How to convert rows to logical

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Simon Allosserie
Simon Allosserie 2022-10-27
评论： Simon Allosserie 2022-11-10
Ran in:
I am filtering a set of values in a column vector in multiple steps. In the end, I get the row numbers of the values that should be kept, eg.:
A = 1:10;
%fitering happens
r = [2 7 8]; %the rows to keep
Now for subsequent calculations, I need A with only the filtered values, and 0's in the other values. At the moment I do it like this:
filter = ~min(logical(abs(A-r(:))),[],1)
filter = 1×10 logical array
0 1 0 0 0 0 1 1 0 0
Afiltered = A.*filter
Afiltered = 1×10
0 2 0 0 0 0 7 8 0 0
However this method isn't that elegant so I was wondering if there are some standard functions / quicker ways to get the same?
Due to the nature of the filters, it is not possible to work with masks that have the same length as A. r will always be shorter than A.

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采纳的回答

Star Strider
Star Strider 2022-10-27
Ran in:
Why not just:
A = 1:10;
%fitering happens
r = [2 7 8]; %the rows to keep
Afiltered = zeros(size(A));
Afiltered(r) = A(r)
Afiltered = 1×10
0 2 0 0 0 0 7 8 0 0
.
  2 个评论
Cris LaPierre
Cris LaPierre 2022-10-28
This is likely the better answer. Here, Afiltered is assigned the values in A. In my answer, it is assigned the row numbers. It only looks correct because, in this simplified example, the elements in A are the row numbers.
Simon Allosserie
Simon Allosserie 2022-11-10
Okay I will look into the differences between the two approaches and see which renders the results the fastest for the data that I use. Thanks for coming back to this!

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更多回答（1 个）

Cris LaPierre
Cris LaPierre 2022-10-27
Ran in:
I'm not aware of a special function for this, but perhaps a more readable way would be this?
A = 1:10;
r = [2 7 8];
Afiltered=zeros(1,length(A));
Afiltered(r)=r
Afiltered = 1×10
0 2 0 0 0 0 7 8 0 0
  1 个评论
Simon Allosserie
Simon Allosserie 2022-10-27
Yes this is indeed much more simple, I was making it way too complicated. Thanks for helping me out of my thinking rut!

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