Suppose I have a matrix A and I sort the rows of this matrix. How do I replicate the same ordering on a matrix B (same size of course)?
E.g.
A = rand(3,4);
[val ind] = sort(A,2);
B = rand(3,4);
%// Reorder the elements of B according to the reordering of A
This is the best I've come up with
m = size(A,1);
B = B(bsxfun(@plus,(ind-1)*m,(1:m)'));
Out of curiosity, any alternatives?
Update: Jonas' excellent solution profiled on 2008a (XP):
n = n
0.048524 1.4632 1.4791 1.195 1.0662 1.108 1.0082 0.96335 0.93155 0.90532 0.88976
n = 2m
0.63202 1.3029 1.1112 1.0501 0.94703 0.92847 0.90411 0.8849 0.8667 0.92098 0.85569
It just goes to show that loops aren't anathema to MATLAB programmers anymore thanks to JITA (perhaps).
Answer
A somewhat clearer way to do this is to use a loop
A = rand(3,4);
B = rand(3,4);
[sortedA,ind] = sort(A,2);
for r = 1:size(A,1)
B(r,:) = B(r,ind(r,:));
end
Interestingly, the loop version is faster for small (<12 rows) and large (>~700 rows) square arrays (r2010a, OS X). The more columns there are relative to rows, the better the loop performs.
Here's the code I quickly hacked up for testing:
siz = 10:100:1010;
tt = zeros(100,2,length(siz));
for s = siz
for k = 1:100
A = rand(s,1*s);
B = rand(s,1*s);
[sortedA,ind] = sort(A,2);
tic;
for r = 1:size(A,1)
B(r,:) = B(r,ind(r,:));
end,tt(k,1,s==siz) = toc;
tic;
m = size(A,1);
B = B(bsxfun(@plus,(ind-1)*m,(1:m).'));
tt(k,2,s==siz) = toc;
end
end
m = squeeze(mean(tt,1));
m(1,:)./m(2,:)
For square arrays
ans =
0.7149 2.1508 1.2203 1.4684 1.2339 1.1855 1.0212 1.0201 0.8770 0.8584 0.8405
For twice as many columns as there are rows (same number of rows)
ans =
0.8431 1.2874 1.3550 1.1311 0.9979 0.9921 0.8263 0.7697 0.6856 0.7004 0.7314