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!

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
```

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)
```

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