当输入是对角线元素时，将输入转换为矩阵

Question

我的列表如下：

[[1],[2,1],[3,2,1],[3,1,2],[1,2],[1]]

以上是输入内容。

这些是原始清单的对角线元素：

[[3,3,1,1],[2,2,1,2],[1,1,1,2]]

也就是说，

3 3 1 1
2 2 1 2
1 1 1 2

所以问题是要从对角元素的输入中重建原始列表。

如您所见，对角线元素遵循主对角线的方向。对于任意尺寸的矩阵，该如何实现？

Answer 1

您可以按照以下步骤进行操作：

• 从输入中填充零背景矩阵（以便剪切感兴趣的值）
• 应用剪切（甚至是转置），以便所有零最终出现在边缘的某些行/列中
• 切出背景零点

这可以通过NumPy数组或纯Python list s来完成。

使用NumPy

import numpy as np


l = [[1],[2,1],[3,2,1],[3,1,2],[1,2],[1]]
m = np.array([[3,3,1,1],[2,2,1,2],[1,1,1,2]])

rows = max(len(x) for x in l)
cols = len(l) - rows + 1

# fill a shear matrix from the input
a = np.zeros((rows, len(l)), dtype=m.dtype)
for j, x in enumerate(l):
    k = rows - j - 1
    a[:, j] = [0] * k + x + [0] * (-k - 1)
print(a)
# [[0 0 3 3 1 1]
#  [0 2 2 1 2 0]
#  [1 1 1 2 0 0]]

# apply the shearing
for i in range(rows):
    a[i, :] = np.roll(a[i, :], i)
print(a)
# [[0 0 3 3 1 1]
#  [0 0 2 2 1 2]
#  [0 0 1 1 1 2]]

# slice out the zeros
a = a[:, len(l) - cols:]
print(a)
# [[3 3 1 1]
#  [2 2 1 2]
#  [1 1 1 2]]

np.all(a == m)
# True

使用list s

def transpose(seq):
    return type(seq)(map(type(seq), zip(*seq)))


def rotate(seq, i):
    return seq[-i:] + seq[:-i]


def shear(seq):
    return type(seq)(rotate(x, i) for i, x in enumerate(seq))


l = [[1],[2,1],[3,2,1],[3,1,2],[1,2],[1]]
m = [[3,3,1,1],[2,2,1,2],[1,1,1,2]]

rows = max(len(x) for x in l)
cols = len(l) - rows + 1

# fill zero-background matrix from the input
a = [([0] * (rows - j - 1) + x + [0] * (j - rows)) for j, x in enumerate(l)]
print(a)
# [[0, 0, 1], [0, 2, 1], [3, 2, 1], [3, 1, 2], [1, 2, 0], [1, 0, 0]]

# apply the transposition and the shearing
print(transpose(a))
# [[0, 0, 3, 3, 1, 1], [0, 2, 2, 1, 2, 0], [1, 1, 1, 2, 0, 0]]
print(shear(transpose(a)))
# [[0, 0, 3, 3, 1, 1], [0, 0, 2, 2, 1, 2], [0, 0, 1, 1, 1, 2]]

# slice out the zeros
a = [x[len(a) - cols:] for x in shear(transpose(a))]
print(a)
# [[3, 3, 1, 1], [2, 2, 1, 2], [1, 1, 1, 2]]
print(a == m)
# True

或者，不将逻辑操作分开而更加简洁：

l = [[1],[2,1],[3,2,1],[3,1,2],[1,2],[1]]
m = [[3,3,1,1],[2,2,1,2],[1,1,1,2]]

rows = max(len(x) for x in l)
cols = len(l) - rows + 1

a = [([0] * (rows - j - 1) + x + [0] * (j - rows)) for j, x in enumerate(l)]
a = [list(x[len(a) - cols - i:len(a) - i]) for i, x in enumerate(zip(*a))]
print(a)
# [[3, 3, 1, 1], [2, 2, 1, 2], [1, 1, 1, 2]]
print(a == m)
# True