列表只包含第二对元素
我是 python 的新手，所以我正在做一些实验，但我现在有一个小问题。
我有一个包含n 个数字的列表，我想创建一个新列表，其中只包含每对第二对 数字。
所以基本上如果我有这样的清单
oldlist = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
那么我希望新列表看起来像这样
newlist = [3, 4, 7, 8]
我已经尝试过该slice()功能，但我没有找到任何方法可以将我的列表分成两部分。然后我想我可以使用两个slice()以四为单位移动并以一为单位移动的函数，但是如果我合并这两个新列表，它们的顺序将不正确。
回答
如果您枚举列表，您将采用那些索引为 2 或 3 作为除以 4 的余数的条目：
>>> [val for j, val in enumerate(old_list) if j % 4 in (2, 3)]
[3, 4, 7, 8]
- I might suggest `if ((j // 2) % 2) == 1` where now the first `2` is the size of the grouping, the second `2` is the interval of groups, and `1` is the index to select from the interval. `if ((j // size) % interval) == index` This allows you to parameterize the group size and index. E.g. for `if ((j // 3) % 2) == 0` you can get groups sized 3, with an interval of 2 groups and take the first index, `[1, 2, 3, 7, 8, 9]`.
- @flakes big +1 to the suggestion! Can I include your suggestion in my [answer](https://stackoverflow.com/a/68122252/3896008) as well? Also, not sure if I should delete my answer because Mustafa's answer does the same thing but is more pythonic since no explicit indexing is used in this answer.
回答
a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
b = [a[i] for i in range(len(a)) if i%4 in (2,3)]
# Output: b = [3, 4, 7, 8]
在这里，我们使用第 3、第 4、第 7、第 8 次……等等的想法。索引除以 4 时的余数为 2 或 3。
回答
first_part = oldList[2::4] # every 4th item, starting from the 3rd item
second_part = oldList[3::4] # every 4th item starting from the 4th item
pairs = zip(first_part, second_part)
final_result = chain.from_iterable(pairs)