# How to find and replace stand alone values in a logical array

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Teddy Fisher on 19 Dec 2019
Commented: Teddy Fisher on 26 Feb 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 :)

the cyclist on 19 Dec 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;

Teddy Fisher on 19 Dec 2019
this hasn't worked. when I try as you have written, i get an error message:
"error using horzcat
dimensions of matrices being concatenated are not consistent."
which i assume is because my input is a matrix. however, when i try to fix this with:
x(strfind([1 1:numel(x) 1],[1 0 1]))=1
OR
x(strfind(1:numel(x),[1 0 1]))=1
nothing happens
the cyclist on 19 Dec 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
Teddy Fisher on 26 Feb 2020
Thanks! that totally works. i dont know how or why, but it does exactly what i was trying to