在Python中更快实现ReLu派生工具? [英] Faster implementation for ReLu derivative in python?

问题描述

我将ReLu派生实现为:

I have implemented ReLu derivative as:

def relu_derivative(x):
     return (x>0)*np.ones(x.shape)

我也尝试过:

def relu_derivative(x):
   x[x>=0]=1
   x[x<0]=0
   return x

X的大小=(3072,10000). 但这需要花费大量时间进行计算.还有其他优化的解决方案吗?

Size of X=(3072,10000). But it's taking much time to compute. Is there any other optimized solution?

推荐答案

方法1:使用numexpr

在处理大数据时，我们可以使用 numexpr模块支持多核处理.在这里，一种方法是-

Approach #1 : Using numexpr

When working with large data, we can use numexpr module that supports multi-core processing if the intended operations could be expressed as arithmetic ones. Here, one way would be -

(X>=0)+0

因此，要解决我们的问题，应该是-

Thus, to solve our case, it would be -

import numexpr as ne

ne.evaluate('(X>=0)+0')

方法2:使用NumPy views

另一个技巧是使用views，方法是将比较掩码作为int数组查看，就像这样-

Approach #2 : Using NumPy views

Another trick would be to use views by viewing the mask of comparisons as an int array, like so -

(X>=0).view('i1')

关于性能，它应该与创建X>=0相同.

On performance, it should be identical to creating X>=0.

时间

在随机数组上比较所有已发布的解决方案-

Comparing all posted solutions on a random array -

In [14]: np.random.seed(0)
    ...: X = np.random.randn(3072,10000)

In [15]: # OP's soln-1
    ...: def relu_derivative_v1(x):
    ...:      return (x>0)*np.ones(x.shape)
    ...: 
    ...: # OP's soln-2     
    ...: def relu_derivative_v2(x):
    ...:    x[x>=0]=1
    ...:    x[x<0]=0
    ...:    return x

In [16]: %timeit ne.evaluate('(X>=0)+0')
10 loops, best of 3: 27.8 ms per loop

In [17]: %timeit (X>=0).view('i1')
100 loops, best of 3: 19.3 ms per loop

In [18]: %timeit relu_derivative_v1(X)
1 loop, best of 3: 269 ms per loop

In [19]: %timeit relu_derivative_v2(X)
1 loop, best of 3: 89.5 ms per loop

基于numexpr的线程具有8线程.因此，随着更多线程可用于计算，它应该进一步改进. Related post 有关如何控制多核功能的信息.

The numexpr based one was with 8 threads. Thus, with more number of threads available for compute, it should improve further. Related post on how to control multi-core functionality.

将两种方法混合使用，以获得大型阵列的最佳选择-

Mix both of those for the most optimal one for large arrays -

In [27]: np.random.seed(0)
    ...: X = np.random.randn(3072,10000)

In [28]: %timeit ne.evaluate('X>=0').view('i1')
100 loops, best of 3: 14.7 ms per loop

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