Logistic Regression
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。
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import numpy.random
import time
import os
path = 'LogiReg_data.txt'
pdData = pd.read_csv(path, header=None, names=['Exam 1', 'Exam 2', 'Admitted'])
pdData.head()
orig_data = pdData.as_matrix()
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')
def sigmoid(z):
return 1 / (1 + np.exp(-z))
函数图像为:
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')
def model(X, theta):
return sigmoid(np.dot(X, theta.T))
关于X,Y数据的处理:
orig_data = pdData.as_matrix() #原始数据
X矩阵的构造:
cols = orig_data.shape[1]
X = orig_data[:,0:cols-1]
len1 = np.shape(X)[0]
X = np.c_[np.ones(len1),X]
y的构造：
y = orig_data[:,cols-1:cols]
theta的初始化:
theta = np.zeros([1, 3])
X.shape, y.shape, theta.shape
((100, 3), (100, 1), (1, 3))
损失函数(造价函数)：
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)))
#print(type(np.sum(left - right)))
return float(np.sum(left - right)) / (X.shape[0]) * 1.0
cost(X, y, theta)
返回结果:0.6931471805599453
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:
STOP_ITER = 0
STOP_COST = 1
STOP_GRAD = 2
theta = np.zeros([1, 3])
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
数据进行洗牌:
#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
完整的函数:
n=100
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)
print(X)
grad = np.zeros(theta.shape) # 计算的梯度
len1 = np.shape(X)[0]
X = np.c_[np.ones(len1), X]
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
len1 = np.shape(X)[0]
X = np.c_[np.ones(len1), X]
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
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')
plt.show()
return theta
#选择的梯度下降方法是基于所有样本的
runExpe(orig_data, theta, n, STOP_ITER, thresh=5000, alpha=0.000001)
根据损失值停止
设定阈值 1E-6, 差不多需要110 000次迭代
runExpe(orig_data, theta, 1, STOP_COST, thresh=0.000001, alpha=0.001)
根据梯度变化停止
设定阈值 0.05,差不多需要40 000次迭代
runExpe(orig_data, theta, n, STOP_GRAD, thresh=0.05, alpha=0.001)
Stochastic descent
runExpe(orig_data, theta, 1, STOP_ITER, thresh=5000, alpha=0.001)
有点爆炸。。。很不稳定,再来试试把学习率调小一些
runExpe(orig_data, theta, 1, STOP_ITER, thresh=15000, alpha=0.000002)
Mini-batch descent
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
from sklearn import preprocessing as pp
scaled_data = orig_data.copy()
scaled_data[:, 1:3] = pp.scale(orig_data[:, 1:3])
print(orig_data)
print('-------------------------')
runExpe(scaled_data, theta, n, STOP_ITER, thresh=5000, alpha=0.001)
runExpe(scaled_data, theta, n, STOP_GRAD, thresh=0.02, alpha=0.001)
theta = runExpe(scaled_data, theta, 1, STOP_GRAD, thresh=0.002/5, alpha=0.001)
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！！！
runExpe(scaled_data, theta, 16, STOP_GRAD, thresh=0.002*2, alpha=0.001)
精度:
#设定阈值
def predict(X, theta):
return [1 if x >= 0.5 else 0 for x in model(X, theta)]
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))
