Find the index of the first digit in a string

Question

I have a string like

\"xdtwkeltjwlkejt7wthwk89lk\"

how can I get the index of the first digit in the string?

Answer 1

In Python 3.8+ you can use re.search to look for the first \d (for digit) character class like this:

import re

my_string = "xdtwkeltjwlkejt7wthwk89lk"

if first_digit := re.search(r"\d", my_string):
    print(first_digit.start())

Answer 2

instr = 'nkfnkjbvhbef0njhb h2konoon8ll'
numidx = next((i for i, s in enumerate(instr) if s.isdigit()), None)
print numidx

output:

12

numidx will be the index of the first occurrence of a digit in instr. If there are no digits in instr, numidx will be None.

I didn't see this solution here, and thought it should be.

Answer 3

import re
first_digit = re.search('\d', 'xdtwkeltjwlkejt7wthwk89lk')
if first_digit:
    print(first_digit.start())

Answer 4

Here is another regex-less way, more in a functional style. This one finds the position of the first occurrence of each digit that exists in the string, then chooses the lowest. A regex is probably going to be more efficient, especially for longer strings (this makes at least 10 full passes through the string and up to 20).

haystack = "xdtwkeltjwlkejt7wthwk89lk"
digits   = "012345689"
found    = [haystack.index(dig) for dig in digits if dig in haystack]
firstdig = min(found) if found else None

Answer 5

import re
result = "  Total files:...................     90"
match = re.match(r".*[^\d](\d+)$", result)
if match:
    print(match.group(1))

will output

90

Answer 6

Seems like a good job for a parser:

>>> from simpleparse.parser import Parser
>>> s = 'xdtwkeltjwlkejt7wthwk89lk'
>>> grammar = """
... integer := [0-9]+
... <alpha> := -integer+
... all     := (integer/alpha)+
... """
>>> parser = Parser(grammar, 'all')
>>> parser.parse(s)
(1, [('integer', 15, 16, None), ('integer', 21, 23, None)], 25)
>>> [ int(s[x[1]:x[2]]) for x in parser.parse(s)[1] ]
[7, 89]