机器学习算法整理（二）梯度下降求解逻辑回归 python实现

逻辑回归(Logistic regression) 

以下均为自己看视频做的笔记，自用，侵删！

还参考了：http://www.ai-start.com/ml2014/

 用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 [7]:

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

In [8]:

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[8]:

	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 [10]:

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[10]:

Text(0,0.5,'Exam 2 Score')

 

The logistic regression 

目标：建立分类器（求解出三个参数 θ0θ1θ2）

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

sigmoid 函数

In [11]:

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

In [12]:

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[12]:

[<matplotlib.lines.Line2D at 0x15117048>]

 

 

Sigmoid

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

In [13]:

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

In [14]:

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 seful for further computations  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 [15]:

X[:5]

Out[15]:

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 [16]:

y[:5]

Out[16]:

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

In [17]:

theta

Out[17]:

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

In [18]:

X.shape, y.shape, theta.shape

Out[18]:

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

损失函数

In [19]:

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 [20]:

cost(X, y, theta)

Out[20]:

0.69314718055994529

计算梯度

In [21]:

def gradient(X, y, theta): grad = np.zeros(theta.shape) # （1,3） 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 [22]:

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 [23]:

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 [24]:

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 [25]:

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 [26]:

#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: 1.52s

Out[26]:

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: 39.86s

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: 13.44s

Out[28]:

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

 

对比不同的梯度下降方法

Stochastic descent

In [29]:

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

***Original data - learning rate: 0.001 - Stochastic descent - Stop: 5000 iterations
Theta: [[-0.38585397  0.09042018 -0.01044445]] - Iter: 5000 - Last cost: 1.53 - Duration: 0.48s

Out[29]:

array([[-0.38585397,  0.09042018, -0.01044445]])

 

 

有点爆炸。。。很不稳定,再来试试把学习率调小一些

In [30]:

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

***Original data - learning rate: 2e-06 - Stochastic descent - Stop: 15000 iterations
Theta: [[-0.00201963  0.01014321  0.00107125]] - Iter: 15000 - Last cost: 0.63 - Duration: 1.70s

Out[30]:

array([[-0.00201963,  0.01014321,  0.00107125]])

 

 

速度快，但稳定性差，需要很小的学习率

Mini-batch descent

In [31]:

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: [[-1.032863    0.03624659  0.02571257]] - Iter: 15000 - Last cost: 0.97 - Duration: 2.11s

Out[31]:

array([[-1.032863  ,  0.03624659,  0.02571257]])

 

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 [32]:

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: 1.92s

Out[32]:

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 [33]:

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: [[ 1.0707921   2.63030842  2.41079787]] - Iter: 59422 - Last cost: 0.22 - Duration: 21.58s

Out[33]:

array([[ 1.0707921 ,  2.63030842,  2.41079787]])

 

更多的迭代次数会使得损失下降的更多！

In [34]:

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: [[ 1.14904527  2.7920262   2.56725991]] - Iter: 72624 - Last cost: 0.22 - Duration: 9.14s

 

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 [35]:

 

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: [[ 1.16033549  2.81496841  2.59589695]] - Iter: 2393 - Last cost: 0.22 - Duration: 0.40s

Out[35]:

array([[ 1.16033549,  2.81496841,  2.59589695]])

 

精度

In [36]:

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

 

In [37]:

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%