有发电机版本的吗`字符串。拆分()`在Python中?

re.finditer 使用相当小的内存开销。

def split_iter(string):
    return (x.group(0) for x in re.finditer(r"[A-Za-z']+", string))

演示:

>>> list( split_iter("A programmer's RegEx test.") )
['A', "programmer's", 'RegEx', 'test']

编辑: 我刚刚确认了在python3.2.1中这需要恒定的内存,假设我的测试方法是正确的。我创建了一个非常大的字符串(1GB左右),然后用 for

更通用的版本:

str.split “,下面是一个更通用的版本:

def splitStr(string, sep="\s+"):
    # warning: does not yet work if sep is a lookahead like `(?=b)`
    if sep=='':
        return (c for c in string)
    else:
        return (_.group(1) for _ in re.finditer(f'(?:^|{sep})((?:(?!{sep}).)*)', string))

    # alternatively, more verbosely:
    regex = f'(?:^|{sep})((?:(?!{sep}).)*)'
    for match in re.finditer(regex, string):
        fragment = match.group(1)
        yield fragment

我们的想法是 ((?!pat).)* 通过确保组贪婪地匹配直到模式开始匹配(lookaheads不使用regex有限状态机中的字符串),来“否定”组。在伪代码中:重复使用( begin-of-string 异或 {sep} ) + as much as possible until we would be able to begin again (or hit end of string)

演示:

>>> splitStr('.......A...b...c....', sep='...')
<generator object splitStr.<locals>.<genexpr> at 0x7fe8530fb5e8>

>>> list(splitStr('A,b,c.', sep=','))
['A', 'b', 'c.']

>>> list(splitStr(',,A,b,c.,', sep=','))
['', '', 'A', 'b', 'c.', '']

>>> list(splitStr('.......A...b...c....', '\.\.\.'))
['', '', '.A', 'b', 'c', '.']

>>> list(splitStr('   A  b  c. '))
['', 'A', 'b', 'c.', '']

(应当指出 str.split 有一个丑陋的行为:它有特殊情况 sep=None 作为第一件事 str.strip "\s+"

r'(.*?)($|,)' ',,,a,,b,c' 退货 ['', '', '', 'a', '', 'b', 'c', '']

(如果您想自己实现它以获得更高的性能(尽管它们很重,regex最重要的是在C中运行),您应该编写一些代码(使用ctypes?不知道如何让发电机与它一起工作?),使用以下用于固定长度分隔符的伪代码:哈希长度为L的分隔符。在使用运行哈希算法扫描字符串时,保留长度为L的运行哈希,O(1)更新时间。每当散列可能等于您的分隔符时,手动检查过去的几个字符是否是分隔符;如果是,则从上次yield开始生成子字符串。字符串开头和结尾的特殊情况。这将是教科书算法的一个生成器版本,用于做O(N)文本搜索。多处理版本也是可能的。他们可能看起来杀伤力过大,但这个问题意味着一个人正在处理非常巨大的字符串。。。在这一点上,您可能会考虑一些疯狂的事情,比如缓存字节偏移量(如果它们很少的话),或者在磁盘上使用一些由testring view object支持的磁盘,购买更多的RAM等等。)

我能想到的最有效的方法就是用 offset str.find() 方法。这避免了大量的内存使用,并在不需要时依赖regexp的开销。

[编辑2016-8-2:更新此选项以可选地支持regex分隔符]

def isplit(source, sep=None, regex=False):
    """
    generator version of str.split()

    :param source:
        source string (unicode or bytes)

    :param sep:
        separator to split on.

    :param regex:
        if True, will treat sep as regular expression.

