Matrix and find
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Hello actually i have a random matrix done of a lot of 0 and some 1. i need to save the position (i,j) of the first 1 for each row/column. If i have a matrix like this one
0 1 0 0 0 1 0 0
0 0 1 0 0 0 0 0
0 0 1 0 0 0 0 0
doing a(i)=find(matrix(i,:),1) i got a=[2,3,3], and this is ok.
Instead if i have this matrix:
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0
doing the same thing with the find i got error.
Improper assignment with rectangular empty matrix.
why?
1 件のコメント
Image Analyst
2012 年 1 月 15 日
What do you want or expect to get when you have no 1's in the row? Maybe -1 or something? It's got to be something.
採用された回答
Jan
2012 年 1 月 15 日
You have to decide, what you want as output, if a row does not contain any 1.
n = size(matrix, 1);
a = zeros(1, n); % Pre-allocate!
for i = 1:n
found = find(matrix(i,:), 1);
if isempty(found)
a(i) = NaN; % Or Inf, or -1, or 0, or whatever?!
else
a(i) = found;
end
end
[EDITED]: You can define the default value in the pre-allocation to make the code 2 lines shorter:
n = size(matrix, 1);
a = nan(1, n); % Pre-allocate!
for i = 1:n
found = find(matrix(i,:), 1);
if ~isempty(found)
a(i) = found;
end
end
その他の回答 (2 件)
the cyclist
2012 年 1 月 15 日
For the first matrix, when i=1, "find(matrix(1,:),1)" looks at the first row of the matrix, and looks for the first column that has a non-zero in it. That's column 2, so you are doing a(1)=2, which is fine.
For the second matrix, when i=1, the same command returns the empty matrix, because there is no non-zero in the first row. Therefore, you are trying to do a(1)=[], which gives the error you see.
Edit in response to your comment:
Here's a different way from Jan's, in which only the rows that actually have non-zeros are displayed:
x = [0 1 0 0 0 1 0 0 ...
0 0 1 0 0 0 0 0 ...
0 0 1 0 0 0 0 0];
[m,n]=find(x);
[rowsWithNonZero,indexToFindColumn] = unique(m,'first');
columnsOfFirstNonZero = n(indexToFindColumn);
disp('row:')
disp(rowsWithNonZero)
disp('column:')
disp(columnsOfFirstNonZero)
0 件のコメント
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