如何计算子列表中某个模式的次数发生在列表中，然后将其添加到Sublist中

Question

挑战是我要计算某些项目在某些索引中发生的某些项目模式的次数。

例如，我想计算在索引0和索引1处发生唯一模式的次数。" a"one_answers" z"出现在 index 0和 index 1下3次，而'1'and'2'在索引0和索引1下方发生两次。我只关心在索引0和1处出现的对-列表。

List = [['a','z','g','g','g'],['a','z','d','d','d'],['a','z','z','z','d'],['1','2','f','f','f'],['1','2','3','f','f'],['1','1','g','g','g']]
Desired_List = [['a','z','g','g','g',3],['a','z','d','d','d',3],['a','z','z','z','d',3],['1','2','f','f','f',2],['1','2','3','f','f',2],['1','1','g','g','g',1]]

目前，我的尝试是：

from collections import Counter
l1 = Counter(map(lambda x: (x[0] + "|" + x[1]),List)
Deduped_Original_List = map(lambda x: Counter.keys().split("|"),l1)
Counts = map(lambda x: Counter.values(),l1)
for ele_a, ele_b in zip(Deduped_Original_List, Counts):
    ele_a.append(ele_b)

这显然不起作用，因为在此过程中，我丢失了索引2,3和4。

Answer 1

您可以使用 collections.Counter：

使用列表理解

from collections import Counter
lst = [['a','z','g','g','g'],['a','z','d','d','d'],['a','z','z','z','d'],['1','2','f','f','f'],['1','2','3','f','f'],['1','1','g','g','g']]
cnt = Counter([tuple(l[:2]) for l in lst])
lst_output = [l + [cnt[tuple(l[:2])]] for l in lst]
print lst_output

ouput：

[['a', 'z', 'g', 'g', 'g', 3], ['a', 'z', 'd', 'd', 'd', 3], ['a', 'z', 'z', 'z', 'd', 3], ['1', '2', 'f', 'f', 'f', 2], ['1', '2', '3', 'f', 'f', 2], ['1', '1', 'g', 'g', 'g', 1]]

Answer 2

>>> import collections
>>> List = [['a','z','g','g','g'],['a','z','d','d','d'],['a','z','z','z','d'],['1','2','f','f','f'],['1','2','3','f','f'],['1','1','g','g','g']]
>>> patterns = ['az', '12']
>>> answer = collections.defaultdict(int)
>>> for subl in List:
...     for pattern in patterns:
...         if all(a==b for a,b in zip(subl, pattern)):
...             answer[pattern] += 1
...             break
... 
>>> for i,subl in enumerate(List):
...     if ''.join(subl[:2]) in answer:
...         List[i].append(answer[''.join(subl[:2])])
... 
>>> List
[['a', 'z', 'g', 'g', 'g', 3], ['a', 'z', 'd', 'd', 'd', 3], ['a', 'z', 'z', 'z', 'd', 3], ['1', '2', 'f', 'f', 'f', 2], ['1', '2', '3', 'f', 'f', 2], ['1', '1', 'g', 'g', 'g']]
>>> 

Answer 3

我喜欢ys-l的Counter方法。这是另一种方法：

>>> List = [['a','z','g','g','g'], ['a','z','d','d','d'], ['a','z','z','z','d'],['1','2','f','f','f'], ['1','2','3','f','f'], ['1','1','g','g','g']]
>>> d = {}
>>> for i in List:
       key = i[0] + i[1]
       if not d.get(key, None): d[key] = 1
       else: d[key] += 1

>>> Desired_List = [li + [d[li[0] + li[1]]] for li in List]
>>> Desired_List
[['a', 'z', 'g', 'g', 'g', 3], ['a', 'z', 'd', 'd', 'd', 3], ['a', 'z', 'z', 'z', 'd', 3], ['1', '2', 'f', 'f', 'f', 2], ['1', '2', '3', 'f', 'f', 2], ['1', '1', 'g', 'g', 'g', 1]]