逻辑回归和梯度下降简单应用案例

Posted 2020-10-29 xuyiqing

实例：

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。

 

data.txt:

34.62365962451697,78.0246928153624,0
30.28671076822607,43.89499752400101,0
35.84740876993872,72.90219802708364,0
60.18259938620976,86.30855209546826,1
79.0327360507101,75.3443764369103,1
45.08327747668339,56.3163717815305,0
61.10666453684766,96.51142588489624,1
75.02474556738889,46.55401354116538,1
76.09878670226257,87.42056971926803,1
84.43281996120035,43.53339331072109,1
95.86155507093572,38.22527805795094,0
75.01365838958247,30.60326323428011,0
82.30705337399482,76.48196330235604,1
69.36458875970939,97.71869196188608,1
39.53833914367223,76.03681085115882,0
53.9710521485623,89.20735013750205,1
69.07014406283025,52.74046973016765,1
67.94685547711617,46.67857410673128,0
70.66150955499435,92.92713789364831,1
76.97878372747498,47.57596364975532,1
67.37202754570876,42.83843832029179,0
89.67677575072079,65.79936592745237,1
50.534788289883,48.85581152764205,0
34.21206097786789,44.20952859866288,0
77.9240914545704,68.9723599933059,1
62.27101367004632,69.95445795447587,1
80.1901807509566,44.82162893218353,1
93.114388797442,38.80067033713209,0
61.83020602312595,50.25610789244621,0
38.78580379679423,64.99568095539578,0
61.379289447425,72.80788731317097,1
85.40451939411645,57.05198397627122,1
52.10797973193984,63.12762376881715,0
52.04540476831827,69.43286012045222,1
40.23689373545111,71.16774802184875,0
54.63510555424817,52.21388588061123,0
33.91550010906887,98.86943574220611,0
64.17698887494485,80.90806058670817,1
74.78925295941542,41.57341522824434,0
34.1836400264419,75.2377203360134,0
83.90239366249155,56.30804621605327,1
51.54772026906181,46.85629026349976,0
94.44336776917852,65.56892160559052,1
82.36875375713919,40.61825515970618,0
51.04775177128865,45.82270145776001,0
62.22267576120188,52.06099194836679,0
77.19303492601364,70.45820000180959,1
97.77159928000232,86.7278223300282,1
62.07306379667647,96.76882412413983,1
91.56497449807442,88.69629254546599,1
79.94481794066932,74.16311935043758,1
99.2725269292572,60.99903099844988,1
90.54671411399852,43.39060180650027,1
34.52451385320009,60.39634245837173,0
50.2864961189907,49.80453881323059,0
49.58667721632031,59.80895099453265,0
97.64563396007767,68.86157272420604,1
32.57720016809309,95.59854761387875,0
74.24869136721598,69.82457122657193,1
71.79646205863379,78.45356224515052,1
75.3956114656803,85.75993667331619,1
35.28611281526193,47.02051394723416,0
56.25381749711624,39.26147251058019,0
30.05882244669796,49.59297386723685,0
44.66826172480893,66.45008614558913,0
66.56089447242954,41.09209807936973,0
40.45755098375164,97.53518548909936,1
49.07256321908844,51.88321182073966,0
80.27957401466998,92.11606081344084,1
66.74671856944039,60.99139402740988,1
32.72283304060323,43.30717306430063,0
64.0393204150601,78.03168802018232,1
72.34649422579923,96.22759296761404,1
60.45788573918959,73.09499809758037,1
58.84095621726802,75.85844831279042,1
99.82785779692128,72.36925193383885,1
47.26426910848174,88.47586499559782,1
50.45815980285988,75.80985952982456,1
60.45555629271532,42.50840943572217,0
82.22666157785568,42.71987853716458,0
88.9138964166533,69.80378889835472,1
94.83450672430196,45.69430680250754,1
67.31925746917527,66.58935317747915,1
57.23870631569862,59.51428198012956,1
80.36675600171273,90.96014789746954,1
68.46852178591112,85.59430710452014,1
42.0754545384731,78.84478600148043,0
75.47770200533905,90.42453899753964,1
78.63542434898018,96.64742716885644,1
52.34800398794107,60.76950525602592,0
94.09433112516793,77.15910509073893,1
90.44855097096364,87.50879176484702,1
55.48216114069585,35.57070347228866,0
74.49269241843041,84.84513684930135,1
89.84580670720979,45.35828361091658,1
83.48916274498238,48.38028579728175,1
42.2617008099817,87.10385094025457,1
99.31500880510394,68.77540947206617,1
55.34001756003703,64.9319380069486,1
74.77589300092767,89.52981289513276,1

 

 

代码：

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

# 读取本地e盘一个记录成绩的文本(随机生成的一堆数字)
path = ‘e:\\data.txt‘
# 两次考试成绩和是否被录取
pdData = pd.read_csv(path, header=None, names=[‘Exam1‘, ‘Exam2‘, ‘Admitted‘])

‘‘‘
print(pdData.shape)
(100, 3)
获得数据的维度（100行3列）
说明现在一共有100条数据
‘‘‘

‘‘‘
# 这里画一张图，大概观察下数据(见图1)
positive = pdData[pdData[‘Admitted‘] == 1]
negative = pdData[pdData[‘Admitted‘] == 0]
fig, ax = plt.subplots(figsize=(10, 5))
ax.scatter(positive[‘Exam1‘], positive[‘Exam2‘], s=30, c=‘b‘, marker=‘o‘, label=‘Admitted‘)
ax.scatter(negative[‘Exam1‘], negative[‘Exam2‘], s=30, c=‘r‘, marker=‘x‘, label=‘Not Admitted‘)
ax.legend()
ax.set_xlabel(‘Exam1 Score‘)
ax.set_ylabel(‘Exam2 Score‘)
#plt.show()
‘‘‘


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


‘‘‘
# 可以画sigmoid函数图形观察(图2):
nums = np.arange(-10, 10, step=1)
fig, ax = plt.subplots(figsize=(12, 4))
ax.plot(nums, sigmoid(nums), ‘b‘)
plt.show()
‘‘‘


# 预测函数
def model(X, theta):
    return sigmoid(np.dot(X, theta.T))


# 添加一列：全为1
pdData.insert(0, ‘Ones‘, 1)

orig_data = pdData.as_matrix()
cols = orig_data.shape[1]
X = orig_data[:, 0:cols - 1]
y = orig_data[:, cols - 1:cols]
theta = np.zeros([1, 3])


# 定义损失函数
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)))
    return np.sum(left - right) / (len(X))


# 计算梯度
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


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


# 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


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


def runExpe(data, theta, batchSize, stopType, thresh, alpha):
    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


# 选择的梯度下降方法是基于所有样本的

n = 100

# 设定迭代次数(图3)
runExpe(orig_data, theta, n, STOP_ITER, thresh=5000, alpha=0.000001)

# 根据损失值停止(图4)
runExpe(orig_data, theta, n, STOP_COST, thresh=0.000001, alpha=0.001)

# 根据梯度变化停止(图5)
runExpe(orig_data, theta, n, STOP_GRAD, thresh=0.05, alpha=0.001)
# plt.show()

 

图1

 

图2

 

图3：

 

图4：

 

图5：

 

 

精度：

from sklearn import preprocessing as pp

scaled_data = orig_data.copy()
scaled_data[:, 1:3] = pp.scale(orig_data[:, 1:3])

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

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))

#最后结论：accuracy = 89%