*j SS jr g )Nc [ U 5 n[ X15 n [ U5 nSR XE5 nU=( d [ U5 e! [ a U$ f = f)a If *iterable* has only one item, return it.
If it has zero items, return *default*.
If it has more than one item, raise the exception given by *too_long*,
which is ``ValueError`` by default.
>>> only([], default='missing')
'missing'
>>> only([1])
1
>>> only([1, 2]) # doctest: +IGNORE_EXCEPTION_DETAIL
Traceback (most recent call last):
...
ValueError: Expected exactly one item in iterable, but got 1, 2,
and perhaps more.'
>>> only([1, 2], too_long=TypeError) # doctest: +IGNORE_EXCEPTION_DETAIL
Traceback (most recent call last):
...
TypeError
Note that :func:`only` attempts to advance *iterable* twice to ensure there
is only one item. See :func:`spy` or :func:`peekable` to check
iterable contents less destructively.
zLExpected exactly one item in iterable, but got {!r}, {!r}, and perhaps more.)iternextformat
ValueError
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