Getting only element from a single-element list in Python?

PythonPython 3.xPython 2.7ListIterable Unpacking

Python Problem Overview


When a Python list is known to always contain a single item, is there a way to access it other than:

mylist[0]

You may ask, 'Why would you want to?'. Curiosity alone. There seems to be an alternative way to do everything in Python.

Python Solutions


Solution 1 - Python

Raises exception if not exactly one item:

Sequence unpacking:

singleitem, = mylist
# Identical in behavior (byte code produced is the same),
# but arguably more readable since a lone trailing comma could be missed:
[singleitem] = mylist

All others silently ignore spec violation, producing first or last item:

Explicit use of iterator protocol:

singleitem = next(iter(mylist))
Destructive pop:

singleitem = mylist.pop()
Negative index:

singleitem = mylist[-1]
Set via single iteration for (because the loop variable remains available with its last value when a loop terminates):
for singleitem in mylist: break
Rampant insanity:
# But also the only way to retrieve a single item and raise an exception on failure
# for too many, not just too few, elements as an expression, rather than a statement,
# without resorting to defining/importing functions elsewhere to do the work
singleitem = (lambda x: x)(*mylist)

There are many others (combining or varying bits of the above, or otherwise relying on implicit iteration), but you get the idea.

Solution 2 - Python

I will add that the more_itertools library has a tool that returns one item from an iterable.

from more_itertools import one


iterable = ["foo"]
one(iterable)
# "foo"

In addition, more_itertools.one raises an error if the iterable is empty or has more than one item.

iterable = []
one(iterable)
# ValueError: not enough values to unpack (expected 1, got 0)

iterable = ["foo", "bar"]
one(iterable)
# ValueError: too many values to unpack (expected 1)

more_itertools is a third-party package > pip install more-itertools

Solution 3 - Python

(This is an adjusted repost of my answer to a similar question related to sets.)

One way is to use reduce with lambda x: x.

from functools import reduce

> reduce(lambda x: x, [3]})
3

> reduce(lambda x: x, [1, 2, 3])
TypeError: <lambda>() takes 1 positional argument but 2 were given

> reduce(lambda x: x, [])
TypeError: reduce() of empty sequence with no initial value

Benefits:

  • Fails for multiple and zero values
  • Doesn't change the original list
  • Doesn't need a new variable and can be passed as an argument

Cons: "API misuse" (see comments).

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionPydermanView Question on Stackoverflow
Solution 1 - PythonShadowRangerView Answer on Stackoverflow
Solution 2 - PythonpylangView Answer on Stackoverflow
Solution 3 - PythonnocibambiView Answer on Stackoverflow