迭代器列表的迭代器 [英] Iterator of list of list of iterables

问题描述

我想迭代Python3中的Iterables列表.

I would like to iterate of a list of list of Iterables in Python3.

换句话说，我有一个可迭代的矩阵，我想遍历并在每次迭代时都得到一个值矩阵.更具体地说，我有几个文件(行)，它们有多个版本(列)，我希望在每次迭代时，得到一个包含所有文件第一行的元组/矩阵，依此类推.

Stated differently, I have a matrix of iterables and I would like to loop through and get at each iteration a matrix of values. More concretely, I have several files (the rows) which have multiple versions of them (the columns) and I would like at each iteration, get a tuple/matrix containing the first line of all my files and so on.

所以，给定这样的东西

a = [
  [iter(range(1,10)), iter(range(11,20)), iter(range(21,30))],
  [iter(range(101,110)), iter(range(111,120)), iter(range(121,130))]
]

我想

for sources_with_their_factors in MAGIC_HERE(a):
  print(sources_with_their_factors)

并获得

((1,11,21), (101,111,121))
((2,12,22), (102,112,122))
…

我尝试了

for b in zip(zip(*zip(*a))):
    ...:     print(b)
    ...:
((<range_iterator object at 0x2b688d65b630>, <range_iterator object at 0x2b688d65b7e0>, <range_iterator object at 0x2b688d65b540>),)
((<range_iterator object at 0x2b688d65ba50>, <range_iterator object at 0x2b688d65b6f0>, <range_iterator object at 0x2b688d65b0c0>),)

但这并不是在重复我的范围.

But it isn’t iterating my ranges.

推荐答案

很显然，您可以将每个子列表中的迭代器一起zip在一起，只是缺少了如何将结果迭代器一起zip在一起.我将解压缩一个生成器表达式:

Obviously you can zip the iterators in each sublist together, you're just missing how to zip the resulting iterators together. I would unpack a generator expression:

for t in zip(*(zip(*l) for l in a)):
    print(t)

((1, 11, 21), (101, 111, 121))
((2, 12, 22), (102, 112, 122))
...

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