11
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Bug introduced in 11.0 and fixed in 11.3.0

I am trying to reshape a $2 \times 2 \times 2$ array into a $n \times m$ array. It seems that if the matrix is defined via SparseArray, the ArrayReshape command doesn't care what $n$ and $m$ I choose, it always reshapes it to a $4 \times 2$ array.

In this code I try to reshape to $1 \times 8$:

Clear["Global`*"]
m = SparseArray[{i_, i_, i_} -> 1., {2, 2, 2}];
mt = ArrayReshape[m, {1, 8}];
MatrixForm[mt]

$\begin{pmatrix}1.&0.\\ 0.&0.\\0.&0.\\0.&1.\end{pmatrix}$

Why does it work like that? Am I doing something wrong?

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4
  • $\begingroup$ Seems to have something to do with SparseArray, despite what the docs say. Dimensions@ArrayReshape[m, {1, 8}] gives {4, 2}, but Dimensions@ArrayReshape[Normal@m, {1, 8}] gives {1, 8}. $\endgroup$
    – aardvark2012
    Oct 28, 2017 at 12:42
  • $\begingroup$ In Mathematica 11 Miguels code produces a list with 8 entries ({1,8} dimensional) $\endgroup$
    – M. Stern
    Oct 28, 2017 at 14:51
  • $\begingroup$ Mi edition is 11.2 ... $\endgroup$
    – MBolin
    Oct 28, 2017 at 15:59
  • $\begingroup$ bug introduced in 11.2? $\endgroup$
    – M. Stern
    Oct 28, 2017 at 16:37

3 Answers 3

Reset to default
8
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ArrayReshape[SparseArray`SparseArrayFlatten[m], {1, 8}]
(* or ArrayReshape[Flatten[m], {1, 8}] *)

SparseArray[<2>,{1,8}]

Normal @ %

{{1, 0, 0, 0, 0, 0, 0, 1}}

For the special case where the desired shape is a list with Length equal to Times@@Dimensions[m], you can also use SparseArray`SparseArrayFlatten:

SparseArray`SparseArrayFlatten[m]

SparseArray[<2>,{8}]

Normal @ %

{1, 0, 0, 0, 0, 0, 0, 1}

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4
  • $\begingroup$ That's right @kglr but my aim is not a $1 \times 8$ array, but any $n \times m$ array. The $1 \times 8$ in the code was just an example. $\endgroup$
    – MBolin
    Oct 28, 2017 at 16:01
  • $\begingroup$ @Miguel, this approach works for any n and m. For example, ArrayReshape[SparseArray`SparseArrayFlatten[m], {5, 3}, x] gives SparseArray[<9>,{5,3}] as it should $\endgroup$
    – kglr
    Oct 28, 2017 at 16:29
  • $\begingroup$ Sorry, you're right. Thank you @kglr $\endgroup$
    – MBolin
    Oct 29, 2017 at 2:07
  • $\begingroup$ @MiguelBolín, my pleasure. Thank you for the accept. And welcome to mma.se. $\endgroup$
    – kglr
    Oct 29, 2017 at 2:08
5
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It seems one have to make SparseArray to Normal first to reshape it. 

m=SparseArray[{i_,i_,i_}->1,{2,2,2}];
ArrayReshape[m,{1,8}]//Normal

Mathematica graphics

But now

ArrayReshape[Normal@m,{1,8}]

Mathematica graphics

Not sure if this is by design or not.

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6
  • 1
    $\begingroup$ Okay but then I loose the computation saving of SparseArray, right? When you make it Normal it adds all the zeros and you carry them along $\endgroup$
    – MBolin
    Oct 28, 2017 at 13:29
  • $\begingroup$ @MiguelBolín that is right. THat was the only way I could find to get it to work. I suggest you use Kglr solution using SparseArray`SparseArrayFlatten which I did not know about. $\endgroup$
    – Nasser
    Oct 28, 2017 at 15:21
  • 1
    $\begingroup$ mt = SparseArray@ArrayReshape[Normal@m, {1, 8}][[1]] $\endgroup$
    – Bob Hanlon
    Oct 28, 2017 at 15:22
  • $\begingroup$ Or just mt = SparseArray@ArrayReshape[Normal@m, {1, 8}] for the 1 x 8 $\endgroup$
    – Bob Hanlon
    Oct 28, 2017 at 15:37
  • $\begingroup$ I guess that has the same problem as Nasser's, right @BobHanlon ? You are adding the zeros and carrying them along. $\endgroup$
    – MBolin
    Oct 28, 2017 at 16:04
3
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Comparing the timings for the approaches proposed by kglr and Nasser (modified):

$HistoryLength = 0;

m = SparseArray[{i_, i_, i_} -> 1., {16, 16, 16}];

(kglr = ArrayReshape[SparseArray`SparseArrayFlatten[m], #] & /@ {{1, 
      16^3}, {16^3, 1}, {128, 32}, {32, 128}}) // RepeatedTiming

enter image description here

(nasser = SparseArray@ArrayReshape[Normal@m, #] & /@ {{1, 16^3}, {16^3, 
      1}, {128, 32}, {32, 128}}) // RepeatedTiming

enter image description here

Verifying their equivalence

kglr === nasser

True
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