Sort list by frequency

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]

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]

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