梯度下降求解逻辑回归

 Logistic Regression

 

The data

 

我们将建立一个逻辑回归模型来预测一个学生是否被大学录取。假设你是一个大学系的管理员，你想根据两次考试的结果来决定每个申请人的录取机会。你有以前的申请人的历史数据，你可以用它作为逻辑回归的训练集。对于每一个培训例子，你有两个考试的申请人的分数和录取决定。为了做到这一点，我们将建立一个分类模型，根据考试成绩估计入学概率。

In [ ]:

 

#三大件
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
%matplotlib inline

 

In [2]:

 

import os
path = 'data' + os.sep + 'LogiReg_data.txt'
pdData = pd.read_csv(path, header=None, names=['Exam 1', 'Exam 2', 'Admitted'])
pdData.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]:

 

pdData.shape

 

Out[3]:

(100, 3)

In [4]:

 

positive = pdData[pdData['Admitted'] == 1] # returns the subset of rows such Admitted = 1, i.e. the set of *positive* examples
negative = pdData[pdData['Admitted'] == 0] # returns the subset of rows such Admitted = 0, i.e. the set of *negative* examples
​
fig, ax = plt.subplots(figsize=(10,5))
ax.scatter(positive['Exam 1'], positive['Exam 2'], s=30, c='b', marker='o', label='Admitted')
ax.scatter(negative['Exam 1'], negative['Exam 2'], s=30, c='r', marker='x', label='Not Admitted')
ax.legend()
ax.set_xlabel('Exam 1 Score')
ax.set_ylabel('Exam 2 Score')

 

Out[4]:

<matplotlib.text.Text at 0x1bab4215208>

 

 

The logistic regression

 

目标：建立分类器（求解出三个参数 𝜃0𝜃1𝜃2θ0θ1θ2）

设定阈值，根据阈值判断录取结果

要完成的模块

• sigmoid : 映射到概率的函数

• model : 返回预测结果值

• cost : 根据参数计算损失

• gradient : 计算每个参数的梯度方向

• descent : 进行参数更新

• accuracy: 计算精度

 

 

 

In [5]:

 

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

 

In [6]:

 

nums = np.arange(-10, 10, step=1) #creates a vector containing 20 equally spaced values from -10 to 10
fig, ax = plt.subplots(figsize=(12,4))
ax.plot(nums, sigmoid(nums), 'r')

 

Out[6]:

[<matplotlib.lines.Line2D at 0x1bab41d49b0>]

 

 

Sigmoid

• 𝑔:ℝ→[0,1]g:R→[0,1]
• 𝑔(0)=0.5g(0)=0.5
• 𝑔(−∞)=0g(−∞)=0
• 𝑔(+∞)=1g(+∞)=1

In [7]:

 

def model(X, theta):
    return sigmoid(np.dot(X, theta.T))

 

 

 

In [8]:

 

pdData.insert(0, 'Ones', 1) # in a try / except structure so as not to return an error if the block si executed several times
​
​
# set X (training data) and y (target variable)
orig_data = pdData.as_matrix() # convert the Pandas representation of the data to an array useful for further computations
cols = orig_data.shape[1]
X = orig_data[:,0:cols-1]
y = orig_data[:,cols-1:cols]
​
# convert to numpy arrays and initalize the parameter array theta
#X = np.matrix(X.values)
#y = np.matrix(data.iloc[:,3:4].values) #np.array(y.values)
theta = np.zeros([1, 3])

 

In [10]:

 

X[:5]

 

Out[10]:

array([[  1.        ,  34.62365962,  78.02469282],
       [  1.        ,  30.28671077,  43.89499752],
       [  1.        ,  35.84740877,  72.90219803],
       [  1.        ,  60.18259939,  86.3085521 ],
       [  1.        ,  79.03273605,  75.34437644]])

In [11]:

 

y[:5]

 

Out[11]:

array([[ 0.],
       [ 0.],
       [ 0.],
       [ 1.],
       [ 1.]])

In [12]:

 

theta

 

Out[12]:

array([[ 0.,  0.,  0.]])

