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

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

其他回答

字符串。标点符号包含以下字符:

'!"#$%&\'()*+,-./:;<=>?@[\]^_`{|}~'

可以使用translate和maketrans函数将标点符号映射为空值(替换)

import string

'This, is. A test!'.translate(str.maketrans('', '', string.punctuation))

输出:

'This is A test'

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

这将删除字符串中的所有特殊字符、标点符号和空格,只包含数字和字母。

import re

sample_str = "Hel&&lo %% Wo$#rl@d"

# using isalnum()
print("".join(k for k in sample_str if k.isalnum()))


# using regex
op2 = re.sub("[^A-Za-z]", "", sample_str)
print(f"op2 = ", op2)


special_char_list = ["$", "@", "#", "&", "%"]

# using list comprehension
op1 = "".join([k for k in sample_str if k not in special_char_list])
print(f"op1 = ", op1)


# using lambda function
op3 = "".join(filter(lambda x: x not in special_char_list, sample_str))
print(f"op3 = ", op3)

使用翻译:

import string

def clean(instr):
    return instr.translate(None, string.punctuation + ' ')

警告:仅适用于ascii字符串。

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"))