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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
  32 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
  32 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