In [13]:

 

X.shape, y.shape, theta.shape

 

Out[13]:

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

 

 

In [14]:

 

def cost(X, y, theta):
    left = np.multiply(-y, np.log(model(X, theta)))
    right = np.multiply(1 - y, np.log(1 - model(X, theta)))
    return np.sum(left - right) / (len(X))

 

In [15]:

 

cost(X, y, theta)

 

Out[15]:

0.69314718055994529

 

 

In [23]:

 

def gradient(X, y, theta):
    grad = np.zeros(theta.shape)
    error = (model(X, theta)- y).ravel()
    for j in range(len(theta.ravel())): #for each parmeter
        term = np.multiply(error, X[:,j])
        grad[0, j] = np.sum(term) / len(X)
    return grad

 

Gradient descent

 

比较3中不同梯度下降方法

In [16]:

 

STOP_ITER = 0
STOP_COST = 1
STOP_GRAD = 2
​
def stopCriterion(type, value, threshold):
    #设定三种不同的停止策略
    if type == STOP_ITER:        return value > threshold
    elif type == STOP_COST:      return abs(value[-1]-value[-2]) < threshold
    elif type == STOP_GRAD:      return np.linalg.norm(value) < threshold

 

In [17]:

 

import numpy.random
#洗牌
def shuffleData(data):
    np.random.shuffle(data)
    cols = data.shape[1]
    X = data[:, 0:cols-1]
    y = data[:, cols-1:]
    return X, y

 

In [18]:

 

import time
​
def descent(data, theta, batchSize, stopType, thresh, alpha):
    #梯度下降求解
    init_time = time.time()
    i = 0 # 迭代次数
    k = 0 # batch
    X, y = shuffleData(data)
    grad = np.zeros(theta.shape) # 计算的梯度
    costs = [cost(X, y, theta)] # 损失值
​
    while True:
        grad = gradient(X[k:k+batchSize], y[k:k+batchSize], theta)
        k += batchSize #取batch数量个数据
        if k >= n: 
            k = 0 
            X, y = shuffleData(data) #重新洗牌
        theta = theta - alpha*grad # 参数更新
        costs.append(cost(X, y, theta)) # 计算新的损失
        i += 1 
​
        if stopType == STOP_ITER:       value = i
        elif stopType == STOP_COST:     value = costs
        elif stopType == STOP_GRAD:     value = grad
        if stopCriterion(stopType, value, thresh): break
    return theta, i-1, costs, grad, time.time() - init_time

 

In [2]:

 

def runExpe(data, theta, batchSize, stopType, thresh, alpha):
    #import pdb; pdb.set_trace();
    theta, iter, costs, grad, dur = descent(data, theta, batchSize, stopType, thresh, alpha)
    name = "Original" if (data[:,1]>2).sum() > 1 else "Scaled"
    name += " data - learning rate: {} - ".format(alpha)
    if batchSize==n: strDescType = "Gradient"
    elif batchSize==1:  strDescType = "Stochastic"
    else: strDescType = "Mini-batch ({})".format(batchSize)
    name += strDescType + " descent - Stop: "
    if stopType == STOP_ITER: strStop = "{} iterations".format(thresh)
    elif stopType == STOP_COST: strStop = "costs change < {}".format(thresh)
    else: strStop = "gradient norm < {}".format(thresh)
    name += strStop
    print ("***{}\nTheta: {} - Iter: {} - Last cost: {:03.2f} - Duration: {:03.2f}s".format(
        name, theta, iter, costs[-1], dur))
    fig, ax = plt.subplots(figsize=(12,4))
    ax.plot(np.arange(len(costs)), costs, 'r')
    ax.set_xlabel('Iterations')
    ax.set_ylabel('Cost')
    ax.set_title(name.upper() + ' - Error vs. Iteration')
    return theta

 

不同的停止策略

 

设定迭代次数

In [29]:

 

#选择的梯度下降方法是基于所有样本的
n=100
runExpe(orig_data, theta, n, STOP_ITER, thresh=5000, alpha=0.000001)

 

***Original data - learning rate: 1e-06 - Gradient descent - Stop: 5000 iterations
Theta: [[-0.00027127  0.00705232  0.00376711]] - Iter: 5000 - Last cost: 0.63 - Duration: 1.18s

Out[29]:

array([[-0.00027127,  0.00705232,  0.00376711]])

 

 

根据损失值停止

 

设定阈值 1E-6, 差不多需要110 000次迭代

In [27]:

 

 

 

 

 

runExpe(orig_data, theta, n, STOP_COST, thresh=0.000001, alpha=0.001)

 

***Original data - learning rate: 0.001 - Gradient descent - Stop: costs change < 1e-06
Theta: [[-5.13364014  0.04771429  0.04072397]] - Iter: 109901 - Last cost: 0.38 - Duration: 24.47s

Out[27]:

array([[-5.13364014,  0.04771429,  0.04072397]])

 

 

根据梯度变化停止

 

设定阈值 0.05,差不多需要40 000次迭代

In [28]:

 

 

runExpe(orig_data, theta, n, STOP_GRAD, thresh=0.05, alpha=0.001)

 

