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

I would probably implement this recursively:

def generate(max, digits):
    for d in range(max + 1):
        if digits == 1:
            yield d
        else:
            first = d * 10**(digits-1)
            for n in generate(d, digits - 1):
                yield first + n

The output:

In : list(generate(3, 3))
Out:
[0,
 100,
 110,
 111,
 200,
 210,
 211,
 220,
 221,
 222,
 300,
 310,
 311,
 320,
 321,
 322,
 330,
 331,
 332,
 333]

Answer 2

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]<a[1]<...<a[n-1] of {0,1,...,n+8}, we pick the number with the i^th 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.

Answer 3

I implemented @iFlo's suggestion as commented originally. It's not hyper efficient but it certainly doesn't take ages.

def digit_test(n):
    while n > 9:
        if (n % 100 / 10) < (n % 10): return False
        n /= 10
    return True

# under a second to construct a list of all numbers below 1000000 meeting the criteria
candidates = [x for x in xrange(1,1000000) if digit_test(x)]

# should be 8001 elements, consistent with other algorithms
print len(candidates)

Answer 4

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

Answer 5

this an approach using itertools:

from itertools import combinations_with_replacement

N = 3

for kji in combinations_with_replacement((str(i) for i in range(10)), N):
    print(''.join(reversed(kji)))

note that the order is not the same as in your original approach.

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