17 
value = 'ad.41.bd'

if len(value) == 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?

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7 Answers 7

Reset to default
19

What about this?

import re
if not re.search('\d+', value):
    # no numbers
else:
    # numbers present
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9 
>>> 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.

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8 Comments

It would be interesting to see how good (or bad) this performs compared to a regex.
I like any, but map generates the entire list first. If value is huge and the first character is a digit, this code still processes the whole thing.
I believe any stops when it finds the first True. However map doesn't return a generator (apart from I think in python 3) so a whole list is created in memory (even if the first character is a digit). Otherwise I'm guessing it would be fairly similar.
@robert king, w/o the parens you create a list of methods. This will always evaluate to True! Try it with value = 'abc'.
any(c.isdigit() for c in value) uses a generator expression and dispenses with the lambda and the map.
|
4 
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).

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2 Comments

I played with your timings and found that as the value increases in length the regexes perform better and better.
@Steven: That seems reasonable as it shifts a lot of the work to C. I changed value to "abcd" * 1000 + "9" and decreased number to 10000. any(imap) took 7.588s, re.search (stop after first match) took 0.377s, and any(c in digits) took 4.283s. Rewriting the last as any(d in value for d in digits) took only .310s (and .073s when the last digit was "1" instead of "9"), again I suppose because it pushed most of the workload to off to core.
3

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
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4 Comments

You don't need to anchor the expression with ^ and would probably be better off with re.search. Also, .* matches zero or more characters so ? is redundant.
@Kirk: I thought search might look for all locations that match this pattern (by I need only one), whereas match will only match at the beginning (so yes, I could omit ^ here). I want the shortest possible match, hence the .*? (not greedy).
Good points. After I've slept a little and had some coffee, I'll re-evaluate. :-)
search give only the first match.
1 
if not any(c.isdigit() for c in value)
    # no numbers
else:
    # numbers present
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Comments

1

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

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

0

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
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