在Python中,如何拆分字符串并保留分隔符？

Question

这是解释这个问题的最简单方法.这是我正在使用的:

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

这就是我想要的:

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

原因是我想将一个字符串拆分成标记,操纵它,然后再将它重新组合在一起.

Answer 1

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

Answer 2

如果要拆分换行符,请使用splitlines(True).

>>> 'line 1\nline 2\nline without newline'.splitlines(True)
['line 1\n', 'line 2\n', 'line without newline']

(不是一般的解决方案,但是如果有人来这里没有意识到这种方法存在的话,可以在这里添加它.)

Answer 3

另一个在Python 3上运行良好的非正则表达式解决方案

# Split strings and keep separator
test_strings = ['<Hello>', 'Hi', '<Hi> <Planet>', '<', '']

def split_and_keep(s, sep):
   if not s: return [''] # consistent with string.split()

   # Find replacement character that is not used in string
   # i.e. just use the highest available character plus one
   # Note: This fails if ord(max(s)) = 0x10FFFF (ValueError)
   p=chr(ord(max(s))+1) 

   return s.replace(sep, sep+p).split(p)

for s in test_strings:
   print(split_and_keep(s, '<'))


# If the unicode limit is reached it will fail explicitly
unicode_max_char = chr(1114111)
ridiculous_string = '<Hello>'+unicode_max_char+'<World>'
print(split_and_keep(ridiculous_string, '<'))

Answer 4

如果您只有一个分隔符,则可以使用列表推导:

text = 'foo,bar,baz,qux'  
sep = ','

附加/预先分隔符:

result = [x+sep for x in text.split(sep)]
#['foo,', 'bar,', 'baz,', 'qux,']
# to get rid of trailing
result[-1] = result[-1].strip(sep)
#['foo,', 'bar,', 'baz,', 'qux']

result = [sep+x for x in text.split(sep)]
#[',foo', ',bar', ',baz', ',qux']
# to get rid of trailing
result[0] = result[0].strip(sep)
#['foo', ',bar', ',baz', ',qux']

分隔符作为它自己的元素:

result = [u for x in text.split(sep) for u in (x, sep)]
#['foo', ',', 'bar', ',', 'baz', ',', 'qux', ',']
results = result[:-1]   # to get rid of trailing

Answer 5

另一个示例，拆分非字母数字并保留分隔符

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

输出：

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

说明

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

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

Answer 6

一种懒惰且简单的解决方案

假设您的正则表达式模式是split_pattern = r'(!|\?)'

首先，添加一些与新分隔符相同的字符，例如“[cut]”

new_string = re.sub(split_pattern, '\\1[cut]', your_string)

然后拆分新的分隔符new_string.split('[cut]')。

Answer 7

这是一个无需正则表达式即可工作的简单.split解决方案。

这是Python split() 的答案，没有删除分隔符，所以不完全是原始帖子所要求的，但另一个问题作为这个问题的重复项被关闭。

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

Answer 8

您还可以使用字符串数组而不是正则表达式来分割字符串，如下所示：

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 = ["'''", '+=', '+', "/", "*", "\\'", '\\"', "-=", "-", " ", '"""', "(", ")"]))