Question

我遇到使用选定的元素上的枚举器形成可迭代的（即序列或迭代器< / em>或类似的）并且希望返回 原始索引而不是默认count，从0开始，然后转到len(iterable) - 1。

一种非常天真的方法是声明一个新的生成器对象 ，称为_enumerate()

>>> def _enumerate(iterable, offset = 0, step = 1): index = offset for element in iterable: yield index, element index += step

...一个新的列表对象months 。

>>> months = ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"]

使用Pythons 内置enumerate函数会为[5::2]切片产生输出：

>>> for index, element in enumerate(months[5::2]): print(index, element) 0 June 1 August 2 October 3 December

我们自己的枚举器的 预期输出 _enumerate再次针对[5::2]切片：

>>> for index, element in _enumerate(months[5::2], offset = 5, step = 2): print(index, element) 5 June 7 August 9 October 11 December

你知道更好，更pythonic和更易读的解决方案吗？：）

Answer 1

这里我的评论作为答案;）

months = ["January", "February", "March", "April", "May", "June",
          "July", "August", "September", "October", "November", "December"]

offset = 5
step = 2
for index, element in enumerate(months[offset::step]):

    # recalculate original index
    index = offset + index*step

    # actually repetition of the month is trivial,
    # but I put it just to show that the index is right
    print(index, element, months[index])

打印：

5 June June
7 August August
9 October October
11 December December

Answer 2

我不喜欢使用range(len(...，但也许在这里没问题。

>>> months = ["January", "February", "March", "April", "May", "June",
...           "July", "August", "September", "October", "November", "December"]
>>> print(*('{:>2} {}'.format(i, months[i]) for i in range(len(months))[5::2]), sep='\n')
 5 June
 7 August
 9 October
11 December

扩展单线：

offset = 5
step = 2
r = range(len(months))

for i in r[offset::step]:
    print('{:>2} {}'.format(i, months[i]))

Answer 3

import itertools as it

months = [
    "January", "February", "March", "April", "May", "June", "July",
    "August", "September", "October", "November", "December"
]

print list(
    it.islice(it.izip(it.count(1), months), 5, len(months), 2)
)

回答评论中的其他问题：

普通切片无法在itertools.izip上工作，因为它不支持__getitem__方法。
是的，itertools.izip已从Python 3中删除，而常规内置zip使用相同的生成器语义。

Answer 4

您可以使用itertools.islice() + enumerate()来选择具有原始索引的元素：

>>> import calendar
>>> from itertools import islice
>>> for i, month in islice(enumerate(calendar.month_abbr), 6, None, 2):
...     print(i, month)
... 
6 Jun
8 Aug
10 Oct
12 Dec

它不会复制切片信息，并使用熟悉行为的现有函数。

或者如果你知道原始的可迭代性很小;你可以致电list()：

>>> list(enumerate(calendar.month_abbr))[6::2]
[(6, 'Jun'), (8, 'Aug'), (10, 'Oct'), (12, 'Dec')]

如何枚举可迭代保留原始索引中的选定元素？

4 个答案: