将矩阵转换为字典值列表的Python程序

给定 Matrix，任务是编写一个Python程序，将每一列的值映射为另一个列表中的自定义键。

Input : test_list = [[4, 5, 6], [1, 3, 5], [3, 8, 1], [10, 3, 5]], map_list = [4, 5, 6]

Output : {4: [4, 1, 3, 10], 5: [5, 3, 8, 3], 6: [6, 5, 1, 5]}

Explanation : 4 is mapped with all the 0th index of lists, 4, 1 ,3, 10.

Input : test_list = [[4, 5, 6], [1, 3, 5], [3, 8, 1]], map_list = [4, 5, 6]

Output : {4: [4, 1, 3], 5: [5, 3, 8], 6: [6, 5, 1]}

Explanation : 4 is mapped with all the 0th index of lists, 4, 1 ,3.

方法 1：使用字典理解+ zip()

在这种情况下，使用 zip() 完成列与自定义列表索引元素的映射，完成字典理解以将提取的键分配给映射值。

Python3

# Python3 code to demonstrate working of
# Convert Matrix to Dictionary Value List
# Using dictionary comprehension + zip()
from collections import defaultdict
  
# initializing list
test_list = [[4, 5, 6], [1, 3, 5], [3, 8, 1], [10, 3, 5]]
               
# printing original list
print("The original list is : " + str(test_list))
  
# initializing map list 
map_list = [4, 5, 6]
  
# mapping column using zip(), dictionary comprehension for key
# converts to list of dictionary
temp = [{key : val for key, 
         val in zip(map_list, idx)} for idx in test_list]
  
# convert to dictionary value list 
res = defaultdict(list) 
{res[key].append(sub[key]) for sub in temp for key in sub}
  
# printing result
print("Converted Dictionary : " + str(dict(res)))

输出：

The original list is : [[4, 5, 6], [1, 3, 5], [3, 8, 1], [10, 3, 5]]

Converted Dictionary : {4: [4, 1, 3, 10], 5: [5, 3, 8, 3], 6: [6, 5, 1, 5]}

方法 2：使用dict() +列表推导+ zip()

在这种情况下，将值映射到字典键和转换的任务是使用 dict() 和 zip() 以及字典理解完成的。休息功能类似于上述方法。

蟒蛇3

# Python3 code to demonstrate working of
# Convert Matrix to Dictionary Value List
# Using dict() + list comprehension + zip()
from collections import defaultdict
  
# initializing list
test_list = [[4, 5, 6], [1, 3, 5], [3, 8, 1], [10, 3, 5]]
               
# printing original list
print("The original list is : " + str(test_list))
  
# initializing map list 
map_list = [4, 5, 6]
  
# mapping column using zip() and conversion using using dict()
# converts to list of dictionary
temp = [dict(zip(map_list, sub)) for sub in test_list]
  
# convert to dictionary value list 
res = defaultdict(list) 
{res[key].append(sub[key]) for sub in temp for key in sub}
  
# printing result
print("Converted Dictionary : " + str(dict(res)))

输出：

The original list is : [[4, 5, 6], [1, 3, 5], [3, 8, 1], [10, 3, 5]]

Converted Dictionary : {4: [4, 1, 3, 10], 5: [5, 3, 8, 3], 6: [6, 5, 1, 5]}