使用regex搜索最后一个单词具有最大字符数的短语。

Question

因此，我有以下几个词的清单：

list = ['cyber attacks, 28','cyber attacks. A', 'cyber attacks, intrusions', 'cyber attacks; 3','cyber attack. Our', 'cyber intrusions, data']

我想要做的是，如果第三个单词有超过三个字符，就删除列表中的短语。因此，最后的清单是：

new_list = ['cyber attacks, 28','cyber attacks. A', 'cyber attacks; 3','cyber attack. Our']

这是我到目前为止所掌握的，但也包括了最后一个单词超过三个字符的短语：

new_list = []

for phrase in list:
    max_three_char = re.match('cyber\s\w{1,}(\.|,|;|\)|\/|:|"|])\s\w{,3}', phrase)
    if max_three_char:
        new_list.append(phrase)

Answer 1

我会这样做：

import re 

li = ['cyber attacks, 28','cyber attacks. A', 'cyber attacks, intrusions', 'cyber attacks; 3','cyber attack. Our', 'cyber intrusions, data']

>>> [s for s in li if re.search(r'(?<=\W)\w{1,3}$', s)]
['cyber attacks, 28', 'cyber attacks. A', 'cyber attacks; 3', 'cyber attack. Our']

或者，如果您可以指望有一个空格分隔符：

>>> [s for s in li if len(s.split()[-1])<=3]
# same

Answer 2

您可以使用列表理解，如

import re

lst = ['cyber attacks, 28','cyber attacks. A', 'cyber attacks, intrusions', 'cyber attacks; 3','cyber attack. Our', 'cyber intrusions, data']
pattern = re.compile(r'[, ]+')

new_lst = [item
           for item in lst
           for splitted in [pattern.split(item)]
           if not (len(splitted) > 2 and len(splitted[2]) > 3)]

print(new_lst)

就会屈服

['cyber attacks, 28', 'cyber attacks. A', 'cyber attacks; 3', 'cyber attack. Our']

不要以内置的东西(如list等)来命名变量。

Answer 3

不需要regex，您可以使用string.split

if len(my_phrase.split()[2]) <=3:
    //process my_phrase

这是因为单词之间有空格。