Sorting a List by frequency of occurrence in a list

Question

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]

Here the element 5 appears 4 times in the list, 4 appears 3 times. So the output sorted list would be :

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

I tried using a.count(), but it gives the number of occurrence of the element. I would like to sort it. Any idea how to do it ?

Thanks

Answer 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
print sorted(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)

Answer 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]

Answer 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 in sorted(counts, key=counts.__getitem__, reverse=True):
    answer.extend([k for _ in range(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 in sorted(counts, key=counts.__getitem__, reverse=True)]))

Answer 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]

Answer 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 in sorted(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), [])

Answer 6

Occurrence in array and within a sets of equal size:

rev=True

arr = [6, 6, 5, 2, 9, 2, 5, 9, 2, 5, 6, 5, 4, 6, 9, 1, 2, 3, 4, 7 ,8 ,8, 8, 2]
print arr

arr.sort(reverse=rev)

ARR = {}
for n in arr:
  if n not in ARR:
    ARR[n] = 0
  ARR[n] += 1

arr=[]
for k,v in sorted(ARR.iteritems(), key=lambda (k,v): (v,k), reverse=rev):
  arr.extend([k]*v)
print arr

Results:

[6, 6, 5, 2, 9, 2, 5, 9, 2, 5, 6, 5, 4, 6, 9, 1, 2, 3, 4, 7, 8, 8, 8, 2]
[2, 2, 2, 2, 2, 6, 6, 6, 6, 5, 5, 5, 5, 9, 9, 9, 8, 8, 8, 4, 4, 7, 3, 1]

Answer 7

Dart Solution

String sortedString = '';
Map map = {};
for (int i = 0; i < s.length; i++) {
   map[s[i]] = (map[s[i]] ?? 0) + 1;
   // OR 
   //  map.containsKey(s[i])
   //   ? map.update(s[i], (value) => ++value)
   //   : map.addAll({s[i]: 1});
}
var sortedByValueMap = Map.fromEntries(
    map.entries.toList()..sort((e1, e2) => e1.value.compareTo(e2.value)));
sortedByValueMap.forEach((key, value) {
  sortedString += key * value;
});
return sortedString.split('').reversed. Join();