Question

此程序在我的程序中运行超过200万次。任何人都可以建议优化这个？ x和y是元组。

def distance(x,y):

    return sqrt((x[0]-y[0])*(x[0]-y[0])+(x[1]-y[1])*(x[1]-y[1])+(x[2]-y[2])*(x[2]-y[2]))

我的尝试：我尝试使用math.sqrt，pow和x **。但是没有太多的性能提升。

Answer 1

你可以通过不访问相同的x [i]元素并在本地绑定它来削减一些周期。

def distance(x,y):
    x0, x1, x2 = x
    y0, y1, y2 = y
    return sqrt((x0-y0)*(x0-y0)+(x1-y1)*(x1-y1)+(x2-y2)*(x2-y2))

例如，比较：

>>> import timeit
>>> timer = timeit.Timer(setup='''
... from math import sqrt
... def distance(x,y):
...    return sqrt((x[0]-y[0])*(x[0]-y[0])+(x[1]-y[1])*(x[1]-y[1])+(x[2]-y[2])*(x[2]-y[2]))
... ''', stmt='distance((0,0,0), (1,2,3))')
>>> timer.timeit(1000000)
1.2761809825897217

使用：

>>> import timeit
>>> timer = timeit.Timer(setup='''
... from math import sqrt
... def distance(x,y):
...    x0, x1, x2 = x
...    y0, y1, y2 = y
...    return sqrt((x0-y0)*(x0-y0)+(x1-y1)*(x1-y1)+(x2-y2)*(x2-y2))
... ''', stmt='distance((0,0,0), (1,2,3))')
>>> timer.timeit(1000000)
0.8375771045684814

还有更多performance tips on the python wiki.

Answer 2

原件：

>>> timeit.timeit('distance2((0,1,2),(3,4,5))', '''
... from math import sqrt
... def distance2(x,y):
...     return sqrt((x[0]-y[0])*(x[0]-y[0])+(x[1]-y[1])*(x[1]-y[1])+(x[2]-y[2])*(x[2]-y[2]))
... ''')
1.1989610195159912

消除常见的Subexpression：

>>> timeit.timeit('distance((0,1,2),(3,4,5))', '''
... def distance(x,y):
...     d1 = x[0] - y[0]
...     d2 = x[1] - y[1]
...     d3 = x[2] - y[2]
...     return (d1 * d1 + d2 * d2 + d3 * d3) ** .5''')
0.93855404853820801

优化解包：

>>> timeit.timeit('distance((0,1,2),(3,4,5))', '''
... def distance(x,y):
...     x1, x2, x3 = x
...     y1, y2, y3 = y
...     d1 = x1 - y1
...     d2 = x2 - y2
...     d3 = x3 - y3
...     return (d1 * d1 + d2 * d2 + d3 * d3) ** .5''')
0.90851116180419922

图书馆职能：

>>> timeit.timeit('distance((0,1,2),(3,4,5))', '''
... import math
... def distance(x,y):
...     x1, x2, x3 = x
...     y1, y2, y3 = y
...     d1 = x1 - y1
...     d2 = x2 - y2
...     d3 = x3 - y3
...     return math.sqrt(d1 * d1 + d2 * d2 + d3 * d3)
... ''')
0.78318595886230469

虚线：

>>> timeit.timeit('distance((0,1,2),(3,4,5))', '''
... from math import sqrt
... def distance(x,y):
...     x1, x2, x3 = x
...     y1, y2, y3 = y
...     d1 = x1 - y1
...     d2 = x2 - y2
...     d3 = x3 - y3
...     return sqrt(d1 * d1 + d2 * d2 + d3 * d3)
... ''')
0.75629591941833496

Answer 3

scipy有一个欧几里德距离函数。你可能不会比这更快。

http://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.distance.euclidean.html#scipy.spatial.distance.euclidean

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

# x and y are 1 x 3 vectors
x = np.random.rand(1,3) 
y = np.random.rand(1,3)
euclidean(x,y)

编辑：实际上，通过timeit对OP的pure-python distance（）函数运行，实际上这在python浮点数上变慢了。我认为scipy版本浪费了一些时间将花车投射到numpy dtypes。至少可以说我很惊讶。

Python优化用于计算欧氏距离

3 个答案: