# How to sort Counter by value? – python

## Question:

Other than doing list comprehensions of reversed list comprehension, is there a pythonic way to sort Counter by value? If so, it is faster than this:

``````>>> from collections import Counter
>>> x = Counter({'a':5, 'b':3, 'c':7})
>>> sorted(x)
['a', 'b', 'c']
>>> sorted(x.items())
[('a', 5), ('b', 3), ('c', 7)]
>>> [(l,k) for k,l in sorted([(j,i) for i,j in x.items()])]
[('b', 3), ('a', 5), ('c', 7)]
>>> [(l,k) for k,l in sorted([(j,i) for i,j in x.items()], reverse=True)]
[('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)
[('b', 3), ('a', 5), ('c', 7)]
>>> sorted(x.items(), key=lambda i: i, 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)]
``````

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, reverse=True)
[('c', 7), ('a', 5), ('b', 3)]
``````

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)  # Ascending
[('c', 7), ('a', 5), ('b', 3)]
``````