Question

我的问题可能很简单，但我无法找到一种方法来加快这项操作

  print a[(b==c[i]) for i in arange(0,len(c))]

其中a，b和c是三个numpy数组。我正在处理数百万条目的数组，上面的代码片段是我程序的瓶颈。

Answer 1

您是否尝试将a的值设为b==c？

如果是这样，您可以a[b==c]：

from numpy import *

a = arange(11)
b = 11*a
c = b[::-1]

print a        # [  0   1   2   3   4   5   6   7   8   9  10]
print b        # [  0  11  22  33  44  55  66  77  88  99 110]
print c        # [110  99  88  77  66  55  44  33  22  11   0]
print a[b==c]  # [5]

Answer 2

你应该考虑广播。我假设您正在寻找以下内容？

>>> b=np.arange(5)
>>> c=np.arange(6).reshape(-1,1)
>>> b
array([0, 1, 2, 3, 4])
>>> c
array([[0],
       [1],
       [2],
       [3],
       [4],
       [5]])
>>> b==c
array([[ True, False, False, False, False],
       [False,  True, False, False, False],
       [False, False,  True, False, False],
       [False, False, False,  True, False],
       [False, False, False, False,  True],
       [False, False, False, False, False]], dtype=bool)
>>> np.any(b==c,axis=1)
array([ True,  True,  True,  True,  True, False], dtype=bool)

对于大型阵列，您可以尝试：

import timeit

s="""
import numpy as np
array_size=500
a=np.random.randint(500, size=(array_size))
b=np.random.randint(500, size=(array_size))
c=np.random.randint(500, size=(array_size))
"""

ex1="""
a[np.any(b==c.reshape(-1,1),axis=0)]
"""

ex2="""
a[np.in1d(b,c)]
"""

print 'Example 1 took',timeit.timeit(ex1,setup=s,number=100),'seconds.'
print 'Example 2 took',timeit.timeit(ex2,setup=s,number=100),'seconds.'

当array_size为50时：

Example 1 took 0.00323104858398 seconds.
Example 2 took 0.0125901699066 seconds.

当array_size为500时：

Example 1 took 0.142632007599 seconds.
Example 2 took 0.0283041000366 seconds.

当array_size为5,000时：

Example 1 took 16.2110910416 seconds.
Example 2 took 0.170011043549 seconds.

当array_size为50,000（number = 5）时：

Example 1 took 33.0327301025 seconds.
Example 2 took 0.0996031761169 seconds.

注意我必须更改np.any（）的轴，以便结果相同。 np.in1d的逆序或np.any的切换轴为期望的效果。您可以从示例1中重新塑造，但重塑非常快。切换以获得所需的效果。真的很有趣 - 我将来必须使用它。

Answer 3

np.where()：

怎么样？

>>> a  = np.array([2,4,8,16])
>>> b  = np.array([0,0,0,0])
>>> c  = np.array([1,0,0,1])
>>> bc = np.where(b==c)[0] #indices where b == c
>>> a[bc]
array([4,8])

这应该可以解决问题。不确定时间是否适合您的目的

>>> a = np.random.randint(0,10000,1000000)
>>> b = np.random.randint(0,10000,1000000)
>>> c = np.random.randint(0,10000,1000000)
>>> %timeit( a[ np.where( b == c )[0] ]   )
100 loops, best of 3: 11.3 ms per loop

python中的慢数组操作

3 个答案: