Python 使用numpy获取多维数组的所有对角线（包括小对角线）

我试图得到多维对象的对角线（和反对角线）元素

这些形状类似于

（2,2）

，

（3,3,3）

，

（4,4,4,4）

，

（5,5,5,5）

等等。不过，我认为这并不太相关

我用

ndarray

的

.diagonal

方法找到了获得对角线元素的方法，但是我找不到任何可以得到反对角线的东西

那么我必须用手来做吗

[编辑] 所以

所以我想要“水平”对角线，比如：

[54, 45, 49],
[ 4. 20, 75],
etc.

但从某种意义上说，它们也是水平的

[ 6, 45, 31],
[39, 20, 18]

然后是“垂直”的，比如：

[54, 33, 28],
[81, 20, 52],
etc.

但这些也是垂直的：

[6, 33, 38],
[11, 20, 32]

然后这个，不管你怎么称呼它

[54, 20, 63]

然后这些也是“更长”的对角线，就像前一条一样（如果你把矩阵看作一个三维几何结构，在几何意义上更长，数字放在立方体的顶点上，在它们之间的线的中间）

然后，在这个矩阵中，小对角线是从右到左或从下到上（但不是同时从右到左）的对角线，类似于：

[31, 45, 6],
[31, 83, 38]  # this is the first classical anti-diagonal in the first matrix

当然,我没有把所有的对角线都放在这里,但这是我的要求。我不需要从任何主/反对角线偏移的对角线

---

如果您也知道这是不可能的，请告诉我，因为我会手工完成。

如果您想要从

corner1

到

corner2

的对角线，并以

（0,0,0，…，0），（0,0,0，…，1），（1,1,1，…，1）

其中0表示“0处的此维度”，1表示“1处的此维度/结束”

然后这将返回从

corner1

到

corner2

得到的值，假设数组在每个维度上都具有相同的大小

import numpy
def diagonal(arr,corner1,corner2):
    arr=numpy.array(arr)
    #Change values to fit array
    corner1Copy=(len(arr)-1)*numpy.array(corner1)
    corner2Copy=(len(arr)-1)*numpy.array(corner2)

    #create return array by running from corner1 to corner2 and returning the values
    return [arr[tuple((i*corner2Copy+(len(arr)-i-1)*corner1Copy)/(len(arr)-1))] for i in range(len(arr))]

这里有两个小测试用例，但我建议再创建一些，以防我遗漏了什么：

arr=[[[i+j+k for i in range(5)]for j in range(5)] for k in range(5)]
corner1=[0,0,0]
corner2=[1,1,1]

#returns arr[0,0,0],arr[1,1,1],....,arr[-1,-1,-1]
print(diagonal(arr,corner1,corner2))
print([arr[i][i][i] for i in range(len(arr))])

arr2=[[i+j for i in range(5)]for j in range(5)]

corner12=[0,1]
corner22=[1,1]
#return arr[0,-1],arr[1,-1],....,arr[-1,-1]
print(diagonal(arr2,corner12,corner22))
print([arr2[i][-1] for i in range(len(arr2))])

---

这应该可以通过使用numpy数组实现。它提供了一个生成器，其中包含与数组维数相同的所有对角线。 还提供原始阵列的视图（而不是副本）。 代码下面有解释

import numpy as np
def get_diagonals_np(arr):
    if arr.ndim == 1:
        yield arr
    else:
        yield from get_diagonals_np(arr.diagonal())
        yield from get_diagonals_np(np.flip(arr, 0).diagonal())

# The function is recursive. How it works is best shown by example.
# 1d: arr = [0, 1] then the diagonal is also [0, 1].

# 2d: arr = [[0, 1],
#            [2, 3]]
# The numpy diagonal method gives the main diagonal = [0, 3], a 1d array
# which is recursively passed to the function.
# To get the opposite diagonal we first use the numpy flip function to
# reverse the order of the elements along the given dimension, 0 in this case.
# This gives [[2, 3],
#              0, 1]]
# The numpy diagonal method gives the main diagonal = [2, 1], a 2d array
# which is recursively passed to the function.

