python3 sort object list by attr same values

Question

I have a problem with my sorting method.

This is my object contained in list: Address class, with city attribute

My list looklike (simplified) :

[Address('Paris'), Address('Denver'), Address('Paris'), Address('Test'), Address('Denver')]

For this example, i have two duplicates cities: Paris and Denver,

I want to have a result like:

[Address('Devenr'), Address('Denver'), Address('Paris'), Address('Paris'), Address('Test')]

Sorted by duplicates count, and in case of same number, by alphanumeric order.

I tried:

self.dictionnary.sort(key=lambda address: len([x for x in self.dictionnary if address.city == x.city]))

By this don't work...

Can anyone help me?

Thank you in advance !

Mark Ransom · Accepted Answer · 2017-07-07 18:47:12Z

1

import collections
counts = collections.Counter(address.city for address in self.dictionnary)
self.dictionnary.sort(key=lambda address: (-counts[address.city], address.city))

By using Counter to count the duplicates in a separate step, you save the overhead of scanning the list each time you need a new key. This can make a big difference in the run time for a long list. The key then becomes a tuple; by taking the negative of the count, the larger counts will come first in the sort order. The second part of the tuple, the city name itself, will only be considered when the counts are equal.

edited Jul 7, 2017 at 18:47

answered Jul 7, 2017 at 18:11

Mark Ransom

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3 Comments

Mark Ransom Over a year ago

Could the downvoter please explain themselves? I just checked this code and it works perfectly. Plus it's more efficient than counting the entire list every time you need a new key.

vaultah Over a year ago

I can see why someone might downvote this: it's a code-only answer. In general, including an explanation really helps to improve the quality of your post.

Mark Ransom Over a year ago

@vaultah for me that's worth a comment, not a downvote. I reserve my downvotes for things that are actually incorrect.

AltF4 · Accepted Answer · 2017-07-07 19:01:30Z

1

The problem is that paris and denver both have count 2 so they don't get sorted.

If you add the string to the sort so ties are broken lexically it should work

Example:

from collections import Counter

l = ['a', 'b', 'a', 'b', 'c']
c = Counter(l)
l.sort(key=lambda x : -c[x])
# l is unchanged
l.sort(key=lambda x : (-c[x],x))
# l is ['a', 'a', 'b', 'b', 'c']

edit: Mark's solution uses a counter which is much better than recounting every time. I am going to steal that Idea

edited Jul 7, 2017 at 19:01

answered Jul 7, 2017 at 18:18

AltF4

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