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

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 2

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.