NumPy first and last element from array

Question

I am trying to dynamically get the first and last element from an array.

So, let us suppose the array has 6 elements.

test = [1,23,4,6,7,8] 

If I am trying to get the first and last = 1,8, 23,7 and 4,6. Is there a way to get elements in this order? I looked at a couple of questions Link Link2. I took help of these links and I came up with this prototype..

#!/usr/bin/env python  import numpy  test = [1,23,4,6,7,8] test1 = numpy.array([1,23,4,6,7,8]) len_test = len(test) first_list = [0,1,2] len_first = len(first_list) second_list = [-1,-2,-3] len_second = len(second_list)  for a in range(len_first):         print numpy.array(test)[[first_list[a] , second_list[a]]]         print test1[[first_list[a], second_list[a]]] 

But this prototype won't scale for if you have more than 6 elements. So, I was wondering if there is way to dynamically get the pair of elements.

Thanks!

Answer 1

I ended here, because I googled for "python first and last element of array", and found everything else but this. So here's the answer to the title question:

a = [1,2,3] a[0] # first element (returns 1) a[-1] # last element (returns 3) 

Answer 2

How about:

In [10]: arr = numpy.array([1,23,4,6,7,8])  In [11]: [(arr[i], arr[-i-1]) for i in range(len(arr) // 2)] Out[11]: [(1, 8), (23, 7), (4, 6)] 

Depending on the size of arr, writing the entire thing in NumPy may be more performant:

In [41]: arr = numpy.array([1,23,4,6,7,8]*100)  In [42]: %timeit [(arr[i], arr[-i-1]) for i in range(len(arr) // 2)] 10000 loops, best of 3: 167 us per loop  In [43]: %timeit numpy.vstack((arr, arr[::-1]))[:,:len(arr)//2] 100000 loops, best of 3: 16.4 us per loop