Avoiding unpacking an array when altering its dimensionMultidimensional array reduction through summation...
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Avoiding unpacking an array when altering its dimension
Multidimensional array reduction through summation over one of its dimensionsFastest way to test if a numerical array has complex elementsFiltering the elements of an array to split them into two categoriesQuickly pruning elements in one structured array that exist in a separate unordered arrayHow to implement the general array broadcasting method from NumPy?Table with List iterator return unpacked listMemory optimization for exporting “VideoFrames” with many large Array plots2D array summation operationsPacking and unpacking of arrays with Map/TableConditionally replace element value to 0. in blocks of small matrice
$begingroup$
I am trying to flatten a matrix from 3-dimensional to 2-dimensional one.
The original one has dimension for example {10,120,120}:
list1 = RandomReal[1, {10, 120, 120}];
I want to convert it to a new matrix with {10,120*120} dimension. I am using:
On["Packing"] (*For checking if exist unpacked arrays *)
Flatten[Map[Flatten, {list1}, {-3}], 1];
But this will generate unpacked array. How should I avoid this issue?
list-manipulation packed-arrays
edited 1 hour ago
m_goldberg
87.2k872197
asked 2 hours ago
cj9435042cj9435042
35416
$endgroup$
add a comment |
$begingroup$
I am trying to flatten a matrix from 3-dimensional to 2-dimensional one.
The original one has dimension for example {10,120,120}:
list1 = RandomReal[1, {10, 120, 120}];
I want to convert it to a new matrix with {10,120*120} dimension. I am using:
On["Packing"] (*For checking if exist unpacked arrays *)
Flatten[Map[Flatten, {list1}, {-3}], 1];
But this will generate unpacked array. How should I avoid this issue?
list-manipulation packed-arrays
edited 1 hour ago
m_goldberg
87.2k872197
asked 2 hours ago
cj9435042cj9435042
35416
$endgroup$
2
$begingroup$
TryFlatten[list1, {{1}, {2, 3}}].
$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
1 hour ago
$begingroup$
Flatten /@ list1
$endgroup$
– Okkes Dulgerci
33 mins ago
add a comment |
$begingroup$
I am trying to flatten a matrix from 3-dimensional to 2-dimensional one.
The original one has dimension for example {10,120,120}:
list1 = RandomReal[1, {10, 120, 120}];
I want to convert it to a new matrix with {10,120*120} dimension. I am using:
On["Packing"] (*For checking if exist unpacked arrays *)
Flatten[Map[Flatten, {list1}, {-3}], 1];
But this will generate unpacked array. How should I avoid this issue?
list-manipulation packed-arrays
edited 1 hour ago
m_goldberg
87.2k872197
asked 2 hours ago
cj9435042cj9435042
35416
$endgroup$
I am trying to flatten a matrix from 3-dimensional to 2-dimensional one.
The original one has dimension for example {10,120,120}:
list1 = RandomReal[1, {10, 120, 120}];
I want to convert it to a new matrix with {10,120*120} dimension. I am using:
On["Packing"] (*For checking if exist unpacked arrays *)
Flatten[Map[Flatten, {list1}, {-3}], 1];
But this will generate unpacked array. How should I avoid this issue?
list-manipulation packed-arrays
list-manipulation packed-arrays
edited 1 hour ago
m_goldberg
87.2k872197
asked 2 hours ago
cj9435042cj9435042
35416
edited 1 hour ago
m_goldberg
87.2k872197
asked 2 hours ago
cj9435042cj9435042
35416
edited 1 hour ago
m_goldberg
87.2k872197
edited 1 hour ago
m_goldberg
87.2k872197
edited 1 hour ago
m_goldberg
87.2k872197
87.2k872197
asked 2 hours ago
cj9435042cj9435042
35416
asked 2 hours ago
cj9435042cj9435042
35416
asked 2 hours ago
cj9435042cj9435042
35416
35416
2
$begingroup$
TryFlatten[list1, {{1}, {2, 3}}].
