Python – 从前后键值构建字典

给定一个列表，使用键作为前半部分值和从最后开始的值构造字典。

Input : test_list = [4, 10, 5, 3]
Output : {4: 3, 10: 5}
Explanation : First (4) is paired with rear (3) and so on.

Input : test_list = [5, 3]
Output : {5: 3}
Explanation : First (5) is paired with rear (3).

Input : test_dict = {“Apple” : 2, “Mango” : 2, “Grapes” : 2}, {“Apple” : 2, “Mango” : 2, “Grapes” : 2}
Output : 12
Explanation : (2*2) + (2*2) + (2*2) = 12.

Input : test_dict = {“Apple” : 3, “Mango” : 2, “Grapes” : 3}, {“Apple” : 2, “Mango” : 2, “Grapes” : 2}
Output : 16
Explanation : The summation of product leads to 16 as above.

方法#1：使用循环

这是可以执行此任务的粗暴方式。在此，我们从头开始运行一个循环，从 beg-end 中提取值，并相应地构造键值映射。

Python3

# Python3 code to demonstrate working of 
# Dictionary construction from front-rear key values
# Using loop
  
# initializing list
test_list = [4, 6, 3, 10, 5, 3]
  
# printing original list
print("The original list : " + str(test_list))
  
  
# initializing size and empty Dictionary
n = len(test_list)
res = dict()
  
# running loop till mid
for idx in range(n // 2):
  
    # mapping as required
    res.__setitem__(test_list[idx], test_list[n - idx - 1]) 
  
# printing result 
print("The mapped dictionary : " + str(res))

输出

The original list : [4, 6, 3, 10, 5, 3]
The mapped dictionary : {4: 3, 6: 5, 3: 10}

方法#2：使用 zip() + dict()

这是可以执行此任务的另一种方式。在此，我们使用 zip() 执行压缩键和值元素的任务，并使用 dict() 将结果转换为字典。

Python3

# Python3 code to demonstrate working of 
# Dictionary construction from front-rear key values
# Using zip() + dict()
  
# initializing list
test_list = [4, 6, 3, 10, 5, 3]
  
# printing original list
print("The original list : " + str(test_list))
  
# using zip to cut first and second half 
n = len(test_list)
res = dict(zip(test_list[:n // 2], test_list[n // 2:][::-1]))
  
# printing result 
print("The mapped dictionary : " + str(res))

输出

The original list : [4, 6, 3, 10, 5, 3]
The mapped dictionary : {4: 3, 6: 5, 3: 10}