sorting - How to sort Counter by value? - python

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Top 5 Answer for sorting - How to sort Counter by value? - python

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Use the Counter.most_common() method, it'll sort the items for you:

>>> from collections import Counter >>> x = Counter({'a':5, 'b':3, 'c':7}) >>> x.most_common() [('c', 7), ('a', 5), ('b', 3)] 

It'll do so in the most efficient manner possible; if you ask for a Top N instead of all values, a heapq is used instead of a straight sort:

>>> x.most_common(1) [('c', 7)] 

Outside of counters, sorting can always be adjusted based on a key function; .sort() and sorted() both take callable that lets you specify a value on which to sort the input sequence; sorted(x, key=x.get, reverse=True) would give you the same sorting as x.most_common(), but only return the keys, for example:

>>> sorted(x, key=x.get, reverse=True) ['c', 'a', 'b'] 

or you can sort on only the value given (key, value) pairs:

>>> sorted(x.items(), key=lambda pair: pair[1], reverse=True) [('c', 7), ('a', 5), ('b', 3)] 

See the Python sorting howto for more information.

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A rather nice addition to @MartijnPieters answer is to get back a dictionary sorted by occurrence since Collections.most_common only returns a tuple. I often couple this with a json output for handy log files:

from collections import Counter, OrderedDict  x = Counter({'a':5, 'b':3, 'c':7}) y = OrderedDict(x.most_common()) 

With the output:

OrderedDict([('c', 7), ('a', 5), ('b', 3)]) {   "c": 7,    "a": 5,    "b": 3 } 
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>>> from collections import Counter >>> x = Counter({'a':5, 'b':3, 'c':7}) 

Using the sorted keyword key and a lambda function:

>>> sorted(x.items(), key=lambda i: i[1]) [('b', 3), ('a', 5), ('c', 7)] >>> sorted(x.items(), key=lambda i: i[1], reverse=True) [('c', 7), ('a', 5), ('b', 3)] 

This works for all dictionaries. However Counter has a special function which already gives you the sorted items (from most frequent, to least frequent). It's called most_common():

>>> x.most_common() [('c', 7), ('a', 5), ('b', 3)] >>> list(reversed(x.most_common()))  # in order of least to most [('b', 3), ('a', 5), ('c', 7)] 

You can also specify how many items you want to see:

>>> x.most_common(2)  # specify number you want [('c', 7), ('a', 5)] 
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More general sorted, where the key keyword defines the sorting method, minus before numerical type indicates descending:

>>> x = Counter({'a':5, 'b':3, 'c':7}) >>> sorted(x.items(), key=lambda k: -k[1])  # Ascending [('c', 7), ('a', 5), ('b', 3)] 
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