Sorting list containing value:None, results to TypeError: '<' not supported between instances of 'NoneType' and 'str'

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I've dictionary containing list of key:value pairs. How to sort the list containing key:value pairs (where value is None)

from operator import itemgetter
test = {"test":[{ "name" : "Nandini", "age" : 20}, { "name" : "Manjeet", "age" : 20 }, { "name" : None , "age" : 19 }] }
print(sorted(test["test"], key=itemgetter('name')) )

Results to TypeError: '<' not supported between instances of 'NoneType' and 'str'

Output should include None values and look like: [{ "name" : None , "age" : 19 }, { "name" : "Manjeet", "age" : 20 },{ "name" : "Nandini","age" : 20}]

nonetype python-3.x sorting operators
1 Answers

You can use the "or" trick:

# python 2 you can use with the itemgetter('name') 
# python3 needs a lambda, see further below
from operator import itemgetter
test = {"test":[{ "name" : "Nandini", "age" : 20}, 
                { "name" : "Manjeet", "age" : 20 }, 
                { "name" : None , "age" : 19 }] }


# supply a default value that can be compared - here "" is a good one
print(sorted(test["test"], key=itemgetter('name') or "") )

to get:

[{'age': 19, 'name': None}, 
 {'age': 20, 'name': 'Manjeet'}, 
 {'age': 20, 'name': 'Nandini'}]

You essentially supply a default value for None`s (and empty strings - it operates on the Truthy-ness of the value).

For python 3 you would need to use a lambda instead:

print(sorted(test["test"], key=lambda x: x['name'] or ""  ))

More infos:

Here the lambda is kindof like the itemgetter() - x is each inner dict of the list, x["name"] is the value you want as key and if that is Falsy (None, "") it will use whatever you supply after the or:

print(None or "works")
print("" or "works as well")
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