How can I find one index for each row of a 2d matrix where a condition is met?

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funkadelala
funkadelala 2015-2-13
评论： Geoff Hayes 2015-2-21
A is an n by n matrix. B is an n by 1 matrix.
I would like to find the last instance of an element in each row of A being greater than the element in the corresponding row of B, and return an n by 1 list of indices.
For example:
A = [1 3 2;
2 5 -1;
0 2 3]
B = [1;
0;
2]
idx = [3;
2;
3]
I'm sure I could do this with a for loop, but would like to avoid it as the matrices are potentially very large.
Many thanks in advance.

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回答（1 个）

Geoff Hayes
Geoff Hayes 2015-2-13
Funkadelala - try using the following which assumes that A and B have been defined as above
cell2mat(arrayfun(@(k)find(A(k,:)>B(k),1,'last'),1:size(A,1),'UniformOutput',false)')
We use arrayfun to apply the anonymous function
@(k)find(A(k,:)>B(k),1,'last')
to the kth row of A, using find to find the last index which is greater than the kth element of B. The result from arrayfun is a cell array (row), so we transpose it and convert to a matrix using cell2mat to produce the desired result of
ans =
3
2
3
Try the above and see what happens!
  2 个评论
Geoff Hayes
Geoff Hayes 2015-2-21
funkadelala's answer moved here
Thanks for the response, and sorry for the delay (I've been trying to sort out errors myself).
The above suggestion returns the following error:
Error using cat
Dimensions of matrices being concatenated are not consistent.
Error in cell2mat (line 83)
m{n} = cat(1,c{:,n});
Error in untitled (line 60)
ans = cell2mat(arrayfun(@(k)find(A(k,:)>B(k),1,'last'),1:size(A,1),'UniformOutput',false)');
And simply using:
ans = arrayfun(@(k)find(A(k,:)>B(k,:),1,'last'),A);
Returns the error:
Attempted to access A(-0,:); index must be a positive integer or logical.
Error in @(k)find(A(k,:)>B(k,:),1,'last')
Error in untitled (line 61)
zindex = arrayfun(@(k)find(A(k,:)>B(k,:),1,'last'),A)';
Any ideas?
Geoff Hayes
Geoff Hayes 2015-2-21
funkadelala - are you using a different A and B than the ones described from above, and if so, what are they? Or, what is the output of just
arrayfun(@(k)find(A(k,:)>B(k),1,'last'),1:size(A,1),'UniformOutput',false)

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