Python程序在矩阵中查找连续行差异

给定一个矩阵，任务是编写一个Python程序，根据存在的元素执行与前一行的差异。

Input : test_list = [[5, 6, 3, 1], [7, 5, 3, 1], [3, 2], [7, 3, 3, 2], [2, 3], [9, 8, 1]]

Output : [[], [7], [2], [7], [], [8, 9, 1]]

Explanation : Comparing 1st and 2nd row, only 7 exists in 2nd row and not in 1st hence [7] is output.

Input : test_list = [[5, 6], [7, 5, 3, 1], [3, 4], [7], [2, 9], [9, 8, 1]]

Output : [[], [1, 3, 7], [4], [7], [9, 2], [8, 1]]

Explanation : Comparing 2nd and 3rd list, only 4 is present in 3rd list and not in 2nd, hence output.

示例：使用set.difference() +循环

在这里，保持了前一组，保持了前一组的轨迹以获得差异。这将从当前列表中删除已经存在于前一个列表中的所有元素。

Python3

# Python3 code to demonstrate working of
# Sucessive row difference in Matrix
# Using set.difference + loop
  
# initializing list
test_list = [[5, 6, 3, 1], [7, 5, 3, 1],
             [3, 2], [7, 3, 3, 2],
             [2, 3], [9, 8, 1]]
               
# printing original list
print("The original list is : " + str(test_list))
  
res = []
prev = set(test_list[0])
for ele in test_list:
      
    # appending difference set diff with previous
    # element
    res.append(list(set(ele).difference(prev)))
      
    # updating prev. for comparison
    prev = set(ele)
  
# printing result
print("Sucessive Row difference : " + str(res))

输出：

The original list is : [[5, 6, 3, 1], [7, 5, 3, 1], [3, 2], [7, 3, 3, 2], [2, 3], [9, 8, 1]]

Sucessive Row difference : [[], [7], [2], [7], [], [8, 9, 1]]