Sorting A List By Frequency Of Occurrence In A List

I have a list of integers(or could be even strings), which I would like to sort by the frequency of occurrences in Python, for instance: a = [1, 1, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5

Solution 1:

from collections import Counter
print [item for items, c in Counter(a).most_common() for item in [items] * c]
# [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]

Or even better (efficient) implementation

from collections import Counter
from itertools import repeat, chain
print list(chain.from_iterable(repeat(i, c) for i,c in Counter(a).most_common()))
# [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]

Or

from collections import Counter
printsorted(a, key=Counter(a).get, reverse=True)
# [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]

If you prefer in-place sort

a.sort(key=Counter(a).get, reverse=True)

Solution 2:

Using Python 3.3 and the built in sorted function, with the count as the key:

>>>a = [1,1,2,3,3,3,4,4,4,5,5,5,5]>>>sorted(a,key=a.count)
[2, 1, 1, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5]
>>>sorted(a,key=a.count,reverse=True)
[5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]

Solution 3:

In [15]: a = [1,1,2,3,3,3,4,4,4,5,5,5,5]

In [16]: counts = collections.Counter(a)

In [17]: list(itertools.chain.from_iterable([[k for _ in range(counts[k])] for k in sorted(counts, key=counts.__getitem__, reverse=True)]))
Out[17]: [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]

Alternatively:

answer = []
for k insorted(counts, key=counts.__getitem__, reverse=True):
    answer.extend([k for _ inrange(counts[k])])

Of course, [k for _ in range(counts[k])] can be replaced with [k]*counts[k]. So line 17 becomes

list(itertools.chain.from_iterable([[k]*counts[k] for k insorted(counts, key=counts.__getitem__, reverse=True)]))

Solution 4:

If you happen to be using numpy already, or if using it is an option, here's another alternative:

In [309]: import numpy as np

In [310]: a = [1, 2, 3, 3, 1, 3, 5, 4, 4, 4, 5, 5, 5]

In [311]: vals, counts = np.unique(a, return_counts=True)

In [312]: order= np.argsort(counts)[::-1]

In [313]: np.repeat(vals[order], counts[order])
Out[313]: array([5, 5, 5, 5, 4, 4, 4, 3, 3, 3, 1, 1, 2])

That result is a numpy array. If you want to end up with a Python list, call the array's tolist() method:

In [314]: np.repeat(vals[order], counts[order]).tolist()
Out[314]: [5, 5, 5, 5, 4, 4, 4, 3, 3, 3, 1, 1, 2]

Solution 5:

Not interesting way...

a = [1,1,2,3,3,3,4,4,4,5,5,5,5]

from collections import Counter
result = []
for v, times insorted(Counter(a).iteritems(), key=lambda x: x[1], reverse=True):
    result += [v] * times

One liner:

reduce(lambda a, b: a + [b[0]] * b[1], sorted(Counter(a).iteritems(), key=lambda x: x[1], reverse=True), [])