Python 如何检查字符串列表是否在另一个字符串列表中？

范例

我想检查一下我清单上的单词是否在描述中，如果是，请不要做任何事情。如果我的列表不在描述中，我希望返回字符串“Keywords not found”

---

我该如何编写此代码？

您可以使用双列表理解的

all

：

description = ['This is a random sentence. I like to travel and stuff','Hello world', 'Things on my bucket list, travel']

my_list = ['travel','fire']

您可以使用从

description

中的句子中提取所有单词，并检查其中是否存在

my_list

中的任何单词：

description = ['This is a random sentence. I like to travel and stuff','Hello world', 'Things on my bucket list, travel']
my_list = ['travel','fire']
def response():
   return "found" if any(i in b for i in my_list for b in description) else "Keywords not found"

哪些产出：

import re

def find_words(words, text):
    desc_words = re.findall(r'[^\s,.]+', "".join(text))

    for word in words:
        if word in desc_words:
            return "found"

    return "Keywords not found"

或者您可以使用此

set（）

方法：

>>> description = ['This is a random sentence. I like to travel and stuff','Hello world', 'Things on my bucket list, travel']
>>> my_list = ['travel','fire']
>>> find_words(my_list, description)
found

---

注意：如果您遇到除

、、
之外的标点符号不同的句子，则必须更新正则表达式。
 将单词保存在集合中，并检查
my_list
中的单词是否在集合中。只有当
我的列表中没有短语时，此才起作用。i、 e.
my_list
中的所有单词都是单字

def find_words(words, text):
    return "found" if set(words).intersection(re.findall(r'[^\s,.]+', "".join(text))) else "Keywords not found"

使用
isdisjoint

description = ['This is a random sentence. I like to travel and stuff','Hello world', 'Things on my bucket list, travel']
my_list = ['travel','fire']
set_words_in_description = set()
for s in description:  
    # add new words in set_words_in_description
    set_words_in_description.update(set(w for w in s.split())) 

你可以试试这样的设置吗

def find_word_method_disjoint(my_list, set_words_in_description):
    # check if my_list is disjoint with set_wrods_in_description
    return not set_words_in_description.isdisjoint(my_list)

%timeit find_word_method_disjoint(my_list, set_words_in_description)
189 ns ± 1.53 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

%timeit response()  # function given by the accepted answer.
572 ns ± 9.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

输出：

description = ['This is a random sentence. I like to travel and stuff','Hello world', 'Things on my bucket list, travel']

my_list = ['travel','fire']

flat=[k for i in description for k in i.split()]


print(any({i}.issubset(set(flat))for i in my_list))

“通常的方式”对你有帮助吗？你在
my_list
中的单词是否包含短语？可能重复的可能重复的可能重复的@ayodhyankittpaul是的。我不理解那些不发表评论的反对者。事实上，不用担心，兄弟，我们不能改变他们的行为：）是的，我对其他答案投了高票，因为把所有答案都投了反对票看起来很愚蠢。
True