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}]
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")