机器学习作业5-偏差和方差

公号sumsmile

发布于 2021-03-12 11:57:17

43500

代码可运行

文章被收录于专栏：音视频技术学习笔记音视频技术学习笔记

运行总次数：0

代码可运行

基于偏差和方差的优化

模型训练完成，怎么直观的观察模型的好坏呢？观察训练集的方差和交叉验证的方差，可以大致判断是欠拟合还是过拟合

方差的评价

观察训练集的损失图和交叉样本集的损失图，判断优化方向

过拟合-增加正则	欠拟合-需增加多项式拟合
过拟合	欠拟合

减轻过拟合	lambda太大造成欠拟合
lambda=1	lambda=100

代码说明

这个案例中，输入是水位，输出是流速。即研究水库中水位是如何影响水的流速的。

加载数据

import numpy as np
import scipy.io as sio
import scipy.optimize as opt
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns

def load_data():
    """for ex5
    d['X'] shape = (12, 1)
    pandas has trouble taking this 2d ndarray to construct a dataframe, so I ravel
    the results
    pandas 通过二维数组构造dataframe的功能有点问题，这里将结果摊开，原始数据是列形式，转成行
    """
    d = sio.loadmat('ex5data1.mat')
    return map(np.ravel, [d['X'], d['y'], d['Xval'], d['yval'], d['Xtest'], d['ytest']])

X, y, Xval, yval, Xtest, ytest = load_data()
df = pd.DataFrame({'water_level':X, 'flow':y})

sns.set(context="notebook", style="darkgrid", palette=sns.color_palette("RdBu", 2))
sns.lmplot(x = 'water_level', y = 'flow', data=df, fit_reg=False, height=7)
plt.show()

# 增加一列"1"，即截距
X, Xval, Xtest = [np.insert(x.reshape(x.shape[0], 1), 0, np.ones(x.shape[0]), axis=1) for x in (X, Xval, Xtest)]

复习下代价函数

代码函数

def cost(theta, X, y):
    """
    X: R(m*n), m records, n features
    y: R(m)
    theta : R(n), linear regression parameters
    """
    m = X.shape[0]

    inner = X @ theta - y  # R(m*1)

    # 1*m @ m*1 = 1*1 in matrix multiplication
    # but you know numpy didn't do transpose in 1d array, so here is just a
    # vector inner product to itselves
    # 既然一维向量相乘不做转置处理，求的是内积，那这里的转置有什么意义呢？看起来一维向量不需要转置
    square_sum = inner.T @ inner
    cost = square_sum / (2 * m)

    return cost

theta = np.ones(X.shape[1])
cost(theta, X, y)
303.9515255535976

复习下梯度 偏导求梯度

求解运算

def gradient(theta, X, y):
    m = X.shape[0]

    inner = X.T @ (X @ theta - y)  # (m,n).T @ (m, 1) -> (n, 1)

    return inner / m

gradient(theta, X, y)
array([-15.30301567, 598.16741084])
# 这里的theta向量是两个参数，即一维的线性拟合

正则化梯度

def regularized_gradient(theta, X, y, l=1):
    m = X.shape[0]

    regularized_term = theta.copy()  # same shape as theta
    regularized_term[0] = 0  # don't regularize intercept theta

    regularized_term = (l / m) * regularized_term

    return gradient(theta, X, y) + regularized_term
regularized_gradient(theta, X, y)
array([-15.30301567, 598.25074417])
# l = 1时，求第一次的梯度差别不大

拟合数据

正则项

def linear_regression_np(X, y, l=1):
    """linear regression
    args:
        X: feature matrix, (m, n+1) # with incercept x0=1
        y: target vector, (m, )
        l: lambda constant for regularization

    return: trained parameters
    """
    # init theta
    theta = np.ones(X.shape[1])

    # train it
    res = opt.minimize(fun=regularized_cost,
                       x0=theta,
                       args=(X, y, l),
                       method='TNC',
                       jac=regularized_gradient,
                       options={'disp': True})
    return res

def regularized_cost(theta, X, y, l=1):
    m = X.shape[0]

    regularized_term = (l / (2 * m)) * np.power(theta[1:], 2).sum()

    return cost(theta, X, y) + regularized_term

theta = np.ones(X.shape[0])

final_theta = linear_regression_np(X, y, l=0).get('x')

b = final_theta[0] # intercept
m = final_theta[1] # slope

plt.scatter(X[:,1], y, label="Training data")
plt.plot(X[:, 1], X[:, 1]*m + b, label="Prediction")
plt.legend(loc=2)
plt.show()

