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2

$begingroup$

Consider a data having the form

data = {{1,7,4,6},{1,6,4,8},{2,4,9,2},{E,...},{1,4,6,3},{4,4,6,2},{E,...},...}

i.e., some number $n_{1}$ of rows followed by row ${E,...}$, then some number $n_{2}$ of rows followed by row ${E,...}$ and so on.

Could you please tell me how to leave only the last rows before ${E,}$, i.e. to obtain

subdata= {{2,4,9,2},{4,4,6,2},...}?

list-manipulation data

share|improve this question

asked 3 hours ago

John TaylorJohn Taylor

787211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  e.g. SequenceCases[data, {x_List, {E, ___}} :> x]
  $endgroup$
  – C. E.
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Consider a data having the form

data = {{1,7,4,6},{1,6,4,8},{2,4,9,2},{E,...},{1,4,6,3},{4,4,6,2},{E,...},...}

i.e., some number $n_{1}$ of rows followed by row ${E,...}$, then some number $n_{2}$ of rows followed by row ${E,...}$ and so on.

Could you please tell me how to leave only the last rows before ${E,}$, i.e. to obtain

subdata= {{2,4,9,2},{4,4,6,2},...}?

list-manipulation data

share|improve this question

asked 3 hours ago

John TaylorJohn Taylor

787211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  e.g. SequenceCases[data, {x_List, {E, ___}} :> x]
  $endgroup$
  – C. E.
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider a data having the form

data = {{1,7,4,6},{1,6,4,8},{2,4,9,2},{E,...},{1,4,6,3},{4,4,6,2},{E,...},...}

i.e., some number $n_{1}$ of rows followed by row ${E,...}$, then some number $n_{2}$ of rows followed by row ${E,...}$ and so on.

Could you please tell me how to leave only the last rows before ${E,}$, i.e. to obtain

subdata= {{2,4,9,2},{4,4,6,2},...}?

list-manipulation data

share|improve this question

asked 3 hours ago

John TaylorJohn Taylor

787211

$endgroup$

data = {{1,7,4,6},{1,6,4,8},{2,4,9,2},{E,...},{1,4,6,3},{4,4,6,2},{E,...},...}

subdata= {{2,4,9,2},{4,4,6,2},...}?

share|improve this question

asked 3 hours ago

John TaylorJohn Taylor

787211

share|improve this question

asked 3 hours ago

John TaylorJohn Taylor

787211

asked 3 hours ago

John TaylorJohn Taylor

787211

asked 3 hours ago

John TaylorJohn Taylor

787211

2 Answers
2

active

oldest

votes

2

$begingroup$

Try SequenceCases:

data = {{1, 7, 4, 6}, {1, 6, 4, 8}, {2, 4, 9, 2}, {E, 1, 2, 3}, 

        {1, 4, 6, 3}, {4, 4, 6, 2}, {E, 4, 5, 6}}

SequenceCases[data, {p_, {E, ___}} :> p]

yields

{{2, 4, 9, 2}, {4, 4, 6, 2}}

share|improve this answer

answered 3 hours ago

sakrasakra

2,8231429

$endgroup$

add a comment | 

2

$begingroup$

The most idiomatic solution to this problem is, in my opinion, pattern matching (as Sakra has also answered):

SequenceCases[data, {x_List, {E, ___}} :> x]

{{2, 4, 9, 2}, {4, 4, 6, 2}}

But the problem also lends itself to functional solutions, e.g.:

pairs = Partition[data, 2, 1];

If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs

{{2, 4, 9, 2}, {4, 4, 6, 2}}

Or in one go:

BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]

