如何反转itertools.chain对象?

问题描述

我的函数创建了一系列的生成器:

My function creates a chain of generators:

def bar(num):
    import itertools
    some_sequence = (x*1.5 for x in range(num))
    some_other_sequence = (x*2.6 for x in range(num))
    chained = itertools.chain(some_sequence, some_other_sequence)
    return chained

我的函数有时需要以相反的顺序返回chained.从概念上讲，以下是我想做的事情:

My function sometimes needs to return chained in reversed order. Conceptually, the following is what I would like to be able to do:

if num < 0:
    return reversed(chained)
return chained

不幸的是:

>>> reversed(chained)
TypeError: argument to reversed() must be a sequence

我有什么选择?

这是一些实时图形渲染代码，因此我不想使其变得太复杂/太慢.

This is in some realtime graphic rendering code so I don't want to make it too complicated/slow.

编辑:当我第一次提出这个问题时，我还没有想到发电机的可逆性.正如许多人指出的那样，生成器是不可逆的.

EDIT:When I first posed this question I hadn't thought about the reversibility of generators. As many have pointed out, generators can't be reversed.

事实上，我确实想颠倒链中扁平化的内容；不只是生成器的顺序.

I do in fact want to reverse the flattened contents of the chain; not just the order of the generators.

基于响应，我没有一个可用于反转itertools.chain的调用，因此，我认为这里唯一的解决方案是使用列表，至少对于反转情况，也许对两者都使用. >

Based on the responses, there is no single call I can use to reverse an itertools.chain, so I think the only solution here is to use a list, at least for the reverse case, and perhaps for both.

推荐答案

if num < 0:
    lst = list(chained)
    lst.reverse()
    return lst
else:
    return chained

reversed()需要一个实际的序列，因为它会按索引向后迭代，这对于生成器(仅具有"next"项的概念)不起作用.

reversed() needs an actual sequence, because it iterates it backwards by index, and that wouldn't work for a generator (which only has the notion of "next" item).

由于无论如何都需要展开整个生成器以进行反转，所以最有效的方法是将其读取到列表中，并使用.reverse()方法就地反转该列表.

Since you will need to unroll the whole generator anyway for reversing, the most efficient way is to read it to a list and reverse the list in-place with the .reverse() method.

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