Question

我需要一种方法，在类似矩阵结构中的每一行的下降索引位置附加或插入/。我发生的事情是我有一个程序，它接受列表的起始值和长度^ 2然后从这些值创建列表I.E。：

answer(17, 4) = [17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]

然后我将其转换为矩阵或嵌套列表：

matrix = [id_arr[i:i + length] for i in range(0, len(id_arr), length)]

输出：[[17, 18, 19, 20], [21, 22, 23, 24], [25, 26, 27, 28], [29, 30, 31, 32]]

[17, 18, 19, 20]  # matrix[0]
[21, 22, 23, 24]  # matrix[1]
[25, 26, 27, 28]  # matrix[2]
[29, 30, 31, 32]  # matrix[3]

现在我想要一种方法在矩阵的每一行的下降索引位置插入或附加/ .E：

17 18 19 20 /
21 22 23 / 24
25 26 / 27 28
29 / 30 31 32

完整代码：

def answer(start, length):
    # Generate the list of given start value and length
    id_arr = list(range(start, start + length ** 2))

    # Create a matrix of the list
    matrix = [id_arr[i:i + length] for i in range(0, len(id_arr), length)]

    # Attempt to try and insert the /
    for row in matrix:
        row.insert(len(matrix), '/')
        print row
    print len(matrix)

answer(17, 4)

输出：

[17, 18, 19, 20, '/']
[21, 22, 23, 24, '/']
[25, 26, 27, 28, '/']
[29, 30, 31, 32, '/']

正如您所看到的，我当前的代码只会在每行的末尾附加/。

Answer 1

使用枚举并获取每一行的索引并插入/每行的长度减去索引

 def answer(start, length):
    # Generate the list of given start value and length
    id_arr = list(range(start, start + length ** 2))

    # Create a matrix of the list
    matrix = [id_arr[i:i + length] for i in range(0, len(id_arr), length)]

    # Attempt to try and insert the /
    for i,row in enumerate(matrix):
        row.insert(len(matrix)-i, '/')
        print row
    print len(matrix)

 answer(17, 4)

输出：

[17, 18, 19, 20, '/']
[21, 22, 23, '/', 24]
[25, 26, '/', 27, 28]
[29, '/', 30, 31, 32]
4

Answer 2

您应该减少插入/

的索引

def answer(start, length):
    # Generate the list of given start value and length
    id_arr = list(range(start, start + length ** 2))

    # Create a matrix of the list
    matrix = [id_arr[i:i + length] for i in range(0, len(id_arr), length)]

    # Attempt to try and insert the /
    l = len(matrix)
    for row in matrix:
        row.insert(l, '/')
        l -= 1
        print row
    print len(matrix)

answer(17, 4)

在嵌套列表中的降序索引位置插入字符

2 个答案: