Wondering why scipy.spatial.distance.sqeuclidean is twice slower than numpy.sum((y1-y2)**2)

问题

Here is my code

import numpy as np
import time
from scipy.spatial import distance

y1=np.array([0,0,0,0,1,0,0,0,0,0])
y2=np.array([0. , 0.1, 0. , 0. , 0.7, 0.2, 0. , 0. , 0. , 0. ])

start_time = time.time()
for i in range(1000000):
    distance.sqeuclidean(y1,y2)
print("--- %s seconds ---" % (time.time() - start_time))

---15.212640523910522 seconds---

start_time = time.time()
for i in range(1000000):
    np.sum((y1-y2)**2)
print("--- %s seconds ---" % (time.time() - start_time))

---8.381187438964844--- seconds

I supposed that the Scipy is kind of optimized so it should be faster.

Any comments will be appreciated.

回答1:

Here is a more comprehensive comparison (credit to @Divakar's benchit package):

def m1(y1,y2):
  return distance.sqeuclidean(y1,y2)

def m2(y1,y2):
  return np.sum((y1-y2)**2)

in_ = {n:[np.random.rand(n), np.random.rand(n)] for n in [10,100,1000,10000,20000]}

scipy gets more efficient for larger arrays. For smaller arrays, the overhead of calling the function most likely outweighs its benefit. According to source, scipy calculates np.dot(y1-y2,y1-y2).

And if you want an even faster solution, use np.dot directly without the overhead of extra lines and function calling:

def m3(y1,y2):
  y_d = y1-y2
  return np.dot(y_d,y_d)

来源：https://stackoverflow.com/questions/63592683/wondering-why-scipy-spatial-distance-sqeuclidean-is-twice-slower-than-numpy-sum