How to find and replace stand alone values in a logical array
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Teddy Fisher
le 19 Déc 2019
Commenté : Teddy Fisher
le 26 Fév 2020
Hello,
I have a logical aray, a 370x29 matrix. I am trying to find all 1s and 0s that are non-consecuitve and replace them with the other one, in each column. So in a series: 00000100000 I want to be able to find and replace that stand alone 1 with a 0, and in a series: 1111101111 I want to find and replace that 0 with a 1. So far, I have been able to find and replace these with this script, where a is my 370x29 matrix:
x=find(a(diff(a)==1),1,1);
y=find(a(diff(a)==-1),1,1);
a(x)=0;
a(y)=1
The only problem with this is that it also finds and replaces the first 1 in a string of 1s and the first 0 in a string of 0s.
Is there any way to find and replace these stand alone 1s and 0s?
Thanks :)
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the cyclist
le 19 Déc 2019
I think this does what you want:
a(strfind([1 a 1],[1 0 1])) = 1;
a(strfind([0 a 0],[0 1 0])) = 0;
3 commentaires
the cyclist
le 19 Déc 2019
Sorry I was not clear. My code is applied to each row. You'd need to use a loop.
I also see now that you wanted to apply to each column, not row, which makes my code slightly awkward because strfind doesn't handle column data well. However, the following should work:
% Make up an input matrix
v = [0 1 1 0 1 1 1 0 1 1 0]';
a = [v 1-v v];
for nc = 1:size(a,2)
a(strfind([1 a(:,nc)' 1],[1 0 1]),nc) = 1;
a(strfind([0 a(:,nc)' 0],[0 1 0]),nc) = 0;
end
Use your input matrix a, instead of the data I made up.
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