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

Use re.search():

>>> import re
>>> s1 = "thishasadigit4here"
>>> m = re.search(r"\d", s1)
>>> if m:
...     print("Digit found at position", m.start())
... else:
...     print("No digit in that string")
... 
Digit found at position 13

Answer 2

Here is a better and more flexible way, regex is overkill here.

s = 'xdtwkeltjwlkejt7wthwk89lk'

for i, c in enumerate(s):
    if c.isdigit():
        print(i)
        break

output:

15

To get all digits and their positions, a simple expression will do

>>> [(i, c) for i, c in enumerate('xdtwkeltjwlkejt7wthwk89lk') if c.isdigit()]
[(15, '7'), (21, '8'), (22, '9')]

Or you can create a dict of digit and its last position

>>> {c: i for i, c in enumerate('xdtwkeltjwlkejt7wthwk89lk') if c.isdigit()}
{'9': 22, '8': 21, '7': 15}

Answer 3

Thought I'd toss my method on the pile. I'll do just about anything to avoid regex.

sequence = 'xdtwkeltjwlkejt7wthwk89lk'
i = [x.isdigit() for x in sequence].index(True)

To explain what's going on here:

• [x.isdigit() for x in sequence] is going to translate the string into an array of booleans representing whether each character is a digit or not
• [...].index(True) returns the first index value that True is found in.

Answer 4

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

Answer 5

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]

Answer 6

To get all indexes do:

idxs = [i for i in range(0, len(string)) if string[i].isdigit()]

Then to get the first index do:

if len(idxs):
    print(idxs[0])
else:
    print('No digits exist')

Answer 7

As the other solutions say, to find the index of the first digit in the string we can use regular expressions:

>>> s = 'xdtwkeltjwlkejt7wthwk89lk'
>>> match = re.search(r'\d', s)
>>> print match.start() if match else 'No digits found'
15
>>> s[15] # To show correctness
'7'

While simple, a regular expression match is going to be overkill for super-long strings. A more efficient way is to iterate through the string like this:

>>> for i, c in enumerate(s):
...     if c.isdigit():
...         print i
...         break
... 
15

In case we wanted to extend the question to finding the first integer (not digit) and what it was:

>>> s = 'xdtwkeltjwlkejt711wthwk89lk'
>>> for i, c in enumerate(s):
...     if c.isdigit():
...         start = i
...         while i < len(s) and s[i].isdigit():
...             i += 1
...         print 'Integer %d found at position %d' % (int(s[start:i]), start)
...         break
... 
Integer 711 found at position 15

Answer 8

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 9

I'm sure there are multiple solutions, but using regular expressions you can do this:

>>> import re
>>> match = re.search("\d", "xdtwkeltjwlkejt7wthwk89lk")
>>> match.start(0)
15

Answer 10

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 11

def first_digit_index(iterable):
    try:
        return next(i for i, d in enumerate(iterable) if d.isdigit())
    except StopIteration:
        return -1

This does not use regex and will stop iterating as soon as the first digit is found.

Answer 12

you can use regular expression

import re
y = "xdtwkeltjwlkejt7wthwk89lk"

s = re.search("\d",y).start()

Answer 13

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

will output

90

Answer 14

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.