如何检查字符串中的特定字符？

Question

如何使用Python2检查字符串值中是否包含确切的字符？具体来说，我正在寻找是否有美元符号（“ $”），逗号（“，”）和数字。

Answer 1

假设您的字符串是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

对于其他字符，依此类推。

... 要么

pattern = re.compile(r'\d\$,')

if pattern.findall(s):

    print('Found')

else

    print('Not found')

... 要么

chars = set('0123456789$,')

if any((c in chars) for c in s):

    print('Found')

else:

    print('Not Found')

Answer 2

它应该工作：

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

“ 1”，“ 2”等应替换为您要查找的字符。

有关字符串的一些信息，请参阅Python 2.7文档中的此页面，包括有关使用in运算符进行子字符串测试的信息。

更新：这与我上面的建议做的工作相同，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

Answer 3

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，任何代码都更紧凑，而条件代码则更快。