How to sort Counter by value? - python
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.
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
}
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)]
Yes:
>>> 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)]