTiming[dist15;][[1]], 5]



            {0.097359, 0.094737, 0.092589, 0.08872, 0.087478}







            share|improve this answer











            $endgroup$



            I will choose a bit better GraphLayout for a tree:



            net = Graph[graph, VertexLabels -> "Name", GraphLayout -> "RadialEmbedding"];


            I suggest don't just count directly - get an object - a subgraph - of your query, so you can then run various computations on it and don't need count all over again based on different criteria w/ a different code.



            nei[v_, d_] := NeighborhoodGraph[net, v, d]


            Take distance 1:



            nei[19, 1]


            enter image description here



            and see it is right:



            HighlightGraph[net, nei[19, 1]]


            enter image description here



            Now you can compute whatever you need:



            VertexList[nei[19, 1]]
            Length[%] - 1



            {19, 15, 22, 23, 39, 80, 83}



            6




            For the distance 2:



            VertexList[nei[19, 1]]
            VertexList[nei[19, 2]]
            Complement[%, %%]
            Length[%]



            {19, 15, 22, 23, 39, 80, 83}



            {19, 13, 15, 22, 23, 39, 49, 51, 59, 80, 82, 83, 96, 98}



            {13, 49, 51, 59, 82, 96, 98}



            7




            Timings for large graphs



            net = RandomGraph[BarabasiAlbertGraphDistribution[20000, 1]];

            nei[v_, d_] := NeighborhoodGraph[net, v, d]

            dist15:=Length[Complement[VertexList[nei[#,15]],VertexList[nei[#,14]]]&@RandomInteger[1000]]

            Table[AbsoluteTiming[dist15;][[1]], 5]



            {0.097359, 0.094737, 0.092589, 0.08872, 0.087478}








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 10 hours ago

























            answered 14 hours ago









            Vitaliy KaurovVitaliy Kaurov

            57.6k6162282




            57.6k6162282












            • $begingroup$
              Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15}).
              $endgroup$
              – ralph
              10 hours ago










            • $begingroup$
              @ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
              $endgroup$
              – Vitaliy Kaurov
              10 hours ago










            • $begingroup$
              Please forgive me. I meant about 200,000 no 20,000 nodes.
              $endgroup$
              – ralph
              10 hours ago










            • $begingroup$
              @Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
              $endgroup$
              – Vitaliy Kaurov
              9 hours ago










            • $begingroup$
              @VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
              $endgroup$
              – Szabolcs
              7 hours ago




















            • $begingroup$
              Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15}).
              $endgroup$
              – ralph
              10 hours ago










            • $begingroup$
              @ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
              $endgroup$
              – Vitaliy Kaurov
              10 hours ago










            • $begingroup$
              Please forgive me. I meant about 200,000 no 20,000 nodes.
              $endgroup$
              – ralph
              10 hours ago










            • $begingroup$
              @Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
              $endgroup$
              – Vitaliy Kaurov
              9 hours ago










            • $begingroup$
              @VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
              $endgroup$
              – Szabolcs
              7 hours ago


















            $begingroup$
            Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15}).
            $endgroup$
            – ralph
            10 hours ago




            $begingroup$
            Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15}).
            $endgroup$
            – ralph
            10 hours ago












            $begingroup$
            @ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
            $endgroup$
            – Vitaliy Kaurov
            10 hours ago




            $begingroup$
            @ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
            $endgroup$
            – Vitaliy Kaurov
            10 hours ago












            $begingroup$
            Please forgive me. I meant about 200,000 no 20,000 nodes.
            $endgroup$
            – ralph
            10 hours ago




            $begingroup$
            Please forgive me. I meant about 200,000 no 20,000 nodes.
            $endgroup$
            – ralph
            10 hours ago












            $begingroup$
            @Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
            $endgroup$
            – Vitaliy Kaurov
            9 hours ago




            $begingroup$
            @Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
            $endgroup$
            – Vitaliy Kaurov
            9 hours ago












            $begingroup$
            @VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
            $endgroup$
            – Szabolcs
            7 hours ago






            $begingroup$
            @VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
            $endgroup$
            – Szabolcs
            7 hours ago













