我想使用quantreg包从分位数回归中提取系数和上限和下限。以下是帮助文件中的示例。
data(engel)
attach(engel)
taus <- c(.05,.1,.25,.75,.9,.95)
f <- rq((foodexp)~(income),tau=taus)
sf <- summary(f)
sf[1]
#[[1]]
#Call: rq(formula = (foodexp) ~ (income), tau = taus)
#tau: [1] 0.05
#Coefficients:
# coefficients lower bd upper bd
#(Intercept) 124.88004 98.30212 130.51695
#income 0.34336 0.34333 0.38975
我知道我可以使用coefficients()来获取系数。
cf <- t(data.frame(coefficients(f))) # transpose for better arrangement
cf
# (Intercept) income
#tau..0.05 124.88004 0.3433611
#tau..0.10 110.14157 0.4017658
#tau..0.25 95.48354 0.4741032
#tau..0.75 62.39659 0.6440141
#tau..0.90 67.35087 0.6862995
#tau..0.95 64.10396 0.7090685
但我无法弄清楚如何获得summary()中出现的上限/下限。我看了str(sf)，但我没看到如何提取。
最后，我想在数据帧中加上taus，coefficient和upper / lower bound进行进一步处理。
答案 0 :(得分：2)
我假设你只想要非截距词的系数。这个怎么样
sapply(sf, function(x) c(tau=x$tau, x$coefficients[-1, ]))
这将迭代tau的不同级别并提取系数的间隔
[,1] [,2] [,3] [,4] [,5] [,6]
tau 0.0500000 0.1000000 0.2500000 0.7500000 0.9000000 0.9500000
coefficients 0.3433611 0.4017658 0.4741032 0.6440141 0.6862995 0.7090685
lower bd 0.3433270 0.3420992 0.4203298 0.5801552 0.6493680 0.6739000
upper bd 0.3897500 0.4507941 0.4943288 0.6904127 0.7422294 0.7344405
答案 1 :(得分：1)
您可以将函数coef与summary返回的对象一起使用来提取值。
library(quantreg)
f <- rq(stack.loss ~ stack.x,.5)
sf <- summary(f)
sf
# Call: rq(formula = stack.loss ~ stack.x, tau = 0.5)
# tau: [1] 0.5
# Coefficients:
# coefficients lower bd upper bd
# (Intercept) -39.68986 -41.61973 -29.67754
# stack.xAir.Flow 0.83188 0.51278 1.14117
# stack.xWater.Temp 0.57391 0.32182 1.41090
# stack.xAcid.Conc. -0.06087 -0.21348 -0.02891
coef(sf)
# coefficients lower bd upper bd
# (Intercept) -39.68985507 -41.6197317 -29.67753515
# stack.xAir.Flow 0.83188406 0.5127787 1.14117115
# stack.xWater.Temp 0.57391304 0.3218235 1.41089812
# stack.xAcid.Conc. -0.06086957 -0.2134829 -0.02891341
这里，coef返回一个矩阵。下限和上限分别位于第二列和第三列。
答案 2 :(得分：1)
这是整洁解决方案，需要扫帚：
library("broom")
tidy(rq(data = engel, foodexp ~ income, tau = c(.05,.1,.25,.75,.9,.95)))
term estimate conf.low conf.high tau
1 (Intercept) 124.8800408 96.9260199 131.1526572 0.05
2 income 0.3433611 0.3256521 0.3957653 0.05
3 (Intercept) 110.1415742 74.7176192 151.7365540 0.10
4 income 0.4017658 0.3399444 0.4542858 0.10
5 (Intercept) 95.4835396 67.5662860 134.9199638 0.25
6 income 0.4741032 0.4168149 0.5112239 0.25
7 (Intercept) 62.3965855 27.8317452 120.0708811 0.75
8 income 0.6440141 0.5738877 0.7051561 0.75
9 (Intercept) 67.3508721 34.4846868 114.7983532 0.90
10 income 0.6862995 0.6408633 0.7430180 0.90
11 (Intercept) 64.1039632 44.4751575 107.6600046 0.95
12 income 0.7090685 0.6444755 0.7381036 0.95
请参阅：https://cran.r-project.org/web/packages/broom/vignettes/broom.html
答案 3 :(得分：0)
这可能会有所帮助。
matrix(unlist(lapply(sf, function(x) data.frame(unlist(x)[3:8]))), nrow=6, byrow=TRUE)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 124.88004 0.3433611 98.30212 0.3433270 130.51695 0.3897500
[2,] 110.14157 0.4017658 79.88753 0.3420992 146.18875 0.4507941
[3,] 95.48354 0.4741032 73.78608 0.4203298 120.09847 0.4943288
[4,] 62.39659 0.6440141 32.74488 0.5801552 107.31362 0.6904127
[5,] 67.35087 0.6862995 37.11802 0.6493680 103.17399 0.7422294
[6,] 64.10396 0.7090685 46.26495 0.6739000 83.57896 0.7344405
或
matrix(unlist(lapply(sf, function(x) data.frame(unlist(x)[3:8]))), nrow=6, byrow=FALSE)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 124.8800408 110.1415742 95.4835396 62.3965855 67.3508721 64.1039632
[2,] 0.3433611 0.4017658 0.4741032 0.6440141 0.6862995 0.7090685
[3,] 98.3021170 79.8875277 73.7860765 32.7448768 37.1180206 46.2649456
[4,] 0.3433270 0.3420992 0.4203298 0.5801552 0.6493680 0.6739000
[5,] 130.5169478 146.1887545 120.0984745 107.3136214 103.1739898 83.5789592
[6,] 0.3897500 0.4507941 0.4943288 0.6904127 0.7422294 0.7344405