NumPy first and last element from array

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!

Solution 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)

Solution 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

Solution 3:

Using Numpy's fancy indexing:

>>> test
array([ 1, 23,  4,  6,  7,  8])

>>> test[::-1]  # test, reversed
array([ 8,  7,  6,  4, 23,  1])

>>> numpy.vstack([test, test[::-1]])  # stack test and its reverse
array([[ 1, 23,  4,  6,  7,  8],
       [ 8,  7,  6,  4, 23,  1]])

>>> # transpose, then take the first half;
>>> # +1 to cater to odd-length arrays
>>> numpy.vstack([test, test[::-1]]).T[:(len(test) + 1) // 2]
array([[ 1,  8],
       [23,  7],
       [ 4,  6]])

vstack copies the array, but all the other operations are constant-time pointer tricks (including reversal) and hence are very fast.