从交替的边循环列表 [英] Cycle a list from alternating sides
问题描述
给出一个清单
a = [0,1,2,3,4,5,6,7,8,9]
我怎样才能获得
b = [0,9,1,8,2,7,3,6,4,5]
也就是说,产生一个新的列表,其中每个连续的元素交替地从原始列表的两边取出?
That is, produce a new list in which each successive element is alternately taken from the two sides of the original list?
推荐答案
>>> [a[-i//2] if i % 2 else a[i//2] for i in range(len(a))]
[0, 9, 1, 8, 2, 7, 3, 6, 4, 5]
说明:
此代码从头开始选择数字( a [i // 2]
)并从最后选择( a [-i // 2]
) a
,交替地(,如果i%2其他
)。选择总计 len(a)
的数字,因此即使 len(a)
,也不会产生任何不良影响奇数。
[ - i // 2 for i in range(len(a))]
yield 0,-1 ,-1,-2,-2,-3,-3,-4,-4,-5
,
[i // 2 for i in range(len(a))]
yield 0,0,1,1,2,2,3,3,4,4
,
和 i%2
在 False
和之间交替显示True
,
所以我们从 a
中提取的索引是: 0,-1,1, -2,2,3,3,4,4,5
。
Explanation:
This code picks numbers from the beginning (a[i//2]
) and from the end (a[-i//2]
) of a
, alternatingly (if i%2 else
). A total of len(a)
numbers are picked, so this produces no ill effects even if len(a)
is odd.
[-i//2 for i in range(len(a))]
yields 0, -1, -1, -2, -2, -3, -3, -4, -4, -5
,
[ i//2 for i in range(len(a))]
yields 0, 0, 1, 1, 2, 2, 3, 3, 4, 4
,
and i%2
alternates between False
and True
,
so the indices we extract from a
are: 0, -1, 1, -2, 2, -3, 3, -4, 4, -5
.
我对pythonicness的评估:
关于这个单线程的好处是它很短并且显示对称性( + i // 2
和 -i // 2
)。
但不好的是,这种对称性具有欺骗性:
认为 -i // 2
与 i // 2
相同,并且标志被翻转。 但在Python中,整数除法返回发言权结果的a>而不是截断为零。所以 -1 // 2 == -1
。
另外,我发现通过索引访问列表元素比pythonic更少迭代。
My assessment of pythonicness:
The nice thing about this one-liner is that it's short and shows symmetry (+i//2
and -i//2
).
The bad thing, though, is that this symmetry is deceptive:
One might think that -i//2
were the same as i//2
with the sign flipped. But in Python, integer division returns the floor of the result instead of truncating towards zero. So -1//2 == -1
.
Also, I find accessing list elements by index less pythonic than iteration.
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