Pythonic方法获取序列的第一个和最后一个元素

What is the easiest and cleanest way to get the first AND the last elements of a sequence? E.g., I have a sequence [1, 2, 3, 4, 5], and I'd like to get [1, 5] via some kind of slicing magic. What I have come up with so far is:

获取序列的第一个和最后一个元素的最简单，最简洁的方法是什么？例如，我有一个序列[1,2,3,4,5]，我想通过某种切片魔法获得[1,5]。到目前为止我想出的是：

l = len(s)
result = s[0:l:l-1]

I actually need this for a bit more complex task. I have a 3D numpy array, which is cubic (i.e. is of size NxNxN, where N may vary). I'd like an easy and fast way to get a 2x2x2 array containing the values from the vertices of the source array. The example above is an oversimplified, 1D version of my task.

我实际上需要这个更复杂的任务。我有一个3D numpy数组，它是立方的（即大小为NxNxN，其中N可能不同）。我想要一种简单快捷的方法来获取包含源数组顶点值的2x2x2数组。上面的示例是我的任务的过度简化的1D版本。

6 个解决方案

#1

Since you're using a numpy array, you may want to use fancy indexing:

由于您使用的是numpy数组，因此您可能希望使用花哨的索引：

a = np.arange(27)
indices = [0, -1]
b = a[indices] # array([0, 26])

For the 3d case:

对于3d案例：

vertices = [(0,0,0),(0,0,-1),(0,-1,0),(0,-1,-1),(-1,-1,-1),(-1,-1,0),(-1,0,0),(-1,0,-1)]
indices = list(zip(*vertices))  #Can store this for later use.
a = np.arange(27).reshape((3,3,3)) #dummy array for testing.  Can be any shape size :)
vertex_values = a[indices].reshape((2,2,2))

I first write down all the vertices (although I am willing to bet there is a clever way to do it using itertools which would let you scale this up to N dimensions ...). The order you specify the vertices is the order they will be in the output array. Then I "transpose" the list of vertices (using zip) so that all the x indices are together and all the y indices are together, etc. (that's how numpy likes it). At this point, you can save that index array and use it to index your array whenever you want the corners of your box. You can easily reshape the result into a 2x2x2 array (although the order I have it is probably not the order you want).

我首先写下所有的顶点（虽然我愿意打赌，有一种聪明的方法可以使用itertools来实现它，这可以让你将它扩展到N维...）。指定顶点的顺序是它们在输出数组中的顺序。然后我“转置”顶点列表（使用zip），以便所有x索引在一起，所有y索引在一起等等（这就是numpy喜欢它的方式）。此时，您可以保存该索引数组，并在需要方框的角落时使用它来索引数组。您可以轻松地将结果重新整形为2x2x2阵列（尽管我拥有它的顺序可能不是您想要的顺序）。

#2

Use this:

用这个：

result = [s[0], s[-1]]

#3

This would give you a list of the first and last element in your sequence:

这将为您提供序列中第一个和最后一个元素的列表：

 result = [s[0], s[-1]]

Alternatively, this would give you a tuple

或者，这会给你一个元组

 result = s[0], s[-1]

#4

With the particular case of a (N,N,N) ndarray X that you mention, would the following work for you?

对于你提到的（N，N，N）ndarray X的特殊情况，下面的工作会对你有用吗？

s = slice(0,N,N-1)
X[s,s,s]

Example

例

>>> N = 3
>>> X = np.arange(N*N*N).reshape(N,N,N)
>>> s = slice(0,N,N-1)
>>> print X[s,s,s]
[[[ 0  2]
  [ 6  8]]
 [[18 20]
  [24 26]]]

#5

>>> from operator import itemgetter
>>> first_and_last = itemgetter(0, -1)
>>> first_and_last([1, 2, 3, 4, 5])
(1, 5)

#6

Why do you want to use a slice? Getting each element with

为什么要使用切片？获得每个元素

result = [s[0], s[-1]]

is better and more readable.

更好，更可读。

If you really need to use the slice, then your solution is the simplest working one that I can think of.

如果你真的需要使用切片，那么你的解决方案是我能想到的最简单的工作方式。

This also works for the 3D case you've mentioned.

这也适用于您提到的3D案例。

#1