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

A simple recursive approach:

def ordered_digits_generator(numDigits,min=1,max=9):
    for first in range(min,max+1):
        if numDigits == 1:
             yield first
        else:
             addend = first*10**(numDigits-1)
             for rest in ordered_digits(numDigits-1,min=0,max=first):
                 yield addend+rest

Then called via:

for number in ordered_digits_generator(10):
    print number

works as expected.

The mathematician's approach

The itertools package already has logic which essentially already implements this recursion. Presumably better than we can, with significant testing. So we can use it as follows:

import itertools
def ordered_digits_combo(numDigits):
    exponent = [10**i for i in range(0,numDigits)]

    for subset in itertools.combinations(range(0,numDigits+9),numDigits):
        if subset[numDigits-1]>numDigits-1:
            v = 0
            for i in range(0,numDigits):
                v += exponent[i]*(subset[i]-i)
            yield v

Given an ordered subset a[0] of {0,1,...,n+8}, we pick the number with the ith digit from the right equal to a[i]-i. We have to exclude the case a[n-1]==n-1 because that consists of he number with all zeros.