将矩阵转换为字典列表的Python程序

给定一个矩阵，通过映射相似的索引值将其转换为字典列表。

Input : test_list = [[“Gfg”, [1, 2, 3]], [“best”, [9, 10, 11]]] 
Output : [{‘Gfg’: 1, ‘best’: 9}, {‘Gfg’: 2, ‘best’: 10}, {‘Gfg’: 3, ‘best’: 11}] 

Input : test_list = [[“Gfg”, [1, 2, 3]]] 
Output : [{‘Gfg’: 1}, {‘Gfg’: 2}, {‘Gfg’: 3}] 

执行此任务的粗暴方式是使用循环。在此，我们迭代 Matrix 中的所有元素，并将字典绑定键更新为字典中的适当值。

Python3

# initializing list
test_list = [["Gfg", [1, 2, 3]],
             ["is", [6, 5, 4]], ["best", [9, 10, 11]]]
  
# printing original list
print("The original list : " + str(test_list))
  
# using loop to bind Matrix elements to dictionary
res = []
for key, val in test_list:
  
    for idx, val in enumerate(val):
  
        # append values according to rows structure
        if len(res) - 1 < idx:
            res.append({key: val})
  
        else:
            res[idx].update({key: val})
  
# printing result
print("Converted dictionary : " + str(res))

输出：

The original list : [[‘Gfg’, [1, 2, 3]], [‘is’, [6, 5, 4]], [‘best’, [9, 10, 11]]]
Converted dictionary : [{‘Gfg’: 1, ‘is’: 6, ‘best’: 9}, {‘Gfg’: 2, ‘is’: 5, ‘best’: 10}, {‘Gfg’: 3, ‘is’: 4, ‘best’: 11}]