            3












            $begingroup$

            You could build it using BreadthFirstScan:



            net = RandomGraph[BarabasiAlbertGraphDistribution[200000, 1]];

            distance =
            GroupBy[Reap[
            BreadthFirstScan[net,
            19, {"DiscoverVertex" -> (Sow[#3 -> #1] &)}]][[2, 1]],
            First -> Last, Association[{"length" -> Length[#], "set" -> #}] &];


            Get length:



            distance[3, "length"]



            1194




            distance[[All, "length"]]



            <|0 -> 1, 1 -> 214, 2 -> 1194, 3 -> 3058, 4 -> 5826, 5 -> 10069, 6
            -> 15110, 7 -> 19992, 8 -> 23821, 9 -> 24910, 10 -> 24767, 11 -> 21459, 12 -> 17869, 13 -> 13525, 14 -> 9119, 15 -> 5146, 16 -> 2406,
            17 -> 1025, 18 -> 337, 19 -> 106, 20 -> 34, 21 -> 11, 22 -> 1|>




            and set
            distance[21, "set"]




            {182224, 145742, 171910, 124658, 125540, 128520, 196392, 166986,
            159530, 196846, 144772}







            share|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Amazingly fast, halmir! Any idea why this solution is so much faster than GraphDistance, which I would have thought works by BreadthFirstScan internally?
              $endgroup$
              – Roman
              7 hours ago






            • 1




              $begingroup$
              @Roman I had the conviction that GraphDistance compute the entire GraphDistanceMatrix even if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.
              $endgroup$
              – Szabolcs
              6 hours ago










            • $begingroup$
              @Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity of GraphDistance is quadratic in the graph size even when given just one vertex. That should not be so.
              $endgroup$
              – Szabolcs
              6 hours ago












            • $begingroup$
              @halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
              $endgroup$
              – Szabolcs
              6 hours ago












            • $begingroup$
              @Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
              $endgroup$
              – Szabolcs
              6 hours ago
















            3












            $begingroup$

            You could build it using BreadthFirstScan:



            net = RandomGraph[BarabasiAlbertGraphDistribution[200000, 1]];

            distance =
            GroupBy[Reap[
            BreadthFirstScan[net,
            19, {"DiscoverVertex" -> (Sow[#3 -> #1] &)}]][[2, 1]],
            First -> Last, Association[{"length" -> Length[#], "set" -> #}] &];


            Get length:



            distance[3, "length"]



            1194




            distance[[All, "length"]]



            <|0 -> 1, 1 -> 214, 2 -> 1194, 3 -> 3058, 4 -> 5826, 5 -> 10069, 6
            -> 15110, 7 -> 19992, 8 -> 23821, 9 -> 24910, 10 -> 24767, 11 -> 21459, 12 -> 17869, 13 -> 13525, 14 -> 9119, 15 -> 5146, 16 -> 2406,
            17 -> 1025, 18 -> 337, 19 -> 106, 20 -> 34, 21 -> 11, 22 -> 1|>




            and set
            distance[21, "set"]




            {182224, 145742, 171910, 124658, 125540, 128520, 196392, 166986,
            159530, 196846, 144772}







            share|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Amazingly fast, halmir! Any idea why this solution is so much faster than GraphDistance, which I would have thought works by BreadthFirstScan internally?
              $endgroup$
              – Roman
              7 hours ago






            • 1




              $begingroup$
              @Roman I had the conviction that GraphDistance compute the entire GraphDistanceMatrix even if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.
              $endgroup$
              – Szabolcs
              6 hours ago










            • $begingroup$
              @Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity of GraphDistance is quadratic in the graph size even when given just one vertex. That should not be so.
              $endgroup$
              – Szabolcs
              6 hours ago












            • $begingroup$
              @halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
              $endgroup$
              – Szabolcs
              6 hours ago












            • $begingroup$
              @Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
              $endgroup$
              – Szabolcs
              6 hours ago














            3












            3








            3





            $begingroup$

            You could build it using BreadthFirstScan:



            net = RandomGraph[BarabasiAlbertGraphDistribution[200000, 1]];

            distance =
            GroupBy[Reap[
            BreadthFirstScan[net,
            19, {"DiscoverVertex" -> (Sow[#3 -> #1] &)}]][[2, 1]],
            First -> Last, Association[{"length" -> Length[#], "set" -> #}] &];


