# How to sort Counter by value? - python

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)
```

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)]
```