使用列表推导删除与模式匹配的词典列表中的词典

Question

我有一个以文本为值的字典列表，我想删除文本中包含某些单词的字典。

df = [{'name':'jon','text':'the day is light'},{'name':'betty','text':'good night'},{'name':'shawn','text':'good afternoon'}]

我要删除包含“文本”键的单词“ light”和“ night”的字典：

words = ['light','night']
pattern = re.compile(r"|".join(words))

预期结果：

df = [{'name':'shawn','text':'good afternoon'}]

Answer 1

[x for x in df if not any(w in x['text'] for w in words)]

Answer 2

您已经关闭。您所需要做的就是编写列表理解并应用搜索模式：

result = [x for x in df if not re.search(pattern, x['text'])]

完整示例：

import re

df = [{'name':'jon','text':'the day is light'},{'name':'betty','text':'good night'},{'name':'shawn','text':'good afternoon'}]
words = ['light','night']
pattern = re.compile(r"|".join(words))
result = [x for x in df if not re.search(pattern, x['text'])]
print(result) # => [{'name': 'shawn', 'text': 'good afternoon'}]

Answer 3

我找到了答案：

[x for x in df if not pattern.search(x['text'])]