从 numpy 中的一维数组创建方形 nxn 矩阵的最快方法

Question

Suppose the following numpy array:

arr = np.array([0, 1, 2, 3, 4]) # can be any array

I want to know the fastest way to generate the following operation:

n = arr.shape[0]
result = np.tile(arr, (n, 1)) - arr.reshape((-1, 1))
print(result):

array([[ 0,  1,  2,  3,  4],
       [-1,  0,  1,  2,  3],
       [-2, -1,  0,  1,  2],
       [-3, -2, -1,  0,  1],
       [-4, -3, -2, -1,  0]])

(1) How to efficiently create matrix "result" (because n >> 0 can be very large)?

(2) Does this matrix have a particular name?

Answer 1

This is a bit faster:

result = arr-arr[:,None]

cursory benchmarks, nothing scientific. (timeit 100 times with arr):

          5 items (arr) 100 times   10,000 items (np.arange) once
OP:       0.0006383560000000066     0.7902513520000001
This one: 0.0001735200000000381     0.3640661519999999
Kelly's:  0.00027326299999996806    0.36036748900000015 (see comments)

Answer 2

Comparing the broadcast to scipy.linalg.toeplitz and OP:

N = 1000
arr = np.arange(N)

%timeit np.tile(arr, (N, 1)) - arr.reshape((-1, 1))
1.83 ms ± 53.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit arr - arr[:, None]
1.11 ms ± 25 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit scipy.linalg.toeplitz(-arr, arr)
727 µs ± 21.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

N = 10000
arr = np.arange(N)

%timeit np.tile(arr, (N, 1)) - arr.reshape((-1, 1))
184 ms ± 9.27 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit arr - arr[:, None]
85.3 ms ± 597 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit scipy.linalg.toeplitz(-arr, arr)
70.6 ms ± 573 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

It seems that the scipy solution is generally faster.