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"

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))

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)));

flog = open('test.txt', 'r')
flogLines = flog.readlines()
strlist = ['SUCCESS', 'Done','SUCCESSFUL']
res = False
for line in flogLines:
     for fstr in strlist:
         if line.find(fstr) != -1:
            print('found') 
            res = True


if res:
    print('res true')
else: 
    print('res 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'。