如何检查数组中的任何字符串是否存在于另一个字符串中?
例如:
a = ['a', 'b', 'c']
s = "a123"
if a in s:
print("some of the strings found in s")
else:
print("no strings found in s")
我如何替换如果a在s:行得到适当的结果?
如何检查数组中的任何字符串是否存在于另一个字符串中?
例如:
a = ['a', 'b', 'c']
s = "a123"
if a in s:
print("some of the strings found in s")
else:
print("no strings found in s")
我如何替换如果a在s:行得到适当的结果?
当前回答
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'。
其他回答
python文档中推荐的regex模块支持这一点
words = {'he', 'or', 'low'}
p = regex.compile(r"\L<name>", name=words)
m = p.findall('helloworld')
print(m)
输出:
['he', 'low', 'or']
实现的一些细节:link
a = ['a', 'b', 'c']
str = "a123"
a_match = [True for match in a if match in str]
if True in a_match:
print "some of the strings found in str"
else:
print "no strings found in str"
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"
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']
>>>
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