从字符串中删除Python中两个特定字符之间包含的所有字符

Posted 2021-03-19

Python中的两个特定字符之间包含所有字符的快速方法是什么？

答案

您可以使用此正则表达式：\(.*?\)。在这里演示：https://regexr.com/3jgmd

然后你可以用这段代码删除零件：

import re
test_string = 'This is a string (here is a text to remove), and here is a text not to remove'
new_string = re.sub(r" \(.*?\)", "", test_string)

这个正则表达式（正则表达式）将在空格前面的括号中查找任何文本（没有换行符）

另一答案

你很可能会使用像这样的正则表达式

\s*\([^()]*\)\s*

为此（见a demo on regex101.com）。 该表达式删除括号和周围空格中的所有内容。

---

In Python this could be:

import re
test_string = 'This is a string (here is a text to remove), and here is a text not to remove'
new_string = re.sub(r'\s*\([^()]*\)\s*', '', test_string)
print(new_string)
# This is a string, and here is a text not to remove

---

However, for learning purposes, you could as well go with the builtin methods:

test_string = 'This is a string (here is a text to remove), and here is a text not to remove'
left = test_string.find('(')
right = test_string.find(')', left)

if left and right:
    new_string = test_string[:left] + test_string[right+1:]
    print(new_string)
    # This is a string , and here is a text not to remove

后者的问题：它没有考虑多次出现并且不会删除空格，但肯定会更快。

---

Executing this a 100k times each, the measurements yield:

0.578398942947 # regex solution
0.121736049652 # non-regex solution

另一答案

要删除（和）中的所有文本，您可以使用findall()中的re方法并使用replace()删除它们：

import re
test_string = 'This is a string (here is a text to remove), and here is a (second one) text not to remove'
remove = re.findall(r" \(.*?\)",test_string)
for r in remove:
    test_string = test_string.replace(r,'')
print(test_string)
#result: This is a string , and here is a  text not to remove