I want to calculate the determinant of mm subarrays of a mm*n dimensional arrays, and would like to do this in a fast/more elegant way. The brute-force approach works:
import numpy as n
array=n.array([[[0.,1.,2.,3.],[2,1,1,0]],[[0.5, 0.5,2,2],[0.5,1,0,2]]])
detarray=n.zeros(4)
for i in range(4):
detarray[i]= n.linalg.det(array[:,:,i])
I would have tried doing this with apply_along_axis, but I know this is only for 1D arguments to the function, and so I presume I can't get this to work.
However, I thought apply_over_axes should also work:
n.apply_over_axes(n.linalg.det, array, [0,1])
but this gives me an error: "det() takes exactly 1 argument (2 given)"
Does anyone know why this doesn't work? If this type of calculation is really not possible with apply_over_axes, is there a better way to do this rather than the for-loop?
Utilizing the transpose semantics of NumPy for 3D arrays, you can simply pass the transposed array to numpy.linalg.det() as in:
In [13]: arr = np.array([[[0.,1.,2.,3.],
[2, 1, 1, 0]],
[[0.5, 0.5,2,2],
[0.5, 1, 0, 2]]])
In [14]: np.linalg.det(arr.T)
Out[14]: array([-1. , 0.5, -2. , 6. ])
Performance wise this approach seems to be twice as fast as the other approach of manually moving the axes using numpy.moveaxis
In [29]: %timeit np.linalg.det(np.moveaxis(arr, 2, 0))
12.9 µs ± 192 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [30]: %timeit np.linalg.det(arr.T)
6.2 µs ± 136 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)