[英]NumPy: construct squares along diagonal of matrix / expand diagonal matrix
Suppose you have either two arrays:假设您有两个数组:
index = [1, 2, 3]
counts = [2, 3, 2]
or a singular array或奇异数组
arr = [1, 1, 2, 2, 2, 3, 3]
How can I efficiently construct the matrix我怎样才能有效地构造矩阵
[
[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]
]
with NumPy?使用 NumPy?
I know that我知道
square = np.zeros((7, 7))
np.fill_diagnol(square, arr) # see arr above
produces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]
]
How do I "expand" the diagonal by n
where n
is counts[index-1]
for the values specified by index[I]
如何将对角线“扩展”
n
,其中n
是index[I]
指定的值的counts[index-1]
]
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.]])
This gets what I want, but probably doesn't scale that well?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
yield屈服
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.]])
You can use scipy.linalg.block_diag to make that work: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.]]
if you want to get there from a list of values and a list of counts you can do this: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)
Here is a generic solution.这是一个通用的解决方案。
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)
output: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]])
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)
output: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]])
Try broadcasting:尝试广播:
idx = np.repeat(np.arange(len(counts)), counts)
np.where(idx==idx[:,None], arr, 0)
# or
# arr * (idx==idx[:,None])
Output;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]])
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