Detect numbers in string

value = 'ad.41.bd'

if len(v开发者_StackOverflow社区alue) == len(value.strip({0,1,2,3,4,5,6,7,8,9})):
    # no numbers
else:
    # numbers present

There a cleaner way of detecting numbers in a string in Python?

What about this?

import re
if not re.search('\d+', value):
    # no numbers
else:
    # numbers present

>>> value="ab3asdf"
>>> any(c.isdigit() for c in value)
True
>>> value="asf"
>>> any(c.isdigit() for c in value)
False




>>> value = 'ad.41.bd'
>>> any(map(lambda c:c.isdigit(),value))
True

EDIT:

>>> value="1"+"a"*10**6
>>> any(map(lambda c:c.isdigit(),value))
True
>>> from itertools import imap
>>> any(imap(lambda c:c.isdigit(),value))
True

map took 1 second (on old python) imap was instant because imap returns a generator. note often in the real world there is a higher probability of the number being at the end of the file name.

from string import digits
def containsnumbers(value):
    return any(char in digits for char in value)

EDIT:

And just for thoroughness:

any(c.isdigit()):

>>> timeit.timeit('any(c.isdigit() for c in value)', setup='value = "abcd1"')
1.4080650806427002

any(c in digits):

>>> timeit.timeit('any(c in digits for c in value)', setup='from string import digits; value = "abcd1"')
1.392179012298584

re.search (1 or more digits):

>>> timeit.timeit("re.search('\d+', value)", setup='import re; value = "abcd1"')
1.8129329681396484

re.search (stop after one digit):

>>> timeit.timeit("re.search('\d', value)", setup='import re; value = "abcd1"')
1.599431037902832

re.match (non-greedy):

>>> timeit.timeit("re.match(r'^.*?\d', value)", setup='import re; value = "abcd1"')
1.6654980182647705

re.match(greedy):

>>> timeit.timeit("re.match(r'^.*\d', value)", setup='import re; value = "abcd1"')
1.5637178421020508

any(map()):

>>> timeit.timeit("any(map(lambda c:c.isdigit(),value))", setup='value = "abcd1"')
1.9165890216827393

any(imap()):

>>> timeit.timeit("any(imap(lambda c:c.isdigit(),value))", setup='from itertools import imap;value = "abcd1"')
1.370448112487793

Generally, the less complex regexps ran more quickly. c.isdigit() and c in digits are almost equivalent. re.match is slightly faster than re.search. map() is the slowest solution, but imap() was the fastest (but within rounding error of any(c.isdigit) and any(c in digits).

You can use a regular expression:

import re
# or if re.search(r'\d', value):
if re.match(r'^.*?\d', value):
    # numbers present
else:
    # no numbers

if not any(c.isdigit() for c in value)
    # no numbers
else:
    # numbers present

To detect signs in the numbers, use the ? operator.

import re
if not re.search('-?\d+', value):
    # no numbers
else:
    # numbers present

If you want to know how big is the difference, you can use re.sub()

import re
digits_num = len(value) - len(re.sub(r'\d','',value))
if not digits_num:
    #without numbers
else:
    #with numbers - or elif digist_num == 3