how to efficiently find the indices?

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Pinpress
Pinpress 2012-5-2
Hi,
I have a matrix of N-by-M with integers. I need to efficiently find the indices for all of the unique elements in the matrix. The solution I have is via a "for" loop:
uM = unique (M(:));
for i = 1 : length(uM)
I(i) = find(M == uM(i));
end
This works fine, but with a large matrix, this is slow. I wonder if there are better solutions. thanks very much!
  1 个评论
Geoff
Geoff 2012-5-3
One way to make this code faster without changing anything fundamental would be to preallocate the cell array before your loop:
I = cell(length(uM),1);

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采纳的回答

Richard Brown
Richard Brown 2012-5-2
sort is your friend:
First we sort the entries of A so that like entries are adjacent
[S, iSorted] = sort(A(:));
Then we find the markers - where the value changes, plus the endpoints
markers = [0; find(diff(S)); numel(A)];
All we need to do is find the entries of iSorted corresponding to each block.
for i = 1:numel(markers)-1
uM(i) = S(markers(i)+1);
I{i} = iSorted(markers(i)+1:markers(i+1));
end
  2 个评论
Pinpress
Pinpress 2012-5-3
Great! This turns out to be working fine for me. It's still a "for" loop, but the computational efficiency is OK.
Richard Brown
Richard Brown 2012-5-3
for loops get a bit of an unnecessarily bad rap sometimes

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更多回答（6 个）

Leah
Leah 2012-5-2
Matlab has a nice function built in for this
[B,I,J] = UNIQUE(...) also returns index vectors I and J such
that B = A(I) and A = B(J) (or B = A(I,:) and A = B(J,:)).
Andrei Bobrov
Andrei Bobrov 2012-5-3
[uM,n,n] = unique (M(:));
I = accumarray(n,1:numel(n),[],@(x){sort(x)});
  2 个评论
Richard Brown
Richard Brown 2012-5-3
I think even @(x) {x} would meet the brief for your function

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Richard Brown
Richard Brown 2012-5-2
Depending on whether you want the first or the last occurence
[uM, I] = unique(M, 'first')
[uM, I] = unique(M)
Walter Roberson
Walter Roberson 2012-5-2
Your suggested code will not work if there are any duplicates, as the find() would return multiple values in that case and multiple values cannot be stored into a single numeric array element.
Have you considered using the second or third return value from unique() ?
Pinpress
Pinpress 2012-5-2
Thanks guys -- forgot to mention that I need to find the indices for all of the elements if there are multiple occurrences. So simply using "unique" doesn't seem to work.
And yes, the original code I had will not work when more than one indices returns (which was precisely the reason I did not use [A, I, J] = unique (B);].
Any other thoughts?
  2 个评论
Oleg Komarov
Oleg Komarov 2012-5-2
A and J will give you what you want.
Richard Brown
Richard Brown 2012-5-2
Not really - J is just A(:), but with the unique elements replaced with 1:nUnique. So it's no better

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Geoff
Geoff 2012-5-2
This comes straight out of some of my own code... I guess it's the reverse of what you want though.
uM = unique(M);
I = arrayfun(@(x) find(uM==x,1), M);
For every element in M, it gives an index into uM. I use this to reduce columns of data in a matrix that are common to multiple targets.
So you seem to want: for every element in uM an array of indices into M. The result of course would be a cell array.
This would be:
I = arrayfun(@(x) find(M==x), uM, 'UniformOutput', false);
  2 个评论
Richard Brown
Richard Brown 2012-5-2
I think that's what he was trying before, but found the repeated calls to find to be too slow
Pinpress
Pinpress 2012-5-3
Calling "find" thousands of times will be slow.

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