这是最简单的解释。这是我正在使用的:

re.split('\W', 'foo/bar spam\neggs')
>>> ['foo', 'bar', 'spam', 'eggs']

这是我想要的:

someMethod('\W', 'foo/bar spam\neggs')
>>> ['foo', '/', 'bar', ' ', 'spam', '\n', 'eggs']

原因是我想把一个字符串分割成令牌,操作它,然后再把它组合在一起。


当前回答

另一个例子,在非字母数字上进行分割,并保留分隔符

import re
a = "foo,bar@candy*ice%cream"
re.split('([^a-zA-Z0-9])',a)

输出:

['foo', ',', 'bar', '@', 'candy', '*', 'ice', '%', 'cream']

解释

re.split('([^a-zA-Z0-9])',a)

() <- keep the separators
[] <- match everything in between
^a-zA-Z0-9 <-except alphabets, upper/lower and numbers.

其他回答

你也可以用字符串数组而不是正则表达式分割字符串,就像这样:

def tokenizeString(aString, separators):
    #separators is an array of strings that are being used to split the string.
    #sort separators in order of descending length
    separators.sort(key=len)
    listToReturn = []
    i = 0
    while i < len(aString):
        theSeparator = ""
        for current in separators:
            if current == aString[i:i+len(current)]:
                theSeparator = current
        if theSeparator != "":
            listToReturn += [theSeparator]
            i = i + len(theSeparator)
        else:
            if listToReturn == []:
                listToReturn = [""]
            if(listToReturn[-1] in separators):
                listToReturn += [""]
            listToReturn[-1] += aString[i]
            i += 1
    return listToReturn
    

print(tokenizeString(aString = "\"\"\"hi\"\"\" hello + world += (1*2+3/5) '''hi'''", separators = ["'''", '+=', '+', "/", "*", "\\'", '\\"', "-=", "-", " ", '"""', "(", ")"]))

将所有分隔符:(\W)替换为分隔符+ new_分隔符:(\W;) 由new_separator分隔符拆分:(;)

def split_and_keep(seperator, s):
  return re.split(';', re.sub(seperator, lambda match: match.group() + ';', s))

print('\W', 'foo/bar spam\neggs')

这里有一个简单的.split解决方案,不需要regex。

这是一个没有删除分隔符的Python split()的答案,所以不完全是最初的帖子所要求的,但另一个问题被关闭为这个问题的副本。

def splitkeep(s, delimiter):
    split = s.split(delimiter)
    return [substr + delimiter for substr in split[:-1]] + [split[-1]]

随机测试:

import random

CHARS = [".", "a", "b", "c"]
assert splitkeep("", "X") == [""]  # 0 length test
for delimiter in ('.', '..'):
    for _ in range(100000):
        length = random.randint(1, 50)
        s = "".join(random.choice(CHARS) for _ in range(length))
        assert "".join(splitkeep(s, delimiter)) == s

另一个例子,在非字母数字上进行分割,并保留分隔符

import re
a = "foo,bar@candy*ice%cream"
re.split('([^a-zA-Z0-9])',a)

输出:

['foo', ',', 'bar', '@', 'candy', '*', 'ice', '%', 'cream']

解释

re.split('([^a-zA-Z0-9])',a)

() <- keep the separators
[] <- match everything in between
^a-zA-Z0-9 <-except alphabets, upper/lower and numbers.

之前发布的一些答案,会重复分隔符,或者有一些我在自己的情况下遇到的其他错误。你可以使用这个函数:

def split_and_keep_delimiter(input, delimiter):
    result      = list()
    idx         = 0
    while delimiter in input:
        idx     = input.index(delimiter);
        result.append(input[0:idx+len(delimiter)])
        input = input[idx+len(delimiter):]
    result.append(input)
    return result