[英]nth item in List to dictionary python
I am trying to create a dictionary from every nth element of a list with list's original index as key. 我正在尝试使用列表的原始索引作为键,从列表的每n个元素创建一个字典。 So for example:
因此,例如:
l = [1,2,3,4,5,6,7,8,9]
Now running 正在运行
dict(enumerate(l)).items()
gives me: 给我:
dict_items([(0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 9)])
which is what I want. 这就是我想要的。 However the problem begins when I want to now select every second value from l to do this, so I try
但是,当我现在想从l中选择第二个值来执行此操作时,问题就开始了,所以我尝试
dict(enumerate(l[::2])).items()
which gives me 这给了我
dict_items([(0, 1), (1, 3), (2, 5), (3, 7), (4, 9)])
but I do not want that, I want to preserve the original index when making a dictionary. 但是我不想要那样,我想在制作字典时保留原始索引。 What is the best way to do this?
做这个的最好方式是什么?
I want the following output 我想要以下输出
dict_items([(0, 1), (2, 3), (4, 5), (6, 7), (8, 9)])
Use itertools.islice()
on the enumerate()
object: 在
enumerate()
对象上使用itertools.islice()
:
from itertools import islice
dict(islice(enumerate(l), None, None, 2)).items()
islice()
gives you a slice on any iterator ; islice()
为您提供了任何迭代器的切片; the above takes every second element: 以上是第二个要素:
>>> from itertools import islice
>>> l = [1,2,3,4,5,6,7,8,9]
>>> dict(islice(enumerate(l), None, None, 2)).items()
dict_items([(0, 1), (8, 9), (2, 3), (4, 5), (6, 7)])
(note that the output is as expected, but order is, as always, determined by the hash table ). (请注意,输出与预期的一样,但顺序通常由哈希表确定 )。
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