Is there any way in Python, wherein I can sort a list by its frequency?
For example,
[1,2,3,4,3,3,3,6,7,1,1,9,3,2]
the above list
In case you want to use a double comparator.
For example: Sort the list by frequency in descending order and in case of a clash the smaller one comes first.
import collections
def frequency_sort(a):
f = collections.Counter(a)
a.sort(key = lambda x:(-f[x], x))
return a
from collections import Counter
a = [2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8]
count = Counter(a)
a = []
while len(count) > 0:
c = count.most_common(1)
for i in range(c[0][1]):
a.append(c[0][0])
del count[c[0][0]]
print(a)
I think this would be a good job for a collections.Counter
:
counts = collections.Counter(lst)
new_list = sorted(lst, key=lambda x: -counts[x])
Alternatively, you could write the second line without a lambda:
counts = collections.Counter(lst)
new_list = sorted(lst, key=counts.get, reverse=True)
If you have multiple elements with the same frequency and you care that those remain grouped, we can do that by changing our sort key to include not only the counts, but also the value:
counts = collections.Counter(lst)
new_list = sorted(lst, key=lambda x: (counts[x], x), reverse=True)
l = [1,2,3,4,3,3,3,6,7,1,1,9,3,2]
print sorted(l,key=l.count,reverse=True)
[3, 3, 3, 3, 3, 1, 1, 1, 2, 2, 4, 6, 7, 9]
Was practising this one for fun. This solution use less time complexity.
from collections import defaultdict
lis = [1,2,3,4,3,3,3,6,7,1,1,9,3,2]
dic = defaultdict(int)
for num in lis:
dic[num] += 1
s_list = sorted(dic, key=dic.__getitem__, reverse=True)
new_list = []
for num in s_list:
for rep in range(dic[num]):
new_list.append(num)
print(new_list)
You can use a Counter
to get the count of each item, use its most_common
method to get it in sorted order, then use a list comprehension to expand again
>>> lst = [1,2,3,4,3,3,3,6,7,1,1,9,3,2]
>>>
>>> from collections import Counter
>>> [n for n,count in Counter(lst).most_common() for i in range(count)]
[3, 3, 3, 3, 3, 1, 1, 1, 2, 2, 4, 6, 7, 9]