Using a column vector of indices to replace values in a matrix

Hello,

I have a column vector, C, (55x1 double) which are the indices of my values of interest in Matrix, M which is 370x29 double.

I want to replace all the values in M that are not these values of interest with 0.

I am trying to use C to index into M and replace ~C with 0.

However when I try

M(~C)=0

all it does is change all the values in M to just different values and I'm not sure why.

When I try to create a new Matrix

N=M(~C==0) it returns a columnn vector with just the first 55 values in the first column of M.

Does anyone know how I can replace all the elements in M that are not listed in C with 0? I want my output to be another 370x29 double matrix.

Thanks!

[Guillaume]

If C is a vector of indices, it's very unclear why you think that ~C would useful for anything ~0 is logical 1 and ~anythingelse is logical 0.

So is C a vector of indices (values came go from 1 to numel(M) == 10730) or is C something else? It would be useful if you told us how you obtained C.

[Teddy Fisher]

I had another Matrix, R, that was 0s and 1s, also 370x29 double

I used

C=find(R) to get the indexes of the 1s in R (which correspond to the values I want in M)

So the values in C are numbers (between 1-10730), not [row,column] format

Is there any way I can use C to index into M ?

[Fangjun Jiang]

with this comments, the answer below should provide one solution.

Answer 1

[Fangjun Jiang]

C=[1,2,3]';
M=magic(5);
index=setdiff(1:numel(M),C);
M(index)=0

[Teddy Fisher]

Thank you so much!! I can't tell you how much this helped

[sruizpp]

c00L

Answer 2

[Guillaume]

Well, now you have explained how you got C, a much simpler method than the accepted one is not to create C and use:

M(~R) = 0;  %that's all that is needed!

As usual in matlab, find is completely unnecessary and what you were trying to do initially would have worked had you kept your original logical vector.

[Fangjun Jiang]

indeed!