r - 增强回归树 - 偏差值

Question

我正在使用 R 中的 gbm 包为以下模型拟合 BRT 模型:

离地高度~年龄+季节+栖息地+一天中的时间

离地高度是一个连续变量，一天中的时间也是如此。季节和栖息地是二项式变量。

我的偏差非常高，但我不知道为什么...... 有人可以帮我设置参数吗？

> M1 <- gbm.step(data=data, gbm.x = 2:5, gbm.y = 1,
+                family = "gaussian", tree.complexity = 4,
+                learning.rate = 0.01, bag.fraction = 0.50,
+                tolerance.method = "fixed",
+                tolerance = 0.01)


 GBM STEP - version 2.9 

Performing cross-validation optimisation of a boosted regression tree model 
for HAG and using a family of gaussian 
Using 15439 observations and 4 predictors 
creating 10 initial models of 50 trees 

 folds are unstratified 
total mean deviance =  55368.22 
tolerance is fixed at  0.01 
ntrees resid. dev. 
50    51050.65 
now adding trees... 
100   48935.65 
150   47805.14 
200   47193.43 
250   46841.71 
300   46631.33 
350   46498.56 
400   46418.58 
450   46371.7 
500   46336.54 
550   46317.53 
600   46309.25 
650   46300.57 
700   46296.82 
750   46297 
800   46299.11 
850   46297.7 
900   46298.34 
950   46292.32 
1000   46297.62 
1050   46295.78 
1100   46301.32 
1150   46306.59 
1200   46312.55 
1250   46314.67 
1300   46318.64 
1350   46321.38 
1400   46324.33 
1450   46322.9 
fitting final gbm model with a fixed number of 950 trees for HAG

mean total deviance = 55368.21 
mean residual deviance = 45913.34 

estimated cv deviance = 46292.32 ; se = 1366.501 

training data correlation = 0.413 
cv correlation =  0.406 ; se = 0.008 

elapsed time -  0.02 minutes 

Answer 1

gbm 中的偏差是均方误差，它将取决于因变量所处的范围。

例如:

library(dismo)
library(mlbench)
data(BostonHousing)
idx=sample(nrow(BostonHousing),400)
TrnData = BostonHousing[idx,]
TestData = BostonHousing[-idx,]

因变量是最后一列“medv”，因此我们对原始数据运行 gbm:

gbm_0 = gbm.step(data=TrnData,gbm.x=1:13,gbm.y=14,family="gaussian")

mean total deviance = 84.02 
mean residual deviance = 7.871 

estimated cv deviance = 13.959 ; se = 1.909 

training data correlation = 0.952 
cv correlation =  0.916 ; se = 0.012 

您可以看到平均偏差也可以根据残差计算(即 y - y 预测):

mean(gbm_0$residuals^2)
[1] 7.871158

使用 testData(模型尚未接受过训练)总是好的。您还可以使用相关性或 MAE(平均绝对误差)检查它与实际数据的接近程度:

pred = predict(gbm_0,TestData,1000)    
# or pearson if you like
cor(pred,TestData$medv,method="spearman")
[1] 0.8652737
# MAE
mean(abs(TestData$medv-pred))
[1] 2.75325

想象一下，良好的相关性意味着您的预测平均偏差 3。

现在，如果您更改因变量的规模，则根据您对相关性或 MAE 的解释而产生的偏差将保持不变:

TrnData$medv = TrnData$medv*2
TestData$medv = TestData$medv*2
gbm_2 = gbm.step(data=TrnData,gbm.x=1:13,gbm.y=14,family="gaussian")

mean total deviance = 336.081 
mean residual deviance = 30.983 

estimated cv deviance = 57.52 ; se = 10.254 

training data correlation = 0.953 
cv correlation =  0.911 ; se = 0.019 

elapsed time -  0.2 minutes

pred = predict(gbm_2,TestData,1000)    
cor(pred,TestData$medv,method="spearman")
[1] 0.8676821
mean(abs(TestData$medv-pred))
[1] 5.47673