Python程序通过切片行和列总和对矩阵进行排序

给定一个矩阵和一系列索引，任务是编写一个Python程序，该程序可以根据每行和列的给定索引范围的总和对矩阵进行排序，即行和列要从给定的开始到结束索引，此外，矩阵仅使用来自每行或每列的那些切片总和进行排序。

Input : test_list = [[1, 4, 3, 1, 3], [3, 4, 5, 2, 4], [23, 5, 5, 3], [2, 3, 5, 1, 6]], i, j = 1, 3
Output : [[1, 4, 3, 1, 3], [2, 3, 5, 1, 6], [3, 4, 5, 2, 4], [23, 5, 5, 3]]
Explanation : 7 < 8 < 9 < 10, is summation of 1st and 2nd element.

Input : test_list = [[1, 4, 3, 1, 3], [23, 5, 5, 3], [2, 3, 5, 1, 6]], i, j = 1, 3
Output : [[1, 4, 3, 1, 3], [2, 3, 5, 1, 6], [23, 5, 5, 3]]
Explanation : 7 < 8 < 10, is summation of 1st and 2nd element.

编程需要懂一点英语

行求和

方法 1：使用sort() 、 slice和sum()

在这里，我们使用 sort() 执行就地排序的任务，使用 sum() 完成求和，使用列表切片完成切片操作，它们一起形成排序的关键函数。

例子：

Python3

# get sliced summation
def get_sliced_sum(row):
    return sum(row[i:j])
  
  
# initializing list
test_list = [[1, 4, 3, 1, 3], [3, 4, 5, 2, 4],
             [23, 5, 5, 3], [2, 3, 5, 1, 6]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing range
i, j = 1, 3
  
# performing sort
test_list.sort(key=get_sliced_sum)
  
# printing result
print("Sorted List : " + str(test_list))

Python3

# initializing list
test_list = [[1, 4, 3, 1, 3], [3, 4, 5, 2, 4],
             [23, 5, 5, 3], [2, 3, 5, 1, 6]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing range
i, j = 1, 3
  
# performing sort using sorted()
# filter util. using lambda fnc.
res = sorted(test_list, key=lambda row: sum(row[i:j]))
  
# printing result
print("Sorted List : " + str(res))

Python3

# get sliced summation
def get_sliced_sum(row):
    return sum(row[i:j])
  
  
# initializing list
test_list = [[1, 4, 3, 1], [3, 4, 5, 2],
             [23, 5, 5, 3], [2, 3, 5, 1]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing range
i, j = 1, 3
  
# transposing matrix
test_list = list(zip(*test_list))
  
# performing sort
test_list.sort(key=get_sliced_sum)
  
# performing transpose again to get 
# result.
test_list = zip(*test_list)
  
# converting list of tuples to list of 
# lists
res = [list(sub) for sub in test_list]
  
# printing result
print("Sorted List Columnwise : " + str(res))

Python3

# initializing list
test_list = [[1, 4, 3, 1], [3, 4, 5, 2], 
             [23, 5, 5, 3], [2, 3, 5, 1]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing range
i, j = 1, 3
  
# transposing matrix
test_list = zip(*test_list)
  
# performing sort using sorted()
# filter util. using lambda fnc.
res = sorted(test_list, key=lambda row: sum(row[i:j]))
  
# performing transpose again to get result.
res = zip(*res)
  
# converting list of tuples to list of lists
res = [list(sub) for sub in res]
  
# printing result
print("Sorted List Columnwise : " + str(list(res)))

输出：

The original list is : [[1, 4, 3, 1, 3], [3, 4, 5, 2, 4], [23, 5, 5, 3], [2, 3, 5, 1, 6]]

Sorted List : [[1, 4, 3, 1, 3], [2, 3, 5, 1, 6], [3, 4, 5, 2, 4], [23, 5, 5, 3]]

编程需要懂一点英语

方法 2：使用sorted() 、 lambda 、 sum()和切片

在这种情况下，执行排序的任务是使用 sorted() 完成的，并且 lambda函数用于获取使用单语句渲染的切片的总和，无需外部函数调用。

例子：

蟒蛇3

# initializing list
test_list = [[1, 4, 3, 1, 3], [3, 4, 5, 2, 4],
             [23, 5, 5, 3], [2, 3, 5, 1, 6]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing range
i, j = 1, 3
  
# performing sort using sorted()
# filter util. using lambda fnc.
res = sorted(test_list, key=lambda row: sum(row[i:j]))
  
# printing result
print("Sorted List : " + str(res))

输出：

The original list is : [[1, 4, 3, 1, 3], [3, 4, 5, 2, 4], [23, 5, 5, 3], [2, 3, 5, 1, 6]]

Sorted List : [[1, 4, 3, 1, 3], [2, 3, 5, 1, 6], [3, 4, 5, 2, 4], [23, 5, 5, 3]]

编程需要懂一点英语

列求和

方法 1：使用sort() 、 slice和sum()

在此，我们执行基矩阵的转置，然后使用上述方法 1 执行获取切片的常规任务，排序后，矩阵再次转换为其转置格式。

例子：

蟒蛇3

# get sliced summation
def get_sliced_sum(row):
    return sum(row[i:j])
  
  
# initializing list
test_list = [[1, 4, 3, 1], [3, 4, 5, 2],
             [23, 5, 5, 3], [2, 3, 5, 1]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing range
i, j = 1, 3
  
# transposing matrix
test_list = list(zip(*test_list))
  
# performing sort
test_list.sort(key=get_sliced_sum)
  
# performing transpose again to get 
# result.
test_list = zip(*test_list)
  
# converting list of tuples to list of 
# lists
res = [list(sub) for sub in test_list]
  
# printing result
print("Sorted List Columnwise : " + str(res))

输出：

The original list is : [[1, 4, 3, 1], [3, 4, 5, 2], [23, 5, 5, 3], [2, 3, 5, 1]]

Sorted List Columnwise : [[1, 4, 3, 1], [2, 4, 5, 3], [3, 5, 5, 23], [1, 3, 5, 2]]

编程需要懂一点英语

方法 2：使用sorted() 、 lambda 、 sum()和切片

在此，我们执行基矩阵的转置，然后使用上述方法 2 执行获取切片的常规任务，排序后，矩阵再次转换为其转置格式。

例子：

蟒蛇3

# initializing list
test_list = [[1, 4, 3, 1], [3, 4, 5, 2], 
             [23, 5, 5, 3], [2, 3, 5, 1]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing range
i, j = 1, 3
  
# transposing matrix
test_list = zip(*test_list)
  
# performing sort using sorted()
# filter util. using lambda fnc.
res = sorted(test_list, key=lambda row: sum(row[i:j]))
  
# performing transpose again to get result.
res = zip(*res)
  
# converting list of tuples to list of lists
res = [list(sub) for sub in res]
  
# printing result
print("Sorted List Columnwise : " + str(list(res)))

输出：

The original list is : [[1, 4, 3, 1], [3, 4, 5, 2], [23, 5, 5, 3], [2, 3, 5, 1]]

Sorted List Columnwise : [[1, 4, 3, 1], [2, 4, 5, 3], [3, 5, 5, 23], [1, 3, 5, 2]]

编程需要懂一点英语