Andrew Ng -- machine learning --Logistic Regression --- ex2/吴恩达机器学习ex2

这个项目包含了吴恩达机器学习ex2的python实现，主要知识点为逻辑回归、正则化，题目内容可以查看数据集中的ex2.pdf
代码来自网络（原作者黄广海的github），添加了部分对于题意的中文翻译，以及修改成与习题一致的结构，方便大家理解

另，原来代码中的高次项有些问题，以及缺少作图部分，已经修改补全

其余练习的传送门

1 逻辑回归¶

在训练的初始阶段，我们将要构建一个逻辑回归模型来预测，某个学生是否被大学录取。
设想你是大学相关部分的管理者，想通过申请学生两次测试的评分，来决定他们是否被录取。
现在你拥有之前申请学生的可以用于训练逻辑回归的训练样本集。对于每一个训练样本，你有他们两次测试的评分和最后是被录取的结果。

1.1 数据可视化¶

In [1]:

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

/opt/conda/lib/python3.6/importlib/_bootstrap.py:219: RuntimeWarning: numpy.dtype size changed, may indicate binary incompatibility. Expected 96, got 88
  return f(*args, **kwds)
/opt/conda/lib/python3.6/importlib/_bootstrap.py:219: RuntimeWarning: numpy.dtype size changed, may indicate binary incompatibility. Expected 96, got 88
  return f(*args, **kwds)

In [2]:

path = '/home/kesci/input/andrew_ml_ex22391/ex2data1.txt'
data = pd.read_csv(path, header=None, names=['Exam 1', 'Exam 2', 'Admitted'])
data.head()

Out[2]:

	Exam 1	Exam 2	Admitted
0	34.623660	78.024693	0
1	30.286711	43.894998	0
2	35.847409	72.902198	0
3	60.182599	86.308552	1
4	79.032736	75.344376	1

In [3]:

positive = data[data['Admitted'].isin([1])]
negative = data[data['Admitted'].isin([0])]

fig, ax = plt.subplots(figsize=(12,8))
ax.scatter(positive['Exam 1'], positive['Exam 2'], s=50, c='b', marker='o', label='Admitted')
ax.scatter(negative['Exam 1'], negative['Exam 2'], s=50, c='r', marker='x', label='Not Admitted')
ax.legend()
ax.set_xlabel('Exam 1 Score')
ax.set_ylabel('Exam 2 Score')
plt.show()

1.2 实现¶

1.2.1 sigmoid 函数¶

逻辑回归函数为

h_{θ} = g (θ^{T} x)

g代表一个常用的逻辑函数（logistic function）为S形函数（Sigmoid function），公式为：

g (z) = \frac{1}{1 + e^{- z}}

合起来，我们得到逻辑回归模型的假设函数：

h_{θ} (x) = \frac{1}{1 + e^{- θ^{T} x}}

In [4]:

# 实现sigmoid函数
def sigmoid(z):
    return 1 / (1 + np.exp(-z))

1.2.2 代价函数和梯度¶

代价函数：

J (θ) = \frac{1}{m} \sum_{i = 1}^{m} [- y^{(i)} \log (h_{θ} (x^{(i)})) - (1 - y^{(i)}) \log (1 - h_{θ} (x^{(i)}))]

梯度：

\frac{\partial J (θ)}{\partial θ_{j}} = \frac{1}{m} \sum_{i = 1}^{m} (h_{θ} (x^{(i)}) - y^{(i)}) x_{_{j}}^{(i)}

虽然这个梯度和前面线性回归的梯度很像，但是要记住$h_\theta(x)$是不一样的
实现完成后，用初始$\theta$代入计算，结果应该是0.693左右

In [5]:

# 实现代价函数
def cost(theta, X, y):
    theta = np.matrix(theta)
    X = np.matrix(X)
    y = np.matrix(y)
    first = np.multiply(-y, np.log(sigmoid(X * theta.T)))
    second = np.multiply((1 - y), np.log(1 - sigmoid(X * theta.T)))
    return np.sum(first - second) / (len(X))

初始化X，y，$\theta$

In [6]:

