I need help with permutations.
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for example: There are matrix m and r:
m = [1.2000 2.4000 1.3000
2.3000 1.5000 1.0000];
r = [0.2000 0.4000 0.3000];
rr = triu(ones(3),1);
rr(rr > 0) = r;
mm = bsxfun(@times,permute(m,[2,3,1]),permute(m,[3,2,1]));
K = bsxfun(@times,mm,rr + rr.' + eye(3));
This code gives me the result I want.
The values of the matrix m can vary. for example:
m = [1.2000 2.4000 1.3000
2.3000 1.5000 1.0000
1.2000 2.4000 3.0000];
r = [0.2000 0.4000 0.3000 0.4444];
So how can I write code like "[row col] = size (m)"?
I mean;
rr = triu(ones(row),1);
rr(rr > 0) = r;
mm = bsxfun(@times,permute(m,[?,?,?]),permute(m,[?,?,?]));
K = bsxfun(@times,mm,rr + rr.' + eye(row));
5 个评论
Adam
2017-5-5
What is your question?
Muhendisleksi
2017-5-5
Guillaume
2017-5-5
It's fairly clear that you aren't the author of the code you're showing (since it does not appear that you understand it). Can you provide a link to the question where you got this code so we have context.
Muhendisleksi
2017-5-5
Andrei Bobrov
2017-5-5
" m = [1.2000 2.4000 1.3000
2.3000 1.5000 1.0000
1.2000 2.4000 3.0000];
r = [0.2000 0.4000 0.3000 0.4444]; "
What result do you expect, what is the expression for K ?
回答（1 个）
What's wrong with
[r,c]=size(m);
mm = bsxfun(@times,permute(m,[r,c,1]),permute(m,[c,r,1]));
ADDENDUM
Which, if you haven't recognized it, can be written as
permute(m,[size(m),1]),permute(m,[flip(size(m)),1])
ERRATUM
As Guillaume astutely points out, the permute indices are NOT size(m)-dependent, but reflect the dimensionality of m as 2D array. The above, while cute, is nonsensical for the purpose.
5 个评论
Muhendisleksi
2017-5-5
dpb
2017-5-5
Well, that's dynamic using the size() returned????
Muhendisleksi
2017-5-5
My aim is to create these tables. However, as the values of "m" and "r" increase, the result will change.
Guillaume
2017-5-5
No idea what the original purpose of the code is, and I'm not going to try to decrypt some obscure image with no associated explanation, but it's clear to me that in the original code, the dimensions to permute do not depend on the size of m. It's always going to be [2 3 1] and [3 2 1] respectively.
The only thing that would need to change in the original code if the size of m changes is the size of rr, with
rr = triu(ones(size(m, 2)));
You would of course need more elements in r to accodomate the change in size of rr.
dpb
2017-5-5
Good catch, Guillaume! The permutations are not size-dependent but dimensional.
I didn't really try to read the image, either, and while the above will respond to OP's plea, it isn't useful. I'll scratch it altho I was kinda' proud of the flip(size()) thingie; I'd not thunk a' it before and can recall at least one instance could have used it to eliminate a temporary...
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