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
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
其他回答
re.split的文档中提到:
根据出现的模式拆分字符串。如果捕获 括号是在模式中使用的,然后是文本中的所有组 模式也作为结果列表的一部分返回。
所以你只需要用一个捕获组来包装分隔符:
>>> re.split('(\W)', 'foo/bar spam\neggs')
['foo', '/', 'bar', ' ', 'spam', '\n', 'eggs']
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 = ["'''", '+=', '+', "/", "*", "\\'", '\\"', "-=", "-", " ", '"""', "(", ")"]))
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
另一个在Python 3上工作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, '<'))
如果你想拆分字符串,同时通过regex保留分隔符,而不捕获组:
def finditer_with_separators(regex, s):
matches = []
prev_end = 0
for match in regex.finditer(s):
match_start = match.start()
if (prev_end != 0 or match_start > 0) and match_start != prev_end:
matches.append(s[prev_end:match.start()])
matches.append(match.group())
prev_end = match.end()
if prev_end < len(s):
matches.append(s[prev_end:])
return matches
regex = re.compile(r"[\(\)]")
matches = finditer_with_separators(regex, s)
如果假设regex被封装到捕获组中:
def split_with_separators(regex, s):
matches = list(filter(None, regex.split(s)))
return matches
regex = re.compile(r"([\(\)])")
matches = split_with_separators(regex, s)
这两种方法也将删除空组,在大多数情况下是无用和恼人的。