bsxfun seems to return too few values
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In Matlab 2016a, the following command behaves as I expect:
>> bsxfun (@(x,y)((y-x)), [1], [3, 4])
ans =
2 3
However, this very similar command returns a scalar, instead of a 1x2 matrix:
>> bsxfun (@(x,y)(min(y-x)), [1], [3, 4])
ans =
2
Am I right in assuming this is a bug in bsxfun? If so, how can I report it. If not, could someone please explain this behaviour?
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KSSV
le 28 Juil 2016
bsxfun (@(x,y)((y-x)), [1], [3, 4])
It's output is 2, 3. The minimum value of (2,3) is 2..The second line
bsxfun (@(x,y)(min(y-x)), [1], [3, 4])
is giving you minimum value of the output....
I am using MATLAB2015a and I got the same result as said.
2 commentaires
Stephen23
le 28 Juil 2016
"I am using MATLAB2015a and I got the same result as said"
This does not answer the question "is this a bug?"
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Stephen23
le 28 Juil 2016
Modifié(e) : Stephen23
le 28 Juil 2016
@Lachlan: this is a bug (or an undocumented "feature"), and you should report it. You can read how here:
When you read the bsxfun documentation it clearly states that "A binary element-wise function of the form C = fun(A,B) accepts arrays A and B of arbitrary, but equal size and returns output of the same size. Each element in the output array C is the result of an operation on the corresponding elements of A and B only. fun must also support scalar expansion, such that if A or B is a scalar, C is the result of applying the scalar to every element in the other input array."
The function you provided does not fulfill these conditions, because it returns scalar when provided with a vector and a scalar:
>> fun = @(x,y)min(y-x);
>> fun(1, [3, 4])
ans =
2
Note that scalar x was not expanded, as the documentation requires. So you are already using bsxfun outside of its documented functionality. The question is, what should bsxfun do when you provide an incorrect function that returns a scalar and not a vector?
By accepting this function and returning an undocumented output is at least misleading and at worst totally erroneous.
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