11

我正在尝试有效地实现块引导技术来获得回归系数的分布。主要大纲如下。

我有一个面板数据集,并说公司和年份是指数。对于引导程序的每次迭代,我希望对 n 个主题进行替换抽样。从这个样本中,我需要构建一个新的数据框,它是rbind()每个采样对象的所有观察值的堆栈,运行回归并提取系数。重复一堆迭代,比如 100 次。

  • 每个公司都可能被多次选择,所以我需要在每次迭代的数据集中多次包含它的数据。
  • 使用循环和子集方法,如下所示,在计算上似乎很繁重。
  • 请注意,对于我的真实数据框,n 和迭代次数远大于下面的示例。

我最初的想法是使用命令将现有数据框按主题分解为列表split()。从那里,使用

sample(unique(df1$subject),n,replace=TRUE)

获取新列表,然后可能quickdfplyr包中实现以构造一个新的数据框。

慢代码示例:

require(plm)
data("Grunfeld", package="plm")

firms = unique(Grunfeld$firm)
n = 10
iterations = 100
mybootresults=list()

for(j in 1:iterations){

  v = sample(length(firms),n,replace=TRUE)
  newdata = NULL

  for(i in 1:n){
    newdata = rbind(newdata,subset(Grunfeld, firm == v[i]))
  }

  reg1 = lm(value ~ inv + capital, data = newdata)
  mybootresults[[j]] = coefficients(reg1)

}

mybootresults = as.data.frame(t(matrix(unlist(mybootresults),ncol=iterations)))
names(mybootresults) = names(reg1$coefficients)
mybootresults

  (Intercept)      inv    capital
1    373.8591 6.981309 -0.9801547
2    370.6743 6.633642 -1.4526338
3    528.8436 6.960226 -1.1597901
4    331.6979 6.239426 -1.0349230
5    507.7339 8.924227 -2.8661479
...
...
4

5 回答 5

16

像这样的东西怎么样:

myfit <- function(x, i) {
   mydata <- do.call("rbind", lapply(i, function(n) subset(Grunfeld, firm==x[n])))
   coefficients(lm(value ~ inv + capital, data = mydata))
}

firms <- unique(Grunfeld$firm)

b0 <- boot(firms, myfit, 999)
于 2012-08-12T05:47:16.663 回答
5

您还可以使用带有固定块重采样方案的包中的tsboot功能。boot

require(plm)
require(boot)
data(Grunfeld)

### each firm is of length 20
table(Grunfeld$firm)
##  1  2  3  4  5  6  7  8  9 10 
## 20 20 20 20 20 20 20 20 20 20


blockboot <- function(data) 
{
 coefficients(lm(value ~ inv + capital, data = data))

}

### fixed length (every 20 obs, so for each different firm) block bootstrap
set.seed(321)
boot.1 <- tsboot(Grunfeld, blockboot, R = 99, l = 20, sim = "fixed")

boot.1    
## Bootstrap Statistics :
##      original     bias    std. error
## t1* 410.81557 -25.785972    174.3766
## t2*   5.75981   0.451810      2.0261
## t3*  -0.61527   0.065322      0.6330

dim(boot.1$t)
## [1] 99  3

head(boot.1$t)
##        [,1]   [,2]      [,3]
## [1,] 522.11 7.2342 -1.453204
## [2,] 626.88 4.6283  0.031324
## [3,] 479.74 3.2531  0.637298
## [4,] 557.79 4.5284  0.161462
## [5,] 568.72 5.4613 -0.875126
## [6,] 379.04 7.0707 -1.092860
于 2012-08-12T18:35:17.750 回答
3

这是一种通常应该比接受的答案更快的方法,返回相同的结果并且不依赖于其他包(除了boot)。这里的关键是使用which和整数索引来构造每个 data.frame 复制而不是split/subsetand do.call/rbind

# get function for boot
myIndex <- function(x, i) {
  # select the observations to subset. Likely repeated observations
  blockObs <- unlist(lapply(i, function(n) which(x[n] == Grunfeld$firm)))
  # run regression for given replicate, return estimated coefficients
  coefficients(lm(value~ inv + capital, data=Grunfeld[blockObs,]))
}

现在,引导

# get result
library(boot)
set.seed(1234)
b1 <- boot(firms, myIndex, 200)

运行接受的答案

set.seed(1234)
b0 <- boot(firms, myfit, 200)

让我们来个对比

使用索引

b1

ORDINARY NONPARAMETRIC BOOTSTRAP


Call:
boot(data = firms, statistic = myIndex, R = 200)


Bootstrap Statistics :
       original      bias    std. error
t1* 410.8155650 -6.64885086 197.3147581
t2*   5.7598070  0.37922066   2.4966872
t3*  -0.6152727 -0.04468225   0.8351341

