我收集了这里提出的不同方法,以及另外两个问题 ,并测量了不同方法的速度:
import numpy as np
import scipy.spatial
import sklearn.metrics
def dist_direct(x, y):
d = np.expand_dims(x, -2) - y
return np.sum(np.square(d), axis=-1)
def dist_einsum(x, y):
d = np.expand_dims(x, -2) - y
return np.einsum('ijk,ijk->ij', d, d)
def dist_scipy(x, y):
return scipy.spatial.distance.cdist(x, y, "sqeuclidean")
def dist_sklearn(x, y):
return sklearn.metrics.pairwise.pairwise_distances(x, y, "sqeuclidean")
def dist_layers(x, y):
res = np.zeros((x.shape[0], y.shape[0]))
for i in range(x.shape[1]):
res += np.subtract.outer(x[:, i], y[:, i])**2
return res
# inspired by the excellent https://github.com/droyed/eucl_dist
def dist_ext1(x, y):
nx, p = x.shape
x_ext = np.empty((nx, 3*p))
x_ext[:, :p] = 1
x_ext[:, p:2*p] = x
x_ext[:, 2*p:] = np.square(x)
ny = y.shape[0]
y_ext = np.empty((3*p, ny))
y_ext[:p] = np.square(y).T
y_ext[p:2*p] = -2*y.T
y_ext[2*p:] = 1
return x_ext.dot(y_ext)
# https://stackoverflow.com/a/47877630/648741
def dist_ext2(x, y):
return np.einsum('ij,ij->i', x, x)[:,None] + np.einsum('ij,ij->i', y, y) - 2 * x.dot(y.T)
我timeit
用来比较不同方法的速度。为了比较,我使用长度为 10 的向量,第一组有 100 个向量,第二组有 1000 个向量。
import timeit
p = 10
x = np.random.standard_normal((100, p))
y = np.random.standard_normal((1000, p))
for method in dir():
if not method.startswith("dist_"):
continue
t = timeit.timeit(f"{method}(x, y)", number=1000, globals=globals())
print(f"{method:12} {t:5.2f}ms")
在我的笔记本电脑上,结果如下:
dist_direct 5.07ms
dist_einsum 3.43ms
dist_ext1 0.20ms <-- fastest
dist_ext2 0.35ms
dist_layers 2.82ms
dist_scipy 0.60ms
dist_sklearn 0.67ms
虽然 和 两种方法dist_ext1
都dist_ext2
基于写(x-y)**2
为的思想,x**2 - 2*x*y + y**2
速度非常快,但也有一个缺点:当 和 之间的距离x
非常y
小时,由于消除误差,数值结果有时可能(非常轻微)为负。