Variable Number Of Nested For Loops With Fixed Range

I'm looking for a method to have a variable number of nested for loops instead of the following code. For example if the variable n represents the number of nested for loops and n

Solution 1:

It's important for one to develop the skills to reason about these problems. In this case, Python includes itertools.product but what happens the next time you need to write a behaviour specific to your program? Will there be another magical built-in function? Maybe someone else will have published a 3rd party library to solve your problem?

Below, we design product as a simple recursive function that accepts 1 or more lists.

defproduct (first, *rest):
  ifnot rest:
    for x in first:
      yield (x,)
  else:
    for p in product (*rest):
      for x in first:
        yield (x, *p)

for p in product (range(2), range(2), range(2)):
  print ('x: %d, y: %d z: %d' % p)

# x: 0, y: 0 z: 0# x: 1, y: 0 z: 0# x: 0, y: 1 z: 0# x: 1, y: 1 z: 0# x: 0, y: 0 z: 1# x: 1, y: 0 z: 1# x: 0, y: 1 z: 1# x: 1, y: 1 z: 1

Assuming you want a more conventional iteration ordering, you can accomplish do so by using an auxiliary loop helper

defproduct (first, *rest):
  defloop (acc, first, *rest):
    ifnot rest:
      for x in first:
        yield (*acc, x)
    else:
      for x in first:
        yieldfrom loop ((*acc, x), *rest)
  return loop ((), first, *rest)

for p in product (range(2), range(2), range(2)):
  print ('x: %d, y: %d z: %d' % p)

# x: 0, y: 0 z: 0# x: 0, y: 0 z: 1# x: 0, y: 1 z: 0# x: 0, y: 1 z: 1# x: 1, y: 0 z: 0# x: 1, y: 0 z: 1# x: 1, y: 1 z: 0# x: 1, y: 1 z: 1

Solution 2:

Something like this should work:

import itertools

n=3
fixed=26
p = list(itertools.product(range(fixed), repeat=n))

This solution uses the optimized functions of itertools, so it should be quite fast.

Mind that itertools.product returns an iterator, so one needs to transform it to get an array.