返回列表的python筛选器字符串

使用正则表达式--> re.findall .

前任:

import re

s = "Welcome to 'Jungle', is a song by American rock band 'Guns N Roses' released in 1987."
print(re.findall(r"'(.*?)'", s))

['Jungle', 'Guns N Roses']

你也可以这样试试。这是不使用任何库的最简单的编程方法。你可以试一试:

full_string = input("Enter String: ")
quoted_strings = []

start = 0
quoted_string = ""
for letter in full_string:
    if letter == "'" and start == 0:
        start = 1
    elif letter == "'" and start == 1:
        quoted_strings.append(quoted_string)
        quoted_string = ""
        start = 0
    elif start == 1:
        quoted_string += letter
    else:
        pass

print("Entered Full String: " + full_string)
print("Quoted Strings: ", quoted_strings)

IMHO,使用regex的最佳方法是避免“贪婪”。

import re
a = "'Jungle', is a song by American rock band 'Guns N Roses'"]
re.findall(r"'(.+?)'", a)

这将寻找一个被引语包围的词。e、 它将跳过空的报价。

我们用“?”来减少搜索的贪婪。