# 加一列常数列
data.insert(0, 'Ones', 1)

# 初始化X，y，θ
cols = data.shape[1]
X = data.iloc[:,0:cols-1]
y = data.iloc[:,cols-1:cols]
theta = np.zeros(3)

# 转换X，y的类型
X = np.array(X.values)
y = np.array(y.values)

In [7]:

# 检查矩阵的维度
X.shape, theta.shape, y.shape

Out[7]:

((100, 3), (3,), (100, 1))

In [8]:

# 用初始θ计算代价
cost(theta, X, y)

Out[8]:

0.6931471805599453

In [9]:

# 实现梯度计算的函数（并没有更新θ）
def gradient(theta, X, y):
    theta = np.matrix(theta)
    X = np.matrix(X)
    y = np.matrix(y)
    
    parameters = int(theta.ravel().shape[1])
    grad = np.zeros(parameters)
    
    error = sigmoid(X * theta.T) - y
    
    for i in range(parameters):
        term = np.multiply(error, X[:,i])
        grad[i] = np.sum(term) / len(X)
    
    return grad

1.2.3 用工具库计算θ的值¶

在此前的线性回归中，我们自己写代码实现的梯度下降（ex1的2.2.4的部分）。当时我们写了一个代价函数、计算了他的梯度，然后对他执行了梯度下降的步骤。这次，我们不自己写代码实现梯度下降，我们会调用一个已有的库。这就是说，我们不用自己定义迭代次数和步长，功能会直接告诉我们最优解。
andrew ng在课程中用的是Octave的“fminunc”函数，由于我们使用Python，我们可以用scipy.optimize.fmin_tnc做同样的事情。
（另外，如果对fminunc有疑问的，可以参考下面这篇百度文库的内容https://wenku.baidu.com/view/2f6ce65d0b1c59eef8c7b47a.html ）
如果一切顺利的话，最有θ对应的代价应该是0.203

In [10]:

import scipy.optimize as opt
result = opt.fmin_tnc(func=cost, x0=theta, fprime=gradient, args=(X, y))
result

Out[10]:

(array([-25.16131863,   0.20623159,   0.20147149]), 36, 0)

让我们看看在这个结论下代价函数计算结果是什么个样子~

In [11]:

# 用θ的计算结果代回代价函数计算
cost(result[0], X, y)

Out[11]:

0.20349770158947458

画出决策曲线

In [12]:

plotting_x1 = np.linspace(30, 100, 100)
plotting_h1 = ( - result[0][0] - result[0][1] * plotting_x1) / result[0][2]

fig, ax = plt.subplots(figsize=(12,8))
ax.plot(plotting_x1, plotting_h1, 'y', label='Prediction')
ax.scatter(positive['Exam 1'], positive['Exam 2'], s=50, c='b', marker='o', label='Admitted')
ax.scatter(negative['Exam 1'], negative['Exam 2'], s=50, c='r', marker='x', label='Not Admitted')
ax.legend()
ax.set_xlabel('Exam 1 Score')
ax.set_ylabel('Exam 2 Score')
plt.show()

1.2.4 评价逻辑回归模型¶

在确定参数之后，我们可以使用这个模型来预测学生是否录取。如果一个学生exam1得分45，exam2得分85，那么他录取的概率应为0.776

In [13]:

# 实现hθ
def hfunc1(theta, X):
    return sigmoid(np.dot(theta.T, X))
hfunc1(result[0],[1,45,85])

Out[13]:

0.7762906238162848

另一种评价θ的方法是看模型在训练集上的正确率怎样。写一个predict的函数，给出数据以及参数后，会返回“1”或者“0”。然后再把这个predict函数用于训练集上，看准确率怎样。

In [14]:

# 定义预测函数
def predict(theta, X):
    probability = sigmoid(X * theta.T)
    return [1 if x >= 0.5 else 0 for x in probability]

In [15]:

