The data

咱們將創建一個邏輯迴歸模型來預測一個學生是否被大學錄取。假設你是一個大學系的管理員，你想根據兩次考試的結果來決定每一個申請人的錄取機會。你有之前的申請人的歷史數據，你能夠用它做爲邏輯迴歸的訓練集。對於每個培訓例子，你有兩個考試的申請人的分數和錄取決定。爲了作到這一點，咱們將創建一個分類模型，根據考試成績估計入學機率。html

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

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

(100, 3)

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')

<matplotlib.text.Text at 0x1bab4215208>

The logistic regression

目標：創建分類器（求解出三個參數 python

設定閾值，根據閾值判斷錄取結果app

要完成的模塊

sigmoid : 映射到機率的函數dom
model : 返回預測結果值函數
cost : 根據參數計算損失學習
gradient : 計算每一個參數的梯度方向atom
descent : 進行參數更新spa
accuracy: 計算精度code

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')

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

Sigmoid

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

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])

X[:5]

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]])

y[:5]

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

theta

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

X.shape, y.shape, theta.shape

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

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))

cost(X, y, theta)

0.69314718055994529

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3中不一樣梯度降低方法orm

STOP_ITER = 0
STOP_COST = 1
STOP_GRAD = 2
​
def stopCriterion(type, value, threshold):
    #設定三種不一樣的中止策略
    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

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):
    #梯度降低求解
    init_time = time.time()
    i = 0 # 迭代次數
    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數量個數據
        if k >= n: 
            k = 0 
            X, y = shuffleData(data) #從新洗牌
        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

不一樣的中止策略

設定迭代次數

#選擇的梯度降低方法是基於全部樣本的
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

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

根據損失值中止

設定閾值 1E-6, 差很少須要110 000次迭代

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

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

根據梯度變化中止

設定閾值 0.05,差很少須要40 000次迭代

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

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

對比不一樣的梯度降低方法

Stochastic descent

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

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

有點爆炸。。。很不穩定,再來試試把學習率調小一些

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

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

速度快，但穩定性差，須要很小的學習率

Mini-batch descent

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

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

浮動仍然比較大，咱們來嘗試下對數據進行標準化將數據按其屬性(按列進行)減去其均值，而後除以其方差。最後獲得的結果是，對每一個屬性/每列來講全部數據都彙集在0附近，方差值爲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)

***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

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

它好多了！原始數據，只能達到達到0.61，而咱們獲得了0.38個在這裏！因此對數據作預處理是很是重要的

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

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

更多的迭代次數會使得損失降低的更多！

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

隨機梯度降低更快，可是咱們須要迭代的次數也須要更多，因此仍是用batch的比較合適！！！

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

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

精度

#設定閾值
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%

梯度降低求解邏輯迴歸

Logistic Regression