            Get length:



            distance[3, "length"]



            1194




            distance[[All, "length"]]



            <|0 -> 1, 1 -> 214, 2 -> 1194, 3 -> 3058, 4 -> 5826, 5 -> 10069, 6
            -> 15110, 7 -> 19992, 8 -> 23821, 9 -> 24910, 10 -> 24767, 11 -> 21459, 12 -> 17869, 13 -> 13525, 14 -> 9119, 15 -> 5146, 16 -> 2406,
            17 -> 1025, 18 -> 337, 19 -> 106, 20 -> 34, 21 -> 11, 22 -> 1|>




            and set
            distance[21, "set"]




            {182224, 145742, 171910, 124658, 125540, 128520, 196392, 166986,
            159530, 196846, 144772}







            share|improve this answer









            $endgroup$



            You could build it using BreadthFirstScan:



            net = RandomGraph[BarabasiAlbertGraphDistribution[200000, 1]];

            distance =
            GroupBy[Reap[
            BreadthFirstScan[net,
            19, {"DiscoverVertex" -> (Sow[#3 -> #1] &)}]][[2, 1]],
            First -> Last, Association[{"length" -> Length[#], "set" -> #}] &];


            Get length:



            distance[3, "length"]



            1194




            distance[[All, "length"]]



            <|0 -> 1, 1 -> 214, 2 -> 1194, 3 -> 3058, 4 -> 5826, 5 -> 10069, 6
            -> 15110, 7 -> 19992, 8 -> 23821, 9 -> 24910, 10 -> 24767, 11 -> 21459, 12 -> 17869, 13 -> 13525, 14 -> 9119, 15 -> 5146, 16 -> 2406,
            17 -> 1025, 18 -> 337, 19 -> 106, 20 -> 34, 21 -> 11, 22 -> 1|>




            and set
            distance[21, "set"]




            {182224, 145742, 171910, 124658, 125540, 128520, 196392, 166986,
            159530, 196846, 144772}








            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 8 hours ago









            halmirhalmir

            10.6k2544




            10.6k2544








            • 1




              $begingroup$
              Amazingly fast, halmir! Any idea why this solution is so much faster than GraphDistance, which I would have thought works by BreadthFirstScan internally?
              $endgroup$
              – Roman
              7 hours ago






            • 1




              $begingroup$
              @Roman I had the conviction that GraphDistance compute the entire GraphDistanceMatrix even if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.
              $endgroup$
              – Szabolcs
              6 hours ago










            • $begingroup$
              @Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity of GraphDistance is quadratic in the graph size even when given just one vertex. That should not be so.
              $endgroup$
              – Szabolcs
              6 hours ago












            • $begingroup$
              @halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
              $endgroup$
              – Szabolcs
              6 hours ago












            • $begingroup$
              @Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
              $endgroup$
              – Szabolcs
              6 hours ago














            • 1




              $begingroup$
              Amazingly fast, halmir! Any idea why this solution is so much faster than GraphDistance, which I would have thought works by BreadthFirstScan internally?
              $endgroup$
              – Roman
              7 hours ago






            • 1




              $begingroup$
              @Roman I had the conviction that GraphDistance compute the entire GraphDistanceMatrix even if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.
              $endgroup$
              – Szabolcs
              6 hours ago










            • $begingroup$
              @Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity of GraphDistance is quadratic in the graph size even when given just one vertex. That should not be so.
              $endgroup$
              – Szabolcs
              6 hours ago












            • $begingroup$
              @halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
              $endgroup$
              – Szabolcs
              6 hours ago












            • $begingroup$
              @Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
              $endgroup$
              – Szabolcs
              6 hours ago








            1




            1




            $begingroup$
            Amazingly fast, halmir! Any idea why this solution is so much faster than GraphDistance, which I would have thought works by BreadthFirstScan internally?
            $endgroup$
            – Roman
            7 hours ago




            $begingroup$
            Amazingly fast, halmir! Any idea why this solution is so much faster than GraphDistance, which I would have thought works by BreadthFirstScan internally?
            $endgroup$
            – Roman
            7 hours ago




            1




            1




            $begingroup$
            @Roman I had the conviction that GraphDistance compute the entire GraphDistanceMatrix even if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.
            $endgroup$
            – Szabolcs
            6 hours ago