# 统计预测正确率
theta_min = np.matrix(result[0])
predictions = predict(theta_min, X)
correct = [1 if ((a == 1 and b == 1) or (a == 0 and b == 0)) else 0 for (a, b) in zip(predictions, y)]
accuracy = (sum(map(int, correct)) % len(correct))
print ('accuracy = {0}%'.format(accuracy))

accuracy = 89%

画出对应曲线

2 正则化逻辑回归¶

在训练的第二部分，我们将实现加入正则项提升逻辑回归算法。
设想你是工厂的生产主管，你有一些芯片在两次测试中的测试结果，测试结果决定是否芯片要被接受或抛弃。你有一些历史数据，帮助你构建一个逻辑回归模型。

2.1 数据可视化¶

In [16]:

path =  '/home/kesci/input/andrew_ml_ex22391/ex2data2.txt'
data_init = pd.read_csv(path, header=None, names=['Test 1', 'Test 2', 'Accepted'])
data_init.head()

Out[16]:

	Test 1	Test 2	Accepted
0	0.051267	0.69956	1
1	-0.092742	0.68494	1
2	-0.213710	0.69225	1
3	-0.375000	0.50219	1
4	-0.513250	0.46564	1

In [17]:

positive2 = data_init[data_init['Accepted'].isin([1])]
negative2 = data_init[data_init['Accepted'].isin([0])]

fig, ax = plt.subplots(figsize=(12,8))
ax.scatter(positive2['Test 1'], positive2['Test 2'], s=50, c='b', marker='o', label='Accepted')
ax.scatter(negative2['Test 1'], negative2['Test 2'], s=50, c='r', marker='x', label='Rejected')
ax.legend()
ax.set_xlabel('Test 1 Score')
ax.set_ylabel('Test 2 Score')
plt.show()

以上图片显示，这个数据集不能像之前一样使用直线将两部分分割。而逻辑回归只适用于线性的分割，所以，这个数据集不适合直接使用逻辑回归。

2.2 特征映射¶

一种更好的使用数据集的方式是为每组数据创造更多的特征。所以我们为每组$x_1,x_2$添加了最高到6次幂的特征

In [18]:

degree = 6
data2 = data_init
x1 = data2['Test 1']
x2 = data2['Test 2']

data2.insert(3, 'Ones', 1)

for i in range(1, degree+1):
    for j in range(0, i+1):
        data2['F' + str(i-j) + str(j)] = np.power(x1, i-j) * np.power(x2, j)
#此处原答案错误较多，已经更正

data2.drop('Test 1', axis=1, inplace=True)
data2.drop('Test 2', axis=1, inplace=True)

data2.head()

Out[18]:

	Accepted	Ones	F10	F01	F20	F11	F02	F30	F21	F12	...	F23	F14	F05	F60	F51	F42	F33	F24	F15	F06
0	1	1	0.051267	0.69956	0.002628	0.035864	0.489384	0.000135	0.001839	0.025089	...	0.000900	0.012278	0.167542	1.815630e-08	2.477505e-07	0.000003	0.000046	0.000629	0.008589	0.117206
1	1	1	-0.092742	0.68494	0.008601	-0.063523	0.469143	-0.000798	0.005891	-0.043509	...	0.002764	-0.020412	0.150752	6.362953e-07	-4.699318e-06	0.000035	-0.000256	0.001893	-0.013981	0.103256
2	1	1	-0.213710	0.69225	0.045672	-0.147941	0.479210	-0.009761	0.031616	-0.102412	...	0.015151	-0.049077	0.158970	9.526844e-05	-3.085938e-04	0.001000	-0.003238	0.010488	-0.033973	0.110047
3	1	1	-0.375000	0.50219	0.140625	-0.188321	0.252195	-0.052734	0.070620	-0.094573	...	0.017810	-0.023851	0.031940	2.780914e-03	-3.724126e-03	0.004987	-0.006679	0.008944	-0.011978	0.016040
4	1	1	-0.513250	0.46564	0.263426	-0.238990	0.216821	-0.135203	0.122661	-0.111283	...	0.026596	-0.024128	0.021890	1.827990e-02	-1.658422e-02	0.015046	-0.013650	0.012384	-0.011235	0.010193

