Python - 按无频率对矩阵进行排序

给定一个矩阵，根据 None 元素频率排序。

Input : test_list = [[None, None, 3, None], [12, 4, 5], [None, 3, 4]] 
Output : [[12, 4, 5], [None, 3, 4], [None, None, 3, None]] 
Explanation : 0, 1, 3 counts of None respectively.

Input : test_list = [[None, None, 3, None], [None, 3, 4]] 
Output : [[12, 4, 5], [None, None, 3, None]] 
Explanation : 0, 3 counts of None respectively. 

方法#1：使用sort()

在此，我们使用 sort() 执行排序，并使用外部函数在每行中提取 None 元素频率的任务。这将执行就地排序。

Python3

# Python3 code to demonstrate working of 
# Sort Matrix by None frequency
# Using sort()
  
# external sort function
def get_None_freq(row):
      
    # getting length of None characters
    return len([ele for ele in row if not ele])
  
# initializing list
test_list = [[None, None, 4], [None, None, 3, None],
             [12, 4, 5], [None, 3, 4]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# sorting using sort()
test_list.sort(key = get_None_freq)
  
# printing result 
print("List after sorting : " + str(test_list))

输出：

The original list is : [[None, None, 4], [None, None, 3, None], [12, 4, 5], [None, 3, 4]]
List after sorting : [[12, 4, 5], [None, 3, 4], [None, None, 4], [None, None, 3, None]]

方法 #2：使用 sorted() + lambda

在这里，不是使用外部函数，而是使用 lambda函数来解决这个问题。 sorted() 用于执行排序任务。

蟒蛇3

# Python3 code to demonstrate working of
# Sort Matrix by None frequency
# Using sorted() + lambda
  
# initializing list
test_list = [[None, None, 4], [None, None, 3, None],
             [12, 4, 5], [None, 3, 4]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# sorting using sorted()
# lambda function for None frequency logic
res = sorted(test_list, key=lambda row: len([ele for ele in row if not ele]))
  
# printing result
print("List after sorting : " + str(res))

输出：

The original list is : [[None, None, 4], [None, None, 3, None], [12, 4, 5], [None, 3, 4]]
List after sorting : [[12, 4, 5], [None, 3, 4], [None, None, 4], [None, None, 3, None]]