Python程序对元组列表中的值进行唯一键计数

给定双元组，获取元组中每个值的唯一键计数。

Input : test_list = [(3, 4), (1, 2), (2, 4), (8, 2), (7, 2), (8, 1), (9, 1), (8, 4), (10, 4)] 
Output : {4: 4, 2: 3, 1: 2} 
Explanation : 3, 2, 8 and 10 are keys for value 4.

Input : test_list = [(3, 4), (1, 2), (8, 1), (9, 1), (8, 4), (10, 4)] 
Output : {4: 3, 2: 1, 1: 2} 
Explanation : 3, 8 and 10 are keys for value 4. 
 

方法 #1：使用循环 + defaultdict()

在此，我们迭代每个元组元素，将键作为元组值，并使用字典值列表中遇到的每个不同键递增它。接下来，通过获取映射值列表的长度，使用另一个迭代计算频率，转换为集合以获得唯一计数。

Python3

# Python3 code to demonstrate working of 
# Unique keys count for Value in Tuple List
# Using loop + defaultdict()
from collections import defaultdict
  
# initializing list
test_list = [(3, 4), (1, 2), (2, 4), (8, 2), (7, 2), (8, 1), (9, 1), (8, 4), (10, 4)]
  
# printing original list
print("The original list is : " + str(test_list))
  
  
res = defaultdict(list)
for sub in test_list:
      
    # getting all keys to values
    res[sub[1]].append(sub[0])
      
  
res = dict(res)
  
  
res_dict = dict()
for key in res:
      
    # getting unique key counts for each value
    res_dict[key] = len(list(set(res[key])))
  
  
# printing result 
print("Unique keys for values : " + str(res_dict))

输出：

The original list is : [(3, 4), (1, 2), (2, 4), (8, 2), (7, 2), (8, 1), (9, 1), (8, 4), (10, 4)]

Unique keys for values : {4: 4, 2: 3, 1: 2}

方法 #2：使用循环 + defaultdict() + Counter()

在这种情况下，为了减少计算循环，列表是动态构建的，只有唯一值。然后 Counter() 用于获取唯一值字典。

蟒蛇3

# Python3 code to demonstrate working of 
# loop + defaultdict() + Counter()
# Using loop + defaultdict() + Counter()
from collections import defaultdict, Counter
  
# initializing list
test_list = [(3, 4), (1, 2), (2, 4), (8, 2), (7, 2),
             (8, 1), (9, 1), (8, 4), (10, 4)]
  
# printing original list
print("The original list is : " + str(test_list))
  
mem_dict = defaultdict(list)
res = []
for sub in test_list:
      
    # if not in dict, add value 
    if sub[0] not in mem_dict[sub[1]]:
        mem_dict[sub[1]].append(sub[0])
        res.append(sub[1])
          
# getting frequency 
res = dict(Counter(res))
  
# printing result 
print("Unique keys for values : " + str(res))

输出：

原列表为：[(3, 4), (1, 2), (2, 4), (8, 2), (7, 2), (8, 1), (9, 1), (8) , 4), (10, 4)]
值的唯一键：{4: 4, 2: 3, 1: 2}

蟒蛇3

# Python3 code to demonstrate working of 
# Unique keys count for Value in Tuple List
# Using loop + defaultdict()
from collections import defaultdict
  
# initializing list
test_list = [(3, 4), (1, 2), (2, 4), (8, 2), (7, 2), 
             (8, 1), (9, 1), (8, 4), (10, 4)]
  
# printing original list
print("The original list is : " + str(test_list))
  
  
res = defaultdict(list)
for sub in test_list:
      
    # getting all keys to values
    res[sub[1]].append(sub[0])
      
  
res = dict(res)
  
  
res_dict = dict()
for key in res:
      
    # getting unique key counts for each value
    res_dict[key] = len(list(set(res[key])))
  
  
# printing result 
print("Unique keys for values : " + str(res_dict))