Python – 列出矩阵中的元素分组

给定一个矩阵，根据列表元素进行分组，即每个组应包含列表中的所有元素。

Input : test_list = [[2, 6], [7, 8], [1, 4]], check_list = [1, 2, 4, 6]
Output : [[7, 8], [[1, 2], [4, 6]]]
Explanation : 1, 2, 4, 6 elements rows are grouped.

Input : test_list = [[2, 7], [7, 8], [1, 4]], check_list = [1, 2, 4, 6]
Output : [[2, 7], [7, 8], [1, 4]]
Explanation : No grouping possible.

方法：使用循环 + 列表理解 + Try-Except

在此，对于矩阵中的每一行，从列表中获取缺失的元素，获取元素后，如果我们找到缺失的元素，则与每一行匹配，如果找到，则建立新的组。

Python3

# Python3 code to demonstrate working of 
# List Elements Grouping in Matrix
# Using loop 
  
# initializing list
test_list = [[4, 6], [1, 2], [2, 6], [7, 8], [1, 4]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing check_list
check_list = [1, 2, 4, 6]
  
  
res = []
while test_list:
      
    # getting row
    sub1 = test_list.pop()
      
    # getting elements not in row
    sub2 = [ele for ele in check_list if ele not in sub1]
    try:
          
        # testing if we have list of removed elements
        test_list.remove(sub2)
          
        # grouping if present
        res.append([sub1, sub2])
    except ValueError:
          
        # ungrouped. 
        res.append(sub1)
  
# printing result 
print("The Grouped rows : " + str(res))

输出

The original list is : [[4, 6], [1, 2], [2, 6], [7, 8], [1, 4]]
The Grouped rows : [[[1, 4], [2, 6]], [7, 8], [[1, 2], [4, 6]]]