线性拟合

线性拟合

1.使用训练集的子集来拟合应模型 2.在计算训练代价和交叉验证代价时，没有用正则化 3.记住使用相同的训练集子集来计算训练代价

m = X.shape[0]
for i in range(1, m+1):
#     print('i={}'.format(i))
    res = linear_regression_np(X[:i, :], y[:i], l=0)
    
    tc = regularized_cost(res.x, X[:i, :], y[:i], l=0)
    cv = regularized_cost(res.x, Xval, yval, l=0)
#     print('tc={}, cv={}'.format(tc, cv))
    
    training_cost.append(tc)
    cv_cost.append(cv)

plt.plot(np.arange(1, m+1), training_cost, label='training cost')
plt.plot(np.arange(1, m+1), cv_cost, label='cv cost')
plt.legend(loc=1)
plt.show()

拟合不太好，欠拟合

多项式拟合

注意asMatrix新版本里已经去掉了，用values替代；

def prepare_poly_data(*args, power):
    """
    args: keep feeding in X, Xval, or Xtest
        will return in the same order
    """
    def prepare(x):
        # expand feature
        df = poly_features(x, power=power)

        # normalization
        ndarr = normalize_feature(df).values

        # add intercept term
        return np.insert(ndarr, 0, np.ones(ndarr.shape[0]), axis=1)

    return [prepare(x) for x in args]

def poly_features(x, power, as_ndarray=False):
    data = {'f{}'.format(i): np.power(x, i) for i in range(1, power + 1)}
    df = pd.DataFrame(data)

    return df.values if as_ndarray else df

X, y, Xval, yval, Xtest, ytest = load_data()
poly_features(X, power=3)

扩展多项式特征

准备多项式回归数据

扩展特征到 8阶,或者你需要的阶数
使用 归一化 来合并

don't forget intercept term

def normalize_feature(df):
    """Applies function along input axis(default 0) of DataFrame."""
    return df.apply(lambda column: (column - column.mean()) / column.std())

X_poly, Xval_poly, Xtest_poly= prepare_poly_data(X, Xval, Xtest, power=8)
X_poly[:3, :]

画出学习曲线

首先，我们没有使用正则化，所以

plt.legend(loc = 1)，loc设置"training cost/cv cost的位置"

def plot_learning_curve(X, y, Xval, yval, l=0):
    training_cost, cv_cost = [], []
    m = X.shape[0]

    for i in range(1, m + 1):
        # regularization applies here for fitting parameters
        res = linear_regression_np(X[:i, :], y[:i], l=l)

        # remember, when you compute the cost here, you are computing
        # non-regularized cost. Regularization is used to fit parameters only
        tc = cost(res.x, X[:i, :], y[:i])
        cv = cost(res.x, Xval, yval)

        training_cost.append(tc)
        cv_cost.append(cv)

    plt.plot(np.arange(1, m + 1), training_cost, label='training cost')
    plt.plot(np.arange(1, m + 1), cv_cost, label='cv cost')
    plt.legend(loc=1)

plot_learning_curve(X_poly, y, Xval_poly, yval, l=0)
plt.show()

训练的代价太低，不真实，这是过拟合了，即泛化的能力很差

plot_learning_curve(X_poly, y, Xval_poly, yval, l=1)
plt.show()

训练代价增加了些，不再是0了。也就是说我们减轻过拟合

plot_learning_curve(X_poly, y, Xval_poly, yval, l=100)
plt.show()

过度正则化，欠拟合

找到最佳的

l_candidate = [0, 0.001, 0.003, 0.01, 0.03, 0.1, 0.3, 1, 3, 10]
training_cost, cv_cost = [], []

for l in l_candidate:
    res = linear_regression_np(X_poly, y, l)
    
    tc = cost(res.x, X_poly, y)
    cv = cost(res.x, Xval_poly, yval)
    
    training_cost.append(tc)
    cv_cost.append(cv)

plt.plot(l_candidate, training_cost, label='training')
plt.plot(l_candidate, cv_cost, label='cross validation')
plt.legend(loc=2)

plt.xlabel('lambda')

plt.ylabel('cost')
plt.show()

# best cv I got from all those candidates
l_candidate[np.argmin(cv_cost)]
1

# use test data to compute the cost
for l in l_candidate:
    theta = linear_regression_np(X_poly, y, l).x
    print('test cost(l={}) = {}'.format(l, cost(theta, Xtest_poly, ytest)))

test cost(l=0) = 10.804375286491785
test cost(l=0.001) = 10.911365745177878
test cost(l=0.003) = 11.265060784108712
test cost(l=0.01) = 10.879143763702967
test cost(l=0.03) = 10.022378551698187
test cost(l=0.1) = 8.631776100446476
test cost(l=0.3) = 7.3365081011786275
test cost(l=1) = 7.466282452677015
test cost(l=3) = 11.643940740451052
test cost(l=10) = 27.715080273166386

调参后，