    :returns:
        generator yielding elements of string.
    """
    if sep is None:
        # mimic default python behavior
        source = source.strip()
        sep = "\\s+"
        if isinstance(source, bytes):
            sep = sep.encode("ascii")
        regex = True
    if regex:
        # version using re.finditer()
        if not hasattr(sep, "finditer"):
            sep = re.compile(sep)
        start = 0
        for m in sep.finditer(source):
            idx = m.start()
            assert idx >= start
            yield source[start:idx]
            start = m.end()
        yield source[start:]
    else:
        # version using str.find(), less overhead than re.finditer()
        sepsize = len(sep)
        start = 0
        while True:
            idx = source.find(sep, start)
            if idx == -1:
                yield source[start:]
                return
            yield source[start:idx]
            start = idx + sepsize

你想怎么用就怎么用。。。

>>> print list(isplit("abcb","b"))
['a','c','']

虽然每次执行find()或切片时字符串中都会有一点开销,但这应该是最小的,因为字符串在内存中表示为连续数组。

这是的生成器版本 split() 通过实施 re.search() 这不存在分配太多子字符串的问题。

import re

def itersplit(s, sep=None):
    exp = re.compile(r'\s+' if sep is None else re.escape(sep))
    pos = 0
    while True:
        m = exp.search(s, pos)
        if not m:
            if pos < len(s) or sep is not None:
                yield s[pos:]
            break
        if pos < m.start() or sep is not None:
            yield s[pos:m.start()]
        pos = m.end()


sample1 = "Good evening, world!"
sample2 = " Good evening, world! "
sample3 = "brackets][all][][over][here"
sample4 = "][brackets][all][][over][here]["

assert list(itersplit(sample1)) == sample1.split()
assert list(itersplit(sample2)) == sample2.split()
assert list(itersplit(sample3, '][')) == sample3.split('][')
assert list(itersplit(sample4, '][')) == sample4.split('][')

编辑:

• str.split (默认值=0.3461570239996945)
• 手动搜索(按字符)(Dave Webb的答案之一)=0.8260340550004912
• re.finditer
• str.find
• itertools.takewhile (伊格纳西奥·巴斯克斯·艾布拉姆斯的回答)=2.023023967998597
• str.split(..., maxsplit=1)

递归回答( string.split 具有 maxsplit = 1 )未能在合理的时间内完成 字符串。拆分

测试使用 timeit 日期:

the_text = "100 " * 9999 + "100"

def test_function( method ):
    def fn( ):
        total = 0

        for x in method( the_text ):
            total += int( x )

        return total

    return fn

这就提出了另一个问题,为什么 尽管它的内存使用率很高,但它的速度要快得多。

我只复制主文件的docstring str_split 功能:

str_split(s, *delims, empty=None)

分开绳子 s 其他的论点,可能省略了 空零件( empty 这是一个生成器函数。

当只提供一个分隔符时,字符串将被它简单地拆分。 那么 True 默认情况下。

str_split('[]aaa[][]bb[c', '[]')
    -> '', 'aaa', '', 'bb[c'
str_split('[]aaa[][]bb[c', '[]', empty=False)
    -> 'aaa', 'bb[c'

如果提供了多个分隔符,则字符串将按最长值拆分 设置为 是的 在这种情况下,分隔符只能是单个字符。

str_split('aaa, bb : c;', ' ', ',', ':', ';')
    -> 'aaa', 'bb', 'c'
str_split('aaa, bb : c;', *' ,:;', empty=True)
    -> 'aaa', '', 'bb', '', '', 'c', ''

如果没有提供分隔符, string.whitespace 与相同 str.split()

str_split('aaa\\t  bb c \\n')
    -> 'aaa', 'bb', 'c'

import string

def _str_split_chars(s, delims):
    "Split the string `s` by characters contained in `delims`, including the \
    empty parts between two consecutive delimiters"
    start = 0
    for i, c in enumerate(s):
        if c in delims:
            yield s[start:i]
            start = i+1
    yield s[start:]

def _str_split_chars_ne(s, delims):
    "Split the string `s` by longest possible sequences of characters \
    contained in `delims`"
    start = 0
    in_s = False
    for i, c in enumerate(s):
        if c in delims:
            if in_s:
                yield s[start:i]
                in_s = False
        else:
            if not in_s:
                in_s = True
                start = i
    if in_s:
        yield s[start:]


