从主题列表中举座(Block bootstrap from subject list)

我试图有效地实现块引导技术来获得回归系数的分布。主要要点如下。

我有一个面板数据集，并说公司和年是指数。自举的每一次迭代，我希望与更换，以样本n科目。从该样品，我需要构造新的数据帧，其为rbind()为每个采样受试者所有观察的叠层，运行回归，并拉出系数。重复一串迭代，说100。

每个企业可能会被多次选择，所以我需要给它的数据多次包括在每个迭代的数据集。
使用循环和子集的方式，如下面，似乎在计算繁重。
请注意，我真正的数据帧，n和迭代次数比下面的例子大得多。

我的想法是最初通过使用受试者打破现有数据帧到一个列表中split()命令。从那里，用

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

拿到新的列表，那么也许实施quickdf从plyr包来构建新的数据帧。

例如缓慢代码：

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
...
...

Answer 1:

怎么样是这样的：

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)

Answer 2:

您也可以使用tsboot功能在boot包固定块重新采样方案。

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

Answer 3:

我发现了一个方法，使用dplyr::left_join更加简明一点，只需要大约60％为长，并给出了相同的结果，在由肖恩答案。这里是一个完全独立的例子。

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

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

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 

summary(b1)
##      R  original bootBias    bootSE   bootMed
## 1 5000 410.81557 14.78272 195.62461 413.70175
## 2 5000   5.75981  0.49301   2.42879   6.00692
## 3 5000  -0.61527 -0.13134   0.78854  -0.76452
summary(b2)
##      R  original bootBias    bootSE   bootMed
## 1 5000 410.81557 14.78272 195.62461 413.70175
## 2 5000   5.75981  0.49301   2.42879   6.00692
## 3 5000  -0.61527 -0.13134   0.78854  -0.76452

Answer 4:

该解决方案需要进行修改，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))

Answer 5:

下面是通常应比接受的答案更快的方法，将返回相同的结果，并且不依赖于其它软件包（除了boot ）。这里的关键是使用which和整数索引来构建每个data.frame复制而不是split/subset和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)

运行该接受的答案

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

让我们的眼球比较

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

蓝本

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

这些看起来相当接近。现在，更多的检查

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

和

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

最后，我们会做一个快速的基准

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

在我的电脑，这将返回

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

所以，直接索引是在分配的各个层面快2倍以上。

注意，缺少固定效应
与大多数的答案，缺少“固定效应”的问题可能出现。通常，固定效应用作对照和研究人员的爱好中的一个或几个的变量将被包含在每一个选择的观察。在这种情况下，占主导地位的，有在限制的返回结果没有（或很少）危害myIndex或myfit功能只包括在返回的矢量关心的变量。

文章来源: Block bootstrap from subject list