Simple but elusive array question

1 次查看（过去 30 天）
Will
Will 2012-4-17
Hi,
I have what must be quite a straightforward question, but I can't find a good solution.
I have a large number of arrays, of dimension 500x40. Each column of the arrays consists of a number of (non-repeating) integers between 1 and 40, with the remaining elements being zeros. An example of this on a smaller array would look like this:
1 1 3
2 0 4
3 0 0
0 0 0
What I want to do, is use this to create an array of the same dimensions, which consists of zeros and ones, such that for each column, the entries in that column are the indices of the "ones", and the remaining elements are zero. So, for the example above, I would get:
1 1 0
1 0 0
1 0 1
0 0 1
The catch is, I was trying to think of a nice way of doing it without using for loops, because I have a large number of arrays and want things to run quickly.
I'm sure there must be a very simple solution to this, but I can't think of it!
Any thoughts? Thanks, Will

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采纳的回答

Oleg Komarov
Oleg Komarov 2012-4-17
[r,c,v] = find(a);
out = accumarray([v,c],1,size(a));

更多回答（2 个）

Andrei Bobrov
Andrei Bobrov 2012-4-17
A = [1 1 3
2 0 4
3 0 0
0 0 0]
s1 = size(A);
out = zeros(size(A));
t = A~=0;
A1 = bsxfun(@times,t,1:s1(2));
out(sub2ind(s1,A(t),A1(t))) = 1
Richard
Richard 2012-4-17
should your question be like 'a' below?
clear all
a = [1,1,3;2,0,4;3,0,0;0,0,0];
b = arrayfun(@find,a,'un',0);
If so, use find and then just replace the empty cells with zeros.
  1 个评论
Will
Will 2012-4-17
No, the ones should correspond to the indices of the values. Oleg has posted a very nice answer which does this. Thanks for your help though!

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