Irregular intervals inside a for loop

Hello, I'm trying to work out how to do two actions (lets call them A and B), where action A is repeated n amount of times after which action B is repeated m amount of times, with n and m being integers.

this all takes place in a for loop with say, index p.

Essentially what I want as my output when n = 2 and m = 3 is:

Action A

Action A

Action B

Action B

Action B

Action A

Action A

...etc.

I tried doing the following:

n=5
for  p = 1:100
    q = floor(p/n);
    if rem(q,2) == 0
       disp('Action A')
    else
       disp('Action B')
    end
end

But ofcourse, this only works for regular intervals (I'm quite a nooby in MATLAB so excuse me if this is a weird way to approach this)

Answer 1

[Voss]

n = 2;
m = 3;
z = n+m;
for p = 1:20
    if ismember(mod(p,z),1:n)
        disp('Action A');
    else
        disp('Action B');
    end
end

Action A
Action A

Action B
Action B
Action B

Action A
Action A

Action B
Action B
Action B

Action A
Action A

Action B
Action B
Action B

Action A
Action A

Action B
Action B
Action B

[Kerr Beeldens]

Appriciate your answer, it works

[Voss]

You're welcome!

Be aware that there is a functional difference between this answer and the other answer, which is how many times any action (Action A or Action B) will occur on each iteration of the "p" loop. In the method used in this answer, one action occurs each time; in the method used in the other answer, n+m actions occur each time.

[Kerr Beeldens]

Thanks for your words of caution. This was exactly what I was refering to in my other comment, where I mentioned that the use of for loops inside the for loop may be undesirable in my code. I will play around with both to see which method I favour :)

[Voss]

Whatever works works! I just wanted you to be aware of the difference.

Answer 2

[David Hill]

for p=1:100
    for a=1:n
      %Action A  
    end
    for b=1:m
      %Action B 
    end
end

[Kerr Beeldens]

I'm not entirely sure if putting these actions into for loops works nicely in my case, but thanks for you answer, it works for now :)

[Image Analyst]

@Kerr Beeldens it does work nicely. And it does what you said in a simpler and more straightforward method than messing with floor() and rem() like you tried to do.

[Kerr Beeldens]

Thanks for the reply, I figured out how to use this in my code. Both suggested methods work well, I prefer this one for visual simplicity.

Answer 3

[John D'Errico]

There are multiple ways to do this. For example:

P = primes(20); % Note that P is a ROW vector
P

P = 1×8

2 3 5 7 11 13 17 19

SumP = 0;
for P_i = P
  SumP = SumP + P_i;
end
SumP

SumP = 77

Do you see there is effectively no index at all? The loop variable is the corresponding element of P.

Or, I might have done this:

Q = (1:5).^2

Q = 1×5

1 4 9 16 25

ProdQ = 1;
for ind = 1:numel(Q)
  Q_ind = Q(ind);
  ProdQ = ProdQ*Q_ind;
end
ProdQ

ProdQ = 14400

In the latter case, I used a uniform index in the form of ind, but then immediately index into the non-uniform vector Q.

Next, we can have a loop on characters.

S = 'The Quick Brown Fox jumped over the lazy dog';
words = split(S).'

words = 1×9 cell array
    {'The'}    {'Quick'}    {'Brown'}    {'Fox'}    {'jumped'}    {'over'}    {'the'}    {'lazy'}    {'dog'}

Again, note that words is again a ROW vector. Now the for loop can access the elements of this row vector.

for W = words
  disp(W{:})
end

The
Quick
Brown
Fox
jumped
over
the
lazy
dog

Be careful, as that trick fails when the vector is a column vector.