Question

如何在不使用正则表达式的情况下从python中的字符串中提取数字？我见过isinstance，但数字几乎可以更改为任何数字。有什么想法吗？

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

Answer 1

这有点冗长，但是我会为此使用url解析。与正则表达式相比，它的优点是您可以免费获得更具可读性的代码进行一些输入验证。

>>> 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

Answer 2

如果url始终采用最后只有数字的格式，则可以执行以下操作：

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

Answer 3

您可以使用以下命令在字符串的任意位置提取连续的数字组：

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]

Answer 4

这假定不存在多个整数块（例如www.something212.com/page=?13）

您可以尝试使用列表推导和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

已编辑：现在可用于9以上的数字

Answer 5

我知道您不需要.p1 { margin-bottom: 10px;} .p2 { margin-bottom: 10px;}，但实际上它非常强大。在后台，大多数库都使用func addGesture() { // Gesture recognizers let leftSwipe = UISwipeGestureRecognizer(target: self, action: #selector(self.swipeAction(swipe:))) leftSwipe.direction = UISwipeGestureRecognizer.Direction.left self.view.addGestureRecognizer(leftSwipe) let rightSwipe = UISwipeGestureRecognizer(target: self, action: #selector(self.swipeAction(swipe:))) rightSwipe.direction = UISwipeGestureRecognizer.Direction.right self.view.addGestureRecognizer(rightSwipe) let downSwipe = UISwipeGestureRecognizer(target: self, action: #selector(self.swipeAction(swipe:))) downSwipe.direction = UISwipeGestureRecognizer.Direction.down self.view.addGestureRecognizer(downSwipe) }。这是我处理这种情况的解决方案：

re

输出为

re

从Python中的字符串中提取数字而无需重新

5 个答案: