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r - 来自 Stata 的 xtpcse - 如何在 R 中重写

转载 作者:行者123 更新时间:2023-12-01 22:44:07 25 4
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我目前正在学习 R。我以前没有 STATA 知识。

我想重新分析在 Stata 中完成的一项研究(xtpcse 线性回归与面板校正标准误差)。我在 Stata 中找不到模型或更详细的代码,也找不到任何其他提示如何在 R 中重写它。我为 R 安装了用于计量经济学的 plm 包。就我所知。

下面复制了 STATA .do 文件的第一行(我刚刚看到它非常不可读。这是我复制 .do 内容的 txt 文件的链接:http://dl.dropbox.com/u/4004629/This%20was%20in%20the%20.do%20file.txt)。我不知道如何以更好的方式解决这个问题。我尝试用谷歌搜索 STATA 和 R 比较等,但没有用。

我要复制的研究的所有数据都在这里:

https://umdrive.memphis.edu/rblanton/public/ISQ_data

---STATA---
Group variable: c_code Number of obs = 265
Time variable: year Number of groups = 27
Panels: correlated (unbalanced) Obs per group: min = 3
Autocorrelation: common AR(1) avg = 9.814815
Sigma computed by pairwise selection max = 14
Estimated covariances = 378 R-squared = 0.8604
Estimated autocorrelations = 1 Wald chi2(11) = 8321.15
Estimated coefficients = 15 Prob > chi2 = 0.0000

------------------------------------------------------------------------------
| Panel-corrected
food | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
lag_food | .8449038 .062589 13.50 0.000 .7222316 .967576
ciri | -.010843 .0222419 -0.49 0.626 -.0544364 .0327504
human_cap | .0398406 .0142954 2.79 0.005 .0118222 .0678591
worker_rts | -.1132705 .0917999 -1.23 0.217 -.2931951 .066654
polity_4 | .0113995 .014002 0.81 0.416 -.0160439 .0388429
market_size | .0322474 .0696538 0.46 0.643 -.1042716 .1687665
income | .0382918 .0979499 0.39 0.696 -.1536865 .2302701
econ_growth | .0145589 .0105009 1.39 0.166 -.0060224 .0351402
log_trade | -.3062828 .1039597 -2.95 0.003 -.5100401 -.1025256
fix_dollar | -.0351874 .1129316 -0.31 0.755 -.2565293 .1861545
fixed_xr | -.4941214 .2059608 -2.40 0.016 -.897797 -.0904457
xr_fluct | .0019044 .0106668 0.18 0.858 -.0190021 .0228109
lab_growth | .0396278 .0277936 1.43 0.154 -.0148466 .0941022
english | -.1594438 .1963916 -0.81 0.417 -.5443641 .2254766
_cons | .4179213 1.656229 0.25 0.801 -2.828227 3.66407
-------------+----------------------------------------------------------------
rho | .0819359
------------------------------------------------------------------------------

. xtpcse fab_metal lag_fab_metal ciri human_cap worker_rts polity_4 market
> income econ_growth log_trade fix_dollar fixed_xr xr_fluct lab_growth
> english, pairwise corr(ar1)

更新:

我刚刚尝试了 Vincent 的代码。我尝试了 pcse2 和 vcovBK 代码,它们都有效(尽管我不确定如何处理 vcocBK 产生的相关矩阵)。

但是,我在重现我正在重新分析的论文中的回归系数估计值时仍然遇到问题。我尽可能地遵循他们的方法,我认为我唯一缺少的步骤是 Stata“自相关:公共(public) AR(1)”中完成的部分。我正在分析的论文说:“OLS 回归使用面板校正标准误差 (Beck/Katz '95),控制每个面板内的一阶相关性(Stata 中的 corr AR1 选项)。”

我如何控制 R 中每个面板内的一阶相关性?

这是我到目前为止对我的数据所做的:

## run lm 
res.lm <- lm(total_FDI ~ ciri + human_cap + worker_rts + polity_4 + lag_total + market_size + income + econ_growth + log_trade + fixed_xr + fix_dollar + xr_fluct + english + lab_growth, data=D)
## run pcse
res.pcse <- pcse2(res.lm,groupN="c_code",groupT="year",pairwise=TRUE)

最佳答案

正如 Ramnath 提到的,pcse package将做 Stata 的 xtpcse 所做的事情。或者,您可以使用 plm package. 中的 vcovBK() 函数如果您选择后一个选项,请确保使用 cluster='time' 选项,这是 Beck & Katz (1995) 文章所建议的,也是 Stata 命令实现的。

pcse 包运行良好,但存在一些问题,使得许多直观的用户输入无法接受,尤其是当您的数据集不平衡时。您可能想尝试重写我刚才编写的函数。只需加载pcse包,加载pcse2函数,按照pcse文档中的说明使用即可。恕我直言,下面粘贴的函数比 pcse 人员提供的函数更清晰、更灵活、更健壮。简单的基准测试还表明我的版本可能比他们的快 5 到 10 倍,这对于大数据集可能很重要。

祝你好运!

library(Matrix)
pcse2 <- function(object, groupN, groupT, pairwise=TRUE){
## Extract basic model info
groupT <- tail(as.character((match.call()$groupT)), 1)
groupN <- tail(as.character((match.call()$groupN)), 1)
dat <- eval(parse(text=object$call$data))

## Sanity checks
if(!"lm" %in% class(object)){stop("Formula object must be of class 'lm'.")}
if(!groupT %in% colnames(dat)){stop(paste(groupT, 'was not found in data', object$call$data))}
if(!groupN %in% colnames(dat)){stop(paste(groupN, 'was not found in data', object$call$data))}
if(anyDuplicated(paste(dat[,groupN], dat[,groupT]))>0){stop(paste('There are duplicate groupN-groupT observations in', object$call$data))}
if(length(dat[is.na(dat[,groupT]),groupT])>0){stop('There are missing unit indices in the data.')}
if(length(dat[is.na(dat[,groupN]),groupN])>0){stop('There are missing time indices in the data.')}

## Expand model frame to include groupT, groupN, resid columns.
f <- as.formula(object$call$formula)
f.expanded <- update.formula(f, paste(". ~ .", groupN, groupT, sep=" + "))
dat.pcse <- model.frame(f.expanded, dat)
dat.pcse$e <- resid(object)

## Extract basic model info (part II)
N <- length(unique(dat.pcse[,groupN]))
T <- length(unique(dat.pcse[,groupT]))
nobs <- nrow(dat.pcse)
is.balanced <- length(resid(object)) == N * T

## If balanced dataset, calculate as in Beck & Katz (1995)
if(is.balanced){
dat.pcse <- dat.pcse[order(dat.pcse[,groupN], dat.pcse[,groupT]),]
X <- model.matrix(f, dat.pcse)
E <- t(matrix(dat.pcse$e, N, T, byrow=TRUE))
Omega <- kronecker((crossprod(E) / T), Matrix(diag(1, T)) )

## If unbalanced and pairwise, calculate as in Franzese (1996)
}else if(pairwise==TRUE){
## Rectangularize
rectangle <- expand.grid(unique(dat.pcse[,groupN]), unique(dat.pcse[,groupT]))
names(rectangle) <- c(groupN, groupT)
rectangle <- merge(rectangle, dat.pcse, all.x=TRUE)
rectangle <- rectangle[order(rectangle[,groupN], rectangle[,groupT]),]
valid <- ifelse(is.na(rectangle$e),0,1)
rectangle[is.na(rectangle)] <- 0
X <- model.matrix(f, rectangle)
X[valid==0,1] <- 0

## Calculate pcse
E <- crossprod(t(matrix(rectangle$e, N, T, byrow=TRUE)))
V <- crossprod(t(matrix(valid, N, T, byrow=TRUE)))
if (length(V[V==0]) > 0){stop("Error! A CS-unit exists without any obs or without any obs in a common period with another CS-unit. You must remove that unit from the data passed to pcse().")}
Omega <- kronecker(E/V, Matrix(diag(1, T)))

## If unbalanced and casewise, caluate based on largest rectangular subset of data
}else{
## Rectangularize
rectangle <- expand.grid(unique(dat.pcse[,groupN]), unique(dat.pcse[,groupT]))
names(rectangle) <- c(groupN, groupT)
rectangle <- merge(rectangle, dat.pcse, all.x=TRUE)
rectangle <- rectangle[order(rectangle[,groupN], rectangle[,groupT]),]
valid <- ifelse(is.na(rectangle$e),0,1)
rectangle[is.na(rectangle)] <- 0
X <- model.matrix(f, rectangle)
X[valid==0,1] <- 0

## Keep only years for which we have the max number of observations
large.panels <- by(dat.pcse, dat.pcse[,groupT], nrow) # How many valid observations per year?
if(max(large.panels) < N){warning('There is no time period during which all units are observed. Consider using pairwise estimation.')}
T.balanced <- names(large.panels[large.panels==max(large.panels)]) # Which years have max(valid observations)?
T.casewise <- length(T.balanced)
dat.balanced <- dat.pcse[dat.pcse[,groupT] %in% T.balanced,] # Extract biggest rectangular subset
dat.balanced <- dat.balanced[order(dat.balanced[,groupN], dat.balanced[,groupT]),]
e <- dat.balanced$e

## Calculate pcse as in Beck & Katz (1995)
E <- t(matrix(dat.balanced$e, N, T.casewise, byrow=TRUE))
Omega <- kronecker((crossprod(E) / T.casewise), Matrix(diag(1, T)))
}

## Finish evaluation, clean and output
salami <- t(X) %*% Omega %*% X
bread <- solve(crossprod(X))
sandwich <- bread %*% salami %*% bread
colnames(sandwich) <- names(coef(object))
row.names(sandwich) <- names(coef(object))
pcse <- sqrt(diag(sandwich))
b <- coef(object)
tstats <- b/pcse
df <- nobs - ncol(X)
pval <- 2*pt(abs(tstats), df, lower.tail=FALSE)
res <- list(vcov=sandwich, pcse=pcse, b=b, tstats=tstats, df=df, pval=pval, pairwise=pairwise,
nobs=nobs, nmiss=(N*T)-nobs, call=match.call())
class(res) <- "pcse"
return(res)
}

关于r - 来自 Stata 的 xtpcse - 如何在 R 中重写,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/5541010/

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