如何使用逻辑索引和切片的组合在Matlab中“切分”矩阵？

Question

I have a matrix M that looks similar to this:

M = [   1, 2, 3, 0, 0;
        1, 2, 0, 0, 0;
        2, 3, 4, 5, 0;
        4, 5, 6, 0, 0;
        1, 2, 3, 4, 5;
    ]

I'm trying to get a column vector with the rightmost non-zero value of each row in A, but ONLY for the rows that have the first column == 1.

I'm able to calculate a filter for the rows:

r = M( :, 1 ) == 1;
> r = [ 1; 1; 0; 0; 1 ]

And I have a set of indices for "the rightmost non-zero value of each row in M":

> c = [ 3, 2, 4, 3, 5 ]

How do I combine these in a slicing of A in order to get what I'm looking for? I'm looking for something like:

A( r, c )
> ans = [ 3; 2; 5 ]

But doing this gets me a 3x3 matrix, for some reason.

Answer 1

The shortest way I can think of is as follows:

% Get the values of the last non-zero entry per row
v = M(sub2ind(size(M), 1:size(M,1), c))

% Filter out the rows that does not begin with 1.
v(r == 1)

Answer 2

This seems to work (I assume other operations defining r,c have been performed):

M(sub2ind(size(A),find(r==1).',c(r==1))).'

Short interpretation of the problem and solution:

M( r, c )

gives a 3 x 5 matrix (not 3 x 1 as desired) due to mixing of logical and subscript indices. The logical indices in r pick out rows in A with r==1 . Meanwhile row array c picks out elements from each row according to the numeric index:

ans =

     3     2     0     3     0
     0     2     0     0     0
     3     2     4     3     5

What you really want are indices into the rightmost nonzero elements in each row starting with 1 . The solution uses linear indices (numeric) to get the correct elements from the matrix.

Answer 3

I think this should do the trick. I wonder if there is more elegant way of doing this though.

% get only rows u want, i.e. with first row == 1
M2 = M(r,:);

% get indices of
% "the rightmost non-zero value of each row in M" 
% for the rows u want 
indicesOfinterest = c(r==1);


noOfIndeciesOfinterest = numel(indicesOfinterest);

% desired output column vector
output = zeros(noOfIndeciesOfinterest, 1);

% iterate through the indeces and select element in M2
% from each row and column indicated by the indice.
for idx = 1:noOfIndeciesOfinterest
    output(idx) = M2(idx, indicesOfinterest(idx));
end

output % it is [3; 2 ; 5]

Answer 4

You can use

arrayfun(@(x) M(x,c(x)), find(r))

But unless you need r and c for other purposes, you can use

arrayfun(@(x) M(x,find(M(x,:),1,'last')), find(M(:,1)==1))

Answer 5

Here is a way to do it using linear indexing:

N = M';
lin_index = (0:size(N,1):prod(size(N))-1) + c;
v = N(lin_index);
v(r)