CODEWITHSUNDEEP
pythonlistpython-2.7dictionary
anvd
anvd

Reputation: 4047

Sort list of dicts by list index of dictionary

I have a list of dicts, where the dict has a list itself. How can I sort using factors index?

For example sort descending the list by factors[0]

[{'score': '2.0', 'id': 686, 'factors': [2.0, 2.25, 2.75, 1.5, 2.25]}, {'score': '1.9', 'id': 863, 'factors': [1.5, 3.0, 1.5, 2.5, 1.5]}, {'score': '2.0', 'id': 55, 'factors': [1.5, 3.0, 2.5, 1.5, 1.5]}, {'score': '1.9', 'id': 756, 'factors': [1.25, 2.25, 2.5, 2.0, 1.75]}]

Upvotes: 1

Views: 593

Answers (2)

Celius Stingher
Celius Stingher

Reputation: 18367

Essentially what you need to do is use sorted. However this factors[0] seems a bit arbitrary, perhaps you want to sort the list first and then sort by its first value?

However this exmple get the jobs done. Keep in mind it won't be usable for ties, so you might want to add a second sorting key:

sort = sorted(ex, key=lambda x: x['factors'][0])
sort

Output:

[{'factors': [1.25, 2.25, 2.5, 2.0, 1.75], 'id': 756, 'score': '1.9'},
 {'factors': [1.5, 3.0, 1.5, 2.5, 1.5], 'id': 863, 'score': '1.9'}, #Tied values
 {'factors': [1.5, 3.0, 2.5, 1.5, 1.5], 'id': 55, 'score': '2.0'}, #Tied values
 {'factors': [2.0, 2.25, 2.75, 1.5, 2.25], 'id': 686, 'score': '2.0'}]

Example of sorting by factors[0] followed by id for tied values. Please note how the order is the other way around, first you sort by id (inner sort) and then factors in the code (outer sort):

sort = sorted(sorted(ex,key=lambda x: x['id']),key=lambda x: x['factors'][0])
sort

Output: 

[{'factors': [1.25, 2.25, 2.5, 2.0, 1.75], 'id': 756, 'score': '1.9'},
 {'factors': [1.5, 3.0, 2.5, 1.5, 1.5], 'id': 55, 'score': '2.0'},
 {'factors': [1.5, 3.0, 1.5, 2.5, 1.5], 'id': 863, 'score': '1.9'},
 {'factors': [2.0, 2.25, 2.75, 1.5, 2.25], 'id': 686, 'score': '2.0'}]

Upvotes: 1

mad_
mad_

Reputation: 8273

You can pass it as a key

my_list=[{'score': '2.0', 'id': 686, 'factors': [2.0, 2.25, 2.75, 1.5, 2.25]}, {'score': '1.9', 'id': 863, 'factors': [1.5, 3.0, 1.5, 2.5, 1.5]}, {'score': '2.0', 'id': 55, 'factors': [1.5, 3.0, 2.5, 1.5, 1.5]}, {'score': '1.9', 'id': 756, 'factors': [1.25, 2.25, 2.5, 2.0, 1.75]}]
sorted(my_list, key=lambda x: x['factors'][0])

Output:

[{'score': '1.9', 'id': 756, 'factors': [1.25, 2.25, 2.5, 2.0, 1.75]},
 {'score': '1.9', 'id': 863, 'factors': [1.5, 3.0, 1.5, 2.5, 1.5]},
 {'score': '2.0', 'id': 55, 'factors': [1.5, 3.0, 2.5, 1.5, 1.5]},
 {'score': '2.0', 'id': 686, 'factors': [2.0, 2.25, 2.75, 1.5, 2.25]}]

Upvotes: 1

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