Replace value with index in 2D array

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

Answer 1

[Star Strider]

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

[Tha saliem]

Thankyou so much. but why original matrix changes in output?

[dpb]

'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]

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]

Yes got it. @dpb & @Star Strider Thank you so much for solution it really helped.

[Star Strider]

Our pleasure!

Answer 2

[Jan]

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)

[Stephen23]

+1 nice.

[Tha saliem]

Good one. Thanks alot @Jan Simon

Answer 3

[dpb]

>> [~,j]=find(A);
>> A(A==1)=j
A =
   0     0     3
   1     0     3
   0     2     0
>>

[Tha saliem]

Thanks Alot.