Split number into groups by magnitude

Question

given an input 34567890

I want an output: [34, 567, 890]

I can do it with some modulo math and floor division

num=34567890
output_list = []
division_tracker = num
while division_tracker > 0:
    output_list.insert(0, division_tracker%1000)
    division_tracker = division_tracker//1000

Is there a better way?

Answer 1

This is kind of hacky, but:

>>> list(map(int, f'{34567890:,}'.split(',')))
[34, 567, 890]

f'{n:,}' generates a string of the integer n with commas as thousands separators, we split the resulting string at comma and cast the parts into int.

Answer 2

credit goes to @fsimonjetz

if you want to use list comprehension instead of map:

[int(x) for x in f'{34567890:,}'.split(',')]
[34, 567, 890]

Answer 3

Using a slice approach over strings:

s = str(345678)
n = len(s)
r = n % 3 # remainder

groups = [int(s[r + 3*i:r + 3*(i+1)]) for i in range(n//3)]
if r != 0:
    groups = [int(s[-r:])] + groups

print(groups)

EDIT

A regex approach: nested gropus are not allowed so the pattern is constructed explicitely, based on the amount of digits.

import re

s = '12345678235156721'

regex = ''
if (r:=len(s) % 3) != 0:
    regex += '(\d{{{}}})'.format(r)
regex += '(\d{3})' * (len(s) // 3)
regex = r'{}'.format(regex)

grps = list(map(int, re.search(regex, s).groups()))

Output

[12, 345, 678, 235, 156, 721]