Starting approach
One approach following an usage of broadcasting, would be like so -
np.arccos(sigIn/np.linalg.norm(sigIn,axis=1,keepdims=1))*180/np.pi
Further optimization - I
We could use np.einsum to replace np.linalg.norm part. Thus :
np.linalg.norm(sigIn,axis=1,keepdims=1)
could be replaced by :
np.sqrt(np.einsum('ij,ij->i',sigIn,sigIn))[:,None]
Further optimization - II
Further boost could be brought in with numexpr module, which works really well with huge arrays and with operations involving trigonometrical functions. In our case that would be arcccos. So, we will use the einsum part as used in the previous optimization section and then use arccos from numexpr on it.
Thus, the implementation would look something like this -
import numexpr as ne
pi_val = np.pi
s = np.sqrt(np.einsum('ij,ij->i',signIn,signIn))[:,None]
out = ne.evaluate('arccos(signIn/s)*180/pi_val')
Runtime test
Approaches -
def original_app(sigIn):
N=len(sigIn)
sigOut=np.empty(shape=(N,3))
sigOut[sigOut==0]=None
i=0
while i<N:
sigOut[i,:] = np.arccos(sigIn[i,:]/np.linalg.norm(sigIn[i,:]))*180/math.pi
i=i+1
return sigOut
def broadcasting_app(signIn):
s = np.linalg.norm(signIn,axis=1,keepdims=1)
return np.arccos(signIn/s)*180/np.pi
def einsum_app(signIn):
s = np.sqrt(np.einsum('ij,ij->i',signIn,signIn))[:,None]
return np.arccos(signIn/s)*180/np.pi
def numexpr_app(signIn):
pi_val = np.pi
s = np.sqrt(np.einsum('ij,ij->i',signIn,signIn))[:,None]
return ne.evaluate('arccos(signIn/s)*180/pi_val')
Timings -
In [115]: a = np.random.rand(180000,3)
In [116]: %timeit original_app(a)
...: %timeit broadcasting_app(a)
...: %timeit einsum_app(a)
...: %timeit numexpr_app(a)
...:
1 loops, best of 3: 1.38 s per loop
100 loops, best of 3: 15.4 ms per loop
100 loops, best of 3: 13.3 ms per loop
100 loops, best of 3: 4.85 ms per loop
In [117]: 1380/4.85 # Speedup number
Out[117]: 284.5360824742268
280x speedup there!