分割字符串用星号蟒蛇

Question

我刚开始学习Python的2天前，对不起，如果我取得了明显的错误

strings: "brake break at * time" --> ["at","time"] 
"strang strange I felt very *" --> ["very",""] 

我想之前得到的字和后*

我尝试：

re.match(r"(?P(first_word)\w+) ('_*_') (?P(first_word)\w+)",strings).group('first_word') 

为获得第一个字

re.match(r"(?P(first_word)\w+) ('_*_') (?P(first_word)\w+)",strings).group('last_word') 

为获得最后的话

错误：没有重复

Answer 1

只需使用string.split('*')。

这样的（适用于1只*）：

>>> s = "brake break at * time" 
>>> def my_func(s): 
    parts = s.split('*') 
    a = parts[0].split()[-1] 
    b = parts[1].split()[0] if parts[1].split() else '' 
    return a,b 
>>> my_func(s) 
('at', ' time') 

或者，如果你想正则表达式：

>>> s = "brake break at * time 123 * blah" 
>>> regex = re.compile("(\w+)\s+\*\s*(\w*)") 
# Run findall 
>>> regex.findall(s) 
[(u'at', u'time'), (u'123', u'blah')] 

Answer 2

尝试：

[x.strip() for x in "test1 * test2".split('*', 1)] 

.strip()摆脱掉空格和.split('*', 1)由星号分割字符串一次。

当你想只有一个字：

words = [x.strip() for x in "test1 * test2".split('*', 1)] 
first = words[0].rsplit(' ', 1)[1] 
last = words[1].split(' ', 1)[0] 

Answer 3

import re 
text1 = "brake break at * time" 
text2 = "strang strange I felt very *" 
compiled = re.compile(r''' 
(\w+) # one or more characters from [_0-9a-zA-Z] saved in group 1 
\s+ # one or more spaces 
\* # literal * 
\s* # zero or more spaces 
(\w*) # zero or more characters from [_0-9a-zA-Z] saved in group 2 
''',re.VERBOSE) 

def parse(text): 
    result = compiled.search(text) 
    return [result.group(1), result.group(2)] 

print(parse(text1)) 
print(parse(text2)) 

输出：

['at', 'time'] 
['very', '']