            $begingroup$
            @Roman I had the conviction that GraphDistance compute the entire GraphDistanceMatrix even if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.
            $endgroup$
            – Szabolcs
            6 hours ago












            $begingroup$
            @Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity of GraphDistance is quadratic in the graph size even when given just one vertex. That should not be so.
            $endgroup$
            – Szabolcs
            6 hours ago






            $begingroup$
            @Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity of GraphDistance is quadratic in the graph size even when given just one vertex. That should not be so.
            $endgroup$
            – Szabolcs
            6 hours ago














            $begingroup$
            @halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
            $endgroup$
            – Szabolcs
            6 hours ago






            $begingroup$
            @halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
            $endgroup$
            – Szabolcs
            6 hours ago














            $begingroup$
            @Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
            $endgroup$
            – Szabolcs
            6 hours ago




            $begingroup$
            @Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
            $endgroup$
            – Szabolcs
            6 hours ago











            2












            $begingroup$

            To count how many nodes there are at every distance (unsorted Association): use this if you want to Lookup a particular distance:



            Counts@GraphDistance[net, g]



            <|4 -> 4, 5 -> 12, 3 -> 7, 6 -> 26, 7 -> 20, 2 -> 7, 8 -> 15, 1 -> 6, 0 -> 1, 9 -> 2|>




            Look them all up in order:



            BinCounts[GraphDistance[net, g], {0, d, 1}]



            {1, 6, 7, 7, 4, 12, 26, 20, 15, 2, 0, 0}







            share|improve this answer











            $endgroup$













            • $begingroup$
              Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15})
              $endgroup$
              – ralph
              10 hours ago










            • $begingroup$
              Yes if you want only short distances then @szabolcs has better tools available. This GraphDistance solution is only good if you want the distances to all nodes in the graph.
              $endgroup$
              – Roman
              9 hours ago
















            2












            $begingroup$

            To count how many nodes there are at every distance (unsorted Association): use this if you want to Lookup a particular distance:



            Counts@GraphDistance[net, g]



            <|4 -> 4, 5 -> 12, 3 -> 7, 6 -> 26, 7 -> 20, 2 -> 7, 8 -> 15, 1 -> 6, 0 -> 1, 9 -> 2|>




            Look them all up in order:



            BinCounts[GraphDistance[net, g], {0, d, 1}]



            {1, 6, 7, 7, 4, 12, 26, 20, 15, 2, 0, 0}







            share|improve this answer











            $endgroup$













            • $begingroup$
              Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15})
              $endgroup$
              – ralph
              10 hours ago










            • $begingroup$
              Yes if you want only short distances then @szabolcs has better tools available. This GraphDistance solution is only good if you want the distances to all nodes in the graph.
              $endgroup$
              – Roman
              9 hours ago














            2












            2








            2





            $begingroup$

            To count how many nodes there are at every distance (unsorted Association): use this if you want to Lookup a particular distance:



            Counts@GraphDistance[net, g]



            <|4 -> 4, 5 -> 12, 3 -> 7, 6 -> 26, 7 -> 20, 2 -> 7, 8 -> 15, 1 -> 6, 0 -> 1, 9 -> 2|>




            Look them all up in order:



            BinCounts[GraphDistance[net, g], {0, d, 1}]



            {1, 6, 7, 7, 4, 12, 26, 20, 15, 2, 0, 0}







            share|improve this answer











            $endgroup$



            To count how many nodes there are at every distance (unsorted Association): use this if you want to Lookup a particular distance:



            Counts@GraphDistance[net, g]



            <|4 -> 4, 5 -> 12, 3 -> 7, 6 -> 26, 7 -> 20, 2 -> 7, 8 -> 15, 1 -> 6, 0 -> 1, 9 -> 2|>




            Look them all up in order:



            BinCounts[GraphDistance[net, g], {0, d, 1}]



            {1, 6, 7, 7, 4, 12, 26, 20, 15, 2, 0, 0}








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 10 hours ago

























            answered 14 hours ago









            RomanRoman

            4,3151027




            4,3151027












            • $begingroup$
              Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15})
              $endgroup$
              – ralph
              10 hours ago










