使用正则表达式拆分此字符串 - python答案

【问题标题】：Split this string using regular expression - python使用正则表达式拆分此字符串 - python
【发布时间】：2012-09-13 09:10:48
【问题描述】：

Input string
---------------
South Africa 109/0 
Australia 100
Sri Lanka 111
Sri Lanka 331/4

Expected Output
---------------
['South Africa', '109', '0']
['Australia', '100']
['Sri Lanka', '111']
['Sri Lanka', '331', '4']

我尝试了几个正则表达式，但不知道写出正确的一个。在这种情况下，空格分隔符对我没有帮助，因为国家名称可能有也可能没有空格（南非、印度）。提前致谢

【问题讨论】：

标签： python regex python-2.7

【解决方案1】：

我们可以使用正则表达式：

r'(\D+)\s(\d+)(?:/(\d+))?'

("很多非数字，后跟一个空格，后跟很多数字，然后可选地跟一个斜线，然后是很多数字。")

这将返回，例如

>>> [re.match(r'(\D+)\s(\d+)(?:/(\d+))?', x).groups() 
...  for x in ['South Africa 109/0', 
...            'Australia 100',
...            'Sri Lanka 111',
...            'Sri Lanka 331/4']]
[('South Africa', '109', '0'), 
 ('Australia', '100', None), 
 ('Sri Lanka', '111', None), 
 ('Sri Lanka', '331', '4')]

注意Nones，您可能需要手动过滤掉它。

【讨论】：

您不应该使用[\w\s] 而不是\D 以在'Au$tralia' 上失败吗？
@PierreGM：如果 OP 希望 Bishop's Stortford 和 Xi'an 成功怎么办？也许Áŭ$t®å£ià 真的被认为是有效的。

【解决方案2】：

试试：

import re
re.split(r"(?<=[a-zA-Z])\s+(?=\d)|(?=\d)\s+(?=[a-zA-Z])|/", "South Africa 109/0")

【讨论】：

【解决方案3】：

re.compile("^([\w\s]+)\s(\d+)\/?(\d+)?")

为您提供三个组。我们可以分解它

在行首的一组只有字母和空格([\w\s]+) (^)
一个空格
一组数字，至少一个(\d+)
/ 与否
一组数字（可能是None）

【讨论】：

这会在第一组中输出Australia 100 和Sri Lanka 111。
不，最后会给你一个空组，就像@KennyTM 版本一样。

【解决方案4】：

这是您需要的正则表达式：

for match in re.finditer(r"(?m)^(?P<Country>.*?)\s*(?P<Number1>\d+)\s*?/?\s*?(?P<Number2>\d*?)\s*?$", inputText):
    country = match.group("Country")
    number1 = match.group("Number1")
    number2 = match.group("Number2")

你可以看到结果here。

下面是对模式的解释：

# ^(?P<Country>.*?)\s*(?P<Number1>\d+)\s*?/?\s*?(?P<Number2>\d*?)\s*?$
# 
# Options: ^ and $ match at line breaks
# 
# Assert position at the beginning of a line (at beginning of the string or after a line break character) «^»
# Match the regular expression below and capture its match into backreference with name “Country” «(?P<Country>.*?)»
#    Match any single character that is not a line break character «.*?»
#       Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
# Match a single character that is a “whitespace character” (spaces, tabs, and line breaks) «\s*»
#    Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
# Match the regular expression below and capture its match into backreference with name “Number1” «(?P<Number1>\d+)»
#    Match a single digit 0..9 «\d+»
#       Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
# Match a single character that is a “whitespace character” (spaces, tabs, and line breaks) «\s*?»
#    Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
# Match the character “/” literally «/?»
#    Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
# Match a single character that is a “whitespace character” (spaces, tabs, and line breaks) «\s*?»
#    Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
# Match the regular expression below and capture its match into backreference with name “Number2” «(?P<Number2>\d*?)»
#    Match a single digit 0..9 «\d*?»
#       Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
# Match a single character that is a “whitespace character” (spaces, tabs, and line breaks) «\s*?»
#    Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
# Assert position at the end of a line (at the end of the string or before a line break character) «$»

【讨论】：

【解决方案5】：

你已经得到了正则表达式的答案，但我建议也考虑可用的内置 str 方法（无论如何对于这个用例）：

s = 'South Africa 109/0'
country, numbers = s.rsplit(' ', 1)
# ('South Africa', '109/0')
new_list = [country] + numbers.split('/')
# ['South Africa', '109', '0']

【讨论】：