Replace value with index in 2D array
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Hi I have a 2D array like this
A=[0 0 1; 1 0 1; 0 1 0]
I want to replace 1 in each row with column index value. e.g new matrix will be like this:
result=[0 0 3 ; 1 0 3 ; 0 2 0]
Thanks in advance
采纳的回答
Star Strider
2017-4-3
This works:
A=[0 0 1; 1 0 1; 0 1 0];
[~,CIV] = find(A); % ‘CIV’ = ‘Column Index Value’
A(A>0) = CIV
result = A
result =
0 0 3
1 0 3
0 2 0
5 个评论
Tha saliem
2017-4-3
dpb
2017-4-3
'Cuz (like mine)
A(address)=assignment
is assigning into the original A; we presumed the idea was to replace elements.
If want new and retain A, just
B(A==1)=j;
will create B using the same logical addressing expression for positions and since A==1 returns a logical array same size as A, B will be the same size as A.
Star Strider
2017-4-3
My pleasure.
The original matrix does not have to change. It has to be duplicated in the ‘result’ matrix if you do not want ‘A’ to change:
A=[0 0 1; 1 0 1; 0 1 0];
result = A;
[~,CIV] = find(result); % ‘CIV’ = ‘Column Index Value’
result(result>0) = CIV
Tha saliem
2017-4-3
Star Strider
2017-4-3
Our pleasure!
更多回答（2 个）
A version without FIND:
A = [0 0 1; 1 0 1; 0 1 0];
R = A .* (1:3); % Auto expanding in >= R2016b
In older Matlab versions:
R = bsxfun(@times, A, 1:3)
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