我需要使用正則表達式和 Collection 找到多個匹配項（包含在列表中）。我試過這段代碼，但它顯示空字典：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)感謝幫助！
1 回答
千巷貓影
TA貢獻1829條經(jīng)驗 獲得超7個贊
您需要另一種語法來重新
也不要忘記在 += 之前初始化字典中的鍵
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：根據(jù)評論，我們需要匹配項目的數(shù)量。像這樣又快又臟的東西是怎么回事？
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)
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