Python:矢量化/广播会提高速度吗?

时间:2018-04-12 13:53:55

标签: python loops numpy vectorization numpy-broadcasting

目标:将数组p1合并到p10以创建一个名为“a”的大型数组,并返回“a”中显示为'a'的所有值,共4次。

问题:由于要完成所有循环,此代码速度非常慢,如何让它更快捷?矢量化和/或广播是否有助于提高效率(是否有可能摆脱所有循环)?或任何其他开箱即用的想法?

import numpy as np
import itertools
from numba import jit

p1 = np.random.randint(0,314000,200000)
p2 = np.random.randint(0,314000,100000)
p3 = np.random.randint(0,314000,300000)
p4 = np.random.randint(0,314000,150000)
p5 = np.random.randint(0,314000,220000)
p6 = np.random.randint(0,314000,320000)
p7 = np.random.randint(0,314000,212100)
p8 = np.random.randint(0,314000,100500)
p9 = np.random.randint(0,314000,300700)
p10 = np.random.randint(0,314000,200300)

@jit
def count(a,n):
 counters=np.zeros(10**6,np.int32)
 for i in a:
  counters[i] += 1
 res=np.empty_like(counters)
 k = 0    
 for i,j in enumerate(counters):
  if j == n:
   res[k] = i
   k += 1
 return res[:k]        


for t in range(0, 20000):
 a = itertools.chain(p1,p2,p3,p4,p5,p6,p7,p8,p9,p10)
 count(a,4)

1 个答案:

答案 0 :(得分:3)

是的,是的。你可以摆脱循环,它会加速事情:

>>> a = np.concatenate([p1,p2,p3,p4,p5,p6,p7,p8,p9,p10])
>>> np.flatnonzero(np.bincount(a, minlength=314000)==4)
array([    29,     33,     38, ..., 313944, 313949, 313973])