Ranking Elements of multiple Lists by their count in Python

Question

I want to rank multiple lists according to their elements how often they appear in each list. Example: list1 = 1,2,3,4 list2 = 4,5,6,7 list3 = 4,1,8,9 result = 4,1,2,3,4,5,6,7,8 (4 is counted three times, 1 two times and the rest once) I've tried the following but i need something more intelligent and something i can do with any ammount of lists. ``` l = [] l.append([ 1, 2, 3, 4, 5]) l.append([ 1, 9, 3, 4, 5]) l.append([ 1, 10, 8, 4, 5]) l.append([ 1, 12, 13, 7, 5]) l.append([ 1, 14, 13, 13, 6]) x1 = set(l[0]) & set(l[1]) & set(l[2]) & set(l[3]) x2 = set(l[0]) & set(l[1]) & set(l[2]) & set(l[4]) x3 = set(l[0]) & set(l[1]) & set(l[3]) & set(l[4]) x4 = set(l[0]) & set(l[2]) & set(l[3]) & set(l[4]) x5 = set(l[1]) & set(l[2]) & set(l[3]) & set(l[4]) set1 = set(x1) | set(x2) | set(x3) | set(x4) | set(x5) a1 = list(set(l[0]) & set(l[1]) & set(l[2]) & set(l[3]) & set(l[4])) a2 = getDifference(list(set1),a1) print a1 print a2 ``` Now here is the problem... i can do it again and again with a3,…

Accepted Answer

```
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)]
```

Ranking Elements of multiple Lists by their count in Python

Solution

Problem