懒惰地在Python中转置列表 [英] Lazily transpose a list in Python

问题描述

因此，我有一个3元组的可迭代生成，它们是惰性生成的.我试图弄清楚如何将其变成3个可迭代对象，分别由元组的第一个，第二个和第三个元素组成.但是，我希望这样做会很懒.

So, I have an iterable of 3-tuples, generated lazily. I'm trying to figure out how to turn this into 3 iterables, consisting of the first, second, and third elements of the tuples, respectively. However, I wish this to be done lazily.

因此，例如，我希望将[(1, 2, 3), (4, 5, 6), (7, 8, 9)]转换为[1, 4, 7]，[2, 5, 8]，[3, 6, 9]. (除了我想要的可迭代对象而不是列表.)

So, for example, I wish [(1, 2, 3), (4, 5, 6), (7, 8, 9)] to be turned into [1, 4, 7], [2, 5, 8], [3, 6, 9]. (Except I want iterables not lists.)

标准的zip(*data)惯用语不起作用，因为参数拆包扩展了整个可迭代项. (您可以通过注意到zip(*((x, x+1, x+2) for x in itertools.count(step=3)))挂起来验证这一点.)

The standard zip(*data) idiom doesn't work, because the argument unpacking expands the entire iterable. (You can verify this by noting that zip(*((x, x+1, x+2) for x in itertools.count(step=3))) hangs.)

到目前为止，我想出的最好的方法是:

The best I've come up with thus far is the following:

def transpose(iterable_of_three_tuples):
    teed = itertools.tee(iterable_of_three_tuples, 3)
    return map(lambda e: e[0], teed[0]), map(lambda e: e[1], teed[1]), map(lambda e: e[2], teed[2])

这似乎有效.但这似乎不像是干净的代码.它做了很多不必要的工作.

This seems to work. But it hardly seems like clean code. And it does a lot of what seems to be unnecessary work.

推荐答案

您的transpose正是您所需要的.

使用您选择的任何解决方案，您都必须缓冲未使用的值(例如，要获得7，则必须读取1-6，并将它们存储在内存中，以便其他可迭代对象请求它们) . tee已经完全进行了这种缓冲，因此无需自己实现.

With any solution you'd choose, you'd have to buffer the unused values (e.g. to get to the 7, you have to read 1-6, and store them in memory for when the other iterables ask for them). tee already does exactly that kind of buffering, so there's no need implementing it yourself.

唯一的(次要)问题是我写的略有不同，避免了map和lambda s:

The only other (minor) thing is that I'd write it slightly differently, avoiding the map and lambdas:

def transpose(iterable_of_three_tuples):
    teed = itertools.tee(iterable_of_three_tuples, 3)
    return ( e[0] for e in teed[0] ),  ( e[1] for e in teed[1] ),  ( e[2] for e in teed[2] )

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