嵌套列表上的增量滚动功能

Question

我还是python的初学者，想把这段用Java写的代码转成Python

// shifts the element  of each row according to the row's index and the element's
// index in the row example row 0  no shifts ( since all the elements would get 
// their original positions from this formula [(i+j) % d] row 1, one shift for all
// the elements, and so on for all the rows in the array using the same formula

public void doSomething(int[][] a) {
    int[] temp = new int[d];
    for (int i = 1; i < a.length; i++) {
        for (int j = 0; j < a[i].length; j++) {
            temp[j] = a[i][(j + i) % a[i].length];
        }
        System.arraycopy(temp, 0, a[i], 0, a[i].length);
    }
}

我尝试了类似的方法：

def do_something(self, arr):
    temp_list = []
    i = 0
    for row in arr:
        j = 0
        for _ in row:
            y = (i + j) % self.d
            z = arr[i, y]
            temp_list.append(z)
            j += 1
        arr[i, :] = temp_list
        i += 1
    return arr

我收到此错误：

arr[i, :] = temp_list  
ValueError: cannot copy sequence with size 8 to array axis with dimension 4

对于输入数组：

0 1 2
3 4 5
6 7 8

应该会得到结果：

0 1 2
4 5 3
8 6 7

或者对于这个输入：

0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15

这个输出

0 1 2 3
5 6 7 4
10 11 8 9
15 12 13 14

Answer 1

这是一个可以完成这项工作的快速小功能。本质上，每个子数组都按其在父数组中的索引进行移位。

设置：

l1 = [[0, 1, 2], [3, 4, 5], [6, 7, 8]]
l2 = [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11], [12, 13, 14], [15, 16, 17]]
l3 = [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15]]

功能：

def spiral(array):
    temp = []
    for i, a in enumerate(array):
        n = i % len(a)
        temp.append(a[n:] + a[:n])
        n += 1
    return temp

输出：

>>> spiral(l1)
[[0, 1, 2], [4, 5, 3], [8, 6, 7]]

>>> spiral(l2)
[[0, 1, 2], [4, 5, 3], [8, 6, 7], [9, 10, 11], [13, 14, 12], [17, 15, 16]]

>>> spiral(l3)
[[0, 1, 2, 3], [5, 6, 7, 4], [10, 11, 8, 9], [15, 12, 13, 14]]

编辑：

根据 OP 的 cmets，更新函数以简化并确保与不同长度的嵌套列表兼容。