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).

Note: Since it must collect all the other elements into _, this is inappropriate for large inputs that are already sequences, doing unnecessary work (O(n), vs. O(1) for direct indexing or slicing approaches). In exchange, on top of the error-checking, it will work with any iterable input, not just lists, tuples, and other sequence types.

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')

Answer 4

first, last = some_list[0], some_list[-1]

Answer 5

Some people are answering the wrong question, it seems. You said you want to do:

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

Ie., you want to extract the first and last elements each into separate variables.

In this case, the answers by Matthew Adams, pemistahl, and katrielalex are valid. This is just a compound assignment:

first_item, last_item = some_list[0], some_list[-1]

But later you state a complication: "I am splitting it in the same line, and that would have to spend time splitting it twice:"

x, y = a.split("-")[0], a.split("-")[-1]

So in order to avoid two split() calls, you must only operate on the list which results from splitting once.

In this case, attempting to do too much in one line is a detriment to clarity and simplicity. Use a variable to hold the split result:

lst = a.split("-")
first_item, last_item = lst[0], lst[-1]

Other responses answered the question of "how to get a new list, consisting of the first and last elements of a list?" They were probably inspired by your title, which mentions slicing, which you actually don't want, according to a careful reading of your question.

AFAIK are 3 ways to get a new list with the 0th and last elements of a list:

>>> s = 'Python ver. 3.4'
>>> a = s.split()
>>> a
['Python', 'ver.', '3.4']

>>> [ a[0], a[-1] ]        # mentioned above
['Python', '3.4']

>>> a[::len(a)-1]          # also mentioned above
['Python', '3.4']

>>> [ a[e] for e in (0,-1) ] # list comprehension, nobody mentioned?
['Python', '3.4']

# Or, if you insist on doing it in one line:
>>> [ s.split()[e] for e in (0,-1) ]
['Python', '3.4']

The advantage of the list comprehension approach, is that the set of indices in the tuple can be arbitrary and programmatically generated.

Answer 6

You can do it like this:

some_list[0::len(some_list)-1]

Answer 7

What about this?

>>> first_element, last_element = some_list[0], some_list[-1]

Answer 8

You can use something like

y[::max(1, len(y)-1)]

if you really want to use slicing. The advantage of this is that it cannot give index errors and works with length 1 or 0 lists as well.

Answer 9

Actually, I just figured it out:

In [20]: some_list[::len(some_list) - 1]
Out[20]: ['1', 'F']

Answer 10

This isn't a "slice", but it is a general solution that doesn't use explicit indexing, and works for the scenario where the sequence in question is anonymous (so you can create and "slice" on the same line, without creating twice and indexing twice): operator.itemgetter

import operator

# Done once and reused
first_and_last = operator.itemgetter(0, -1)

...

first, last = first_and_last(some_list)

You could just inline it as (after from operator import itemgetter for brevity at time of use):

first, last = itemgetter(0, -1)(some_list)

but if you'll be reusing the getter a lot, you can save the work of recreating it (and give it a useful, self-documenting name) by creating it once ahead of time.

Thus, for your specific use case, you can replace:

x, y = a.split("-")[0], a.split("-")[-1]

with:

x, y = itemgetter(0, -1)(a.split("-"))

and split only once without storing the complete list in a persistent name for len checking or double-indexing or the like.

Note that itemgetter for multiple items returns a tuple, not a list, so if you're not just unpacking it to specific names, and need a true list, you'd have to wrap the call in the list constructor.

Answer 11

How about this?

some_list[:1] + some_list[-1:]

Result: ['1', 'F']

Answer 12

More General Case: Return N points from each end of list

The answers work for the specific first and last, but some, like myself, may be looking for a solution that can be applied to a more general case in which you can return the top N points from either side of the list (say you have a sorted list and only want the 5 highest or lowest), i came up with the following solution:

In [1]
def GetWings(inlist,winglen):
    if len(inlist)<=winglen*2:
        outlist=inlist
    else:
        outlist=list(inlist[:winglen])
        outlist.extend(list(inlist[-winglen:]))
    return outlist

and an example to return bottom and top 3 numbers from list 1-10:

In [2]
GetWings([1,2,3,4,5,6,7,8,9,10],3)

#Out[2]
#[1, 2, 3, 8, 9, 10]

Answer 13

Fun new approach to "one-lining" the case of an anonymously split thing such that you don't split it twice, but do all the work in one line is using the walrus operator, :=, to perform assignment as an expression, allowing both:

first, last = (split_str := a.split("-"))[0], split_str[-1]

and:

first, last = (split_str := a.split("-"))[::len(split_str)-1]

Mind you, in both cases it's essentially exactly equivalent to doing on one line:

split_str = a.split("-")

then following up with one of:

first, last = split_str[0], split_str[-1]
first, last = split_str[::len(split_str)-1]

including the fact that split_str persists beyond the line it was used and accessed on. It's just technically meeting the requirements of one-lining, while being fairly ugly. I'd never recommend it over unpacking or itemgetter solutions, even if one-lining was mandatory (ruling out the non-walrus versions that explicitly index or slice a named variable and must refer to said named variable twice).