找到给定矩阵的每一行中最后一个非零元素的索引？

Question

For an arbitrary sized matrix x , how do I find the index of the last non-zero element in each row of a given matrix?

For example, for the matrix

x = [ 0 9 7 0 0 0; 5 0 0 6 0 3; 0 0 0 0 0 0; 8 0 4 2 1 0 ]

the vector [ 3 6 0 5 ] should be obtained.

Answer 1

Here's a shorter version, combining find and accumarray

x = [ 0 9 7 0 0 0; 5 0 0 6 0 3; 0 0 0 0 0 0; 8 0 4 2 1 0 ];
%# get the row and column indices for x
[rowIdx,colIdx] = find(x);
%# with accumarray take the maximum column index for every row
v = accumarray(rowIdx,colIdx,[],@max)'
v =
     3   6   0   5

Answer 2

Here's one version:

x = [ 0 9 7 0 0 0; 5 0 0 6 0 3; 0 0 0 0 0 0; 8 0 4 2 1 0 ];
c = arrayfun(@(k) find(x(k,:)~=0,1,'last'), 1:size(x,1), 'UniformOutput',false);
c( cellfun(@isempty,c) ) = {0};
v = cell2mat(c);

v =
     3     6     0     5

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EDIT : Consider this alternative solution:

[m,v] = max( cumsum(x'~=0) );
v(m==0) = 0;

v =
     3     6     0     5

Answer 3

One-line solution with bsxfun :

result = max(bsxfun(@times, x~=0, 1:size(x,2)).');

Or use the two outputs of max :

[val, result] = max(fliplr(x~=0).',[],1); %'
result = (size(A,2)+1-result).*val;

Answer 4

My answer's a bit twisted but it should work too

x = [ 0 9 7 0 0 0; 5 0 0 6 0 3; 0 0 0 0 0 0; 8 0 4 2 1 0 ];
[~,pos] = max([fliplr(x~=0),ones(size(x,1))],[],2);
v = size(x,2)-pos' +1;