***Original data - learning rate: 0.001 - Gradient descent - Stop: gradient norm < 0.05
Theta: [[-2.37033409  0.02721692  0.01899456]] - Iter: 40045 - Last cost: 0.49 - Duration: 10.79s

Out[28]:

array([[-2.37033409,  0.02721692,  0.01899456]])

 

 

对比不同的梯度下降方法

 

Stochastic descent

In [30]:

 

runExpe(orig_data, theta, 1, STOP_ITER, thresh=5000, alpha=0.001)

 

***Original data - learning rate: 0.001 - Stochastic descent - Stop: 5000 iterations
Theta: [[-0.38504802  0.09357723 -0.01034717]] - Iter: 5000 - Last cost: 1.59 - Duration: 0.42s

Out[30]:

array([[-0.38504802,  0.09357723, -0.01034717]])

 

 

有点爆炸。。。很不稳定,再来试试把学习率调小一些

In [31]:

 

runExpe(orig_data, theta, 1, STOP_ITER, thresh=15000, alpha=0.000002)

 

***Original data - learning rate: 2e-06 - Stochastic descent - Stop: 15000 iterations
Theta: [[-0.00202012  0.01009114  0.00103943]] - Iter: 15000 - Last cost: 0.63 - Duration: 1.10s

Out[31]:

array([[-0.00202012,  0.01009114,  0.00103943]])

 

 

速度快，但稳定性差，需要很小的学习率

 

Mini-batch descent

In [33]:

 

runExpe(orig_data, theta, 16, STOP_ITER, thresh=15000, alpha=0.001)

 

***Original data - learning rate: 0.001 - Mini-batch (16) descent - Stop: 15000 iterations
Theta: [[-1.0352224   0.01668297  0.0124234 ]] - Iter: 15000 - Last cost: 0.57 - Duration: 1.44s

Out[33]:

array([[-1.0352224 ,  0.01668297,  0.0124234 ]])

 

 

浮动仍然比较大，我们来尝试下对数据进行标准化 将数据按其属性(按列进行)减去其均值，然后除以其方差。最后得到的结果是，对每个属性/每列来说所有数据都聚集在0附近，方差值为1

In [34]:

 

from sklearn import preprocessing as pp
​
scaled_data = orig_data.copy()
scaled_data[:, 1:3] = pp.scale(orig_data[:, 1:3])
​
runExpe(scaled_data, theta, n, STOP_ITER, thresh=5000, alpha=0.001)

 

***Scaled data - learning rate: 0.001 - Gradient descent - Stop: 5000 iterations
Theta: [[ 0.3080807   0.86494967  0.77367651]] - Iter: 5000 - Last cost: 0.38 - Duration: 1.13s

Out[34]:

array([[ 0.3080807 ,  0.86494967,  0.77367651]])

 

 

它好多了！原始数据，只能达到达到0.61，而我们得到了0.38个在这里！ 所以对数据做预处理是非常重要的

In [25]:

 

runExpe(scaled_data, theta, n, STOP_GRAD, thresh=0.02, alpha=0.001)

 

***Scaled data - learning rate: 0.001 - Gradient descent - Stop: gradient norm < 0.02
Theta: [[ 1.071  2.63   2.411]] - Iter: 59422 - Last cost: 0.22 - Duration: 12.00s

Out[25]:

array([[ 1.071,  2.63 ,  2.411]])

 

 

更多的迭代次数会使得损失下降的更多！

In [35]:

 

theta = runExpe(scaled_data, theta, 1, STOP_GRAD, thresh=0.002/5, alpha=0.001)

 

***Scaled data - learning rate: 0.001 - Stochastic descent - Stop: gradient norm < 0.0004
Theta: [[ 1.14848169  2.79268789  2.5667383 ]] - Iter: 72637 - Last cost: 0.22 - Duration: 7.05s

 

 

随机梯度下降更快，但是我们需要迭代的次数也需要更多，所以还是用batch的比较合适！！！

In [36]:

 

runExpe(scaled_data, theta, 16, STOP_GRAD, thresh=0.002*2, alpha=0.001)

 

***Scaled data - learning rate: 0.001 - Mini-batch (16) descent - Stop: gradient norm < 0.004
Theta: [[ 1.17096801  2.83171736  2.61095087]] - Iter: 3940 - Last cost: 0.21 - Duration: 0.50s

Out[36]:

array([[ 1.17096801,  2.83171736,  2.61095087]])

 

 

精度

In [38]:

 

#设定阈值
def predict(X, theta):
    return [1 if x >= 0.5 else 0 for x in model(X, theta)]

 

In [40]:

 

scaled_X = scaled_data[:, :3]
y = scaled_data[:, 3]
predictions = predict(scaled_X, theta)
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%