Python: nested 'for' loops

Question

I\'d like to go through all n-digit numbers such that second digit of the number is always lower or equal to the first, third is lower or equal to the second etc. I can get this

Answer 1

Could use itertools:

>>> for comb in itertools.combinations_with_replacement(range(9, -1, -1), 3):
        print comb

(9, 9, 9)
(9, 9, 8)
(9, 9, 7)
(9, 9, 6)
...
(4, 0, 0)
(3, 3, 3)
(3, 3, 2)
(3, 3, 1)
(3, 3, 0)
(3, 2, 2)
(3, 2, 1)
(3, 2, 0)
(3, 1, 1)
(3, 1, 0)
(3, 0, 0)
(2, 2, 2)
(2, 2, 1)
(2, 2, 0)
(2, 1, 1)
(2, 1, 0)
(2, 0, 0)
(1, 1, 1)
(1, 1, 0)
(1, 0, 0)
(0, 0, 0)

---

Or recursively, appending more and more digits until enough, which can more directly produce int objects instead of digit tuples (not sure whether that's what you actually need):

def build(enough, prefix=0):
    if prefix >= enough:
        print(prefix)
        return
    for digit in range(prefix % 10 + 1) if prefix else range(1, 10):
        build(enough, prefix * 10 + digit)

Demo (note it leaves out "000", not sure whether you'd want that anyway):

>>> n = 3
>>> build(10**(n-1))
100
110
111
200
210
211
220
221
222
300
310
311
320
321
322
330
331
332
333
400
410
411
420