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
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)
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]
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)]))
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]
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), [])
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
[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]
