5

我想根据元素在每个列表中出现的频率对多个列表进行排名。例子：

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 计 3 次，1 计 2 次，其余计 1 次）

我已经尝试了以下方法，但我需要一些更智能的东西，并且我可以用任何数量的列表来做一些事情。


 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

现在问题来了……我可以用 a3、a4 和 a5 一次又一次地做到这一点，但它太复杂了，我需要一个函数来解决这个问题……但我不知道如何……我的数学卡住了;)

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。你帮了我所有的忙！泰

4

6 回答 6

7 
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（采用任何可迭代的、宽松的哈希要求），允许键和反向参数（匹配排序），并重命名为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)]

例子：

>>> 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)]
于 2009-12-01T22:54:43.850 回答
2

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.
于 2009-12-01T23:29:44.473 回答
1 
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)
于 2009-12-01T23:01:38.017 回答
0

试试这个：

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)]

使用示例：

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

print rank(a,b,c)

您可以使用任意数量的列表作为输入

于 2009-12-01T23:12:48.320 回答
0

您可以计算每个元素的出现次数（直方图），然后按其排序：

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)
于 2009-12-01T23:37:55.300 回答
0

试试这个代码：

def elementFreq(myList):
    #myList is the list of lists
    from collections import Counter
    tmp = []
    for i in myList: tmp += i        
    return(Counter(tmp))

注意：您的列表应该是可散列类型

于 2018-10-30T13:24:37.577 回答