            • $begingroup$
              Yes if you want only short distances then @szabolcs has better tools available. This GraphDistance solution is only good if you want the distances to all nodes in the graph.
              $endgroup$
              – Roman
              9 hours ago


















            • $begingroup$
              Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15})
              $endgroup$
              – ralph
              10 hours ago










            • $begingroup$
              Yes if you want only short distances then @szabolcs has better tools available. This GraphDistance solution is only good if you want the distances to all nodes in the graph.
              $endgroup$
              – Roman
              9 hours ago
















            $begingroup$
            Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15})
            $endgroup$
            – ralph
            10 hours ago




            $begingroup$
            Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = {1,2,3,4, ..., 15})
            $endgroup$
            – ralph
            10 hours ago












            $begingroup$
            Yes if you want only short distances then @szabolcs has better tools available. This GraphDistance solution is only good if you want the distances to all nodes in the graph.
            $endgroup$
            – Roman
            9 hours ago




            $begingroup$
            Yes if you want only short distances then @szabolcs has better tools available. This GraphDistance solution is only good if you want the distances to all nodes in the graph.
            $endgroup$
            – Roman
            9 hours ago











            2












            $begingroup$


            How to count all neighboring nodes within radius 'r' from the node 'g' ?




            Use IGraph/M.



            IGNeighborhoodSize does precisely this and is probably your fastest bet, but I do not have time to benchmark it against other solutions right now.



            If you want to do it for multiple distances in one go, use IGDistanceCounts,



            IGDistanceCounts[graph, {vertex}]


            This gives you the counts of other vertices found at all (unweighted) distances. You can then simply Accumulate that list to get the result for all r at the same time.



            For weighted distances, use IGDistanceHistogram.






            share|improve this answer









            $endgroup$













            • $begingroup$
              Thanks. And how to count the same as the 'IGDistanceCounts[graph, {vertex}]' formula but for weighted networks?
              $endgroup$
              – ralph
              8 hours ago












            • $begingroup$
              @ralph As I said above, use IGDistanceHistogram
              $endgroup$
              – Szabolcs
              7 hours ago
















            2












            $begingroup$


            How to count all neighboring nodes within radius 'r' from the node 'g' ?




            Use IGraph/M.



            IGNeighborhoodSize does precisely this and is probably your fastest bet, but I do not have time to benchmark it against other solutions right now.



            If you want to do it for multiple distances in one go, use IGDistanceCounts,



            IGDistanceCounts[graph, {vertex}]


            This gives you the counts of other vertices found at all (unweighted) distances. You can then simply Accumulate that list to get the result for all r at the same time.



            For weighted distances, use IGDistanceHistogram.






            share|improve this answer









            $endgroup$













            • $begingroup$
              Thanks. And how to count the same as the 'IGDistanceCounts[graph, {vertex}]' formula but for weighted networks?
              $endgroup$
              – ralph
              8 hours ago












            • $begingroup$
              @ralph As I said above, use IGDistanceHistogram
              $endgroup$
              – Szabolcs
              7 hours ago














            2












            2








            2





            $begingroup$


            How to count all neighboring nodes within radius 'r' from the node 'g' ?




            Use IGraph/M.



            IGNeighborhoodSize does precisely this and is probably your fastest bet, but I do not have time to benchmark it against other solutions right now.



            If you want to do it for multiple distances in one go, use IGDistanceCounts,



            IGDistanceCounts[graph, {vertex}]


            This gives you the counts of other vertices found at all (unweighted) distances. You can then simply Accumulate that list to get the result for all r at the same time.



            For weighted distances, use IGDistanceHistogram.






            share|improve this answer









            $endgroup$




            How to count all neighboring nodes within radius 'r' from the node 'g' ?




            Use IGraph/M.



            IGNeighborhoodSize does precisely this and is probably your fastest bet, but I do not have time to benchmark it against other solutions right now.



            If you want to do it for multiple distances in one go, use IGDistanceCounts,



            IGDistanceCounts[graph, {vertex}]


            This gives you the counts of other vertices found at all (unweighted) distances. You can then simply Accumulate that list to get the result for all r at the same time.



