在python中从字符串中提取数字而不必重新

这假设没有多个整数块(例如。 www.something212.com/page=?13

你可以尝试使用列表理解和 str.isdigit()

url = 'https://www.investopedia.com/articles/retirement/?page=6'

digits = [d for d in url if d.isdigit()]

digit = ''.join(digits)

digit
>>> 6

这有点冗长,但我会使用url解析。与regex相比,它的优点是可以免费获得一些输入验证,而且代码可读性更高。

>>> from urllib.parse import urlparse, parse_qs
>>> url = 'https://www.investopedia.com/articles/retirement/?page=6'
>>> parsed = urlparse(url)
>>> query = parse_qs(parsed.query)
>>> [page] = query['page']
>>> int(page)
6

from itertools import groupby

url = 'https://www.investopedia.com/articles/retirement/?page=6&limit=10&offset=15'
print([int(''.join(group)) for key, group in groupby(iterable=url, key=lambda e: e.isdigit()) if key])

输出

[6, 10, 15]

如果 url

s = 'https://www.investopedia.com/articles/retirement/?page=25'
new = []
k = list(s)
[new.append(i) for i in k if i.isdigit()]
print(''.join(new))

(xenial)vash@localhost:~/python/stack_overflow$ python3.7 isdigit.py
25

我知道你不需要 re ,但它实际上非常强大。在幕后,大多数图书馆利用 重新 . 我的解决方案是:

import re

url = "www.fake888.com/article/?article=123&page=9&group=8"

numbers = re.findall(r'(?<==)(\d+)', url)
print(f'Found: {" ".join(numbers)}')

varval = re.findall(r'(\w+)=(\d+)', url)
urldict = {}
for var in varval:
  urldict[var[0]] = var[1]

print(urldict)

输出为

Found: 123 9 8
{'article': '123', 'page': '9', 'group': '8'}