R 中的监督学习:回归
Nina Zumel and John Mount
Win-Vector, LLC



对正态分布而言:
model <- lm(log(y) ~ x, data = train)
model <- lm(log(y) ~ x, data = train)
logpred <- predict(model, data = test)
model <- lm(log(y) ~ x, data = train)
logpred <- predict(model, data = test)
pred <- exp(logpred)
$log(a) + log(b) = log(ab)$
$log(a) - log(b) = log(a/b)$
降低乘法误差可降低相对误差。
RMS 相对误差 = $\sqrt{ \overline{ (\frac{pred-y}{y})^2 }}$
modIncome <- lm(Income ~ AFQT + Educ, data = train)
AFQT:调查前25年的能力测试得分Educ:至调查时的受教育年限Income:调查时的收入test %>%
+ mutate(pred = predict(modIncome, newdata = test),
+ err = pred - Income) %>%
+ summarize(rmse = sqrt(mean(err^2)),
+ rms.relerr = sqrt(mean((err/Income)^2)))
| RMSE | RMS 相对误差 |
|---|---|
| 36,819.39 | 3.295189 |
modLogIncome <- lm(log(Income) ~ AFQT + Educ, data = train)
test %>%
+ mutate(predlog = predict(modLogIncome, newdata = test),
+ pred = exp(predlog),
+ err = pred - Income) %>%
+ summarize(rmse = sqrt(mean(err^2)),
+ rms.relerr = sqrt(mean((err/Income)^2)))
| RMSE | RMS 相对误差 |
|---|---|
| 38,906.61 | 2.276865 |
log(Income) 模型:RMS 相对误差更小,RMSE 更大
| 模型 | RMSE | RMS 相对误差 |
|---|---|---|
基于 Income |
36,819.39 | 3.295189 |
基于 log(Income) |
38,906.61 | 2.276865 |
R 中的监督学习:回归