Sorting list by an attribute that can be None

Question

I’m trying to sort a list of objects using

my_list.sort(key=operator.attrgetter(attr_name))

but if any of the list items has attr = None instead of attr = 'whatever',

then I get a TypeError: unorderable types: NoneType() < str()

In Py2 it wasn’t a problem. How do I handle this in Py3?

Asked By: AlexVhr

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Source

Answer 1

The ordering comparison operators are stricter about types in Python 3, as described here:

The ordering comparison operators (<, <=, >=, >) raise a TypeError
exception when the operands don’t have a meaningful natural ordering.

Python 2 sorts None before any string (even empty string):

>>> None < None
False

>>> None < "abc"
True

>>> None < ""
True

In Python 3 any attempts at ordering NoneType instances result in an exception:

>>> None < "abc"
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unorderable types: NoneType() < str()

The quickest fix I can think of is to explicitly map None instances into something orderable like "":

my_list_sortable = [(x or "") for x in my_list]

If you want to sort your data while keeping it intact, just give sort a customized key method:

def nonesorter(a):
    if not a:
        return ""
    return a

my_list.sort(key=nonesorter)

Answered By: jsalonen

Answer 2

For a general solution, you can define an object that compares less than any other object:

from functools import total_ordering

@total_ordering
class MinType(object):
    def __le__(self, other):
        return True

    def __eq__(self, other):
        return (self is other)

Min = MinType()

Then use a sort key that substitutes Min for any None values in the list

mylist.sort(key=lambda x: Min if x is None else x)

Answered By: augurar

Answer 3

Since there are other things besides None that are not comparable to a string (ints and lists, for starters), here is a more robust solution to the general problem:

my_list.sort(key=lambda x: x if isinstance(x, str) else "")

This will let strings and any type derived from str to compare as themselves, and bin everything else with the empty string. Or substitute a different default default key if you prefer, e.g. "ZZZZ" or chr(sys.maxunicode) to make such elements sort at the end.

Answered By: alexis

Answer 4

The solutions proposed here work, but this could be shortened further:

mylist.sort(key=lambda x: x or 0)

In essence, we can treat None as if it had value 0.

E.g.:

>>> mylist = [3, 1, None, None, 2, 0]
>>> mylist.sort(key=lambda x: x or 0)
>>> mylist
[None, None, 0, 1, 2, 3]

Answered By: fralau

Answer 5

A general solution to deal with None independent of the types to be sorted:

my_list.sort(key=lambda x: (x is not None, x))

putting None values first.

Note that: my_list.sort(key=lambda x: (x is None, x)) puts None values last.

Answered By: eci

Sorting list by an attribute that can be None

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