删除列表单词组合python 3

Question

我有一个字符串列表,我想搜索一个单词组合.如果组合不存在,则删除列表.是否有一个python列表理解可行？

word_list = ["Dogs love ice cream", "Cats love balls", "Ice cream", "ice cream is good with pizza", "cats hate ice cream"]

keep_words = ["Dogs", "Cats"] 

Delete_word = ["ice cream"]

删除里面有冰淇淋的单词,但如果句子中有狗或猫,请保留它.

Desired_output = ["Dogs love ice cream", "Cats love balls", "cats hate ice cream"] 

尝试此代码也试过AND和OR,但无法正确组合.

output_list = [x for x in  word_list if "ice cream" not in x]

Answer 1

这是一个列表理解解决方案:

[x for x in word_list if any(kw.lower() in x.lower() for kw in keep_words) 
 or all(dw.lower() not in x.lower() for dw in Delete_word)]
# ['Dogs love ice cream', 'Cats love balls', 'cats hate ice cream']

这还为删除单词列表中的多个单词增加了灵活性.

说明

如果以下任何一项出现,请对列表进行迭代并保留该单词True:

• 任何保留字都在x中
• 没有删除字在x中

我从你的例子中假设你希望这个不区分大小写,所以要对这些单词的低级版本进行所有比较.

两个有用的功能是any()和all().

Answer 2

作为一个优化的方法,你可以把你keep_word和delete_words集合以及使用itertools.filterfalse()过滤列表出来:

In [48]: def key(x):
             words = x.lower().split()
             return keep_words.isdisjoint(words) or not delete_words.isdisjoint(words)
   ....: 

In [49]: keep_words = {"dogs", "cats"}

In [51]: delete_words = {"ice cream"}

In [52]: list(filterfalse(key ,word_list))
Out[52]: ['Dogs love ice cream', 'Cats love balls', 'cats hate ice cream']