AlexVhr AlexVhr - 2 years ago 104
Python Question

Sorting list by an attribute that can be None

I'm trying to sort a list of objects using


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?

Answer Source

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

>>> None < "abc"

>>> None < ""

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

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