Python slice first and last element in list

Question

Is there a way to slice only the first and last item in a list?

For example; If this is my list:

>>> some_list
['1', 'B', '3', 'D', '5', 'F']

I want to do this (obviously [0,-1] is not valid syntax):

>>> first_item, last_item = some_list[0,-1]
>>> print first_item
'1'
>>> print last_item
'F'

Some things I have tried:

In [3]: some_list[::-1]
Out[3]: ['F', '5', 'D', '3', 'B', '1']

In [4]: some_list[-1:1:-1]
Out[4]: ['F', '5', 'D', '3']

In [5]: some_list[0:-1:-1]
Out[5]: []
...

Answer 1

One way:

some_list[::len(some_list)-1]

A better way (Doesn't use slicing, but is easier to read):

[some_list[0], some_list[-1]]

Answer 2

Python 3 only answer (that doesn't use slicing or throw away the rest of the list, but might be good enough anyway) is use unpacking generalizations to get first and last separate from the middle:

first, *_, last = some_list

The choice of _ as the catchall for the "rest" of the arguments is arbitrary; they'll be stored in the name _ which is often used as a stand-in for "stuff I don't care about".

Unlike many other solutions, this one will ensure there are at least two elements in the sequence; if there is only one (so first and last would be identical), it will raise an exception (ValueError).

Answer 3

Just thought I'd show how to do this with numpy's fancy indexing:

>>> import numpy
>>> some_list = ['1', 'B', '3', 'D', '5', 'F']
>>> numpy.array(some_list)[[0,-1]]
array(['1', 'F'], 
      dtype='|S1')

Note that it also supports arbitrary index locations, which the [::len(some_list)-1] method would not work for:

>>> numpy.array(some_list)[[0,2,-1]]
array(['1', '3', 'F'], 
      dtype='|S1')

As DSM points out, you can do something similar with itemgetter:

>>> import operator
>>> operator.itemgetter(0, 2, -1)(some_list)
('1', '3', 'F')