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
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")
其他回答
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"
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
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'。
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"
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)));