看排版更好的原文地址
公式显示不出来，可以查看pdf版本
感知器
感知器是神经网络的基础构成组件，可以看做节点组合。
一个简单的直线数据分类示例
对于坐标轴为 (p,q)(p,q) 的点，标签 y，以及等式
$$hat{y} = step(w_1x_1 + w_2x_2 + b) $$
给出的预测
- 如果点分类正确，则什么也不做。
- 如果点分类为正，但是标签为负，则分别减去 $$alpha p$$, $$alpha q$$和 $$alpha$$ 至 $$w_1$$, $$w_2$$和 $$b$$
- 如果点分类为负，但是标签为正，则分别将 $$alpha p$$, $$alpha q$$ 和 $$alpha$$加到 $$w_1$$, $$w_2$$和 $$b$$上。
# perceptron.py
import numpy as np
# Setting the random seed, feel free to change it and see different solutions.
np.random.seed(42)
def stepFunction(t):
if t >= 0:
return 1
return 0
def prediction(X, W, b):
return stepFunction((np.matmul(X,W)+b)[0])
# TODO: Fill in the code below to implement the perceptron trick.
# The function should receive as inputs the data X, the labels y,
# the weights W (as an array), and the bias b,
# update the weights and bias W, b, according to the perceptron algorithm,
# and return W and b.
def perceptronStep(X, y, W, b, learn_rate = 0.01):
# Fill in code
return W, b
# This function runs the perceptron algorithm repeatedly on the dataset,
# and returns a few of the boundary lines obtained in the iterations,
# for plotting purposes.
# Feel free to play with the learning rate and the num_epochs,
# and see your results plotted below.
def trainPerceptronAlgorithm(X, y, learn_rate = 0.01, num_epochs = 25):
x_min, x_max = min(X.T[0]), max(X.T[0])
y_min, y_max = min(X.T[1]), max(X.T[1])
W = np.array(np.random.rand(2,1))
b = np.random.rand(1)[0] + x_max
# These are the solution lines that get plotted below.
boundary_lines = []
for i in range(num_epochs):
# In each epoch, we apply the perceptron step.
W, b = perceptronStep(X, y, W, b, learn_rate)
boundary_lines.append((-W[0]/W[1], -b/W[1]))
return boundary_lines
# data.csv
0.78051,-0.063669,1
0.28774,0.29139,1
0.40714,0.17878,1
0.2923,0.4217,1
0.50922,0.35256,1
0.27785,0.10802,1
0.27527,0.33223,1
0.43999,0.31245,1
0.33557,0.42984,1
0.23448,0.24986,1
0.0084492,0.13658,1
0.12419,0.33595,1
0.25644,0.42624,1
0.4591,0.40426,1
0.44547,0.45117,1
0.42218,0.20118,1
0.49563,0.21445,1
0.30848,0.24306,1
0.39707,0.44438,1
0.32945,0.39217,1
0.40739,0.40271,1
0.3106,0.50702,1
0.49638,0.45384,1
0.10073,0.32053,1
0.69907,0.37307,1
0.29767,0.69648,1
0.15099,0.57341,1
0.16427,0.27759,1
0.33259,0.055964,1
0.53741,0.28637,1
0.19503,0.36879,1
0.40278,0.035148,1
0.21296,0.55169,1
0.48447,0.56991,1
0.25476,0.34596,1
0.21726,0.28641,1
0.67078,0.46538,1
0.3815,0.4622,1
0.53838,0.32774,1
0.4849,0.26071,1
0.37095,0.38809,1
0.54527,0.63911,1
0.32149,0.12007,1
0.42216,0.61666,1
0.10194,0.060408,1
0.15254,0.2168,1
0.45558,0.43769,1
0.28488,0.52142,1
0.27633,0.21264,1
0.39748,0.31902,1
0.5533,1,0
0.44274,0.59205,0
0.85176,0.6612,0
0.60436,0.86605,0
0.68243,0.48301,0
1,0.76815,0
0.72989,0.8107,0
0.67377,0.77975,0
0.78761,0.58177,0
0.71442,0.7668,0
0.49379,0.54226,0
0.78974,0.74233,0
0.67905,0.60921,0
0.6642,0.72519,0
0.79396,0.56789,0
0.70758,0.76022,0
0.59421,0.61857,0
0.49364,0.56224,0
0.77707,0.35025,0
0.79785,0.76921,0
0.70876,0.96764,0
0.69176,0.60865,0
0.66408,0.92075,0
0.65973,0.66666,0
0.64574,0.56845,0
