Python - 按 K 的频率对行进行排序

给定一个矩阵，任务是编写一个Python程序，根据 K 的频率对行进行排序。

Input : test_list = [[10, 2, 3, 2, 3], [5, 5, 4, 7, 7, 4], [1, 2], [1, 1, 2, 2, 2]], K = 2 
Output : [[5, 5, 4, 7, 7, 4], [1, 2], [10, 2, 3, 2, 3], [1, 1, 2, 2, 2]] 
Explanation : 0 < 1 < 2 < 3, count of K in Matrix order.
 

Input : test_list = [[5, 5, 4, 7, 7, 4], [1, 2], [1, 1, 2, 2, 2]], K = 2 
Output : [[5, 5, 4, 7, 7, 4], [1, 2], [1, 1, 2, 2, 2]] 
Explanation : 0 < 1 < 3, count of K in Matrix order. 

方法 #1：使用sort() + count()

在这里，我们使用sort()执行就地排序任务，使用count()完成捕获频率。

Python3

# Python3 code to demonstrate working of
# Sort rows by Frequency of K
# Using sort() + count()
  
  
def get_Kfreq(row):
  
    # return Frequency
    return row.count(K)
  
  
# initializing list
test_list = [[10, 2, 3, 2, 3], [5, 5, 4, 7, 7, 4], [1, 2], [1, 1, 2, 2, 2]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing K
K = 2
  
# performing inplace sort
test_list.sort(key=get_Kfreq)
  
# printing result
print("Sorted List : " + str(test_list))

输出：

The original list is : [[10, 2, 3, 2, 3], [5, 5, 4, 7, 7, 4], [1, 2], [1, 1, 2, 2, 2]]
Sorted List : [[5, 5, 4, 7, 7, 4], [1, 2], [10, 2, 3, 2, 3], [1, 1, 2, 2, 2]]

方法 #2：使用sorted() + lambda + count()

在这里，我们使用sorted()和lambda执行排序任务，消除了用于计算的外部函数调用和lambda函数。

蟒蛇3

# Python3 code to demonstrate working of
# Sort rows by Frequency of K
# Using sorted() + lambda + count()
  
# initializing list
test_list = [[10, 2, 3, 2, 3], [5, 5, 4, 7, 7, 4], [1, 2], [1, 1, 2, 2, 2]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing K
K = 2
  
# performing inplace sort
res = sorted(test_list, key=lambda row: row.count(K))
  
# printing result
print("Sorted List : " + str(res))

输出：

The original list is : [[10, 2, 3, 2, 3], [5, 5, 4, 7, 7, 4], [1, 2], [1, 1, 2, 2, 2]]
Sorted List : [[5, 5, 4, 7, 7, 4], [1, 2], [10, 2, 3, 2, 3], [1, 1, 2, 2, 2]]