Mean a matrix columnwise based on another logical matrix
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I want to mean the matrix a based on the logical matrix b columnwise: In other words this:
mean(a(b(:,1)))
mean(a(b(:,2)))
...
a =[
7 8 5 3
1 1 4 7
9 3 8 7
10 1 8 2
7 1 2 2
8 9 5 5
8 7 5 10
4 4 7 4
7 10 8 6
2 1 8 3];
b =logical([
0 0 1 0
1 0 1 0
1 0 1 1
1 0 1 1
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
1 0 0 1
0 0 0 1]);
1 个评论
Jan
2013-6-18
This is obviously an interesting question, which allows to shed light on different powers of Matlab. +1
采纳的回答
Jan
2013-6-16
Or:
m = sum(a .* b, 1) ./ sum(b, 1);
4 个评论
Giorgos Papakonstantinou
2013-6-17
Jan
2013-6-17
Me too. And this one is the leanest and fastest one :-)
Wayne King
2013-6-17
I agree Jan, +1, silly of me not to think of simply using the b matrix to get the correct number of nonzero elements.
Andrei Bobrov
2013-6-18
+1
更多回答（4 个）
Andrei Bobrov
2013-6-17
编辑：Andrei Bobrov
2013-6-18
b1 = bsxfun(@times,b,1:4);
out = accumarray(b1(b),a(b),[],@mean);
or
[~,jj] = find(b);
out = accumarray(jj,a(b),[],@mean);
and in line with accumarray:
out = accumarray(ceil(find(b)/size(b,1)),a(b),[],@mean);
2 个评论
Giorgos Papakonstantinou
2013-6-17
Jan
2013-6-18
Although I do not think that accumarray is the fastest approach here, I vote for it a sign of my admiration for all successful accumarray masters.
Wayne King
2013-6-16
编辑：Wayne King
2013-6-16
you mean just:
mean(a.*b)
or do you want the mean of just the nonzero entries? In other words, divide by the right number of elements.
C = a.*b;
for nn = 1:size(C,2)
numelements(nn) = nnz(C(:,nn));
end
colsumz = sum(C);
meanz = colsumz./numelements;
Azzi Abdelmalek
2013-6-16
c=a.*b
c(c==0)=nan;
out=nanmean(c)
2 个评论
Giorgos Papakonstantinou
2013-6-17
John Doe
2013-6-18
Just learned about nanmean. I've been looking for a function like that for quite some time. I've always used some workaround. Thanks!
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