compare cell array elements with non zero elements in 2D array

hey i have cell array containing multiple elements e.g.

    x{1,1}={2;1}
    x{2,1}={1;[1;2]}

and a 2D array:

   y=[0,0,-1,1,0,0;-1,0,1,-1,0,0]

for x{1,1}{1,1}=2, it will consider 1st non zero column value (e.g. 3 in 1st row is first non zero column index ) in 1st row of y, and in 2nd row(As x{1,1}{1,1}=2), it will check what value is placed at that non zero index and store it in another cell array. for x{1,1}{2,1} it will consider 2nd non zero element of 1st row.

for x{2,1}{1,1}=1, it will consider 2nd row and check which 1st non zero element.

[Tha saliem]

Result would be like this:

   result{1,1}{1,1}=1
  result{1,1}{2,1}=1
  result{2,1}{1,1}=0
  result{2,1}{2,1}=[-1;1]

Answer 1

[Guillaume]

I assume you've made a mistake with your example result. How can you have 0 in the outptu if you only look at the non-zero values. Also, the first non-zero value in row 1 is -1, so I assume that result{1}{2} should be -1.

If so:

result = cell(size(x));
for row = 1:size(x, 1)
  ynonzeros = nonzeros(y(row, :));
  result{row} = cellfun(@(cols) ynonzeros(cols), x{row}, 'UniformOutput', false);
end

Because of the need of the temporary ynonzeros it's not possible to replace the outer loop with an arrayfun with an anonymous function, as I had in my previous answer.

[Tha saliem]

hey. Thankyou for your help. sorry for not clarifying my question Actually what i want to do is that for example i have y like this:

and elements of x{1,1} are:

    x{1,1}{1,1}=4 
    x{1,1}{2,1}=4

it will take first element of x{1,1} which is 4 and this 4 will indicate the row to be used in y matrix. and also as it is first element of x{1,1} it will consider 1st non zero element of 1st row in y. It means value at column 3 of row 1 will be considered. Now it will check value which is present at 3rd column of 4th row and will display result.

for x{1,1}{2,1}=4, it will consider 2nd non zero element of 1st row in y. and will display value present at 4th column of 4th row.

[Tha saliem]

Similarly if x{2,1}{1,1}=[3,4]

it will consider first non-zero element of 2nd row and which is at column 1, it will then display value present at 1st column of 3rd row and 1st column of 4th row.

[Tha saliem]

your code gives following error im unable to find out whats wrong nothing seems erroneous.

    Index exceeds matrix dimensions.
Error in test>@(cols)ynonzeros(cols)
Error in test (line 61) result{row} = cellfun(@(cols) ynonzeros(cols), x{row}, 'UniformOutput', false);% code

[Guillaume]

result = cell(size(x));
for row = 1:size(x, 1)
   ycols = find(y(row, :));
   result{row} = arrayfun(@(r) y(x{row}{r}, ycols(r)), ....
                          (1:numel(x{row})).', ...
                          'UniformOutput', false);
end

[Tha saliem]

yes its completely accurate. Thank you so much for help.