有效检查字符串是否包含python中的数字

Question

I have a huge amount (GB) of text to process, sentence by sentence. In each sentence I have a costly operation to perform on numbers, so I check that this sentence contains at least one digit. I have done this check using different means and measured those solutions using timeit .

s = 'abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz' # example

• any(c.isdigit() for c in s) 3.61 µs

• re.search('\d', s) 402 ns

• d = re.compile('\d') d.search(s) 126 ns

• '0' in s or '1' in s or '2' in s or '3' in s or '4' in s or '5' in s or '6' in s or '7' in s or '8' in s or '9' in s 60ns

The last way is the fastest one, but it is ugly and probably 10x slower than possible.

Of course I could rewrite this in cython, but it seems overkill.

Is there a better pure python solution? In particular, I wonder why you can use str.startswith() and str.endswith() with a tuple argument, but it does not seem to be possible with in operator.

Answer 1

Actual performance might vary depending on your platform and python version, but on my setup (python 3.9.5 / Ubuntu), it turns out that re.match is significantly faster than re.search , and outperforms the long in series version. Also, compiling the regex with [0-9] instead of \d provides a little improvement.

import re
from timeit import timeit

n = 10_000_000
s = 'abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz'

# reference
timeit(lambda: '0' in s or '1' in s or '2' in s or '3' in s or '4' in s or '5' in s or '6' in s or '7' in s or '8' in s or '9' in s, number=n)
# 2.1005349759998353

# re.search with \d, slower
re.compile('\d')
timeit(lambda: d.search(s), number=n)
# 2.9816031390000717

# re.search with [0-9], better but still slower then reference
d = re.compile('[0-9]')
timeit(lambda: d.search(s), number=n)
# 2.640713582999524

# re.match with [0-9], faster than reference
d = re.compile('[0-9]')
timeit(lambda: d.match(s), number=n)
# 1.5671786130005785

So, on my machine , using re.match with a compiled [0-9] pattern is about 25% faster than the long or ... in chaining. And it looks better too.