A safe max() function for empty lists

Question

Evaluating,

max_val = max(a)

will cause the error,

ValueError: max() arg is an empty sequence

Is there a better way of safeguarding against this error other than a try, except catch?

a = []
try:
    max_val = max(a)
except ValueError:
    max_val = default 

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

Another solution could be by using ternary operators:

nums = []
max_val = max(nums) if nums else 0

or

max val = max(iter(nums) if nums else [0])

Solution 4

Can create simple lambda to do this:

get_max = lambda val_list: max([ val for val in val_list if val is not None ]) if val_list else None

You can call it this way:
get_max(your_list)

Solution 5

_DEFAULT = object()

def max_default(*args, **kwargs):
    """
    Adds support for "default" keyword argument when iterable is empty.
    Works for any iterable, any default value, and any Python version (versions >= 3.4
    support "default" parameter natively).

    Default keyword used only when iterable is empty:

    >>> max_default([], default=42)
    42

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

    All original functionality is preserved:

    >>> max_default([])
    Traceback (most recent call last):
    ValueError: max() arg is an empty sequence

    >>> max_default(3, 42)
    42
    """

    default = kwargs.pop('default', _DEFAULT)
    try:
        return max(*args, **kwargs)
    except ValueError:
        if default is _DEFAULT:
            raise
        return default

Bonus:

def min_default(*args, **kwargs):
    """
    Adds support for "default" keyword argument when iterable is empty.
    Works for any iterable, any default value, and any Python version (versions >= 3.4
    support "default" parameter natively).

    Default keyword used only when iterable is empty:

    >>> min_default([], default=42)
    42

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

    All original functionality is preserved:

    >>> min_default([])
    Traceback (most recent call last):
    ValueError: min() arg is an empty sequence

    >>> min_default(3, 42)
    3
    """

    default = kwargs.pop('default', _DEFAULT)
    try:
        return min(*args, **kwargs)
    except ValueError:
        if default is _DEFAULT:
            raise
        return default

View more solutions

Comments

[vaultah]

The default argument specifies an object to return if the provided iterable is empty.

[Martin Tournoij]

@vaultah Why you no post answer as answer?!

[vaultah]

@Carpetsmoker: because it's too simple.

[Martin Tournoij]

@vaultah ... What's wrong with posting answers that are "too simple"? Now someone else has to...

[Steven Rumbalski]

Another option for pre-3.4 would be to create a my_max function that mimics the newer behavior.

[Alexander McFarlane]

max(iter([]) or 0) is precisely what gave me the grief in the first place... Do you have any further suggestions on what do do with it? Secondly it never occurred to me to check the docs... I assumed such a simple function was a one-in-one-out process

[falsetru]

@AlexanderMcFarlane, Duncan suggested a nice alternative. Check it out.

[falsetru]

@AlexanderMcFarlane, I'm not a native speaker. I don't understand what one-in-one-out process. Does it mean a common task?

[Alexander McFarlane]

@falsetru by one-in-one-out I mean one argument in and one return value :)

[Yaakov Belch]

It's counter-intuitive: the max of an empty list should be infinitely small (so that max(x,list...)=max(x,max(list...))) and the min of an empty list should be infinitely large.

[Gareth McCaughan]

Whoops, yes, I wrote that the wrong way around; my apologies. I will fix my answer and credit you for spotting the mistake. [... Done. Thanks again.]

[MajorInc]

is there any reason to use chain over simple list concatenation e.g. max( [42], [] ) where the default is 42?

[Duncan]

@MajorInc calling max with multiple arguments has a different meaning than calling it with a single argument. With a single iterable argument max will return the largest element it contains, but with multiple arguments it compares them directly (i.e. it won't iterate). e.g. max([42], []) is a list [42] but max(chain([42], [])) and max([], default=42) both give 42.

[Milo]

max_safe(42, 3, default=0) Throws TypeError: max() got an unexpected keyword argument (python 2)