有人可以分享如何训练、调整（超参数）、交叉验证和测试游侠分位数回归模型以及错误评估吗？使用 iris 或波士顿住房数据集？
我问的原因是因为我无法在 Kaggle、随机博客、Youtube 上找到很多使用分位数回归的示例或演练。我遇到的大多数问题都是分类问题。
我目前正在使用分位数回归模型，但我希望看到其他示例，特别是超参数调整
问问题
46 次
1 回答
1
There are a lot of parameters for this function. Since this isn't a forum for what it all means, I really suggest that you hit up Cross Validates with questions on the how and why. (Or look for questions that may already be answered.)
library(tidyverse)
library(ranger)
library(caret)
library(funModeling)
data(iris)
#----------- setup data -----------
# this doesn't include exploration or cleaning which are both necessary
summary(iris)
df_status(iris)
#----------------- create training sample ----------------
set.seed(395280469) # for replicability
# create training sample partition (70/20 split)
tr <- createDataPartition(iris$Species,
p = .8,
list = F)
There are a lot of ways to split the data, but I tend to prefer Caret, because they word to even out factors if that's what you feed it.
#--------- First model ---------
fit.r <- ranger(Sepal.Length ~ .,
data = iris[tr, ],
write.forest = TRUE,
importance = 'permutation',
quantreg = TRUE,
keep.inbag = TRUE,
replace = FALSE)
fit.r
# Ranger result
#
# Call:
# ranger(Sepal.Length ~ ., data = iris[tr, ], write.forest = TRUE,
# importance = "permutation", quantreg = TRUE, keep.inbag = TRUE,
# replace = FALSE)
#
# Type: Regression
# Number of trees: 500
# Sample size: 120
# Number of independent variables: 4
# Mtry: 2
# Target node size: 5
# Variable importance mode: permutation
# Splitrule: variance
# OOB prediction error (MSE): 0.1199364
# R squared (OOB): 0.8336928
p.r <- predict(fit.r, iris[-tr, -1],
type = 'quantiles')
It defaults to .1, .5, and .9:
postResample(p.r$predictions[, 1], iris[-tr, 1])
# RMSE Rsquared MAE
# 0.5165946 0.7659124 0.4036667
postResample(p.r$predictions[, 2], iris[-tr, 1])
# RMSE Rsquared MAE
# 0.3750556 0.7587326 0.3133333
postResample(p.r$predictions[, 3], iris[-tr, 1])
# RMSE Rsquared MAE
# 0.6488991 0.7461830 0.5703333
To see what this looks like in practice:
# this performance is the best so far, let's see what it looks like visually
ggplot(data.frame(p.Q1 = p.r$predictions[, 1],
p.Q5 = p.r$predictions[, 2],
p.Q9 = p.r$predictions[, 3],
Actual = iris[-tr, 1])) +
geom_point(aes(x = Actual, y = p.Q1, color = "P.Q1")) +
geom_point(aes(x = Actual, y = p.Q5, color = "P.Q5")) +
geom_point(aes(x = Actual, y = p.Q9, color = "P.Q9")) +
geom_line(aes(Actual, Actual, color = "Actual")) +
scale_color_viridis_d(end = .8, "Error",
direction = -1)+
theme_bw()
# since Quantile .1 performed the best
ggplot(data.frame(p.Q9 = p.r$predictions[, 3],
Actual = iris[-tr, 1])) +
geom_point(aes(x = Actual, y = p.Q9, color = "P.Q9")) +
geom_segment(aes(x = Actual, xend = Actual,
y = Actual, yend = p.Q9)) +
geom_line(aes(Actual, Actual, color = "Actual")) +
scale_color_viridis_d(end = .8, "Error",
direction = -1)+
theme_bw()
#------------ ranger model with options --------------
# last call used default
# splitrule: variance, use "extratrees" (only 2 for this one)
# mtry = 2, use 3 this time
# min.node.size = 5, using 6 this time
# using num.threads = 15 ** this is the number of cores on YOUR device
# change accordingly --- if you don't know, drop this one
set.seed(326)
fit.r2 <- ranger(Sepal.Length ~ .,
data = iris[tr, ],
write.forest = TRUE,
importance = 'permutation',
quantreg = TRUE,
keep.inbag = TRUE,
replace = FALSE,
splitrule = "extratrees",
mtry = 3,
min.node.size = 6,
num.threads = 15)
fit.r2
# Ranger result
# Type: Regression
# Number of trees: 500
# Sample size: 120
# Number of independent variables: 4
# Mtry: 3
# Target node size: 6
# Variable importance mode: permutation
# Splitrule: extratrees
# Number of random splits: 1
# OOB prediction error (MSE): 0.1107299
# R squared (OOB): 0.8464588
This model produced similarly.
p.r2 <- predict(fit.r2, iris[-tr, -1],
type = 'quantiles')
postResample(p.r2$predictions[, 1], iris[-tr, 1])
# RMSE Rsquared MAE
# 0.4932883 0.8144309 0.4000000
postResample(p.r2$predictions[, 2], iris[-tr, 1])
# RMSE Rsquared MAE
# 0.3610171 0.7643744 0.3100000
postResample(p.r2$predictions[, 3], iris[-tr, 1])
# RMSE Rsquared MAE
# 0.6555939 0.8141144 0.5603333
The prediction was pretty similar overall, as well. This isn't a very large set of data, with few predictors. How much do they contribute?
importance(fit.r2)
# Sepal.Width Petal.Length Petal.Width Species
# 0.06138883 0.71052453 0.22956522 0.18082998
#------------ ranger model with options --------------
# drop a predictor, lower mtry, min.node.size
set.seed(326)
fit.r3 <- ranger(Sepal.Length ~ .,
data = iris[tr, -4], # dropped Sepal.Width
write.forest = TRUE,
importance = 'permutation',
quantreg = TRUE,
keep.inbag = TRUE,
replace = FALSE,
splitrule = "extratrees",
mtry = 2, # has to change (var count lower)
min.node.size = 4, # lowered
num.threads = 15)
fit.r3
# Ranger result
# Type: Regression
# Number of trees: 500
# Sample size: 120
# Number of independent variables: 3
# Mtry: 2
# Target node size: 6
# Variable importance mode: permutation
# Splitrule: extratrees
# Number of random splits: 1
# OOB prediction error (MSE): 0.1050143
# R squared (OOB): 0.8543842
The second most important predictor was removed and it improved.
p.r3 <- predict(fit.r3, iris[-tr, -c(1, 4)],
type = 'quantiles')
postResample(p.r3$predictions[, 1], iris[-tr, 1])
# RMSE Rsquared MAE
# 0.4760952 0.8089810 0.3800000
postResample(p.r3$predictions[, 2], iris[-tr, 1])
# RMSE Rsquared MAE
# 0.3738315 0.7769388 0.3250000
postResample(p.r3$predictions[, 3], iris[-tr, 1])
# RMSE Rsquared MAE
# 0.6085584 0.8032592 0.5170000
importance(fit.r3)
# almost everthing relies on Petal.Length
# Sepal.Width Petal.Length Species
# 0.08008264 0.95440333 0.32570147
于 2022-01-05T01:32:31.083 回答