How can I keep just the arrays with 2 is before 3 after random swap 2 numbers.

with: s=[1,2,3,4,5];

[madhan ravi]

Show how your output should look like.

[Hang Vu]

Thanks for response! the output should be s=[1,4,2,3,5] or [2,4,1,5,3] for example. just to make sure 2 before 3

Answer 1

[Jan]

s = [1,2,3,4,5];

s = s(randperm(numel(s)));     % Random permutation?

% Or swap 2 elements:
index = randperm(numel(s), 2);
s(index) = s(flip(index));

s(s == 2 | s == 3) = [2, 3];   % Re-order the elements 2 and 3

[madhan ravi]

Jan I think you mean

s(s==2|s==3)...

Sorry for editing the comment so many times , it was because I was not able to type the short circuit operator in my mobile phone , I have no idea why.

[Jan]

@madhan: you are right, not || but | is required.

[Hang Vu]

Thank you for the re order idea! I applied for this:

s=[1,2,3,4,5];
L=size(s,2);
a=randi([1 L-1],1);
b=randi([1 L-1],1);
   
temp=s(a);
s(a)=s(b);
s(b)=temp
for i=1:5
    ind1=find(s(:,:)==2);
    ind2=find(s(:,:)==3);
end
if ind1>ind2
    s(s == 2 | s == 3) = [2, 3];
    s
else
    s
end

[Jan]

This is not useful:

for i=1:5
    ind1=find(s(:,:)==2);
    ind2=find(s(:,:)==3);
end

The body of the loop is calculated 5 times with identical values. You can omit the loop:

ind1=find(s(:,:)==2);
ind2=find(s(:,:)==3);

But you can omit this test completely. This is sufficient for all cases:

s(s == 2 | s == 3) = [2, 3];

Why do you use L-1 in randi([1 L-1],1) ? I assume you want L instead.

Answer 2

[Raj]

Try this:

s=[1,2,3,4,5]
%First random swap
x=randi([1,4],1,1);
if s(x)~=2
   s([x x+1])= s([x+1 x]);
else
    s([x x-1])=s([x-1 x]);
end
disp('After first random swap s=')
disp(s);
%second random swap
y=randi([1,4],1,1);
if s(y)~=2
   s([y y+1])= s([y+1 y]);
else
    %Do nothing
end
disp('After second random swap s=')
disp(s);

There may be better and optimized way of doing this also but this also works!

[Hang Vu]

Thank you so much for your time!^^