            For weighted distances, use IGDistanceHistogram.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 9 hours ago









            SzabolcsSzabolcs

            163k14447944




            163k14447944












            • $begingroup$
              Thanks. And how to count the same as the 'IGDistanceCounts[graph, {vertex}]' formula but for weighted networks?
              $endgroup$
              – ralph
              8 hours ago












            • $begingroup$
              @ralph As I said above, use IGDistanceHistogram
              $endgroup$
              – Szabolcs
              7 hours ago


















            • $begingroup$
              Thanks. And how to count the same as the 'IGDistanceCounts[graph, {vertex}]' formula but for weighted networks?
              $endgroup$
              – ralph
              8 hours ago












            • $begingroup$
              @ralph As I said above, use IGDistanceHistogram
              $endgroup$
              – Szabolcs
              7 hours ago
















            $begingroup$
            Thanks. And how to count the same as the 'IGDistanceCounts[graph, {vertex}]' formula but for weighted networks?
            $endgroup$
            – ralph
            8 hours ago






            $begingroup$
            Thanks. And how to count the same as the 'IGDistanceCounts[graph, {vertex}]' formula but for weighted networks?
            $endgroup$
            – ralph
            8 hours ago














            $begingroup$
            @ralph As I said above, use IGDistanceHistogram
            $endgroup$
            – Szabolcs
            7 hours ago




            $begingroup$
            @ralph As I said above, use IGDistanceHistogram
            $endgroup$
            – Szabolcs
            7 hours ago











            2












            $begingroup$

            For weighted network:



            g1 = {4798 <-> 2641, 4798 <-> 2310, 4798 <-> 4721, 2310 <-> 1942,2310 <-> 961, 4721 <-> 4507, 4721 <-> 4779, 4779 <-> 4336, 4779 <-> 3238, 4336 <-> 3277, 4336 <-> 3514, 3277 <-> 2923, 2923 <-> 2772, 2923 <-> 2401, 2772 <-> 2, 2772 <-> 2771, 3514 <-> 3042, 3514 <-> 2739, 3042 <-> 3007, 3042 <-> 1655, 2739 <-> 2277, 2739 <-> 1895, 2 <-> 5, 2 <-> 3, 3277 <-> 100, 5 <-> 6, 5 <-> 7, 5 <-> 8, 5 <-> 9};

            w1 = {10, 20, 20, 4, 35, 3, 4, 6, 17, 7, 13, 2, 2, 7, 2, 1, 3, 5, 3, 6,4, 6, 2, 1, 1, 1, 1, 1, 1};

            w2=Table[1, 29];

            net1 = Graph[g1, EdgeWeight -> w1, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]

            net2 = Graph[g1, EdgeWeight -> w2, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]

            s = RandomSample[VertexList[net1], 15];

            Mr = Table[IGDistanceCounts[net1, {s[[i]]}], {i, 1, Length[s]}] (*for non weighted*)

            Mr2 = IGDistanceHistogram[net1, 9] (*for weighted graph ?*)

            Mr3 = IGDistanceHistogram[net2, 9] (*for non weighted graph ? Mr3==Mr *)





            share|improve this answer









            $endgroup$


















              2












              $begingroup$

              For weighted network:



              g1 = {4798 <-> 2641, 4798 <-> 2310, 4798 <-> 4721, 2310 <-> 1942,2310 <-> 961, 4721 <-> 4507, 4721 <-> 4779, 4779 <-> 4336, 4779 <-> 3238, 4336 <-> 3277, 4336 <-> 3514, 3277 <-> 2923, 2923 <-> 2772, 2923 <-> 2401, 2772 <-> 2, 2772 <-> 2771, 3514 <-> 3042, 3514 <-> 2739, 3042 <-> 3007, 3042 <-> 1655, 2739 <-> 2277, 2739 <-> 1895, 2 <-> 5, 2 <-> 3, 3277 <-> 100, 5 <-> 6, 5 <-> 7, 5 <-> 8, 5 <-> 9};

              w1 = {10, 20, 20, 4, 35, 3, 4, 6, 17, 7, 13, 2, 2, 7, 2, 1, 3, 5, 3, 6,4, 6, 2, 1, 1, 1, 1, 1, 1};

              w2=Table[1, 29];

              net1 = Graph[g1, EdgeWeight -> w1, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]

              net2 = Graph[g1, EdgeWeight -> w2, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]

              s = RandomSample[VertexList[net1], 15];