$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
1 hour ago
$begingroup$
Flatten /@ list1
$endgroup$
– Okkes Dulgerci
33 mins ago
add a comment |
2
$begingroup$
TryFlatten[list1, {{1}, {2, 3}}].
$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
1 hour ago
$begingroup$
Flatten /@ list1
$endgroup$
– Okkes Dulgerci
33 mins ago
2
2
$begingroup$
Try
Flatten[list1, {{1}, {2, 3}}].$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
Try
Flatten[list1, {{1}, {2, 3}}].$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
1 hour ago
$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
1 hour ago
$begingroup$
Flatten /@ list1$endgroup$
– Okkes Dulgerci
33 mins ago
$begingroup$
Flatten /@ list1$endgroup$
– Okkes Dulgerci
33 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An alternative to using Flatten as in the comments is to use ArrayReshape. For large arrays, ArrayReshape should be significantly faster. For example:
list1 = RandomReal[1, {10, 1200, 1200}];
Comparison:
r1 = Flatten[list1, {{1}, {2, 3}}]; //RepeatedTiming
r2 = ArrayReshape[list1, {10, 1200 1200}]; //RepeatedTiming
r1 === r2
{0.120, Null}
{0.025, Null}
True
answered 1 hour ago
Carl WollCarl Woll
68.9k391177
$endgroup$
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An alternative to using Flatten as in the comments is to use ArrayReshape. For large arrays, ArrayReshape should be significantly faster. For example:
list1 = RandomReal[1, {10, 1200, 1200}];
Comparison:
r1 = Flatten[list1, {{1}, {2, 3}}]; //RepeatedTiming
r2 = ArrayReshape[list1, {10, 1200 1200}]; //RepeatedTiming
r1 === r2
{0.120, Null}
{0.025, Null}
True
answered 1 hour ago
Carl WollCarl Woll
68.9k391177
$endgroup$
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
add a comment |
$begingroup$
An alternative to using Flatten as in the comments is to use ArrayReshape. For large arrays, ArrayReshape should be significantly faster. For example:
list1 = RandomReal[1, {10, 1200, 1200}];
Comparison:
r1 = Flatten[list1, {{1}, {2, 3}}]; //RepeatedTiming
r2 = ArrayReshape[list1, {10, 1200 1200}]; //RepeatedTiming
r1 === r2
{0.120, Null}
{0.025, Null}
True
answered 1 hour ago
Carl WollCarl Woll
68.9k391177
$endgroup$
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
add a comment |
$begingroup$
An alternative to using Flatten as in the comments is to use ArrayReshape. For large arrays, ArrayReshape should be significantly faster. For example:
list1 = RandomReal[1, {10, 1200, 1200}];
Comparison:
r1 = Flatten[list1, {{1}, {2, 3}}]; //RepeatedTiming
r2 = ArrayReshape[list1, {10, 1200 1200}]; //RepeatedTiming
r1 === r2
{0.120, Null}
{0.025, Null}
True
answered 1 hour ago
Carl WollCarl Woll
68.9k391177
$endgroup$
An alternative to using Flatten as in the comments is to use ArrayReshape. For large arrays, ArrayReshape should be significantly faster. For example:
list1 = RandomReal[1, {10, 1200, 1200}];
Comparison:
r1 = Flatten[list1, {{1}, {2, 3}}]; //RepeatedTiming
r2 = ArrayReshape[list1, {10, 1200 1200}]; //RepeatedTiming
r1 === r2
{0.120, Null}
{0.025, Null}
True
answered 1 hour ago
Carl WollCarl Woll
68.9k391177
answered 1 hour ago
Carl WollCarl Woll
68.9k391177
answered 1 hour ago
Carl WollCarl Woll
68.9k391177
answered 1 hour ago
Carl WollCarl Woll
68.9k391177
68.9k391177
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
add a comment |
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
add a comment |
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$begingroup$
Try
Flatten[list1, {{1}, {2, 3}}].$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
1 hour ago
$begingroup$
Flatten /@ list1$endgroup$
– Okkes Dulgerci
33 mins ago