执行部分列表项匹配到另一个与捕获

如何解决执行部分列表项匹配到另一个与捕获

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 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'

我尝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]

解决方法

如果可以，您应该考虑为变量赋予有意义的名称，这有助于您编写代码

你想要的是

• 遍历句子
• 每次检查没有来自 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']

,

您必须在条件表达式中使用单独的列表推导式。在您的理解中，您为每个不在 j 中的 for i in list_1 添加了一个 i j，这就是您得到重复的原因。

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

,

您想检查短语中是否出现任何单词，因此 any 是要走的路。在我看来，这比使用 all 和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']。

---

解决方案大致相当于：

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)

,

如何使用 set 交叉口？

list_1 = ['world','rock']
list_2 = ['the world is big','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']