我需要从字符串中删除所有特殊字符,标点符号和空格,以便我只有字母和数字。


当前回答

import re
my_string = """Strings are amongst the most popular data types in Python. We can create the strings by enclosing characters in quotes. Python treats single quotes the 

和双引号一样。”""

# if we need to count the word python that ends with or without ',' or '.' at end

count = 0
for i in text:
    if i.endswith("."):
        text[count] = re.sub("^([a-z]+)(.)?$", r"\1", i)
    count += 1
print("The count of Python : ", text.count("python"))

其他回答

Python 2 . *

我认为只要filter(str。Isalnum,字符串)工作

In [20]: filter(str.isalnum, 'string with special chars like !,#$% etcs.')
Out[20]: 'stringwithspecialcharslikeetcs'

Python 3。*

在Python3中,filter()函数将返回一个可迭代对象(而不是与上面不同的字符串)。从itertable中获取字符串必须返回连接:

''.join(filter(str.isalnum, string)) 

或者在连接中传递列表(不确定,但可以快一点)

''.join([*filter(str.isalnum, string)])

注意:unpacking in [*args] valid from Python >= 3.5

下面是一个正则表达式,用于匹配不是字母或数字的字符串:

[^A-Za-z0-9]+

下面是执行正则表达式替换的Python命令:

re.sub('[^A-Za-z0-9]+', '', mystring)
import re
my_string = """Strings are amongst the most popular data types in Python. We can create the strings by enclosing characters in quotes. Python treats single quotes the 

和双引号一样。”""

# if we need to count the word python that ends with or without ',' or '.' at end

count = 0
for i in text:
    if i.endswith("."):
        text[count] = re.sub("^([a-z]+)(.)?$", r"\1", i)
    count += 1
print("The count of Python : ", text.count("python"))

这将删除除空格外的所有非字母数字字符。

string = "Special $#! characters   spaces 888323"
''.join(e for e in string if (e.isalnum() or e.isspace()))

特殊字符空格888323

import re
abc = "askhnl#$%askdjalsdk"
ddd = abc.replace("#$%","")
print (ddd)

你会看到你的结果是

'Askhnlaskdjalsdk