How to make a series of consecutive numbers using a nested for loop?

Hi, I would like a generic expression to obtain an array of consecutive numbers until reaching the value of the product between the limits of two nested loops. I hope the example below will clarify the question. If I have the nested loop:

for i = 1:3
    for j = 1:5
       a = ????;
       disp(a)
    end
end

Which is the expression for 'a' that permits me to obtain a column array from 1 to 15?

a=
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

Maybe there are easier way to obtain this but I must use a nested loop and the last number must be equal to the product of the two limits of the for loop (3 and 5 => 15 in this example). Many thanks.

[Guillaume]

What you're asking does not make any sense to me. You want an array that goes from 1 to limit_of_loop1 * limit_of_loop2. Therefore, none of the elements in a depend in any way on what happens in the loops. You may as well calculate a outside the loops and if the loops don't do anything get rid of them

Therefore, you could just write:

limit_loop1 = 3;
limit_loop2 = 5;
%for i = 1:limit_loop1
%  for j = 1:limit_loop2
    %we're just wasting time with these loops!
%  end
%end
a = 1:limit_loop1*limit_loop2

[Giovanni Rinaldi]

Hi Guillame, thanks for the funny solution you proposed. I realize the problem does not make any sense if taken alone. And also that there are many easier ways of achieving what I need. However, I need to solve it in the way I described because I need it in a much wider context that I cannot include in my my description.

[Guillaume]

I still don't understand what you want. What part of a do you want to calculate in the loop? Would the following work?

limit_loop1 = 3;
limit_loop2 = 5;
for i = 1:limit_loop1
  for j = 1:limit_loop2
    a = 1:limit_loop1*limit_loop2; %but what's the point?
  end
end

[Giovanni Rinaldi]

Hi, the solution you propose gives as result:

       1

       1     2

       1     2     3

       1     2     3     4

       1     2     3     4     5

       1     2

       1     2     3     4

       1     2     3     4     5     6

       1     2     3     4     5     6     7     8

       1     2     3     4     5     6     7     8     9    10

       1     2     3

       1     2     3     4     5     6

       1     2     3     4     5     6     7     8     9

       1     2     3     4     5     6     7     8     9    10    11    12

       1     2     3     4     5     6     7     8     9    10    11    12    13    14    15

Which is different from what I am looking for.

Answer 1

[Jan]

It's not clear to me, what you want to achieve. So let me guess a little bit:

1. The direct approach:

L1 = 3;
L2 = 5;
a  = (1:(L1*L2)).'

2. With the loops:

a = [];
c = 0;
for i = 1:3
  for j = 1:5
     c = c + 1;
     a = [a; c];   % Prefer a pre-allocation
     disp(a)
  end
end

3. Using the paramters mentioned in my first answer:

a = [];
for i = 1:3
  for j = 1:5
     a = [a; (i - 1) * 5 + j];
     disp(a)
  end
end

But nothing will beat the approach 1. in speed and clarity.

Answer 2

[Jan]

This looks like a homework question. Do you really want us to solve it, or would delivering the solution made by someone else be cheating?

What about defining a=0 at the beginning and increasing a by one inside the loops?

You can find some constants b, c and d manually, such that

b * i + c * j + d = 1   % for i=1 and b=1
b * i + c * j + d = 2   % for i=1 and b=2
etc

[Giovanni Rinaldi]

Hi Jan, thanks for your help. This is not at all a homework but a part of a code I am trying to develop in a much wider context. I am sorry but I do not understand how introducing some constants can help. I would like to keep the problem as simple as possible, without introducing new parameters, also because deciding their value could be tricky as well. However, defining a=0 at the beginning is a useful hint. I will further work on this aspect.

Answer 3

[Guillaume]

Maybe one of these is what you want. It's really not clear what you want to do within the loops:

Possible solution #1:

limit_loop1 = 3;
limit_loop2 = 5;
a = zeros(limit_loop1 * limit_loop2, 1);
for i = 1:limit_loop1
  for j = 1:limit_loop2
    aidx = sub2ind([limit_loop1, limit_loop2], i, j);
    a(aidx) = aidx;
  end
end

Possible solution #2:

limit_loop1 = 3;
limit_loop2 = 5;
a = 1;
for i = 1:limit_loop1
  for j = 1:limit_loop2
    if i>1 || j>1
    a = [a(end); a+1];
  end
end

Neither of which make much sense to me, since as pointed out:

a = 1:limit_loop1*limit_loop2

gives the same result without needing the loops