在另一个字符串列表中查找多个字符串的一种紧凑方法是使用set.intersection。这比大型集或列表中的列表理解执行得快得多。

>>> astring = ['abc','def','ghi','jkl','mno']
>>> bstring = ['def', 'jkl']
>>> a_set = set(astring)  # convert list to set
>>> b_set = set(bstring)
>>> matches = a_set.intersection(b_set)
>>> matches
{'def', 'jkl'}
>>> list(matches) # if you want a list instead of a set
['def', 'jkl']
>>>

的元素上进行迭代。

a = ['a', 'b', 'c']
str = "a123"
found_a_string = False
for item in a:    
    if item in str:
        found_a_string = True

if found_a_string:
    print "found a match"
else:
    print "no match found"

这取决于上下文 假设你想检查单个文字(任何单个单词a,e,w，..等)就足够了

original_word ="hackerearcth"
for 'h' in original_word:
      print("YES")

如果你想检查original_word中的任何一个字符: 利用

if any(your_required in yourinput for your_required in original_word ):

如果您想要original_word中的所有输入，请使用所有输入 简单的

original_word = ['h', 'a', 'c', 'k', 'e', 'r', 'e', 'a', 'r', 't', 'h']
yourinput = str(input()).lower()
if all(requested_word in yourinput for requested_word in original_word):
    print("yes")

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。

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)));