原版

b0

ORDINARY NONPARAMETRIC BOOTSTRAP


Call:
boot(data = firms, statistic = myfit, R = 200)


Bootstrap Statistics :
       original      bias    std. error
t1* 410.8155650 -6.64885086 197.3147581
t2*   5.7598070  0.37922066   2.4966872
t3*  -0.6152727 -0.04468225   0.8351341

这些看起来很接近。现在,多一点检查

identical(b0$t, b1$t)
[1] TRUE

identical(summary(b0), summary(b1))
[1] TRUE

最后,我们将做一个快速基准测试

library(microbenchmark)
microbenchmark(index={b1 <- boot(firms, myIndex, 200)}, 
              rbind={b0 <- boot(firms, myfit, 200)})

在我的电脑上,这会返回

Unit: milliseconds
  expr      min       lq     mean   median       uq      max neval
 index 292.5770 296.3426 303.5444 298.4836 301.1119 395.1866   100
 rbind 712.1616 720.0428 729.6644 724.0777 731.0697 833.5759   100

因此,直接索引在分布的每个级别上都快 2 倍以上。

关于缺少固定效果的说明
与大多数答案一样,可能会出现缺少“固定效果”的问题。通常,固定效应用作对照,研究人员对每个选定观察值中包含的一个或几个变量感兴趣。myIndex在这种主要情况下,将ormyfit函数的返回结果限制为仅包含返回向量中感兴趣的变量没有(或很少)危害。

于 2018-03-21T20:44:18.190 回答
2

需要修改解决方案以管理固定效果。

library(boot)  # for boot
library(plm)   # for Grunfeld
library(dplyr) # for left_join

## Get the Grunfeld firm data (10 firms, each for 20 years, 1935-1954)
data("Grunfeld", package="plm")

## Create dataframe with unique firm identifier (one line per firm)
firms <- data.frame(firm=unique(Grunfeld$firm),junk=1)

## for boot(), X is the firms dataframe; i index the sampled firms
myfit <- function(X, i) {
    ## join the sampled firms to their firm-year data
    mydata <- left_join(X[i,], Grunfeld, by="firm")
    ## Distinguish between multiple resamples of the same firm
    ## Otherwise they have the same id in the fixed effects regression
    ## And trouble ensues
    mydata  <- mutate(group_by(mydata,firm,year),
                      firm_uniq4boot = paste(firm,"+",row_number())
                      )
    ## Run regression with and without firm fixed effects
    c(coefficients(lm(value ~ inv + capital, data = mydata)),
    coefficients(lm(value ~ inv + capital + factor(firm_uniq4boot), data = mydata)))
    }

set.seed(1)
system.time(b <- boot(firms, myfit, 1000))

summary(b)

summary(lm(value ~ inv + capital, data=Grunfeld))
summary(lm(value ~ inv + capital + factor(firm), data=Grunfeld))
于 2017-04-26T23:10:09.260 回答
1

我找到了一种dplyr::left_join更简洁的方法,只需要大约 60% 的时间,并且给出的结果与 Sean 的答案相同。这是一个完整的独立示例。

library(boot)  # for boot
library(plm)   # for Grunfeld
library(dplyr) # for left_join

# First get the data
data("Grunfeld", package="plm")

firms <- unique(Grunfeld$firm)

myfit1 <- function(x, i) {
  # x is the vector of firms
  # i are the indexes into x
  mydata <- do.call("rbind", lapply(i, function(n) subset(Grunfeld, firm==x[n])))
  coefficients(lm(value ~ inv + capital, data = mydata))
}

myfit2 <- function(x, i) {
  # x is the vector of firms
  # i are the indexes into x
  mydata <- left_join(data.frame(firm=x[i]), Grunfeld, by="firm")
  coefficients(lm(value ~ inv + capital, data = mydata))
}

# rbind method
set.seed(1)
system.time(b1 <- boot(firms, myfit1, 5000))
  ##  user  system elapsed 
  ## 13.51    0.01   13.62 

# left_join method
set.seed(1)
system.time(b2 <- boot(firms, myfit2, 5000))
   ## user  system elapsed 
   ## 8.16    0.02    8.26 

b1
##        original     bias    std. error
## t1* 410.8155650  9.2896499 198.6877889
## t2*   5.7598070  0.5748503   2.5725441
## t3*  -0.6152727 -0.1200954   0.7829191

b2
##        original     bias    std. error
## t1* 410.8155650  9.2896499 198.6877889
## t2*   5.7598070  0.5748503   2.5725441
## t3*  -0.6152727 -0.1200954   0.7829191
于 2016-11-30T01:46:32.427 回答