How to replace vector values without loop for?

Hi everyone, I have a vector x(1:10000,1) which elements are all 0. I want to replace 0 with 1 in the case a statement is satified. I used the following loop for to do that:

for i = 1:10000
    if y(i,:) >= z(i,:)
        x(i,:) = 1;
    end
end

The code works properly but I would like to optimize the script process. Is there some way to do the same thing without using a loop for? Thanks for help.

Answer 1

[per isakson]

Hint:

    x( y(:,1)>=z(:,1), 1 ) = 1;

However, I didn't say that this is faster than the loop

[Joseph Cheng]

well to test our your piece here is a quick test

clc
x = zeros(10000,1);
y = rand(size(x));
z = rand(size(x));
tic,
for i = 1:10000
    if y(i,:) >= z(i,:)
        x(i,:) = 1;
    end
end
disp(['loop version time: ' num2str(toc)]) 

x1 = zeros(10000,1);
tic,
 x1 = y>=z;
disp(['compare replace all version time: ' num2str(toc)]) 

x2 = zeros(10000,1);
tic
 x2( y(:,1)>=z(:,1), 1 ) = 1;
disp(['replace only true version time: ' num2str(toc)])

which on my machine gets me

loop version time: 0.0088747
compare replace all version time: 0.00061254
replace only true version time: 0.00018797

Which does seem intuitive as replacing only the ones that need to be replaced is faster than replacing everything.