在 Python 中使用 re 查找多个匹配项（初学者问题）

Question

我需要使用正则表达式和 Collection 找到多个匹配项（包含在列表中）。我试过这段代码，但它显示空字典：some_words_lst = ['caT.', 'Cat', 'Dog', 'paper', 'caty', 'London', 'loNdon','londonS']words_to_find = ['cat', 'london']r = re.compile('(?:.*{})'.format(i for i in words_to_find),re.IGNORECASE)count_dictionary = {}for item in some_words_lst:    if r.match(item):        count_dictionary['i']+=1print(count_dictionary)感谢帮助！

Answer 1

您需要另一种语法来重新

也不要忘记在 += 之前初始化字典中的键

import re

some_words_lst = ['caT.', 'Cat', 'Dog', 'paper', 'caty', 'London', 'loNdon','londonS']

words_to_find = ['cat', 'london']

r = re.compile('|'.join(words_to_find), re.IGNORECASE)

count_dictionary = {"i": 0}

for item in some_words_lst:

    if r.match(item):

        count_dictionary['i']+=1

print(count_dictionary)

UPD：根据评论，我们需要匹配项目的数量。像这样又快又脏的东西是怎么回事？

import re

some_words_lst = ['caT.', 'Cat', 'Dog', 'paper', 'caty', 'London', 'loNdon','londonS']

words_to_find = ['cat', 'london']

r = re.compile('|'.join(words_to_find), re.IGNORECASE)

count_dictionary = {word: 0 for word in words_to_find}

for item in some_words_lst:

    if r.match(item):

        my_match = r.match(item)[0]

        count_dictionary[my_match.lower()]+=1

print(count_dictionary)