在Python中按列表数量对多个列表元素进行排名

问题描述

我想根据它们的元素对多个列表进行排名，以它们在每个列表中出现的频率为准.示例:

I want to rank multiple lists according to their elements how often they appear in each list. Example:

列表1 = 1,2,3,4
list2 = 4,5,6,7
list3 = 4,1,8,9

list1 = 1,2,3,4
list2 = 4,5,6,7
list3 = 4,1,8,9

结果= 4,1,2,3,4,5,6,7,8(4被计算3次，1被计算2次，其余被计算一次)

result = 4,1,2,3,4,5,6,7,8 (4 is counted three times, 1 two times and the rest once)

我尝试了以下方法，但是我需要一些更智能的东西，并且可以使用任何数量的列表来做.

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

现在这是问题所在...我可以用a3，a4和a5一次又一次地做到这一点，但是它太复杂了，我需要一个函数...但是我不知道如何...我的数学陷入困境;)

Now here is the problem... i can do it again and again with a3,a4 and a5 but its too complex then, i need a function for this... But i don't know how... my math got stuck ;)

已解决:非常感谢您的讨论.作为一个新手，我以某种方式喜欢这个系统:快速+信息丰富.你帮我全力以赴！ Ty

SOLVED: thanks alot for the discussion. As a newbee i like this system somehow: fast+informative. You helped me all out! Ty

推荐答案

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)

现在让我们对其进行概括(采取任何可迭代的方法，放宽可散列的要求)，允许键和反向参数(以匹配排序)，然后重命名为 freq_sorted :

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)]

示例:

>>> 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)]

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