def _str_split_word(s, delim):
    "Split the string `s` by the string `delim`"
    dlen = len(delim)
    start = 0
    try:
        while True:
            i = s.index(delim, start)
            yield s[start:i]
            start = i+dlen
    except ValueError:
        pass
    yield s[start:]

def _str_split_word_ne(s, delim):
    "Split the string `s` by the string `delim`, not including empty parts \
    between two consecutive delimiters"
    dlen = len(delim)
    start = 0
    try:
        while True:
            i = s.index(delim, start)
            if start!=i:
                yield s[start:i]
            start = i+dlen
    except ValueError:
        pass
    if start<len(s):
        yield s[start:]


def str_split(s, *delims, empty=None):
    """\
Split the string `s` by the rest of the arguments, possibly omitting
empty parts (`empty` keyword argument is responsible for that).
This is a generator function.

When only one delimiter is supplied, the string is simply split by it.
`empty` is then `True` by default.
    str_split('[]aaa[][]bb[c', '[]')
        -> '', 'aaa', '', 'bb[c'
    str_split('[]aaa[][]bb[c', '[]', empty=False)
        -> 'aaa', 'bb[c'

When multiple delimiters are supplied, the string is split by longest
possible sequences of those delimiters by default, or, if `empty` is set to
`True`, empty strings between the delimiters are also included. Note that
the delimiters in this case may only be single characters.
    str_split('aaa, bb : c;', ' ', ',', ':', ';')
        -> 'aaa', 'bb', 'c'
    str_split('aaa, bb : c;', *' ,:;', empty=True)
        -> 'aaa', '', 'bb', '', '', 'c', ''

When no delimiters are supplied, `string.whitespace` is used, so the effect
is the same as `str.split()`, except this function is a generator.
    str_split('aaa\\t  bb c \\n')
        -> 'aaa', 'bb', 'c'
"""
    if len(delims)==1:
        f = _str_split_word if empty is None or empty else _str_split_word_ne
        return f(s, delims[0])
    if len(delims)==0:
        delims = string.whitespace
    delims = set(delims) if len(delims)>=4 else ''.join(delims)
    if any(len(d)>1 for d in delims):
        raise ValueError("Only 1-character multiple delimiters are supported")
    f = _str_split_chars if empty else _str_split_chars_ne
    return f(s, delims)

这个函数在python3中工作,并且可以应用一个简单但相当难看的修复程序使它在2和3版本中都工作。函数的第一行应更改为:

def str_split(s, *delims, **kwargs):
    """...docstring..."""
    empty = kwargs.get('empty')

itertools.takewhile()

编辑:

非常简单的、半途而废的实现:

import itertools
import string

def isplitwords(s):
  i = iter(s)
  while True:
    r = []
    for c in itertools.takewhile(lambda x: not x in string.whitespace, i):
      r.append(c)
    else:
      if r:
        yield ''.join(r)
        continue
      else:
        raise StopIteration()

split()

如果你想写一个,那就相当容易了:

import string

def gsplit(s,sep=string.whitespace):
    word = []

    for c in s:
        if c in sep:
            if word:
                yield "".join(word)
                word = []
        else:
            word.append(c)

    if word:
        yield "".join(word)

def isplit(string, delimiter = None):
    """Like string.split but returns an iterator (lazy)

    Multiple character delimters are not handled.
    """

    if delimiter is None:
        # Whitespace delimited by default
        delim = r"\s"

    elif len(delimiter) != 1:
        raise ValueError("Can only handle single character delimiters",
                        delimiter)

    else:
        # Escape, incase it's "\", "*" etc.
        delim = re.escape(delimiter)

    return (x.group(0) for x in re.finditer(r"[^{}]+".format(delim), string))

下面是我使用的测试(在Python3和Python2中):

