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']
>>>

这是set的另一个解。使用set.intersection。对于一行代码。

subset = {"some" ,"words"} 
text = "some words to be searched here"
if len(subset & set(text.split())) == len(subset):
   print("All values present in text")

if subset & set(text.split()):
   print("Atleast one values present in text")

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")

为了提高速度，我会使用这样的函数:

def check_string(string, substring_list):
    for substring in substring_list:
        if substring in string:
            return True
    return False

一个惊人的快速方法是使用set:

a = ['a', 'b', 'c']
str = "a123"
if set(a) & set(str):
    print("some of the strings found in str")
else:
    print("no strings found in str")

如果a不包含任何多字符值(在这种情况下使用上面列出的any)，则此方法有效。如果是这样，将a指定为字符串会更简单:a = 'abc'。

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)));