How do I resolve these two lines separately?

So I have matrix A that is 3x5 and a matrix B with random values and same dimensions as matrix A.

I want to find the elements in matrix B where the generated random number is lower than 0.6 and then change the coresponding elements in matrix A from 0 to 1 or from 1 to 0. Is there a way to do this without going into a for loop?

B=rand(3,5)

A=[0 0 0 1 0;1 1 1 0 0;1 0 1 1 0]

A(B<0.6 & A==0)=1

A(B<0.6 & A==1)=0

When I run this code it does what it's supposed to but the last line takes the newly-formed ones and then turnes them into zeros as well (which is not what i want).

[Jon]

Can you please clarify what you are trying to do. What is the role of the original values of A in this. Maybe give a small example.

[Sandi Homolak]

it is an example of mutation in a genetic algorithm where every element has 60% chance to mutate

Answer 1

[Sandi Homolak]

Solved it.

B=rand(3,5)

A=[0 0 0 1 0;1 1 1 0 0;1 0 1 1 0]

T=find(B<0.6 & A==1)

Z=find(B<0.6 & A==0)

A(T)=0;

A(Z)=1;

Answer 2

[Les Beckham]

Another approach

B=rand(3,5)

B = 3×5

0.2205 0.2334 0.8353 0.4513 0.3142 0.6426 0.4411 0.6683 0.8477 0.4905 0.0294 0.5650 0.2777 0.6241 0.4927

A=[0 0 0 1 0;1 1 1 0 0;1 0 1 1 0]

A = 3×5

0 0 0 1 0 1 1 1 0 0 1 0 1 1 0

idx = B > 0.6

idx = 3×5 logical array
   0   0   1   0   0
   1   0   1   1   0
   0   0   0   1   0

A(idx) = ~A(idx)

A = 3×5

0 0 1 1 0 0 1 0 1 0 1 0 1 0 0

Answer 3

[Jon]

It looks like you may have already answered your own question, but I think this is a little cleaner approach to do the same thing

B=rand(3,5)
A=[0 0 0 1 0;1 1 1 0 0;1 0 1 1 0]
Aold = A;
A(B<0.6 & Aold==1) = 0;
A(B<0.6 & Aold==0) = 1;

[Jon]

Actually you can do it in one line

A = double((B < 0.6 & ~A) | (B > 0.6 & A))

I turn the result into a double otherwise you would have a logical array rather than an array of ones and zeros. Not sure if that matters for your application

Answer 4

[Tala]

try this

B=rand(3,5);
A=[0 0 0 1 0;1 1 1 0 0;1 0 1 1 0];

B1=B>0.6;
A(B1)=1;

[Sandi Homolak]

This only changes the zeros in A into ones, but not the ones in A into zeros. I need both to happen.