如果需要可逆操作（保留分隔符），可以使用此函数：

def tokenizeSentence_Reversible(sentence):
    setOfDelimiters = ['.', ' ', ',', '*', ';', '!']
    listOfTokens = [sentence]

    for delimiter in setOfDelimiters:
        newListOfTokens = []
        for ind, token in enumerate(listOfTokens):
            ll = [([delimiter, w] if ind > 0 else [w]) for ind, w in enumerate(token.split(delimiter))]
            listOfTokens = [item for sublist in ll for item in sublist] # flattens.
            listOfTokens = filter(None, listOfTokens) # Removes empty tokens: ''
            newListOfTokens.extend(listOfTokens)

        listOfTokens = newListOfTokens

    return listOfTokens

这是我的看法。。。。

def split_string(source,splitlist):
    splits = frozenset(splitlist)
    l = []
    s1 = ""
    for c in source:
        if c in splits:
            if s1:
                l.append(s1)
                s1 = ""
        else:
            print s1
            s1 = s1 + c
    if s1:
        l.append(s1)
    return l

>>>out = split_string("First Name,Last Name,Street Address,City,State,Zip Code",",")
>>>print out
>>>['First Name', 'Last Name', 'Street Address', 'City', 'State', 'Zip Code']

我正在重新熟悉Python，需要同样的东西。findall解决方案可能更好，但我想到了这个：

tokens = [x.strip() for x in data.split(',')]

这是一个有一些解释的答案。

st = "Hey, you - what are you doing here!?"

# replace all the non alpha-numeric with space and then join.
new_string = ''.join([x.replace(x, ' ') if not x.isalnum() else x for x in st])
# output of new_string
'Hey  you  what are you doing here  '

# str.split() will remove all the empty string if separator is not provided
new_list = new_string.split()

# output of new_list
['Hey', 'you', 'what', 'are', 'you', 'doing', 'here']

# we can join it to get a complete string without any non alpha-numeric character
' '.join(new_list)
# output
'Hey you what are you doing'

或者在一行中，我们可以这样做：

(''.join([x.replace(x, ' ') if not x.isalnum() else x for x in st])).split()

# output
['Hey', 'you', 'what', 'are', 'you', 'doing', 'here']

更新的答案

我最近需要这样做，但需要一个与标准库str.split函数有点匹配的函数，当使用0或1参数调用时，该函数的行为与标准库相同。

def split_many(string, *separators):
    if len(separators) == 0:
        return string.split()
    if len(separators) > 1:
        table = {
            ord(separator): ord(separator[0])
            for separator in separators
        }
        string = string.translate(table)
    return string.split(separators[0])

注意：此函数仅在分隔符由单个字符组成时有用（正如我的用例）。

另一种方式，不使用正则表达式

import string
punc = string.punctuation
thestring = "Hey, you - what are you doing here!?"
s = list(thestring)
''.join([o for o in s if not o in punc]).split()