0.89639,0.7085,0
0.85476,0.63167,0
0.62091,0.80424,0
0.79057,0.56108,0
0.58935,0.71582,0
0.56846,0.7406,0
0.65912,0.71548,0
0.70938,0.74041,0
0.59154,0.62927,0
0.45829,0.4641,0
0.79982,0.74847,0
0.60974,0.54757,0
0.68127,0.86985,0
0.76694,0.64736,0
0.69048,0.83058,0
0.68122,0.96541,0
0.73229,0.64245,0
0.76145,0.60138,0
0.58985,0.86955,0
0.73145,0.74516,0
0.77029,0.7014,0
0.73156,0.71782,0
0.44556,0.57991,0
0.85275,0.85987,0
0.51912,0.62359,0
# solution.py
def perceptronStep(X, y, W, b, learn_rate = 0.01):
for i in range(len(X)):
y_hat = prediction(X[i],W,b)
if y[i]-y_hat == 1:
W[0] += X[i][0]*learn_rate
W[1] += X[i][1]*learn_rate
b += learn_rate
elif y[i]-y_hat == -1:
W[0] -= X[i][0]*learn_rate
W[1] -= X[i][1]*learn_rate
b -= learn_rate
return W, b
误差函数
误差函数（ERROR)可以告诉我们目前的状况有多差，与理想解决方案的差别有多大。
离散型到连续型的转化
梯度下降只能用于连续型函数。对于一些离散型数据，将激活函数由跃迁函数改为s函数。
softmax函数
# softmax.py
import numpy as np
# Write a function that takes as input a list of numbers, and returns
# the list of values given by the softmax function.
def softmax(L):
expL = np.exp(L)
sumExpL = sum(expL)
result = []
for i in expL:
result.append(i*1.0/sumExpL)
return result
# Note: The function np.divide can also be used here, as follows:
# def softmax(L):
# expL(np.exp(L))
# return np.divide (expL, expL.sum())
最大似然法
如在点的分类问题中，将每个点分类正确的概率相乘，得到所有点都分类正确的概率。然后尽可能地增大这个概率。这叫做最大似然法。
交叉熵
对最大似然法得到的概率进行求负对数，然后相加。越好的模型求得的交叉熵越小。
交叉熵公式：
import numpy as np
# Write a function that takes as input two lists Y, P,
# and returns the float corresponding to their cross-entropy.
def cross_entropy(Y, P):
Y = np.float_(Y)
P = np.float_(P)
return -np.sum(Y * np.log(P) + (1 - Y) * np.log(1 - P))
交叉熵公式只要保证只加上实际发生事件的概率负对数。
梯度计算
s型函数的导数：$$σ′(x)=σ(x)(1−σ(x))$$
误差公式是：$$E = -frac{1}{m} sum_{i=1}^m left( y_i ln(hat{y_i}) + (1-y_i) ln (1-hat{y_i}) ight)$$
预测是 $$hat{y_i} = sigma(Wx^{(i)} + b)$$
我们的目标是计算 E,E, 在点 $$x = (x _1, ldots, x_n)$$ 时的梯度（偏导数）
$$ abla E =left(frac{partial}{partial w_1} E, cdots, frac{partial}{partial w_n}E, frac{partial}{partial b}E ight)$$
为此，首先我们要计算 $$frac{partial}{partial w_j} hat{y}.$$
最后得：$$∇E(W,b)=(y−hat y)(x _1,…,x _n,1).$$
梯度实际上是标量乘以点的坐标.
梯度下降实验
- Sigmoid activation function
$$sigma(x) = frac{1}{1+e^{-x}}$$
- Output (prediction) formula
$$hat{y} = sigma(w_1 x_1 + w_2 x_2 + b)$$
- Error function
$$Error(y, hat{y}) = - y log(hat{y}) - (1-y) log(1-hat{y})$$
- The function that updates the weights
$$ w_i longrightarrow w_i + alpha (y - hat{y}) x_i$$
$$ b longrightarrow b + alpha (y - hat{y})$$
代码实现：
# Implement the following functions
# Activation (sigmoid) function
def sigmoid(x):
return 1/(1+np.exp(-x))
# Output (prediction) formula
def output_formula(features, weights, bias):
return sigmoid(np.dot(features, weights) + bias)
# Error (log-loss) formula
def error_formula(y, output):
return - y*np.log(output) - (1 - y) * np.log(1-output)
# Gradient descent step
def update_weights(x, y, weights, bias, learnrate):
output = output_formula(x, weights, bias)
d_error = y - output
weights += learnrate * d_error * x
bias += learnrate * d_error
return weights, bias