Avoiding unpacking an array when altering its dimension












3












$begingroup$


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











$endgroup$








  • 2




    $begingroup$
    Try Flatten[list1, {{1}, {2, 3}}].
    $endgroup$
    – J. M. is computer-less♦
    3 hours ago










  • $begingroup$
    Thanks, it works!
    $endgroup$
    – cj9435042
    3 hours ago
















3












$begingroup$


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











$endgroup$








  • 2




    $begingroup$
    Try Flatten[list1, {{1}, {2, 3}}].
    $endgroup$
    – J. M. is computer-less♦
    3 hours ago










  • $begingroup$
    Thanks, it works!
    $endgroup$
    – cj9435042
    3 hours ago














3












3








3


1



$begingroup$


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











$endgroup$




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 hours ago









m_goldberg

87.2k872197




87.2k872197










asked 3 hours ago









cj9435042cj9435042

35916




35916








  • 2




    $begingroup$
    Try Flatten[list1, {{1}, {2, 3}}].
    $endgroup$
    – J. M. is computer-less♦
    3 hours ago










  • $begingroup$
    Thanks, it works!
    $endgroup$
    – cj9435042
    3 hours ago














  • 2




    $begingroup$
    Try Flatten[list1, {{1}, {2, 3}}].
    $endgroup$
    – J. M. is computer-less♦
    3 hours ago










  • $begingroup$
    Thanks, it works!
    $endgroup$
    – cj9435042
    3 hours ago








2




2




$begingroup$
Try Flatten[list1, {{1}, {2, 3}}].
$endgroup$
– J. M. is computer-less♦
3 hours ago




$begingroup$
Try Flatten[list1, {{1}, {2, 3}}].
$endgroup$
– J. M. is computer-less♦
3 hours ago












$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
3 hours ago




$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
3 hours ago










1 Answer
1






active

oldest

votes


















7












$begingroup$

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









$endgroup$













  • $begingroup$
    I'm too slow..... :)
    $endgroup$
    – Michael E2
    3 hours ago











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









7












$begingroup$

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









$endgroup$













  • $begingroup$
    I'm too slow..... :)
    $endgroup$
    – Michael E2
    3 hours ago
















7












$begingroup$

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









$endgroup$













  • $begingroup$
    I'm too slow..... :)
    $endgroup$
    – Michael E2
    3 hours ago














7












7








7





$begingroup$

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









$endgroup$



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








share|improve this answer












share|improve this answer



share|improve this answer










answered 3 hours ago









Carl WollCarl Woll

69k391177




69k391177












  • $begingroup$
    I'm too slow..... :)
    $endgroup$
    – Michael E2
    3 hours ago


















  • $begingroup$
    I'm too slow..... :)
    $endgroup$
    – Michael E2
    3 hours ago
















$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
3 hours ago




$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
3 hours ago


















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