如何获得自定义的元组的计数对两个列表

Question

请帮助我获得了从集合使用列表SS1列表SS2在Python柜台导入柜台或任何其他最快的方式

SS1 = [(1, 2, 3, 4, 5), (1, 2, 3, 4, 6), (1, 2, 3, 5, 6), (1, 2, 4, 5, 6), (1, 3, 4, 5, 6), (2, 3, 4, 5, 6)] 

SS2=[(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 5), 
(1, 4, 6), (1, 5, 6), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6), (2, 5, 6), 
(3, 4, 5), (3, 4, 6), (3, 5, 6), (4, 5, 6)] 

这里是什么我已经试过，我不知道如何为（1,2,4）的数个的每个元组的元素

SS1=[(1, 2, 3, 4, 5), (1, 2, 3, 4, 6), (1, 2, 3, 5, 6), (1, 2, 4, 5, 6), (1, 3, 4, 5, 6), (2, 3, 4, 5, 6)] 

from collections import Counter 
c = Counter(elem[0:3] for elem in SS1[0:6]) 

for k, v in c.items(): 
    if (v > 0): 
     print(k,v) 

现在，这是运行完美的0：3，但我想要的是得到不是1,2,3的计数，但我想1,2,4个元素为每个元组计数。

对不起球员希望你明白我的问题...对不起，我再次是新的这条巨蟒。

Answer 1

信贷@Chiheb的Nexus

SS1=[(1, 2, 3, 4, 5), (1, 2, 3, 4, 6), (1, 2, 3, 5, 6), (1, 2, 4, 5, 6), (1, 3, 4, 5, 6), (2, 3, 4, 5, 6)] 

from collections import Counter 

def get_new_list(a, pos): 
    # Check if any element in pos is > than the length of the tuples 
    if any(k >= len(min(SS1, key=lambda x: len(x))) for k in pos): 
     return 

    for k in a: 
     yield tuple(k[j] for j in pos) 

def elm_counter(elm): 
    if not len(elm): 
     return 

    c = Counter(elm) 
    for k, v in c.items(): 
     if v > 0: 
      print(k, v) 
elm = list(get_new_list(SS1, (2,))) 
elm_counter(elm) 
print('---')  
elm = list(get_new_list(SS1, (0, 2, 3))) 
elm_counter(elm) 
print('---') 
elm = list(get_new_list(SS1, (1, 3, 4))) 
elm_counter(elm) 

Answer 2

好吧我假设你想要什么作为你的输出这里你并不清楚。所以基本上你想要的是在SS2的SS1中找到项目的计数。

例如这将是在SS1

存在的时间(1,4,5)数即在(1, 2, 3, 4, 5)，(1, 2, 4, 5, 6)，(1, 3, 4, 5, 6)

所以(1, 2, 5)这将是再次对不对？目前在 (1, 2, 3, 4, 5),(1, 2, 3, 5, 6),(1, 2, 4, 5, 6)

我想你可能需要的是。

set(tuple2).issubset(tuple1) 

因此，这里是你的问题代码：

SS1 = [(1, 2, 3, 4, 5), (1, 2, 3, 4, 6), (1, 2, 3, 5, 6), (1, 2, 4, 5, 6), (1, 3, 4, 5, 6), (2, 3, 4, 5, 6)] 
SS2=[(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 5), 
(1, 4, 6), (1, 5, 6), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6), (2, 5, 6), 
(3, 4, 5), (3, 4, 6), (3, 5, 6), (4, 5, 6)] 
count=0 
count_list = [] 
for ss2item in SS2: 
    for ss1item in SS1: 
     if set(ss2item).issubset(ss1item): 
      count+=1 
    count_list.append(count)   
    count=0 
print(count_list) 

它的输出将是在SS2每个项目数的列表：

[3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3]