求解逻辑回归—-梯度下降

X[:5]

array([[  1.        ,  30.28671077,  43.89499752],
       [  1.        ,  35.84740877,  72.90219803],
       [  1.        ,  60.18259939,  86.3085521 ],
       [  1.        ,  79.03273605,  75.34437644],
       [  1.        ,  45.08327748,  56.31637178]])

y[:5]

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

theta

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

X.shape,y.shape,theta.shape

((99, 3), (99, 1), (3, 1))

目的是求平均损失的最小值

def costFunction(X,y,theta):
    left = np.multiply(-y,np.log(model(X,theta))) # 同*，元素级乘法
    right = np.multiply((1-y),np.log(1-model(X,theta)))
    return np.sum(left-right)/(len(X))

costFunction(X,y,theta)

0.69314718055994529

def gradient(X,y,theta):
    grad = np.zeros(theta.shape)
    error = np.matmul(X.T,(model(X,theta)-y))
    grad = error/len(X)
    return grad
gradient(X,y,theta)

array([[ -0.10606061],
       [-12.30538878],
       [-11.77067239]])

# 首先设定三种停止策略【迭代次数、损失值差距、梯度】
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

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) # 计算梯度
    costs = [costFunction(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
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