# 3d: arr = [[[0, 1],
#             [2, 3]],
#            [[4, 5],
#             [6, 7]]]
# The numpy diagonal method gives the main diagonals in the 3rd dimension
# as rows.
#            [[0, 6],
#             [1, 7]]
# Note that the diagonals of this array are [0, 7] and [6, 1] which are
# retrieved by a recurive call to the function.
# We now have 2 of the 4 3-agonals of the orginal 3d arr.
# To get the opposite 3-agonals we first use the numpy flip function which
# gives
#           [[[4, 5],
#             [6, 7]],
#            [[0, 1],
#             [2, 3]]]
# and a call to the numpy diagonal method gives
#            [[4, 2],
#             [5, 3]]
# The diagonals of this array are [4, 3] and [2, 5]
# We now have all four 3-agonals of the original 3d arr.

---

列出多维数组的示例输入和预期输出，如

（3,3,3）

？三维或四维数组的（主对角线和反对角线）是什么？因为有很多them@Divakar是 啊我想要所有的。我想要所有可能的对角线。我正在尝试建立一个多维版本的tic-tac-toeyah。。。我们想要一个例子：）你说的小的是什么意思？@Divakar所以如果一个大的只从上到下，从左到右，一个小的会切换其中一个。我会想出一个关于我需要什么的一般数学规则。但我想我得到了我的答案：这在numpy中可能是不可能的，一旦我有了我的数学公式，用python实现它将是微不足道的。

arr=[[[i+j+k for i in range(5)]for j in range(5)] for k in range(5)]
corner1=[0,0,0]
corner2=[1,1,1]

#returns arr[0,0,0],arr[1,1,1],....,arr[-1,-1,-1]
print(diagonal(arr,corner1,corner2))
print([arr[i][i][i] for i in range(len(arr))])

arr2=[[i+j for i in range(5)]for j in range(5)]

corner12=[0,1]
corner22=[1,1]
#return arr[0,-1],arr[1,-1],....,arr[-1,-1]
print(diagonal(arr2,corner12,corner22))
print([arr2[i][-1] for i in range(len(arr2))])

import numpy as np
def get_diagonals_np(arr):
    if arr.ndim == 1:
        yield arr
    else:
        yield from get_diagonals_np(arr.diagonal())
        yield from get_diagonals_np(np.flip(arr, 0).diagonal())

# The function is recursive. How it works is best shown by example.
# 1d: arr = [0, 1] then the diagonal is also [0, 1].

# 2d: arr = [[0, 1],
#            [2, 3]]
# The numpy diagonal method gives the main diagonal = [0, 3], a 1d array
# which is recursively passed to the function.
# To get the opposite diagonal we first use the numpy flip function to
# reverse the order of the elements along the given dimension, 0 in this case.
# This gives [[2, 3],
#              0, 1]]
# The numpy diagonal method gives the main diagonal = [2, 1], a 2d array
# which is recursively passed to the function.

# 3d: arr = [[[0, 1],
#             [2, 3]],
#            [[4, 5],
#             [6, 7]]]
# The numpy diagonal method gives the main diagonals in the 3rd dimension
# as rows.
#            [[0, 6],
#             [1, 7]]
# Note that the diagonals of this array are [0, 7] and [6, 1] which are
# retrieved by a recurive call to the function.
# We now have 2 of the 4 3-agonals of the orginal 3d arr.
# To get the opposite 3-agonals we first use the numpy flip function which
# gives
#           [[[4, 5],
#             [6, 7]],
#            [[0, 1],
#             [2, 3]]]
# and a call to the numpy diagonal method gives
#            [[4, 2],
#             [5, 3]]
# The diagonals of this array are [4, 3] and [2, 5]
# We now have all four 3-agonals of the original 3d arr.