How do I create a list of every combination of 3 digit numbers in Python?

Question

I'm trying to write a python script that would output:

000

001

002

... etc

but I'm running into difficulties. What I have so far is:

from itertools import product


list = [x for x in range(0, 10) if True]
for x in product(list, repeat=3):
    list3 = list(x)


def convert(l):
    c = [str(i) for i in l]
    list2 = int("".join(c))
    return(list2)

print(convert(list3))

but this only outputs:

999

I'm not sure how to get the full list. If I comment out the convert function it provides multiple lists of the numbers, like so:

[0, 0, 0]

[0, 0, 1]

...

Any help would be appreciated, I'm pretty sure I'm missing something simple.

user_494134 · Answer

You're overthinking it.

for n in range(0, 1000):
    print(f'{n:03}')

user_1534218 · Answer

import itertools
import string

for t in itertools.product(string.digits, repeat=3):
    print("".join(t))

user_18004761 · Answer

if you want to use itertools you can also do:

from itertools import combinations

lst = [x for x in range(0, 10) if True]
combos = list(combinations(lst, 3))

documentation on itertools.combinations is here: https://docs.python.org/3/library/itertools.html#itertools.combinations

user_14541164 · Answer

You can try this

from itertools import permutations

digits = [0, 2, 3] #List of digits from which numbers are to be formed
n = 3 #number of digits in a number

num = [int(''.join(map(str, p))) for p in permutations(digits, n) if p[0] != 0]

print(num) #number formed are without repetition of digits

If repetition of digits are allowed, then you can try this

import itertools
rlist = list(itertools.permutations([1, 2, 3]))
plist = []
for r in rlist:
    num = ''
    for _ in r:
        num += str(_)

    plist.append(int(num))

print(plist)

How do I create a list of every combination of 3 digit numbers in Python?

4 Answers