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('$')==-1 # not founds.find('$')!=-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 charactersany(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 subsringsany(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 timeitdef 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 Falseif __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: 0.0737484362111Func2 Time: 0.0125144964371因此，任何代码都更紧凑，而条件代码则更快。