# Wrapper to make it a list
def helper(*args,  **kwargs):
    return list(isplit(*args, **kwargs))

# Normal delimiters
assert helper("1,2,3", ",") == ["1", "2", "3"]
assert helper("1;2;3,", ";") == ["1", "2", "3,"]
assert helper("1;2 ;3,  ", ";") == ["1", "2 ", "3,  "]

# Whitespace
assert helper("1 2 3") == ["1", "2", "3"]
assert helper("1\t2\t3") == ["1", "2", "3"]
assert helper("1\t2 \t3") == ["1", "2", "3"]
assert helper("1\n2\n3") == ["1", "2", "3"]

# Surrounding whitespace dropped
assert helper(" 1 2  3  ") == ["1", "2", "3"]

# Regex special characters
assert helper(r"1\2\3", "\\") == ["1", "2", "3"]
assert helper(r"1*2*3", "*") == ["1", "2", "3"]

# No multi-char delimiters allowed
try:
    helper(r"1,.2,.3", ",.")
    assert False
except ValueError:
    pass

does "the right thing" 对于unicode空白,但我还没有实际测试过。

也可作为 gist

如果你也想 阅读 返回

import itertools as it

def iter_split(string, sep=None):
    sep = sep or ' '
    groups = it.groupby(string, lambda s: s != sep)
    return (''.join(g) for k, g in groups if k)

用法

>>> list(iter_split(iter("Good evening, world!")))
['Good', 'evening,', 'world!']

more_itertools.split_at 提供模拟到 str.split

>>> import more_itertools as mit


>>> list(mit.split_at("abcdcba", lambda x: x == "b"))
[['a'], ['c', 'd', 'c'], ['a']]

>>> "abcdcba".split("b")
['a', 'cdc', 'a']

more_itertools 是第三方软件包。

我想展示如何使用find iter解决方案为给定的分隔符返回一个生成器,然后使用itertools中的成对配方来构建上一个下一个迭代,该迭代将获得与原始split方法相同的实际单词。

from more_itertools import pairwise
import re

string = "dasdha hasud hasuid hsuia dhsuai dhasiu dhaui d"
delimiter = " "
# split according to the given delimiter including segments beginning at the beginning and ending at the end
for prev, curr in pairwise(re.finditer("^|[{0}]+|$".format(delimiter), string)):
    print(string[prev.end(): curr.start()])

2. 这是相当有效的

最愚蠢的方法,没有regex/itertools:

def isplit(text, split='\n'):
    while text != '':
        end = text.find(split)

        if end == -1:
            yield text
            text = ''
        else:
            yield text[:end]
            text = text[end + 1:]

def str_split(text: str, separator: str) -> Iterable[str]:
    i = 0
    n = len(text)
    while i <= n:
        j = text.find(separator, i)
        if j == -1:
            j = n
        yield text[i:j]
        i = j + 1

def split_generator(f,s):
    """
    f is a string, s is the substring we split on.
    This produces a generator rather than a possibly
    memory intensive list. 
    """
    i=0
    j=0
    while j<len(f):
        if i>=len(f):
            yield f[j:]
            j=i
        elif f[i] != s:
            i=i+1
        else:
            yield [f[j:i]]
            j=i+1
            i=i+1

下面是一个简单的回答

def gen_str(some_string, sep):
    j=0
    guard = len(some_string)-1
    for i,s in enumerate(some_string):
        if s == sep:
           yield some_string[j:i]
           j=i+1
        elif i!=guard:
           continue
        else:
           yield some_string[j:]

def isplit(text, sep=None, maxsplit=-1):
    if not isinstance(text, (str, bytes)):
        raise TypeError(f"requires 'str' or 'bytes' but received a '{type(text).__name__}'")
    if sep in ('', b''):
        raise ValueError('empty separator')

    if maxsplit == 0 or not text:
        yield text
        return

    regex = (
        re.escape(sep) if sep is not None
        else [br'\s+', r'\s+'][isinstance(text, str)]
    )
    yield from re.split(regex, text, maxsplit=max(0, maxsplit))