5 rows × 29 columns

3.2 代价函数和梯度¶

这一部分要实现计算逻辑回归的代价函数和梯度的函数。代价函数公式如下：

J (θ) = \frac{1}{m} \sum_{i = 1}^{m} [- y^{(i)} \log (h_{θ} (x^{(i)})) - (1 - y^{(i)}) \log (1 - h_{θ} (x^{(i)}))] + \frac{λ}{2 m} \sum_{j = 1}^{n} θ_{j}^{2}

记住$\theta_0$是不需要正则化的，下标从1开始。
梯度的第j个元素的更新公式为：

θ_{0} := θ_{0} - a \frac{1}{m} \sum_{i = 1}^{m} [h_{θ} (x^{(i)}) - y^{(i)}] x_{_{0}}^{(i)}

θ_{j} := θ_{j} - a \frac{1}{m} \sum_{i = 1}^{m} [h_{θ} (x^{(i)}) - y^{(i)}] x_{j}^{(i)} + \frac{λ}{m} θ_{j}

对上面的算法中 j=1,2,...,n 时的更新式子进行调整可得：

θ_{j} := θ_{j} (1 - a \frac{λ}{m}) - a \frac{1}{m} \sum_{i = 1}^{m} (h_{θ} (x^{(i)}) - y^{(i)}) x_{j}^{(i)}

把初始$\theta$（所有元素为0）带入，代价应为0.693

In [19]:

# 实现正则化的代价函数
def costReg(theta, X, y, learningRate):
    theta = np.matrix(theta)
    X = np.matrix(X)
    y = np.matrix(y)
    first = np.multiply(-y, np.log(sigmoid(X * theta.T)))
    second = np.multiply((1 - y), np.log(1 - sigmoid(X * theta.T)))
    reg = (learningRate / (2 * len(X))) * np.sum(np.power(theta[:,1:theta.shape[1]], 2))
    return np.sum(first - second) / len(X) + reg

In [20]:

# 实现正则化的梯度函数
def gradientReg(theta, X, y, learningRate):
    theta = np.matrix(theta)
    X = np.matrix(X)
    y = np.matrix(y)
    
    parameters = int(theta.ravel().shape[1])
    grad = np.zeros(parameters)
    
    error = sigmoid(X * theta.T) - y
    
    for i in range(parameters):
        term = np.multiply(error, X[:,i])
        
        if (i == 0):
            grad[i] = np.sum(term) / len(X)
        else:
            grad[i] = (np.sum(term) / len(X)) + ((learningRate / len(X)) * theta[:,i])
    
    return grad

In [21]:

# 初始化X，y，θ
cols = data2.shape[1]
X2 = data2.iloc[:,1:cols]
y2 = data2.iloc[:,0:1]
theta2 = np.zeros(cols-1)

# 进行类型转换
X2 = np.array(X2.values)
y2 = np.array(y2.values)

# λ设为1
learningRate = 1

In [22]:

# 计算初始代价
costReg(theta2, X2, y2, learningRate)

Out[22]:

0.6931471805599454

2.3.1 用工具库求解参数¶

In [23]:

result2 = opt.fmin_tnc(func=costReg, x0=theta2, fprime=gradientReg, args=(X2, y2, learningRate))
result2

Out[23]:

(array([ 1.27271026,  0.62529965,  1.18111686, -2.01987398, -0.91743189,
        -1.43166928,  0.12393228, -0.36553118, -0.35725404, -0.17516291,
        -1.45817009, -0.05098418, -0.61558555, -0.27469165, -1.19271298,
        -0.24217841, -0.20603299, -0.04466178, -0.27778951, -0.29539514,
        -0.45645982, -1.04319155,  0.02779373, -0.2924487 ,  0.0155576 ,
        -0.32742405, -0.1438915 , -0.92467487]), 32, 1)

