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]

对Abbafei的帖子的回应时间的快速比较:

import timeit

def func1():
    phrase = 'Lucky Dog'
    return any(i in 'LD' for i in phrase)

def func2():
    phrase = 'Lucky Dog'
    if ('L' in phrase) or ('D' in phrase):
        return True
    else:
        return False

if __name__ == '__main__': 
    func1_time = timeit.timeit(func1, number=100000)
    func2_time = timeit.timeit(func2, number=100000)
    print('Func1 Time: {0}\nFunc2 Time: {1}'.format(func1_time, func2_time))

输出:

Func1 Time: 0.0737484362111
Func2 Time: 0.0125144964371

因此，使用any时代码更紧凑，但使用条件时速度更快。

编辑:TL;DR—对于长字符串，if-then仍然比任何快得多!

我决定根据评论中提出的一些有效点来比较一个长随机字符串的计时:

# Tested in Python 2.7.14

import timeit
from string import ascii_letters
from random import choice

def create_random_string(length=1000):
    random_list = [choice(ascii_letters) for x in range(length)]
    return ''.join(random_list)

def function_using_any(phrase):
    return any(i in 'LD' for i in phrase)

def function_using_if_then(phrase):
    if ('L' in phrase) or ('D' in phrase):
        return True
    else:
        return False

if __name__ == '__main__':
    random_string = create_random_string(length=2000)
    func1_time = timeit.timeit(stmt="function_using_any(random_string)",
                               setup="from __main__ import function_using_any, random_string",
                               number=200000)
    func2_time = timeit.timeit(stmt="function_using_if_then(random_string)",
                               setup="from __main__ import function_using_if_then, random_string",
                               number=200000)
    print('Time for function using any: {0}\nTime for function using if-then: {1}'.format(func1_time, func2_time))

输出:

Time for function using any: 0.1342546
Time for function using if-then: 0.0201827

If-then几乎比任何一个数量级都快!

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

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]

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'$']

s=input("Enter any character:")   
if s.isalnum():   
   print("Alpha Numeric Character")   
   if s.isalpha():   
       print("Alphabet character")   
       if s.islower():   
         print("Lower case alphabet character")   
       else:   
         print("Upper case alphabet character")   
   else:   
     print("it is a digit")   
elif s.isspace():   
    print("It is space character")   

其他: print(“非空格特殊字符”)