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

조회 수: 3 (최근 30일)
Tha saliem
Tha saliem 2017년 4월 11일
댓글: Tha saliem 2017년 4월 13일
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.
  댓글 수: 1
Tha saliem
Tha saliem 2017년 4월 11일
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]

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Guillaume
Guillaume 2017년 4월 11일
편집: Guillaume 2017년 4월 11일
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.
  댓글 수: 5
Guillaume
Guillaume 2017년 4월 13일
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
Tha saliem 2017년 4월 13일
yes its completely accurate. Thank you so much for help.

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