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

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Teddy Fisher
Teddy Fisher 2019 年 12 月 13 日
コメント済み: sruizpp 2020 年 11 月 18 日
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!
  3 件のコメント
Teddy Fisher
Teddy Fisher 2019 年 12 月 13 日
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
Fangjun Jiang 2019 年 12 月 13 日
with this comments, the answer below should provide one solution.

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採用された回答

Fangjun Jiang
Fangjun Jiang 2019 年 12 月 13 日
編集済み: Fangjun Jiang 2019 年 12 月 13 日
C=[1,2,3]';
M=magic(5);
index=setdiff(1:numel(M),C);
M(index)=0
  2 件のコメント
Teddy Fisher
Teddy Fisher 2019 年 12 月 13 日
Thank you so much!! I can't tell you how much this helped
sruizpp
sruizpp 2020 年 11 月 18 日
c00L

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その他の回答 (1 件)

Guillaume
Guillaume 2019 年 12 月 13 日
編集済み: Guillaume 2019 年 12 月 13 日
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.

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