NumPy：沿矩阵的对角线构造正方形/扩展对角矩阵

Question

假设你有两个数组：

index = [1, 2, 3]
counts = [2, 3, 2]

或单数数组

arr = [1, 1, 2, 2, 2, 3, 3]

如何有效地构造矩阵

[
  [1, 1, 0, 0, 0, 0, 0],
  [1, 1, 0, 0, 0, 0, 0],
  [0, 0, 2, 2, 2, 0, 0],
  [0, 0, 2, 2, 2, 0, 0],
  [0, 0, 2, 2, 2, 0, 0],
  [0, 0, 0, 0, 0, 3, 3],
  [0, 0, 0, 0, 0, 3, 3]
]

使用 NumPy？

我知道

square = np.zeros((7, 7))
np.fill_diagnol(square, arr) # see arr above

生产

[
  [1, 0, 0, 0, 0, 0, 0],
  [0, 1, 0, 0, 0, 0, 0],
  [0, 0, 2, 0, 0, 0, 0],
  [0, 0, 0, 2, 0, 0, 0],
  [0, 0, 0, 0, 2, 0, 0],
  [0, 0, 0, 0, 0, 3, 0],
  [0, 0, 0, 0, 0, 0, 3]
]

我如何通过n“扩​​展”对角线，其中n 是counts[index-1] 对于index[I] 指定的值

tmp = np.array((arr * N)).reshape((len(arr), len(arr)) 
np.floor( (tmp + tmp.T) / 2 ) # <-- this is closer


array([[1., 1., 1., 1., 1., 2., 2.],
       [1., 1., 1., 1., 1., 2., 2.],
       [1., 1., 2., 2., 2., 2., 2.],
       [1., 1., 2., 2., 2., 2., 2.],
       [1., 1., 2., 2., 2., 2., 2.],
       [2., 2., 2., 2., 2., 3., 3.],
       [2., 2., 2., 2., 2., 3., 3.]])

这得到了我想要的，但可能不能很好地扩展？

riffled = list(zip(index, counts))
riffled
# [(1, 2), (2, 3), (3, 2)]

a = np.zeros((len(arr), len(arr))) # 7, 7 square
last = 0 # <-- keep track of current sub square
for i, c in riffled:
    a[last:last+c, last:last+c] = np.ones((c, c)) * i 
    last += c # <-- shift square

产量

array([[1., 1., 0., 0., 0., 0., 0.],
       [1., 1., 0., 0., 0., 0., 0.],
       [0., 0., 2., 2., 2., 0., 0.],
       [0., 0., 2., 2., 2., 0., 0.],
       [0., 0., 2., 2., 2., 0., 0.],
       [0., 0., 0., 0., 0., 3., 3.],
       [0., 0., 0., 0., 0., 3., 3.]])

Answer 1

这是一个通用的解决方案。

从索引/计数开始：

index = [1, 2, 1]
counts = [2, 3, 2]

arr = np.repeat(index, counts)
arr2 = np.repeat(range(len(index)), counts)
np.where(arr2 == arr2[:, None], arr, 0)

输出：

array([[1, 1, 0, 0, 0, 0, 0],
       [1, 1, 0, 0, 0, 0, 0],
       [0, 0, 2, 2, 2, 0, 0],
       [0, 0, 2, 2, 2, 0, 0],
       [0, 0, 2, 2, 2, 0, 0],
       [0, 0, 0, 0, 0, 1, 1],
       [0, 0, 0, 0, 0, 1, 1]])

从数组版本开始：

arr = np.array([1, 1, 2, 2, 2, 1, 2])

arr2 = np.cumsum(np.diff(arr,prepend=np.nan) != 0)
np.where(arr2 == arr2[:, None], arr, 0)

输出：

array([[1, 1, 0, 0, 0, 0, 0],
       [1, 1, 0, 0, 0, 0, 0],
       [0, 0, 2, 2, 2, 0, 0],
       [0, 0, 2, 2, 2, 0, 0],
       [0, 0, 2, 2, 2, 0, 0],
       [0, 0, 0, 0, 0, 1, 0],
       [0, 0, 0, 0, 0, 0, 2]])

Answer 2

尝试广播：

idx = np.repeat(np.arange(len(counts)), counts)
np.where(idx==idx[:,None], arr, 0)
# or
# arr * (idx==idx[:,None])

输出；

array([[1, 1, 0, 0, 0, 0, 0],
       [1, 1, 0, 0, 0, 0, 0],
       [0, 0, 2, 2, 2, 0, 0],
       [0, 0, 2, 2, 2, 0, 0],
       [0, 0, 2, 2, 2, 0, 0],
       [0, 0, 0, 0, 0, 3, 3],
       [0, 0, 0, 0, 0, 3, 3]])

Answer 3

您可以使用 scipy.linalg.block_diag 来完成这项工作：

import numpy as np
import scipy.linalg as linalg


a = 1*np.ones((2,2))
b = 2*np.ones((3,3))
c = 3*np.ones((2,2))

superBlock = linalg.block_diag(a,b,c)

print(superBlock)

#returns
#[[1. 1. 0. 0. 0. 0. 0.]
# [1. 1. 0. 0. 0. 0. 0.]
# [0. 0. 2. 2. 2. 0. 0.]
# [0. 0. 2. 2. 2. 0. 0.]
# [0. 0. 2. 2. 2. 0. 0.]
# [0. 0. 0. 0. 0. 3. 3.]
# [0. 0. 0. 0. 0. 3. 3.]]

如果您想从值列表和计数列表中到达那里，您可以这样做：

values = [1,2,3]
counts = [2,3,2]

mats = []
for v,c in zip(values,counts):
    thisMatrix = v*np.ones((c,c))
    mats.append( thisMatrix )


superBlock = linalg.block_diag(*mats)


print(superBlock)