下面是获取可迭代对象“尾部”项的几个选项:

鉴于

n = 9
iterable = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

期望输出值

[2, 3, 4, 5, 6, 7, 8, 9, 10]

Code

我们使用以下任何选项来获得后一个输出:

from collections import deque
import itertools

import more_itertools


# A: Slicing
iterable[-n:]


# B: Implement an itertools recipe
def tail(n, iterable):
    """Return an iterator over the last *n* items of *iterable*.

        >>> t = tail(3, 'ABCDEFG')
        >>> list(t)
        ['E', 'F', 'G']

    """
    return iter(deque(iterable, maxlen=n))
list(tail(n, iterable))


# C: Use an implemented recipe, via more_itertools
list(more_itertools.tail(n, iterable))


# D: islice, via itertools
list(itertools.islice(iterable, len(iterable)-n, None))


# E: Negative islice, via more_itertools
list(more_itertools.islice_extended(iterable, -n, None))

细节

A. Traditional Python slicing is inherent to the language. This option works with sequences such as strings, lists and tuples. However, this kind of slicing does not work on iterators, e.g. iter(iterable). B. An itertools recipe. It is generalized to work on any iterable and resolves the iterator issue in the last solution. This recipe must be implemented manually as it is not officially included in the itertools module. C. Many recipes, including the latter tool (B), have been conveniently implemented in third party packages. Installing and importing these these libraries obviates manual implementation. One of these libraries is called more_itertools (install via > pip install more-itertools); see more_itertools.tail. D. A member of the itertools library. Note, itertools.islice does not support negative slicing. E. Another tool is implemented in more_itertools that generalizes itertools.islice to support negative slicing; see more_itertools.islice_extended.

我该用哪一个?

视情况而定。在大多数情况下，切片(选项A，正如在其他答案中提到的)是最简单的选项，因为它内置在语言中，并支持大多数可迭代类型。对于更通用的迭代器，可以使用其余任何选项。注意，选项C和E需要安装第三方库，一些用户可能会发现这很有用。

下面是获取可迭代对象“尾部”项的几个选项:

鉴于

n = 9
iterable = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

期望输出值

[2, 3, 4, 5, 6, 7, 8, 9, 10]

Code

我们使用以下任何选项来获得后一个输出:

from collections import deque
import itertools

import more_itertools


# A: Slicing
iterable[-n:]


# B: Implement an itertools recipe
def tail(n, iterable):
    """Return an iterator over the last *n* items of *iterable*.

        >>> t = tail(3, 'ABCDEFG')
        >>> list(t)
        ['E', 'F', 'G']

    """
    return iter(deque(iterable, maxlen=n))
list(tail(n, iterable))


# C: Use an implemented recipe, via more_itertools
list(more_itertools.tail(n, iterable))


# D: islice, via itertools
list(itertools.islice(iterable, len(iterable)-n, None))


# E: Negative islice, via more_itertools
list(more_itertools.islice_extended(iterable, -n, None))

细节

A. Traditional Python slicing is inherent to the language. This option works with sequences such as strings, lists and tuples. However, this kind of slicing does not work on iterators, e.g. iter(iterable). B. An itertools recipe. It is generalized to work on any iterable and resolves the iterator issue in the last solution. This recipe must be implemented manually as it is not officially included in the itertools module. C. Many recipes, including the latter tool (B), have been conveniently implemented in third party packages. Installing and importing these these libraries obviates manual implementation. One of these libraries is called more_itertools (install via > pip install more-itertools); see more_itertools.tail. D. A member of the itertools library. Note, itertools.islice does not support negative slicing. E. Another tool is implemented in more_itertools that generalizes itertools.islice to support negative slicing; see more_itertools.islice_extended.

我该用哪一个?

视情况而定。在大多数情况下，切片(选项A，正如在其他答案中提到的)是最简单的选项，因为它内置在语言中，并支持大多数可迭代类型。对于更通用的迭代器，可以使用其余任何选项。注意，选项C和E需要安装第三方库，一些用户可能会发现这很有用。

切片

Python切片是一种非常快的操作，它是快速访问部分数据的方便方法。

Slice表示法从列表(或任何其他支持它的序列，如字符串)中获取最后9个元素，看起来像这样:

num_list[-9:]

当我看到这个时，我把括号里的部分读为“从末尾到末尾的第9个部分”。(实际上，我把它缩写为“-9,on”)

解释:

完整的符号是

sequence[start:stop:step]

但是冒号告诉Python你给它的是一个切片，而不是一个常规的索引。这就是为什么Python 2中复制列表的惯用方法是

list_copy = sequence[:]

清除它们的方法是:

del my_list[:]

(列表得到列表。复制并列出。在Python 3中清除。)

给你的切片起一个描述性的名字!

您可能会发现将切片的形成与将其传递给列表分开是很有用的。__getitem__方法(这就是方括号的作用)。即使您不是新手，它也可以使您的代码更具可读性，以便其他可能需要阅读您的代码的人可以更容易地理解您在做什么。

但是，不能将一些用冒号分隔的整数赋值给变量。你需要使用slice对象:

last_nine_slice = slice(-9, None)

第二个参数None是必需的，因此第一个参数被解释为开始参数，否则它将是停止参数。

然后你可以将slice对象传递给你的序列:

>>> list(range(100))[last_nine_slice]
[91, 92, 93, 94, 95, 96, 97, 98, 99]

islice

来自itertools模块的Islice是另一种可能实现此功能的性能方式。Islice不接受负参数，所以理想情况下，你的iterable有一个__reversed__特殊方法——list有——所以你必须首先将你的list(或带有__reversed__的iterable)传递给reversed。

>>> from itertools import islice
>>> islice(reversed(range(100)), 0, 9)
<itertools.islice object at 0xffeb87fc>

Islice允许对数据管道进行惰性求值，因此要物化数据，将其传递给构造函数(如list):

>>> list(islice(reversed(range(100)), 0, 9))
[99, 98, 97, 96, 95, 94, 93, 92, 91]

最后9个元素可以使用numlist[-9:]从左到右读取，也可以使用numlist[:-10:-1]从右到左读取。

>>> a=range(17)
>>> print a
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
>>> print a[-9:]
[8, 9, 10, 11, 12, 13, 14, 15, 16]
>>> print a[:-10:-1]
[16, 15, 14, 13, 12, 11, 10, 9, 8]

你可以使用负整数和切片运算符。下面是一个使用python CLI解释器的示例:

>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
>>> a[-9:]
[4, 5, 6, 7, 8, 9, 10, 11, 12]

重要的一行是[-9:]

负索引将从列表的末尾开始计数，因此:

num_list[-9:]