使用 catch 将部分列表项匹配到另一个列表项

Question

我有两个列表：

list_1 = ['world', 'abc', 'bcd', 'ghy', 'car', 'hell', 'rock']
list_2 = ['the world is big', 'i want a car', 'puppies are best', 'you rock the world']

我想检查 list_1 中的单词是否以任何形状或形式存在于 list_2 中，然后简单地从 list_2 中删除整个句子，最后打印 list_2

例如：

the word 'world' from list_1 should take out the sentence 'the world is big' from list_2
the word 'car' from list_2 should take out the sentence 'i want a car'

我试过像这样使用列表理解，但遗憾的是它重复了

output = [j for i in list_1 for j in list_2 if i not in j]

Answer 1

您必须在条件表达式中使用单独的列表理解。在您的理解中，您为 j 中不存在的每个 i 添加了一个 j for i in list_1，这就是为什么您要重复的原因。

output = [j for j in list_2 if all([i not in j for i in list_1])]

Answer 2

您应该考虑尽可能为您的变量命名，这有助于您编写代码

你要的是

遍历句子
每次检查其中没有来自 list_1 的单词

output = [sentence for sentence in list_2
          if all(word not in sentence for word in list_1)]

print(output)  # ['puppies are best']

Answer 3

使用 set 路口怎么样？

list_1 = ['world', 'abc', 'bcd', 'ghy', 'car', 'hell', 'rock']
list_2 = ['the world is big', 'i want a car', 'puppies are best', 'you rock the world']

words = set(list_1)
output = [i for i in list_2 if not words & set(i.split())]
print(output)

输出：

['puppies are best']

Answer 4

您想检查短语中是否出现了任何单词，因此 any 是可行的方法。在我看来，这比使用 all 并否定检查更具可读性。

words = ['world', 'abc', 'bcd', 'ghy', 'car', 'hell', 'rock']
phrases = ['the world is big', 'i want a car', 'puppies are best', 'you rock the world']

result = [phrase for phrase in phrases if not any(word in phrase for word in words)]
print(result)

你得到['puppies are best'].

解决方案大致等同于：

result = []
for phrase in phrases:
    contains_any_word = False
    for word in words:
        if word in phrase:
            contains_any_word = True
            break
    if not contains_any_word:
        result.append(phrase)

使用 catch 将部分列表项匹配到另一个列表项

Perform partial one list items match into another with a catch

python

list-comprehension

list