我可以在Python列表上创建一个“视图”吗?

Python标准库中没有“list slice”类(也没有内置的)。因此,您确实需要一个类,尽管它不需要很大——特别是如果您满足于“只读”和“紧凑”切片的话。例如。:

import collections

class ROListSlice(collections.Sequence):

    def __init__(self, alist, start, alen):
        self.alist = alist
        self.start = start
        self.alen = alen

    def __len__(self):
        return self.alen

    def adj(self, i):
        if i<0: i += self.alen
        return i + self.start

    def __getitem__(self, i):
        return self.alist[self.adj(i)]

这有一些限制(不支持“切片”),但在大多数情况下可能是可以的。

要使这个序列r/w,您需要添加 __setitem__ , __delitem__ insert

class ListSlice(ROListSlice):

    def __setitem__(self, i, v):
        self.alist[self.adj(i)] = v

    def __delitem__(self, i, v):
        del self.alist[self.adj(i)]
        self.alen -= 1

    def insert(self, i, v):
        self.alist.insert(self.adj(i), v)
        self.alen += 1

也许只需要使用numpy数组:

In [19]: import numpy as np

In [20]: l=np.arange(10)

returns a view ,不是副本:

In [21]: lv=l[3:6]

In [22]: lv
Out[22]: array([3, 4, 5])

改变 l lv

In [23]: l[4]=-1

In [24]: lv
Out[24]: array([ 3, -1,  5])

吕 影响 我

In [25]: lv[1]=4

In [26]: l
Out[26]: array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

l = [1,2,3,4,5]
lv = (l[i] for i in range(1,4))

lv.next()   # 2
l[2]=-1
lv.next()   # -1
lv.next()   # 4

然而,作为一个生成器,您只能浏览列表一次,如果您删除的元素比您所请求的多,它将爆炸 range .

子类化 more_itertools.SequenceView

代码

import more_itertools as mit


class SequenceView(mit.SequenceView):
    """Overload assignments in views."""
    def __setitem__(self, index, item):
        self._target[index] = item

演示

>>> seq = list(range(10))
>>> view = SequenceView(seq)
>>> view
SequenceView([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

>>> # Mutate Sequence -> Affect View
>>> seq[6] = -1
>>> view[5:8]
[5, -1, 7]

>>> # Mutate View -> Affect Sequence
>>> view[5] = -2
>>> seq[5:8]
[-2, -1, 7]

more_itertools 是第三方图书馆。通过安装 > pip install more_itertools

https://gist.github.com/mathieucaroff/0cf094325fb5294fb54c6a577f05a2c1

上面的链接是一个基于python3范围的解决方案,可以被切片和 以恒定时间索引。

它支持切片、相等比较、字符串转换( __str__ ),和 复制者( __repr__

创建SliceableSequenceView的SliceableSequenceView不会减慢速度

sequenceView.py

# stackoverflow.com/q/3485475/can-i-create-a-view-on-a-python-list

try:
    from collections.abc import Sequence
except ImportError:
    from collections import Sequence # pylint: disable=no-name-in-module

class SliceableSequenceView(Sequence):
    """
    A read-only sequence which allows slicing without copying the viewed list.
    Supports negative indexes.

    Usage:
        li = list(range(100))
        s = SliceableSequenceView(li)
        u = SliceableSequenceView(li, slice(1,7,2))
        v = s[1:7:2]
        w = s[-99:-93:2]
        li[1] += 10
        assert li[1:7:2] == list(u) == list(v) == list(w)
    """
    __slots__ = "seq range".split()
    def __init__(self, seq, sliced=None):
        """
        Accept any sequence (such as lists, strings or ranges).
        """
        if sliced is None:
            sliced = slice(len(seq))
        ls = looksSliceable = True
        ls = ls and hasattr(seq, "seq") and isinstance(seq.seq, Sequence)
        ls = ls and hasattr(seq, "range") and isinstance(seq.range, range)
        looksSliceable = ls
        if looksSliceable:
            self.seq = seq.seq
            self.range = seq.range[sliced]
        else:
            self.seq = seq
            self.range = range(len(seq))[sliced]

    def __len__(self):
        return len(self.range)

    def __getitem__(self, i):
        if isinstance(i, slice):
            return SliceableSequenceView(self.seq, i)
        return self.seq[self.range[i]]

    def __str__(self):
        r = self.range
        s = slice(r.start, r.stop, r.step)
        return str(self.seq[s])

    def __repr__(self):
        r = self.range
        s = slice(r.start, r.stop, r.step)
        return "SliceableSequenceView({!r})".format(self.seq[s])

    def equal(self, otherSequence):
        if self is otherSequence:
            return True
        if len(self) != len(otherSequence):
            return False
        for v, w in zip(self, otherSequence):
            if v != w:
                return False
        return True

