## How to change 0 to 1 and update a matrix

### Antonio Grivas (view profile)

on 21 Jun 2019
Latest activity Answered by Antonio Grivas

on 22 Jun 2019

### Andrei Bobrov (view profile)

clc
clear all
close all
b1 = [10; 20 ; 30 ; 40];
A = [0 20 0 10 ; 20 0 30 0 ; 0 30 0 10 ; 20 0 10 0];
s = size(b1);
x = randi([0 1],s)
G = [];
[ n m ] = size(x);
for i = 1 : n
for j = 1 : m
G(:,i) = x(:,j)
a = find(G(: , i) == 0 , 1 , 'first')
[x y] = size(G(:,i))
for k = 1:y
G(: , k) = 0;
G(a , k) = 1
end
end
i = i + 1;
j = j + 1;
end
This prob wil work only the first time. I want the matrix to take a column, find where the first 0 is and change it to 1. Then proceed to the next line and do the same but keep the first column and not change it to 0s. Final solution must look like
x =
0
1
0
0
(x changes randomly with 0 and 1 )
G =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
I would appreciate any answer :)

Antonio Grivas

### Antonio Grivas (view profile)

on 21 Jun 2019
I can do it in a 4x4 matrix by copying the same code 4 times but what if a matrix is 500x500 !!! final matrix G is always squared.
Stephen Cobeldick

### Stephen Cobeldick (view profile)

on 21 Jun 2019
@Antonio Grivas: Whenever I run your code it always returns this:
G =
0 4 4 4
0 4 4 4
0 4 4 4
0 4 4 4
It is unclear how this output relates to your examples.
Please explain the logic of what you are trying to achieve.
Antonio Grivas

### Antonio Grivas (view profile)

on 21 Jun 2019
It is completely wrong. I don't know how to make it change the final result whick would be
1
0
0
0
1 0
0 1
0 0
0 0
1 0 0
0 1 0
0 0 1
0 0 0
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
I send the (without the wrong loop) answer where the program runs correctly only the first time
I suppose that i need a
for i = 1:size(whatever)
whatever(: , i) = 0
whatever(: , a) = 1
end
that sould be put somewhere to keep the previous results and not change it back to
0
0
0
0
and change the position of the 1 after adding a column
If you run it just check only the 1st time before it changes it back to
0
0
1
0
becomes
0 1
0 0
1 0
0 0
Then for some reason that i cannot see does not continue but gives
0 4
0 4
0 4
0 4
I am trying to be precise if i can to help you
Thank you

### Andrei Bobrov (view profile)

on 21 Jun 2019
Edited by Andrei Bobrov

### Andrei Bobrov (view profile)

on 22 Jun 2019

n = numel(x1);
y1 = ~x1;
% if MATLAB >= R2016b
k = y1.*eye(n);
% if MATLAB <= R2016a
k = bsxfun(@times,y1,eye(n));
out = [x1, zeros(n,n-1)];
p = find(y1);
out(:,2:numel(p)+1) = k(:,p);

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Antonio Grivas

### Antonio Grivas (view profile)

on 21 Jun 2019
I will run this program as soon as i finish from work and go home.
Antonio Grivas

### Antonio Grivas (view profile)

on 21 Jun 2019
Problem with matrix dimensions at k = y1.*eye(n);
Andrei Bobrov

### Andrei Bobrov (view profile)

on 22 Jun 2019
You use old MATLAB.
Replace this expression with the following:
k = bsxfun(@times,y1,eye(n));