Cannot perform null assignment from variable

Question

Aaron 2023-11-13

1
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2046700-cannot-perform-null-assignment-from-variable

评论： Aaron 2023-11-13

The following code "works" (deletes the 3rd element of A, or the third page, if the first two dimensions are not 1 & 1):

A = nan(1,1,3);

A(3) = [];

This modification breaks the functionality and throws an error: "Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 0-by-0."

A = nan(1,1,3);

empty_scalar = [];

A(3) = empty_scalar;

I don't understand what's going on here -- why can't I do the same null assignment from a variable? Is "= []" special syntax?

I'd like to be able to automate the process of either assigning or deleting elements/columns/pages of some arbitrarily-sized array. How can I do this, given the above behavior?

Thanks.

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Dyuman Joshi 2023-11-13

The first one is deletion.

The second one is assignment (as the error states as well).

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Answer 1

Matt J 2023-11-13

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2046700-cannot-perform-null-assignment-from-variable#answer_1351655

编辑：Matt J 2023-11-13

在 MATLAB Online 中打开

Yes, it is special syntax. I agree it is strange behavior, but one workaround is,

A = nan(1,1,3);
empty_scalar = [];
if isempty(empty_scalar)
  A(3) = [];
else
  A(3)=empty_scalar;
end
A
A = 
A(:,:,1) =

   NaN


A(:,:,2) =

   NaN

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Aaron 2023-11-13

It's interesting to me that '= []' is special behavior (for subscripted assignment only?); until today I always assumed it was somehow assigning a 0-by-0 array to all entries in the given slices.

I'm not sure the subsasgn workaround will be any prettier than what I ended up with, which was just some nested if's with different versions of the assignment statement (which had to assign multiple outputs to slices of different arrays).

Thanks for your help.

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Answer 2

Matt J 2023-11-13

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https://ww2.mathworks.cn/matlabcentral/answers/2046700-cannot-perform-null-assignment-from-variable#answer_1351755

在 MATLAB Online 中打开

Another workaround is to call subsasgn direclty,

A = nan(1,3);
empty_scalar = 4;
A=subsasgn(A,  struct('type','()','subs',{{3}})   ,   empty_scalar)
A = 1×3
   NaN   NaN     4
empty_scalar = [];
A=subsasgn(A,  struct('type','()','subs',{{3}})   ,   empty_scalar)
A = 1×2
   NaN   NaN