I have functioning code but I would like help on two aspects:
- Improve the numpy code so that the code runs faster (i.e. don't use for loops)
I would like to first ask about number 1. Here is the code snippet I am inquiring about:
import numpy as np
k = np.linspace(0,4.76*10,2400)
kx,ky = np.meshgrid(k, k)
rx = [0.4, 0.4, 0.4, 0.4, 0.4, 0.4, 0.4, 0.4]
ry = [0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1, 1.1, 1.2]
tensx = np.tensordot(rx,kx, axes = 0)
tensy = np.tensordot(ry,ky, axes = 0)
z = 0
for i in range(0, len(tensx)):
z = z 0.05*np.cos(2*np.pi*(tensx[i] tensy[i]))
The code is currently not slow but when I start to increase the amount of points in the rx and ry vectors, it will be much slower. So my question: is there a "numpy-er" way to write this code?
Any help is greatly appreciated!
CodePudding user response:
You can re-write the for loop as:
0.05*np.cos(2*np.pi*(tensx tensy[:-1])).sum(axis=0)
Or more generally:
0.05*np.cos(2*np.pi*(tensx tensy[:tensx.shape[0]])).sum(axis=0)
You can also avoid the meshgrid
, avoiding calculating unused elements of tensy
, by doing:
tensx = np.tensordot(rx, k[None, :], axes=0)
tensy = np.tensordot(ry[:len(rx)], k[:, None], axes=0)
Timings:
def op():
k = np.linspace(0,4.76*10,2400)
kx,ky = np.meshgrid(k, k)
rx = [0.4, 0.4, 0.4, 0.4, 0.4, 0.4, 0.4, 0.4]
ry = [0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1, 1.1, 1.2]
tensx = np.tensordot(rx,kx, axes = 0)
tensy = np.tensordot(ry,ky, axes = 0)
z = 0
for i in range(0, len(tensx)):
z = z 0.05*np.cos(2*np.pi*(tensx[i] tensy[i]))
return z
def Nin17():
k = np.linspace(0,4.76*10,2400)
rx = [0.4, 0.4, 0.4, 0.4, 0.4, 0.4, 0.4, 0.4]
ry = [0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1, 1.1, 1.2]
tensx = np.tensordot(rx, k[None, :], axes=0)
tensy = np.tensordot(ry[:len(rx)], k[:, None], axes=0)
return 0.05*np.cos(2*np.pi*(tensx tensy)).sum(axis=0)
assert np.isclose(Nin17(), op()).all()
%timeit Nin17()
%timeit op()
Output:
359 ms ± 2.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
410 ms ± 5.47 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)