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