{{2, 4, 9, 2}, {4, 4, 6, 2}}

share|improve this answer

edited 51 mins ago

answered 2 hours ago

C. E.C. E.

51.4k3101207

$endgroup$

add a comment | 

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2

$begingroup$

Try SequenceCases:

data = {{1, 7, 4, 6}, {1, 6, 4, 8}, {2, 4, 9, 2}, {E, 1, 2, 3}, 

        {1, 4, 6, 3}, {4, 4, 6, 2}, {E, 4, 5, 6}}

SequenceCases[data, {p_, {E, ___}} :> p]

yields

{{2, 4, 9, 2}, {4, 4, 6, 2}}

share|improve this answer

answered 3 hours ago

sakrasakra

2,8231429

$endgroup$

add a comment | 

2

$begingroup$

Try SequenceCases:

data = {{1, 7, 4, 6}, {1, 6, 4, 8}, {2, 4, 9, 2}, {E, 1, 2, 3}, 

        {1, 4, 6, 3}, {4, 4, 6, 2}, {E, 4, 5, 6}}

SequenceCases[data, {p_, {E, ___}} :> p]

yields

{{2, 4, 9, 2}, {4, 4, 6, 2}}

share|improve this answer

answered 3 hours ago

sakrasakra

2,8231429

$endgroup$

add a comment | 

$begingroup$

Try SequenceCases:

data = {{1, 7, 4, 6}, {1, 6, 4, 8}, {2, 4, 9, 2}, {E, 1, 2, 3}, 

        {1, 4, 6, 3}, {4, 4, 6, 2}, {E, 4, 5, 6}}

SequenceCases[data, {p_, {E, ___}} :> p]

yields

{{2, 4, 9, 2}, {4, 4, 6, 2}}

share|improve this answer

answered 3 hours ago

sakrasakra

2,8231429

$endgroup$

data = {{1, 7, 4, 6}, {1, 6, 4, 8}, {2, 4, 9, 2}, {E, 1, 2, 3}, 

        {1, 4, 6, 3}, {4, 4, 6, 2}, {E, 4, 5, 6}}

SequenceCases[data, {p_, {E, ___}} :> p]

{{2, 4, 9, 2}, {4, 4, 6, 2}}

share|improve this answer

answered 3 hours ago

sakrasakra

2,8231429

answered 3 hours ago

sakrasakra

2,8231429

answered 3 hours ago

sakrasakra

2,8231429

2

$begingroup$

The most idiomatic solution to this problem is, in my opinion, pattern matching (as Sakra has also answered):

SequenceCases[data, {x_List, {E, ___}} :> x]

{{2, 4, 9, 2}, {4, 4, 6, 2}}

But the problem also lends itself to functional solutions, e.g.:

pairs = Partition[data, 2, 1];

If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs

{{2, 4, 9, 2}, {4, 4, 6, 2}}

Or in one go:

BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]

{{2, 4, 9, 2}, {4, 4, 6, 2}}

share|improve this answer

edited 51 mins ago

answered 2 hours ago

C. E.C. E.

51.4k3101207

$endgroup$

add a comment | 

2

$begingroup$

The most idiomatic solution to this problem is, in my opinion, pattern matching (as Sakra has also answered):

SequenceCases[data, {x_List, {E, ___}} :> x]

{{2, 4, 9, 2}, {4, 4, 6, 2}}

But the problem also lends itself to functional solutions, e.g.:

pairs = Partition[data, 2, 1];

If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs

{{2, 4, 9, 2}, {4, 4, 6, 2}}

Or in one go:

BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]

{{2, 4, 9, 2}, {4, 4, 6, 2}}

share|improve this answer

edited 51 mins ago

answered 2 hours ago

C. E.C. E.

51.4k3101207

$endgroup$

add a comment | 

$begingroup$

The most idiomatic solution to this problem is, in my opinion, pattern matching (as Sakra has also answered):

SequenceCases[data, {x_List, {E, ___}} :> x]

{{2, 4, 9, 2}, {4, 4, 6, 2}}

But the problem also lends itself to functional solutions, e.g.:

pairs = Partition[data, 2, 1];

If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs

{{2, 4, 9, 2}, {4, 4, 6, 2}}

Or in one go:

BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]

{{2, 4, 9, 2}, {4, 4, 6, 2}}

share|improve this answer

edited 51 mins ago

answered 2 hours ago

C. E.C. E.

51.4k3101207

$endgroup$

SequenceCases[data, {x_List, {E, ___}} :> x]

pairs = Partition[data, 2, 1];

If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs

BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]

share|improve this answer

edited 51 mins ago

answered 2 hours ago

C. E.C. E.

51.4k3101207

answered 2 hours ago

C. E.C. E.

51.4k3101207

answered 2 hours ago

C. E.C. E.

51.4k3101207