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I discovered a difference between version 12.0 and version 14.0 in how they handle SparseArray within the Table function.

For example:

SparseArray[{{1, 1} -> 1, {2, 2} -> 2, {3, 3} -> 3, {1, 3} -> z}];
Normal /@ Table[%, {z, 0, 1}]

I got

{{{1, 0, 0}, {0, 2, 0}, {0, 0, 3}}, {{1, 0, 1}, {0, 2, 0}, {0, 0, 3}}}

in version 12.0, and

{{{1, 0, z}, {0, 2, 0}, {0, 0, 3}}, {{1, 0, z}, {0, 2, 0}, {0, 0, 3}}}

in version 14.0. This results in much of the code written for version 12.0 being incompatible with version 14.0. I want to maintain the previous format, namely using SparseArray within the Table function, and I hope it runs in version 14.0. What should I do?

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1 Answer 1

Reset to default
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What about this?

array[z_] := 
 SparseArray[{{1, 1} -> 1, {2, 2} -> 2, {3, 3} -> 3, {1, 3} -> z}]
Normal /@ Table[array[z], {z, 0, 1}]

(*{{{1,0,0},{0,2,0},{0,0,3}},{{1,0,1},{0,2,0},{0,0,3}}}*)

Or you can change the position of Normal.

SparseArray[{{1, 1} -> 1, {2, 2} -> 2, {3, 3} -> 3, {1, 3} -> z}];
Table[Normal@%, {z, 0, 1}]

(*{{{1,0,0},{0,2,0},{0,0,3}},{{1,0,1},{0,2,0},{0,0,3}}}*)

You can also consider the following code.

s = SparseArray[{{1, 1} -> 1, {1, 3} -> z, {2, 1} -> 5, {2, 2} -> 
    2, {3, 3} -> 3}]; Table[
 SparseArray[ArrayRules[s] /. z -> i], {i, 0, 1}];
Normal /@ %
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  • $\begingroup$ Thank you very much for your reply. While the Normal function can temporarily resolve this error, it also leads to increased memory consumption. $\endgroup$ Commented Jul 29, 2024 at 16:45
  • $\begingroup$ @sidy What does you mean? If you want to get two SparseArray, you can use my first method, or just generate SparseArray in Table. $\endgroup$ Commented Jul 29, 2024 at 16:52
  • $\begingroup$ Oh yes, I see what you mean.You are right. $\endgroup$ Commented Jul 29, 2024 at 17:01
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    $\begingroup$ It's not a bug. Please see the 7th bullet point in Incompatible Changes: The internal structure of SparseArray has been changed. It is now treated as a raw object for purposes of pattern matching. $\endgroup$ Commented Jul 29, 2024 at 18:25
  • $\begingroup$ @Domen You are right. But is this structure really good? Number 1/2 is an atom, but expression 1/a is not. Meanwhile, List is never an atom. So why should a SparseArray(especially one with variables) be a raw object? $\endgroup$ Commented Jul 30, 2024 at 1:16

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