为什么我的tanh激活功能表现如此糟糕?

时间:2016-03-04 18:27:02

标签: python python-3.x numpy matplotlib machine-learning

除激活功能外,我有两个相同的感知器算法。一个使用单步函数1 if u >= 0 else -1,另一个使用tanh函数np.tanh(u)

我预计tanh的表现会超过这一步,但实际上它的表现相当可观。我在这里做错了什么,或者有没有理由说它在问题集上表现不佳?

enter image description here

import numpy as np
import matplotlib.pyplot as plt

# generate 20 two-dimensional training data
# data must be linearly separable

# C1: u = (0,0) / E = [1 0; 0 1]; C2: u = (4,0), E = [1 0; 0 1] where u, E represent centre & covariance matrix of the
# Gaussian distribution respectively


def step(u):
    return 1 if u >= 0 else -1


def sigmoid(u):
    return np.tanh(u)

c1mean = [0, 0]
c2mean = [4, 0]
c1cov = [[1, 0], [0, 1]]
c2cov = [[1, 0], [0, 1]]
x = np.ones((40, 3))
w = np.zeros(3)     # [0, 0, 0]
w2 = np.zeros(3)    # second set of weights to see how another classifier compares
t = []  # target array

# +1 for the first 20 then -1
for i in range(0, 40):
    if i < 20:
        t.append(1)
    else:
        t.append(-1)

x1, y1 = np.random.multivariate_normal(c1mean, c1cov, 20).T
x2, y2 = np.random.multivariate_normal(c2mean, c2cov, 20).T

# concatenate x1 & x2 within the first dimension of x and the same for y1 & y2 in the second dimension
for i in range(len(x)):
    if i >= 20:
        x[i, 0] = x2[(i-20)]
        x[i, 1] = y2[(i-20)]
    else:
        x[i, 0] = x1[i]
        x[i, 1] = y1[i]

errors = []
errors2 = []
lr = 0.0001
n = 10

for i in range(n):
    count = 0
    for row in x:
        dot = np.dot(w, row)
        response = step(dot)
        errors.append(t[count] - response)
        w += lr * (row * (t[count] - response))
        count += 1

for i in range(n):
    count = 0
    for row in x:
        dot = np.dot(w2, row)
        response = sigmoid(dot)
        errors2.append(t[count] - response)
        w2 += lr * (row * (t[count] - response))
        count += 1

print(errors[-1], errors2[-1])

# distribution
plt.figure(1)
plt.plot((-(w[2]/w[0]), 0), (0, -(w[2]/w[1])))
plt.plot(x1, y1, 'x')
plt.plot(x2, y2, 'ro')
plt.axis('equal')
plt.title('Heaviside')

# training error
plt.figure(2)
plt.ylabel('error')
plt.xlabel('iterations')
plt.plot(errors)
plt.title('Heaviside Error')

plt.figure(3)
plt.plot((-(w2[2]/w2[0]), 0), (0, -(w2[2]/w2[1])))
plt.plot(x1, y1, 'x')
plt.plot(x2, y2, 'ro')
plt.axis('equal')
plt.title('Sigmoidal')

plt.figure(4)
plt.ylabel('error')
plt.xlabel('iterations')
plt.plot(errors2)
plt.title('Sigmoidal Error')

plt.show()

编辑:即使从错误图中我已经显示tanh函数显示出一些收敛,所以假设只是增加迭代或降低学习速率将允许它减少其错误是合理的。但是我想我真的在问,考虑到阶梯函数的显着更好的性能,对于什么问题集是否可以使用感知器使用tanh?

1 个答案:

答案 0 :(得分:2)

正如评论中已经提到的,您的学习率太小,因此需要花费大量的迭代才能收敛。为了获得可比较的输出,您可以因此增加n和/或lr

如果将lr增加到例如0.1(也1工作正常)和n到10000,结果看起来几乎相同(见下图)和线

print(errors[-1], errors2[-1])

返回

(0, -8.4289020207961585e-11)

如果再次运行,这些值可能会有所不同,因为没有为随机数设置种子。

以下是我获得的上述值的图表:

enter image description here enter image description here

enter image description here enter image description here