r - 在 plm 中循环子集

Question

我正在尝试在 R 中编写一些非常简单的东西(我认为)，但我似乎无法做到正确。我有一个包含 50 个国家(1 到 50 个)的数据集，每个国家 15 年，每个国家大约 20 个变量。现在我只在我的因变量( OS )上测试一个变量( SMD )。我想通过一个国家/地区的循环来执行此操作，因此我将获得每个国家/地区的输出而不是整体输出。

我认为首先创建一个子集是明智的(以便能够首先查看国家/地区 1，然后我的循环应该增加国家/地区和测试国家/地区 2 的数量)。我相信我在页面底部的回归应该给我国家 1 的输出，而不是整个数据集的总分。但是我不断收到这些错误:

> pdata <- plm.data(newdata, index=c("Country","Date"))
  series    are constants and have been removed
> pooling <- plm(Y ~ X, data=pdata, model= "pooling") 
  series Country, xRegion are constants and have been removed
  Error in model.matrix.pFormula(formula, data, rhs = 1, model = model,  : 
  NA in the individual index variable
> summary(pooling)
  Error in summary(pooling) : object 'pooling' not found

我可能认为这一切都错了，但我相信如果不让它工作，就没有必要进一步对循环本身进行编程。任何关于解决我的错误的建议，或其他编程循环的方法都非常感谢。

我的代码:

rm(list = ls())
mydata <- read.table(file = file.choose(), header = TRUE, dec = ",")
names(mydata)
attach(mydata)

Y <- cbind(SMD)
X <- cbind(OS)

newdata <- subset(mydata, Country %in% c(1))

newdata

pdata <- plm.data(newdata, index=c("Country","Date"))
pooling <- plm(Y ~ X, data=pdata, model= "pooling") 
summary(pooling)

编辑:导致相同错误的前 2 个国家的数据样本

dput(mydata) structure(list(Region = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("NAF", "SAME"), class = "factor"), Country = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), Date = c(1995L, 1996L, 1997L, 1998L, 1999L, 2000L, 2001L, 2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L, 2009L, 2010L, 2011L, 2012L, 2013L, 2014L, 1995L, 1996L, 1997L, 1998L, 1999L, 2000L, 2001L, 2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L, 2009L, 2010L, 2011L, 2012L, 2013L, 2014L ), OS = structure(c(19L, 25L, 27L, 15L, 22L, 20L, 23L, 9L, 7L, 5L, 2L, 1L, 4L, 3L, 6L, 10L, 11L, 13L, 11L, 8L, 26L, 25L, 31L, 29L, 28L, 21L, 30L, 24L, 24L, 16L, 11L, 14L, 12L, 17L, 18L, 29L, 32L, 32L, 33L, 34L), .Label = c("51.5", "52.2", "55.6", "56.4", "56.7", "57.7", "57.8", "58.3", "59", "59.2", "59.6", "59.9", "60.2", "60.4", "61.1", "61.2", "62.2", "62.3", "62.8", "63.2", "63.3", "63.8", "63.9", "64.2", "64.3", "64.5", "64.7", "65.3", "65.5", "65.6", "66.4", "68", "69.6", "70.7"), class = "factor"), SMD = structure(c(7L, 12L, 20L, 21L, 17L, 15L, 13L, 10L, 14L, 22L, 23L, 33L, 1L, 32L, 29L, 34L, 28L, 25L, NA, NA, 9L, 6L, 8L, 4L, 2L, 35L, 3L, 36L, 5L, 11L, 16L, 18L, 24L, 19L, 26L, 31L, 27L, 30L, NA, NA), .Label = c("100.3565662", "13.44788845", "13.45858747", "13.56815534", "15.05892471", "17.63789658", "18.04088718", "18.3101351", "19.34226196", "21.25530884", "21.54423145", "23.75898948", "24.08770926", "26.39817342", "29.44079001", "31.40605191", "34.46667996", "34.52913657", "35.66070947", "36.4419931", "39.16875621", "44.0126137", "45.72949566", "49.13062679", "54.83730247", "56.87886311", "59.80971583", "60.5658962", "69.20148901", "70.91362874", "72.64845214", "73.97139238", "75.20140919", "76.18378138", "9.570435019", "9.867635305"), class = "factor")), .Names = c("Region", "Country", "Date", "OS", "SMD"), class = "data.frame", row.names = c(NA, -40L))

Answer 1

您确定需要使用 plm ??这将生成按国家/地区列出的摘要列表。

# convert factors to numeric
mydata$SMD <- as.numeric(mydata$SMD)
mydata$OS  <- as.numeric(mydata$OS)

# Using lapply(...)
smry <- lapply(unique(mydata$Country),
               function(cntry)
                 summary(lm(SMD~OS,data=mydata[mydata$Country==cntry,])))
# Same thing, using for loop
smry <- list()
for (cntry in unique(mydata$Country)) {
  smry <- list(smry, 
               summary(lm(SMD~OS,data=mydata[mydata$Country==cntry,])))
}

在您的数据集中， SMD和 OS是因子，需要先转换为数字。