如何从分位数回归rq()中提取系数的上下界

I'd like to extract coefficients and upper and lower bounds from a quantile regression using the quantreg package. Here's an example from the help file.

我想从量子回归中提取系数和上下界用量子包。这里有一个来自帮助文件的示例。

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

I know I can use coefficients() to get the coefficients.

我知道我可以用系数()来得到系数。

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

But I can't figure out how to get the upper/lower bounds that appear in summary(). I looked at str(sf), but I did not see how to extract.

但是我不知道如何获得summary()中出现的上界/下界。我看了看str(sf)，但没有看到如何提取。

Ultimately, I'd like to put taus, coefficients, and upper/lower bounds in a dataframe for further processing.

最后，我想在dataframe中放入taus、系数和上界/下界，以便进行进一步的处理。

4 个解决方案

#1

I'm assuming you just want the coefficients on the non-intercept term. How about this

我假设你只需要非截距项的系数。这个怎么样

sapply(sf, function(x) c(tau=x$tau, x$coefficients[-1, ]))

That will iterate over the different levels of tau and extract the intervals for the coefficients

它将遍历不同级别的系数并提取系数的区间

                  [,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

#2

You can use the function coef with the object returned by summary to extract the values.

您可以使用带有summary返回的对象的函数coef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

Here, coef returns a matrix. The lower and upper bounds are in the second and third column, respectively.

这里，coef返回一个矩阵。下界和上界分别位于第二和第三列。

#3

Here's a tidy solution that requires a broom:

这里有一个整洁的解决方案，需要一把扫帚:

    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

See: https://cran.r-project.org/web/packages/broom/vignettes/broom.html

参见:https://cran.r-project.org/web/packages/broom/vignettes/broom.html

#4

This might help.

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

#1