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))
...
这篇关于迭代器列表的迭代器的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!