Avoiding unpacking an array when altering its dimension












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?










share|improve this question











$endgroup$








  • 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
    33 mins ago
















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?










share|improve this question











$endgroup$








  • 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
    33 mins ago














2












2








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?










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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









m_goldberg

87.2k872197




87.2k872197










asked 2 hours ago









cj9435042cj9435042

35416




35416








  • 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
    33 mins ago














  • 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
    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










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









$endgroup$













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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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









$endgroup$













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
















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









$endgroup$













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














4












4








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









$endgroup$



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 1 hour ago









Carl WollCarl Woll

68.9k391177




68.9k391177












  • $begingroup$
    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




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


















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