The object argument must be an object that supports the buffer call interface (such as strings, arrays, and buffers). -所以不,很遗憾。

buffer type 就是你要找的。

从链接页粘贴示例:

>>> s = bytearray(1000000)   # a million zeroed bytes
>>> t = buffer(s, 1)         # slice cuts off the first byte
>>> s[1] = 5                 # set the second element in s
>>> t[0]                     # which is now also the first element in t!
'\x05' 

一旦你从一个列表中选取一个片段,你就会创建一个新的列表。好的,它将包含相同的对象,因此只要涉及列表中的对象,它就将是相同的,但是如果修改切片,则原始列表将保持不变。

collection.MutableSequence

class Sublist(collections.MutableSequence):
    def __init__(self, ls, beg, end):
        self.ls = ls
        self.beg = beg
        self.end = end
    def __getitem__(self, i):
        self._valid(i)
        return self.ls[self._newindex(i)]
    def __delitem__(self, i):
        self._valid(i)
        del self.ls[self._newindex(i)]
    def insert(self, i, x):
        self._valid(i)
        self.ls.insert(i+ self.beg, x)
    def __len__(self):
        return self.end - self.beg
    def __setitem__(self, i, x):
        self.ls[self._newindex(i)] = x
    def _valid(self, i):
        if isinstance(i, slice):
            self._valid(i.start)
            self._valid(i.stop)
        elif isinstance(i, int):
            if i<0 or i>=self.__len__():
                raise IndexError()
        else:
            raise TypeError()
    def _newindex(self, i):
        if isinstance(i, slice):
            return slice(self.beg + i.start, self.beg + i.stop, i.step)
        else:
            return i + self.beg

例子:

>>> a = list(range(10))
>>> s = Sublist(a, 3, 8)
>>> s[2:4]
[5, 6]
>>> s[2] = 15
>>> a
[0, 1, 2, 3, 4, 15, 6, 7, 8, 9]

像这样做

shiftedlist = type('ShiftedList',
                   (list,),
                   {"__getitem__": lambda self, i: list.__getitem__(self, i + 3)}
                  )([1, 2, 3, 4, 5, 6])

编辑: 我迟迟才意识到这是行不通的,因为 list() __len__ . 您需要使用代理类;看见 Mr. Martelli's answer

实际上,自己使用 range *您可以分割一个范围,它会为您完成所有复杂的运算:

>>> range(20)[10:]
range(10, 20)
>>> range(10, 20)[::2]
range(10, 20, 2)
>>> range(10, 20, 2)[::-3]
range(18, 8, -6)

因此,您只需要一个对象类,其中包含对原始序列的引用和一个范围。下面是这样一个类的代码(我希望不是太大):

class SequenceView:

    def __init__(self, sequence, range_object=None):
        if range_object is None:
            range_object = range(len(sequence))
        self.range    = range_object
        self.sequence = sequence

    def __getitem__(self, key):
        if type(key) == slice:
            return SequenceView(self.sequence, self.range[key])
        else:
            return self.sequence[self.range[key]]

    def __setitem__(self, key, value):
        self.sequence[self.range[key]] = value

    def __len__(self):
        return len(self.range)

    def __iter__(self):
        for i in self.range:
            yield self.sequence[i]

    def __repr__(self):
        return f"SequenceView({self.sequence!r}, {self.range!r})"

    def __str__(self):
        if type(self.sequence) == str:
            return ''.join(self)
        elif type(self.sequence) in (list, tuple):
            return str(type(self.sequence)(self))
        else:
            return repr(self)

用法:

>>> p = list(range(10))
>>> q = SequenceView(p)[3:6]
>>> print(q)
[3, 4, 5]
>>> q[1] = -1
>>> print(q)
[3, -1, 5]
>>> print(p)
[0, 1, 2, 3, -1, 5, 6, 7, 8, 9]

如果要按顺序访问“视图”,则可以使用itertools.islice(..) You can see the documentation for more info .

l = [1, 2, 3, 4, 5]
d = [1:3] #[2, 3]
d = itertools.islice(2, 3) # iterator yielding -> 2, 3