只是关于如何在String中获得所有列表元素的更多信息

a = ['a', 'b', 'c']
str = "a123" 
list(filter(lambda x:  x in str, a))

data = "firstName and favoriteFood"
mandatory_fields = ['firstName', 'lastName', 'age']


# for each
for field in mandatory_fields:
    if field not in data:
        print("Error, missing req field {0}".format(field));

# still fine, multiple if statements
if ('firstName' not in data or 
    'lastName' not in data or
    'age' not in data):
    print("Error, missing a req field");

# not very readable, list comprehension
missing_fields = [x for x in mandatory_fields if x not in data]
if (len(missing_fields)>0):
    print("Error, missing fields {0}".format(", ".join(missing_fields)));

为了增加regex的多样性:

import re

if any(re.findall(r'a|b|c', str, re.IGNORECASE)):
    print 'possible matches thanks to regex'
else:
    print 'no matches'

或者如果你的列表太长- any(re.findall(r'|'.join(a)， str, re.IGNORECASE))

只是关于如何在String中获得所有列表元素的更多信息

a = ['a', 'b', 'c']
str = "a123" 
list(filter(lambda x:  x in str, a))

如果a或str中的字符串变长，您应该小心。简单的解决方案是O(S*(A^2))，其中S是str的长度，A是A中所有字符串长度的总和。要获得更快的解决方案，请查看用于字符串匹配的Aho-Corasick算法，该算法在线性时间O(S+A)内运行。

如果您想要单词的精确匹配，那么可以考虑对目标字符串进行单词标记。我使用nltk推荐的word_tokenize:

from nltk.tokenize import word_tokenize

下面是接受答案的标记化字符串:

a_string = "A string is more than its parts!"
tokens = word_tokenize(a_string)
tokens
Out[46]: ['A', 'string', 'is', 'more', 'than', 'its', 'parts', '!']

接受的答案修改如下:

matches_1 = ["more", "wholesome", "milk"]
[x in tokens for x in matches_1]
Out[42]: [True, False, False]

在公认的答案中，单词“more”仍然是匹配的。但是，如果“mo”成为匹配字符串，接受的答案仍然找到匹配。这是我不希望看到的行为。

matches_2 = ["mo", "wholesome", "milk"]
[x in a_string for x in matches_1]
Out[43]: [True, False, False]

使用单词标记化，“mo”不再匹配:

[x in tokens for x in matches_2]
Out[44]: [False, False, False]

这是我想要的附加行为。这个答案也回答了这里的重复问题。