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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

2

$begingroup$

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

share|improve this question

edited 1 hour ago

m_goldberg

87.2k872197

asked 2 hours ago

cj9435042cj9435042

35416

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  34 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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

share|improve this question

edited 1 hour ago

m_goldberg

87.2k872197

asked 2 hours ago

cj9435042cj9435042

35416

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  34 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

share|improve this question

edited 1 hour ago

m_goldberg

87.2k872197

asked 2 hours ago

cj9435042cj9435042

35416

$endgroup$

list1 = RandomReal[1, {10, 120, 120}];

On["Packing"]  (*For checking if exist unpacked arrays *)  

Flatten[Map[Flatten, {list1}, {-3}], 1];

share|improve this question

edited 1 hour ago

m_goldberg

87.2k872197

asked 2 hours ago

cj9435042cj9435042

35416

share|improve this question

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

asked 2 hours ago

cj9435042cj9435042

35416

asked 2 hours ago

cj9435042cj9435042

35416

1 Answer
1

active

oldest

votes

4

$begingroup$

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

share|improve this answer

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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4

$begingroup$

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

share|improve this answer

answered 1 hour ago

Carl WollCarl Woll

68.9k391177

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm too slow..... :)
  $endgroup$
  – Michael E2
  1 hour ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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

share|improve this answer

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

share|improve this answer

answered 1 hour ago

Carl WollCarl Woll

68.9k391177

$endgroup$

list1 = RandomReal[1, {10, 1200, 1200}];

r1 = Flatten[list1, {{1}, {2, 3}}]; //RepeatedTiming

r2 = ArrayReshape[list1, {10, 1200 1200}]; //RepeatedTiming



r1 === r2

share|improve this answer

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