如何检查字符串的特定字符?

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"

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:添加'$' in s答案]

用户Jochen Ritzel在回复用户ddapapwit的问题时说了这句话。 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，请参阅Python 2.7文档中的该页，包括使用in操作符进行子字符串测试。

更新:这与我上面的建议相同，但重复次数较少:

# 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

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: < em > TL;博士< / em >——对于长字符串，if-then是仍然比任何字符串都快!

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!

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")

我的简单，简单，简单的方法!= D

代码

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!

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 ....')

或

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

输出:[True, True, False]

另一种方法，可能是神谕的，是这样的:

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