Finding nth biggest value in a nested dictionary

dez8686 0
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I have been trying to figure out how to return the key which has the nth biggest value in a nested dictionary. What causes problems for me is if there's some missing values like

my_dict = {'0': {'name': 'A', 'price': 50}, '1': {'name': 'B', 'price': 20}, '2': {'name': 'C'}, '3': {'name': 'D', 'price': 10}}

If every price existed, I could use code such as this to get the correct key:

sorted_list = sorted(my_dict.items(), key=lambda item: item[1]['price'])
print(sorted_list[-number_im_using_to_find][1]['name'])

How to account for missing values in an efficient way?

nested dictionary python
2 Answers

Use get with a default value, in this case 0 or even a negative value to put the elements with missing prices all the way down the sorted list.

lambda item: item[1].get('price', 0)
fsimonjetz0

you can use dict.get to achieve this:

get(key[, default])

Return the value for key if key is in the dictionary, else default. If default is not given, it defaults to None, so that this method never raises a KeyError.

Here is a code implementation using dict.get:

default_value = -1
key = lambda item: item[1].get('price', default_value)
sorted_list = sorted(my_dict.items(), key=key)

If you want just sorted values you could remove entirely the index [1]:

default_value = -1
key = lambda value: value.get('price', default_value)
sorted_list = sorted(my_dict.values(), key=key)

EDIT: if you want to be sortable with NoneType in the list, you could use something like:

>>> mylist = [3, 1, None, None, 2, 0]
>>> mylist.sort(key=lambda x: x or -1)
>>> mylist
[None, None, 0, 1, 2, 3]
XxJames07-0
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