              Mr = Table[IGDistanceCounts[net1, {s[[i]]}], {i, 1, Length[s]}] (*for non weighted*)

              Mr2 = IGDistanceHistogram[net1, 9] (*for weighted graph ?*)

              Mr3 = IGDistanceHistogram[net2, 9] (*for non weighted graph ? Mr3==Mr *)





              share|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                For weighted network:



                g1 = {4798 <-> 2641, 4798 <-> 2310, 4798 <-> 4721, 2310 <-> 1942,2310 <-> 961, 4721 <-> 4507, 4721 <-> 4779, 4779 <-> 4336, 4779 <-> 3238, 4336 <-> 3277, 4336 <-> 3514, 3277 <-> 2923, 2923 <-> 2772, 2923 <-> 2401, 2772 <-> 2, 2772 <-> 2771, 3514 <-> 3042, 3514 <-> 2739, 3042 <-> 3007, 3042 <-> 1655, 2739 <-> 2277, 2739 <-> 1895, 2 <-> 5, 2 <-> 3, 3277 <-> 100, 5 <-> 6, 5 <-> 7, 5 <-> 8, 5 <-> 9};

                w1 = {10, 20, 20, 4, 35, 3, 4, 6, 17, 7, 13, 2, 2, 7, 2, 1, 3, 5, 3, 6,4, 6, 2, 1, 1, 1, 1, 1, 1};

                w2=Table[1, 29];

                net1 = Graph[g1, EdgeWeight -> w1, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]

                net2 = Graph[g1, EdgeWeight -> w2, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]

                s = RandomSample[VertexList[net1], 15];

                Mr = Table[IGDistanceCounts[net1, {s[[i]]}], {i, 1, Length[s]}] (*for non weighted*)

                Mr2 = IGDistanceHistogram[net1, 9] (*for weighted graph ?*)

                Mr3 = IGDistanceHistogram[net2, 9] (*for non weighted graph ? Mr3==Mr *)





                share|improve this answer









                $endgroup$



                For weighted network:



                g1 = {4798 <-> 2641, 4798 <-> 2310, 4798 <-> 4721, 2310 <-> 1942,2310 <-> 961, 4721 <-> 4507, 4721 <-> 4779, 4779 <-> 4336, 4779 <-> 3238, 4336 <-> 3277, 4336 <-> 3514, 3277 <-> 2923, 2923 <-> 2772, 2923 <-> 2401, 2772 <-> 2, 2772 <-> 2771, 3514 <-> 3042, 3514 <-> 2739, 3042 <-> 3007, 3042 <-> 1655, 2739 <-> 2277, 2739 <-> 1895, 2 <-> 5, 2 <-> 3, 3277 <-> 100, 5 <-> 6, 5 <-> 7, 5 <-> 8, 5 <-> 9};

                w1 = {10, 20, 20, 4, 35, 3, 4, 6, 17, 7, 13, 2, 2, 7, 2, 1, 3, 5, 3, 6,4, 6, 2, 1, 1, 1, 1, 1, 1};

                w2=Table[1, 29];

                net1 = Graph[g1, EdgeWeight -> w1, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]

                net2 = Graph[g1, EdgeWeight -> w2, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]

                s = RandomSample[VertexList[net1], 15];

                Mr = Table[IGDistanceCounts[net1, {s[[i]]}], {i, 1, Length[s]}] (*for non weighted*)

                Mr2 = IGDistanceHistogram[net1, 9] (*for weighted graph ?*)

                Mr3 = IGDistanceHistogram[net2, 9] (*for non weighted graph ? Mr3==Mr *)






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 5 hours ago









                ralphralph

                1707




                1707






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematica Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194581%2fneighboring-nodes-in-the-network%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    IEEEtran - How to include ORCID in TeX/PDF with PdfLatexIs there a standard way to include ORCID in TeX /...

                    Cicindela nigrior Przypisy | Menu nawigacyjneCicindela varians unicolorManual for the Identification of the...

                    Glossaries-extra: Adding glossaries package to “Clas­sicTh­e­sis” template by Dr. André Miede v. 4.6 ...