在python numpy中实现Relu派生 [英] Implement Relu derivative in python numpy

问题描述

我正在尝试实现一个函数，该函数为矩阵中的每个元素计算Relu导数，然后将结果返回到矩阵中.我正在使用Python和Numpy.

I'm trying to implement a function that computes the Relu derivative for each element in a matrix, and then return the result in a matrix. I'm using Python and Numpy.

根据其他交叉验证帖子，x的Relu导数为 当x> 0时为1，当x <0时为0. 0，未定义，或者当x == 0时为0

Based on other Cross Validation posts, the Relu derivative for x is 1 when x > 0, 0 when x < 0, undefined or 0 when x == 0

目前，到目前为止，我有以下代码:

Currently, I have the following code so far:

def reluDerivative(self, x):
    return np.array([self.reluDerivativeSingleElement(xi) for xi in x])

def reluDerivativeSingleElement(self, xi):
    if xi > 0:
        return 1
    elif xi <= 0:
        return 0

不幸的是，因为x是矩阵，所以xi是一个数组. reluDerivativeSingleElement函数不适用于数组.所以我想知道是否有一种方法可以使用numpy将矩阵中的值映射到另一个矩阵，例如numpy中的exp函数?

Unfortunately, xi is an array because x is an matrix. reluDerivativeSingleElement function doesn't work on array. So I'm wondering is there a way to map values in a matrix to another matrix using numpy, like the exp function in numpy?

非常感谢.

推荐答案

我想这就是您要寻找的东西:

I guess this is what you are looking for:

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

>>> z = np.random.uniform(-1, 1, (3,3))
>>> z
array([[ 0.41287266, -0.73082379,  0.78215209],
       [ 0.76983443,  0.46052273,  0.4283139 ],
       [-0.18905708,  0.57197116,  0.53226954]])
>>> reluDerivative(z)
array([[ 1.,  0.,  1.],
       [ 1.,  1.,  1.],
       [ 0.,  1.,  1.]])

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