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機器學習實戰之梯度下降求解邏輯回歸

機器學習實戰之梯度下降求解邏輯回歸

開發環境

軟體:jupyter-notebook

環境:python3

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

path = 'data' + os.sep +'LogiReg_data.txt'
pdData = pd.read_csv(path, header = None, names=['Exam 1','Exam 2','Admitted'])
pdData.head()
pdData.shape
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')
           
機器學習實戰之梯度下降求解邏輯回歸

The logistic regression

目标:建立分類器(求解出三個參數 θ0θ1θ2)

設定門檻值,根據門檻值判斷錄取結果

要完成的子產品

sigmoid : 映射到機率的函數

model : 傳回預測結果值

cost : 根據參數計算損失

gradient : 計算每個參數的梯度方向

descent : 進行參數更新

accuracy: 計算精度
           

sigmoid 函數

機器學習實戰之梯度下降求解邏輯回歸
def sigmoid(z):
    return 1 / (1 + np.exp(-z))
           
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')
           
機器學習實戰之梯度下降求解邏輯回歸

Sigmoid函數

g:ℝ→[0,1]

g(0)=0.5

g(−∞)=0

g(+∞)=1

def model(X, theta):
    return sigmoid(np.dot(X, theta.T))
           
機器學習實戰之梯度下降求解邏輯回歸
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])
           

損失函數

将對數似然函數去負号

機器學習實戰之梯度下降求解邏輯回歸

求平均

機器學習實戰之梯度下降求解邏輯回歸
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))
           

計算梯度

機器學習實戰之梯度下降求解邏輯回歸
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

比較三種不同的梯度下降方法

STOP_ITER = 0
STOP_COST = 1
STOP_GRAD = 2

def stopCriterion(type, value, threshold):
    #設定三種不同的停止政策
    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
           
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
           
import time

def descent(data, theta, batchSize, stopType, thresh, alpha):
    #梯度下降求解
    
    init_time = time.time()
    i = 0 # 疊代次數
    k = 0 # batch
    X, y = shuffleData(data)
    grad = np.zeros(theta.shape) # 計算的梯度
    costs = [cost(X, y, theta)] # 損失值

    
    while True:
        grad = gradient(X[k:k+batchSize], y[k:k+batchSize], theta)
        k += batchSize #取batch數量個資料
        if k >= n: 
            k = 0 
            X, y = shuffleData(data) #重新洗牌
        theta = theta - alpha*grad # 參數更新
        costs.append(cost(X, y, theta)) # 計算新的損失
        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')
    return theta
           

不同的停止政策 設定疊代次數

選擇的梯度下降方法是基于所有樣本的
n=100
runExpe(orig_data, theta, n, STOP_ITER, thresh=5000, alpha=0.000001)
           
機器學習實戰之梯度下降求解邏輯回歸
機器學習實戰之梯度下降求解邏輯回歸
機器學習實戰之梯度下降求解邏輯回歸

對比不同的梯度下降方法

機器學習實戰之梯度下降求解邏輯回歸
機器學習實戰之梯度下降求解邏輯回歸

Mini-batch descent

機器學習實戰之梯度下降求解邏輯回歸

浮動仍然比較大,我們來嘗試下對資料進行标準化 将資料按其屬性(按列進行)減去其均值,然後除以其方差。最後得到的結果是,對每個屬性/每列來說所有資料都聚集在0附近,方內插補點為1

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)
           
機器學習實戰之梯度下降求解邏輯回歸

精度

#設定門檻值
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))
           

accuracy = 89%