最后，我们可以使用第1部分中的预测函数来查看我们的方案在训练数据上的准确度。

In [24]:

theta_min = np.matrix(result2[0])
predictions = predict(theta_min, X2)
correct = [1 if ((a == 1 and b == 1) or (a == 0 and b == 0)) else 0 for (a, b) in zip(predictions, y2)]
accuracy = (sum(map(int, correct)) % len(correct))
print ('accuracy = {0}%'.format(accuracy))

accuracy = 98%

2.4 画出决策的曲线¶

In [25]:

def hfunc2(theta, x1, x2):
    temp = theta[0][0]
    place = 0
    for i in range(1, degree+1):
        for j in range(0, i+1):
            temp+= np.power(x1, i-j) * np.power(x2, j) * theta[0][place+1]
            place+=1
    return temp

In [26]:

def find_decision_boundary(theta):
    t1 = np.linspace(-1, 1.5, 1000)
    t2 = np.linspace(-1, 1.5, 1000)

    cordinates = [(x, y) for x in t1 for y in t2]
    x_cord, y_cord = zip(*cordinates)
    h_val = pd.DataFrame({'x1':x_cord, 'x2':y_cord})
    h_val['hval'] = hfunc2(theta, h_val['x1'], h_val['x2'])

    decision = h_val[np.abs(h_val['hval']) < 2 * 10**-3]
    return decision.x1, decision.x2

In [27]:

fig, ax = plt.subplots(figsize=(12,8))
ax.scatter(positive2['Test 1'], positive2['Test 2'], s=50, c='b', marker='o', label='Accepted')
ax.scatter(negative2['Test 1'], negative2['Test 2'], s=50, c='r', marker='x', label='Rejected')
ax.set_xlabel('Test 1 Score')
ax.set_ylabel('Test 2 Score')

x, y = find_decision_boundary(result2)
plt.scatter(x, y, c='y', s=10, label='Prediction')
ax.legend()
plt.show()

2.5 改变λ，观察决策曲线¶

$\lambda=0$时过拟合

In [28]:

learningRate2 = 0
result3 = opt.fmin_tnc(func=costReg, x0=theta2, fprime=gradientReg, args=(X2, y2, learningRate2))

Out[28]:

(array([   14.60193336,    21.20326682,     4.60748805,  -150.30636263,
          -70.51421716,   -65.71761632,  -167.22986423,  -100.93094956,
          -58.4583472 ,     9.35117823,   538.72438097,   445.25267052,
          633.43046793,   239.567217  ,    92.6608774 ,   300.20568543,
          362.78215934,   440.45538844,   196.63024035,    52.26698467,
          -13.32416223,  -639.85098768,  -782.82561038, -1230.55113233,
         -846.7737968 ,  -793.91524305,  -273.62741174,   -51.4586635 ]),
 280,
 3)

In [29]:

fig, ax = plt.subplots(figsize=(12,8))
ax.scatter(positive2['Test 1'], positive2['Test 2'], s=50, c='b', marker='o', label='Accepted')
ax.scatter(negative2['Test 1'], negative2['Test 2'], s=50, c='r', marker='x', label='Rejected')
ax.set_xlabel('Test 1 Score')
ax.set_ylabel('Test 2 Score')

x, y = find_decision_boundary(result3)
plt.scatter(x, y, c='y', s=10, label='Prediction')
ax.legend()
plt.show()

$\lambda=100$时欠拟合

In [30]:

learningRate3 = 100
result4 = opt.fmin_tnc(func=costReg, x0=theta2, fprime=gradientReg, args=(X2, y2, learningRate3))

In [31]:

fig, ax = plt.subplots(figsize=(12,8))
ax.scatter(positive2['Test 1'], positive2['Test 2'], s=50, c='b', marker='o', label='Accepted')
ax.scatter(negative2['Test 1'], negative2['Test 2'], s=50, c='r', marker='x', label='Rejected')
ax.set_xlabel('Test 1 Score')
ax.set_ylabel('Test 2 Score')

x, y = find_decision_boundary(result4)
plt.scatter(x, y, c='y', s=10, label='Prediction')
ax.legend()
plt.show()

Ungraded Lab: Logistic Regression using Scikit-Learn¶

Goals¶

In this lab you will:

Train a logistic regression model using scikit-learn.

Dataset¶

Let's start with the same dataset as before.