Getting only element from a single-element list in Python?
Question:
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.
Answers:
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
Rampant insanity, unpack the input to the identity lambda function:
# The only even semi-reasonable 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)
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
There are many others (combining or varying bits of the above, or otherwise relying on implicit iteration), but you get the idea.
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
(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).
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.
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
Rampant insanity, unpack the input to the identity lambda function:
# The only even semi-reasonable 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)
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
There are many others (combining or varying bits of the above, or otherwise relying on implicit iteration), but you get the idea.
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
(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).