Python - 按 K 倍数对行进行排序

给定一个矩阵，按行中存在的 K 的倍数执行行排序。

Input : test_list = [[3, 4, 8, 1], [12, 32, 4, 16], [1, 2, 3, 4], [9, 7, 5]], K = 4 
Output : [[9, 7, 5], [1, 2, 3, 4], [3, 4, 8, 1], [12, 32, 4, 16]] 
Explanation : 0 < 1 < 2 < 4, multiple of 4 occurrence order.

Input : test_list = [[3, 4, 8, 1], [12, 32, 4, 16], [1, 2, 3, 4], [9, 7, 5]], K = 2 
Output : [[9, 7, 5], [1, 2, 3, 4], [3, 4, 8, 1], [12, 32, 4, 16]] 
Explanation : 0 < 2 = 2 < 4, multiple of 2 occurrence order. 

方法 #1：使用 sort() + %运算符+ len()

在此，我们使用 %运算符测试多个，然后通过获取过滤元素的长度来计算计数，提供给 sort() 执行行的就地排序的键。

Python3

# Python3 code to demonstrate working of
# Sort row by K multiples
# Using sort() + % operator + len()
 
# checking for multiples count
 
 
def k_mul(row):
    return len([ele for ele in row if ele % K == 0])
 
 
# initializing list
test_list = [[3, 4, 8, 1], [12, 32, 4, 16], [1, 2, 3, 4], [9, 7, 5]]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 4
 
# performing sort
test_list.sort(key=k_mul)
 
# printing result
print("Sorted result : " + str(test_list))

输出：

The original list is : [[3, 4, 8, 1], [12, 32, 4, 16], [1, 2, 3, 4], [9, 7, 5]] 
Sorted result : [[9, 7, 5], [1, 2, 3, 4], [3, 4, 8, 1], [12, 32, 4, 16]]

方法 #2：使用sorted() + lambda + len()

在这种情况下，排序是使用 sorted() 完成的，len() 用于获取 K 的所有倍数的长度，如上述方法。 lambda函数提供单语句替代来执行逻辑注入。

蟒蛇3

# Python3 code to demonstrate working of
# Sort row by K multiples
# Using sorted() + lambda + len()
 
# initializing list
test_list = [[3, 4, 8, 1], [12, 32, 4, 16], [1, 2, 3, 4], [9, 7, 5]]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 4
 
# performing sort using sorted()
# lambda avoiding external function call
res = sorted(test_list, key=lambda row: len(
    [ele for ele in row if ele % K == 0]))
 
# printing result
print("Sorted result : " + str(res))

输出：

The original list is : [[3, 4, 8, 1], [12, 32, 4, 16], [1, 2, 3, 4], [9, 7, 5]] 
Sorted result : [[9, 7, 5], [1, 2, 3, 4], [3, 4, 8, 1], [12, 32, 4, 16]]