A safe max() function for empty lists

Solution 1:

In Python 3.4+, you can use default keyword argument:

>>> max([], default=99)
99

In lower version, you can use or:

>>> max([] or [99])
99

NOTE: The second approach does not work for all iterables. especially for iterator that yield nothing but considered truth value.

>>> max(iter([]) or 0)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: max() arg is an empty sequence

Solution 2:

In versions of Python older than 3.4 you can use itertools.chain() to add another value to the possibly empty sequence. This will handle any empty iterable but note that it is not precisely the same as supplying the default argument as the extra value is always included:

>>> from itertools import chain
>>> max(chain([42], []))
42

But in Python 3.4, the default is ignored if the sequence isn't empty:

>>> max([3], default=42)
3

Solution 3:

The max of an empty sequence "should" be an infinitely small thing of whatever type the elements of the sequence have. Unfortunately, (1) with an empty sequence you can't tell what type the elements were meant to have and (2) there is, e.g., no such thing as the most-negative integer in Python.

So you need to help max out if you want it to do something sensible in this case. In recent versions of Python there is a default argument to max (which seems to me a misleading name, but never mind) which will be used if you pass in an empty sequence. In older versions you will just have to make sure the sequence you pass in isn't empty -- e.g., by oring it with a singleton sequence containing the value you'd like to use in that case.

[EDITED long after posting because Yaakov Belch kindly pointed out in comments that I'd written "infinitely large" where I should have written "infinitely small".]