Question related to vectorized matrix operation.

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Pappu Murthy
Pappu Murthy 2017-1-22
编辑： Pappu Murthy 2017-1-23
i have two matrices A and B; A is say (Nx3) where N is rather large like 1000+... B is (nx2) where n is rather smaller of the order 10 or so.. but it does not matter it is smaller than A row size. i would like to compute
for i = 1:N
D = ( A(i,1)- B(1:n,1) )^2 + (A(i,2) - B(1:n,2)^2 )
end
for e.g. if N is 4, and n =2 then D = a [4x2] matrix...
I am able to perform this using for loops without much difficulty but would like to try using the vectorized Matrix operations instead for improving the performance and speed. Thanks in advance.

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采纳的回答

Stephen23
Stephen23 2017-1-22
编辑：Andrei Bobrov 2017-1-23
>> A = randi(9,4,3)
A =
4 2 9
9 4 4
5 4 8
4 2 8
>> B = randi(9,3,2)
B =
3 2
6 4
3 1
>> D = bsxfun(@minus,A(:,1),B(:,1).').^2 + bsxfun(@minus,A(:,2),B(:,2).').^2
D =
1 8 2
40 9 45
8 1 13
1 8 2
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Pappu Murthy
Pappu Murthy 2017-1-23
thanks to both of you after the corrections I am able to get the right answer and I found this answer to be the most direct and also verified that many times faster than my answer for large size matrices. Thanks again. I will now go ahead and accept the answer.

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更多回答（1 个）

Andrei Bobrov
Andrei Bobrov 2017-1-23
编辑：Andrei Bobrov 2017-1-23
My variants:
For R2016b and later
>> C2 = squeeze(sum((A(:,1:2) - permute(B,[3,2,1])).^2,2))
C2 =
1 8 2
40 9 45
8 1 13
1 8 2
and with bsxfun:
>> C3 = squeeze(sum(bsxfun(@minus,A(:,1:2),permute(B,[3,2,1])).^2,2))
C3 =
1 8 2
40 9 45
8 1 13
1 8 2
  1 个评论
Pappu Murthy
Pappu Murthy 2017-1-23
编辑：Pappu Murthy 2017-1-23
The first solution did not work in 2016a but worked ok in 2016b. Thanks for suggesting the alternative answers, although After testing a lot the original solution and your interpretation with bsxfun appear to be the best from speed point of view. Your first approach was a bit slower. I have tried on huge matrices to establish the speed correctly.

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