Filter löschen
Filter löschen

How to convert rows to logical

2 Ansichten (letzte 30 Tage)
Simon Allosserie
Simon Allosserie am 27 Okt. 2022
Kommentiert: Simon Allosserie am 10 Nov. 2022
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.

Akzeptierte Antwort

Star Strider
Star Strider am 27 Okt. 2022
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 Kommentare
Cris LaPierre
Cris LaPierre am 28 Okt. 2022
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 am 10 Nov. 2022
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!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Cris LaPierre
Cris LaPierre am 27 Okt. 2022
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 Kommentar
Simon Allosserie
Simon Allosserie am 27 Okt. 2022
Yes this is indeed much more simple, I was making it way too complicated. Thanks for helping me out of my thinking rut!

Melden Sie sich an, um zu kommentieren.

Produkte


Version

R2019b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by