检查字符串中是否有字符:

parse_string = lambda chars, string: [char in string for char in chars]

例子:

parse_string('$,x', 'The criminals stole $1,000,000 in ....') 

or

parse_string(['$', ',', 'x'], '..minals stole $1,000,000 i..')

输出:[True, True, False]

这将测试字符串是否由某些组合或数字、美元符号和逗号组成。这就是你想要的吗?

import re

s1 = 'Testing string'
s2 = '1234,12345$'

regex = re.compile('[0-9,$]+$')

if ( regex.match(s1) ):
   print "s1 matched"
else:
   print "s1 didn't match"

if ( regex.match(s2) ):
   print "s2 matched"
else:
   print "s2 didn't match"

我的简单，简单，简单的方法!= D

Code

string_to_test = "The criminals stole $1,000,000 in jewels."
chars_to_check = ["$", ",", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
for char in chars_to_check:
    if char in string_to_test:
        print("Char \"" + char + "\" detected!")

输出

Char "$" detected!
Char "," detected!
Char "0" detected!
Char "1" detected!

假设你的字符串是s:

'$' in s        # found
'$' not in s    # not found

# original answer given, but less Pythonic than the above...
s.find('$')==-1 # not found
s.find('$')!=-1 # found

其他角色也是如此。

．.． 或

pattern = re.compile(r'\d\$,')
if pattern.findall(s):
    print('Found')
else
    print('Not found')

．.． 或

chars = set('0123456789$,')
if any((c in chars) for c in s):
    print('Found')
else:
    print('Not Found')

[编辑:在s的答案中增加了'$']

另一种可能是python式的方法是:

aString = """The criminals stole $1,000,000 in jewels."""
#
if any(list(map(lambda char: char in aString, '0123456789,$')))
   print(True) # Do something.

用户Jochen Ritzel在回复用户dappawit的问题时这样说。 它应该工作:

('1' in var) and ('2' in var) and ('3' in var) ...

'1'， '2'等应替换为您正在寻找的字符。

有关字符串的一些信息，包括关于在子字符串测试中使用in操作符的信息，请参阅Python 2.7文档中的此页。

更新:这和我上面的建议是一样的，而且重复次数更少:

# When looking for single characters, this checks for any of the characters...
# ...since strings are collections of characters
any(i in '<string>' for i in '123')
# any(i in 'a' for i in '123') -> False
# any(i in 'b3' for i in '123') -> True

# And when looking for subsrings
any(i in '<string>' for i in ('11','22','33'))
# any(i in 'hello' for i in ('18','36','613')) -> False
# any(i in '613 mitzvahs' for i in ('18','36','613')) ->True