Ranking Elements of multiple Lists by their count in Python

Answer 1

import collections

data = [
  [1, 2, 3, 4, 5],
  [1, 9, 3, 4, 5],
  [1, 10, 8, 4, 5],
  [1, 12, 13, 7, 5],
  [1, 14, 13, 13, 6],
]

def sorted_by_count(lists):
  counts = collections.defaultdict(int)
  for L in lists:
    for n in L:
      counts[n] += 1

  return [num for num, count in
          sorted(counts.items(),
                 key=lambda k_v: (k_v[1], k_v[0]),
                 reverse=True)]

print sorted_by_count(data)

Now let's generalize it (to take any iterable, loosen hashable requirement), allow key and reverse parameters (to match sorted), and rename to freq_sorted:

def freq_sorted(iterable, key=None, reverse=False, include_freq=False):
  """Return a list of items from iterable sorted by frequency.

  If include_freq, (item, freq) is returned instead of item.

  key(item) must be hashable, but items need not be.

  *Higher* frequencies are returned first.  Within the same frequency group,
  items are ordered according to key(item).
  """
  if key is None:
    key = lambda x: x

  key_counts = collections.defaultdict(int)
  items = {}
  for n in iterable:
    k = key(n)
    key_counts[k] += 1
    items.setdefault(k, n)

  if include_freq:
    def get_item(k, c):
      return items[k], c
  else:
    def get_item(k, c):
      return items[k]

  return [get_item(k, c) for k, c in
          sorted(key_counts.items(),
                 key=lambda kc: (-kc[1], kc[0]),
                 reverse=reverse)]

Example:

>>> import itertools
>>> print freq_sorted(itertools.chain.from_iterable(data))
[1, 5, 4, 13, 3, 2, 6, 7, 8, 9, 10, 12, 14]
>>> print freq_sorted(itertools.chain.from_iterable(data), include_freq=True)
# (slightly reformatted)
[(1, 5),
 (5, 4),
 (4, 3), (13, 3),
 (3, 2),
 (2, 1), (6, 1), (7, 1), (8, 1), (9, 1), (10, 1), (12, 1), (14, 1)]

Answer 2

def items_ordered_by_frequency(*lists):

    # get a flat list with all the values
    biglist = []
    for x in lists:
        biglist += x

    # sort it in reverse order by frequency
    return sorted(set(biglist), 
                  key=lambda x: biglist.count(x), 
                  reverse=True)

Answer 3

Try this one:

def rank(*lists):
    d = dict()
    for lst in lists:
        for e in lst:
            if e in d: d[e] += 1
            else: d[e] = 1
    return [j[1] for j in sorted([(d[i],i) for i in d], reverse=True)]

Usage example:

a = [1,2,3,4]
b = [4,5,6,7]
c = [4,1,8,9]

print rank(a,b,c)

You can use any number of lists as input

Answer 4

Combining a couple of ideas already posted:

from itertools import chain
from collections import defaultdict

def frequency(*lists):
    counter = defaultdict(int)
    for x in chain(*lists):
        counter[x] += 1
    return [key for (key, value) in 
        sorted(counter.items(), key=lambda kv: (kv[1], kv[0]), reverse=True)]

Notes:

1. In Python 2.7, you can use Counter instead of defaultdict(int).
2. This version takes any number of lists as its argument; the leading asterisk means they'll all be packed into a tuple. If you want to pass in a single list containing all of your lists, omit that leading asterisk.
3. This breaks if your lists contain an unhashable type.

Answer 5

You can count the number of appearances of each element (a histogram), then sort by it:

def histogram(enumerable):
  result = {}
  for x in enumerable:
    result.setdefault(x, 0)
    result[x] += 1
  return result

lists = [ [1,2,3,4], [4,5,6,7], ... ]

from itertools import chain

h = histogram(chain(*lists))
ranked = sorted(set(chain(*lists)), key = lambda x : h[x], reverse = True)