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。
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
我可以把它放在这里吗
s = 'foo/bar spam\neggs'
print(s.replace('/', '+++/+++').replace(' ', '+++ +++').replace('\n', '+++\n+++').split('+++'))
['foo', '/', 'bar', ' ', 'spam', '\n', 'eggs']
其他回答
一个懒惰而简单的解决方案
假设你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,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]')
如果你在换行上分割,使用splitlines(True)。
>>> 'line 1\nline 2\nline without newline'.splitlines(True)
['line 1\n', 'line 2\n', 'line without newline']
(不是一个通用的解决方案,但在这里添加这个,以防有人来到这里没有意识到这个方法的存在。)
在下面的代码中,对这个问题有一个简单、高效且经过测试的答案。代码中有解释其中所有内容的注释。
我保证它并不像看起来那么可怕——它实际上只有13行代码!其余的都是注释、文档和断言
def split_including_delimiters(input: str, delimiter: str):
"""
Splits an input string, while including the delimiters in the output
Unlike str.split, we can use an empty string as a delimiter
Unlike str.split, the output will not have any extra empty strings
Conequently, len(''.split(delimiter))== 0 for all delimiters,
whereas len(input.split(delimiter))>0 for all inputs and delimiters
INPUTS:
input: Can be any string
delimiter: Can be any string
EXAMPLES:
>>> split_and_keep_delimiter('Hello World ! ',' ')
ans = ['Hello ', 'World ', ' ', '! ', ' ']
>>> split_and_keep_delimiter("Hello**World**!***", "**")
ans = ['Hello', '**', 'World', '**', '!', '**', '*']
EXAMPLES:
assert split_and_keep_delimiter('-xx-xx-','xx') == ['-', 'xx', '-', 'xx', '-'] # length 5
assert split_and_keep_delimiter('xx-xx-' ,'xx') == ['xx', '-', 'xx', '-'] # length 4
assert split_and_keep_delimiter('-xx-xx' ,'xx') == ['-', 'xx', '-', 'xx'] # length 4
assert split_and_keep_delimiter('xx-xx' ,'xx') == ['xx', '-', 'xx'] # length 3
assert split_and_keep_delimiter('xxxx' ,'xx') == ['xx', 'xx'] # length 2
assert split_and_keep_delimiter('xxx' ,'xx') == ['xx', 'x'] # length 2
assert split_and_keep_delimiter('x' ,'xx') == ['x'] # length 1
assert split_and_keep_delimiter('' ,'xx') == [] # length 0
assert split_and_keep_delimiter('aaa' ,'xx') == ['aaa'] # length 1
assert split_and_keep_delimiter('aa' ,'xx') == ['aa'] # length 1
assert split_and_keep_delimiter('a' ,'xx') == ['a'] # length 1
assert split_and_keep_delimiter('' ,'' ) == [] # length 0
assert split_and_keep_delimiter('a' ,'' ) == ['a'] # length 1
assert split_and_keep_delimiter('aa' ,'' ) == ['a', '', 'a'] # length 3
assert split_and_keep_delimiter('aaa' ,'' ) == ['a', '', 'a', '', 'a'] # length 5
"""
# Input assertions
assert isinstance(input,str), "input must be a string"
assert isinstance(delimiter,str), "delimiter must be a string"
if delimiter:
# These tokens do not include the delimiter, but are computed quickly
tokens = input.split(delimiter)
else:
# Edge case: if the delimiter is the empty string, split between the characters
tokens = list(input)
# The following assertions are always true for any string input and delimiter
# For speed's sake, we disable this assertion
# assert delimiter.join(tokens) == input
output = tokens[:1]
for token in tokens[1:]:
output.append(delimiter)
if token:
output.append(token)
# Don't let the first element be an empty string
if output[:1]==['']:
del output[0]
# The only case where we should have an empty string in the output is if it is our delimiter
# For speed's sake, we disable this assertion
# assert delimiter=='' or '' not in output
# The resulting strings should be combinable back into the original string
# For speed's sake, we disable this assertion
# assert ''.join(output) == input
return output
我在尝试分割文件路径时遇到了类似的问题,并努力找到一个简单的答案。 这对我来说很有效,并且不需要将分隔符替换回分割文本:
My_path = 'folder1/folder2/folder3/file1'
进口再保险
re.findall ('[^/]+/|[^/]+', my_path)
返回:
['folder1/', 'folder2/', 'folder3/', 'file1']
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