bsxfun(minus) vs normal minus
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i have X=eye(3) and A=magic(3) What is the difference between Result1=A-X and the Result2 with this loop
for i=1:3
Result2=bsxfun(@minus,A,X(i,:));
end
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James Tursa
2018년 7월 3일
편집: James Tursa
2018년 7월 3일
Run it and see. For one, your loop overwrites Result2 with each iteration, so you are not even doing the same calculations and thus you shouldn't expect them to match. And you don't define Y (was this supposed to be X?). What are you really trying to compare?
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Rik
2018년 7월 3일
Input arrays, specified as scalars, vectors, matrices,
or multidimensional arrays. Inputs A and B must have
compatible sizes. For more information, see Compatible
Array Sizes for Basic Operations. Whenever a dimension
of A or B is singleton (equal to one), bsxfun virtually
replicates the array along that dimension to match the
other array. In the case where a dimension of A or B is
singleton, and the corresponding dimension in the other
array is zero, bsxfun virtually diminishes the
singleton dimension to zero.
This means the following:
A=[1 2];% 1 by 2
B=[3;4];% 2 by 1
%A will be replicated along the first dimension to make it match B
%B will be replicated along the second dimension to make it match A
%then an element-wise operation can be performed:
C1=bsxfun(@minus,A,B);
C2=repmat(A,size(B,1),1)-repmat(B,1,size(A,2));
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추가 답변 (1개)
Guillaume
2018년 7월 3일
In your loop, which as James pointed out, wouldn't do anything useful since it overwrites Result2 at each step, for each i,
bsxfun(@minus, A, X(i, :))
is exactly equivalent in term of result to
A - repmat(X(i, :), size(A, 1), 1)
but uses much less memory since it doesn't actually replicate the X row.
Note that since R2016b, which introduced implicit expansion, this is also the same as
A - X(i, :)
Before R2016b, the above would have resulted in an error.
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