And*_*eta 55 python numpy machine-learning neural-network
我想制作一个简单的神经网络,我希望使用ReLU功能.有人能告诉我如何使用numpy实现该功能.谢谢你的时间!
Sid*_*Sid 99
有几种方法.
>>> x = np.random.random((3, 2)) - 0.5
>>> x
array([[-0.00590765, 0.18932873],
[-0.32396051, 0.25586596],
[ 0.22358098, 0.02217555]])
>>> np.maximum(x, 0)
array([[ 0. , 0.18932873],
[ 0. , 0.25586596],
[ 0.22358098, 0.02217555]])
>>> x * (x > 0)
array([[-0. , 0.18932873],
[-0. , 0.25586596],
[ 0.22358098, 0.02217555]])
>>> (abs(x) + x) / 2
array([[ 0. , 0.18932873],
[ 0. , 0.25586596],
[ 0.22358098, 0.02217555]])
如果使用以下代码计时结果:
import numpy as np
x = np.random.random((5000, 5000)) - 0.5
print("max method:")
%timeit -n10 np.maximum(x, 0)
print("multiplication method:")
%timeit -n10 x * (x > 0)
print("abs method:")
%timeit -n10 (abs(x) + x) / 2
我们得到:
max method:
10 loops, best of 3: 239 ms per loop
multiplication method:
10 loops, best of 3: 145 ms per loop
abs method:
10 loops, best of 3: 288 ms per loop
所以乘法似乎是最快的.
Ric*_*öhn 37
如果您不介意x进行修改,请使用np.maximum(x, 0, x).丹尼尔S指出了这一点.它要快得多,因为人们可能会忽略它,我会把它作为答案重新发布.这是比较:
max method:
10 loops, best of 3: 238 ms per loop
multiplication method:
10 loops, best of 3: 128 ms per loop
abs method:
10 loops, best of 3: 311 ms per loop
in-place max method:
10 loops, best of 3: 38.4 ms per loop
Tob*_*ias 20
我找到了一个更快的ReLU方法和numpy.您也可以使用numpy的花式索引功能.
花式指数:
每循环20.3 ms±272μs(平均值±标准偏差,7次运行,每次10次循环)
>>> x = np.random.random((5,5)) - 0.5
>>> x
array([[-0.21444316, -0.05676216, 0.43956365, -0.30788116, -0.19952038],
[-0.43062223, 0.12144647, -0.05698369, -0.32187085, 0.24901568],
[ 0.06785385, -0.43476031, -0.0735933 , 0.3736868 , 0.24832288],
[ 0.47085262, -0.06379623, 0.46904916, -0.29421609, -0.15091168],
[ 0.08381359, -0.25068492, -0.25733763, -0.1852205 , -0.42816953]])
>>> x[x<0]=0
>>> x
array([[ 0. , 0. , 0.43956365, 0. , 0. ],
[ 0. , 0.12144647, 0. , 0. , 0.24901568],
[ 0.06785385, 0. , 0. , 0.3736868 , 0.24832288],
[ 0.47085262, 0. , 0.46904916, 0. , 0. ],
[ 0.08381359, 0. , 0. , 0. , 0. ]])
这是我的基准:
import numpy as np
x = np.random.random((5000, 5000)) - 0.5
print("max method:")
%timeit -n10 np.maximum(x, 0)
print("max inplace method:")
%timeit -n10 np.maximum(x, 0,x)
print("multiplication method:")
%timeit -n10 x * (x > 0)
print("abs method:")
%timeit -n10 (abs(x) + x) / 2
print("fancy index:")
%timeit -n10 x[x<0] =0
max method:
241 ms ± 3.53 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
max inplace method:
38.5 ms ± 4 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
multiplication method:
162 ms ± 3.1 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
abs method:
181 ms ± 4.18 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
fancy index:
20.3 ms ± 272 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
Shi*_*hah 12
你可以用更容易的方式完成它而不需要numpy:
def ReLU(x):
return x * (x > 0)
def dReLU(x):
return 1. * (x > 0)
Richard Möhn 的比较 是不公平的。
正如Andrea Di Biagio 的评论，就地方法np.maximum(x, 0, x)将在第一个循环中修改 x。
所以这是我的基准：
import numpy as np
def baseline():
x = np.random.random((5000, 5000)) - 0.5
return x
def relu_mul():
x = np.random.random((5000, 5000)) - 0.5
out = x * (x > 0)
return out
def relu_max():
x = np.random.random((5000, 5000)) - 0.5
out = np.maximum(x, 0)
return out
def relu_max_inplace():
x = np.random.random((5000, 5000)) - 0.5
np.maximum(x, 0, x)
return x
计时：
print("baseline:")
%timeit -n10 baseline()
print("multiplication method:")
%timeit -n10 relu_mul()
print("max method:")
%timeit -n10 relu_max()
print("max inplace method:")
%timeit -n10 relu_max_inplace()
获取结果：
baseline:
10 loops, best of 3: 425 ms per loop
multiplication method:
10 loops, best of 3: 596 ms per loop
max method:
10 loops, best of 3: 682 ms per loop
max inplace method:
10 loops, best of 3: 602 ms per loop
就地最大值方法只比最大值方法快一点，这可能是因为它省略了“out”的变量赋值。而且还是比乘法慢。
并且由于您正在实施 ReLU 函数。您可能必须通过 relu 为反向传播保存“x”。例如：
def relu_backward(dout, cache):
x = cache
dx = np.where(x > 0, dout, 0)
return dx
所以我建议你使用乘法方法。