问题:如何检查字符串中的特定字符?[关闭]

如何使用Python 2检查字符串中是否包含几个特定字符?

例如,给定以下字符串:

罪犯偷走了1,000,000美元的珠宝。

如何检测它是否带有美元符号(“ $”),逗号(“,”)和数字?

How can I check if a string has several specific characters in it using Python 2?

For example, given the following string:

The criminals stole $1,000,000 in jewels.

How do I detect if it has dollar signs (“$”), commas (“,”), and numbers?


回答 0

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')

[编辑:添加'$' in s答案]

Assuming your string is 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

And so on for other characters.

… or

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

… or

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

[Edit: added the '$' in s answers]


回答 1

用户Jochen Ritzel在评论用户dappawit对此问题的答案时说了这一点。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,请参见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

user Jochen Ritzel said this in a comment to an answer to this question from user dappawit. It should work:

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

‘1’, ‘2’, etc. should be replaced with the characters you are looking for.

See this page in the Python 2.7 documentation for some information on strings, including about using the in operator for substring tests.

Update: This does the same job as my above suggestion with less repetition:

# 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

回答 2

快速比较时间,以回应Abbafei的帖子:

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。


编辑: TL; DR-对于长字符串,if-then 仍然比任何字符串都快得多!

我决定根据评论中提出的一些有效点来比较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

如果-那么几乎比任何一个都快一个数量级!

Quick comparison of timings in response to the post by Abbafei:

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))

Output:

Func1 Time: 0.0737484362111
Func2 Time: 0.0125144964371

So the code is more compact with any, but faster with the conditional.


EDIT : TL;DR — For long strings, if-then is still much faster than any!

I decided to compare the timing for a long random string based on some of the valid points raised in the comments:

# 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))

Output:

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

If-then is almost an order of magnitude faster than any!


回答 3

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。那是您要找的东西吗?

汇入

s1 ='测试字符串'
s2 ='1234,12345 $'

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

如果(regex.match(s1)):
   打印“ s1匹配”
其他:
   打印“ s1不匹配”

如果(regex.match(s2)):
   打印“ s2匹配”
其他:
   打印“ s2不匹配”

This will test if strings are made up of some combination or digits, the dollar sign, and a commas. Is that what you’re looking for?

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"

回答 4

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")   

否则:
print(“非空间特殊字符”)

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")   

else:
print(“Non Space Special Character”)


声明:本站所有文章,如无特殊说明或标注,均为本站原创发布。任何个人或组织,在未征得本站同意时,禁止复制、盗用、采集、发布本站内容到任何网站、书籍等各类媒体平台。如若本站内容侵犯了原著者的合法权益,可联系我们进行处理。