我有这个代码。

n_nodes = len(data_x)
X = np.zeros((n_nodes, n_nodes))

for i in range(n_nodes):
  for j in range(n_nodes):
    X[i, j] = data_x[i] ** j

我想在完全不使用循环的情况下完成相同的任务。我如何使用 NumPy 函数做到这一点？

最佳答案

我建议

data_x[:,None]**np.arange(n_nodes)

支票

In [17]: data_x = np.array((3,5,7,4,6))
    ...: n_nodes = len(data_x)
    ...: X = np.zeros((n_nodes, n_nodes))
    ...: 
    ...: for i in range(n_nodes):
    ...:   for j in range(n_nodes):
    ...:     X[i, j] = data_x[i] ** j
    ...: print(X)
    ...: print('-----------')
    ...: print(data_x[:,None]**np.arange(n_nodes))
[[1.000e+00 3.000e+00 9.000e+00 2.700e+01 8.100e+01]
 [1.000e+00 5.000e+00 2.500e+01 1.250e+02 6.250e+02]
 [1.000e+00 7.000e+00 4.900e+01 3.430e+02 2.401e+03]
 [1.000e+00 4.000e+00 1.600e+01 6.400e+01 2.560e+02]
 [1.000e+00 6.000e+00 3.600e+01 2.160e+02 1.296e+03]]
-----------
[[   1    3    9   27   81]
 [   1    5   25  125  625]
 [   1    7   49  343 2401]
 [   1    4   16   64  256]
 [   1    6   36  216 1296]]

一些时机

In [18]: %timeit data_x[:,None]**np.arange(n_nodes)
2.18 µs ± 7.49 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

In [19]: %%timeit
    ...: for i in range(n_nodes):
    ...:     for j in range(n_nodes):
    ...:         X[i, j] = data_x[i] ** j
10.9 µs ± 107 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

https://stackoverflow.com/questions/74155199/