我使用plm处理固定效应回归模型。
该模型如下所示：
FE.model <-plm(fml, data = data.reg2,
index=c('Site.ID','date.hour'), # cross section ID and time series ID
model='within', #coefficients are fixed
effect='individual')
summary(FE.model)
＆＃34; FML＆＃34;是我之前定义的公式。我有很多自变量，所以这使它更有效率。
我想要做的是获取我的拟合值（我的yhats）并将它们连接到我的基础数据集; data.reg2
我可以使用以下代码获取拟合值：
Fe.model.fitted <- FE.model$model[[1]] - FE.model$residuals
然而，这只给我一个拟合值的一列向量 - 我无法将它连接到我的基础数据集。
或者，我尝试过这样的事情：
Fe.model.fitted <- cbind(data.reg2, resid=resid(FE.model), fitted=fitted(FE.model))
但是，我得到了这个错误：
Error in as.data.frame.default(x[[i]], optional = TRUE) : cannot coerce class ""pseries"" to a data.frame
还有其他方法可以在我的基础数据集中获取拟合值吗？或者有人可以解释我得到的错误，也许可以解决它？
我应该注意，我不想根据我的测试版手动计算yhats。我对该选项有太多自变量，我定义的公式（fml）可能会改变，因此选项效率不高。
非常感谢!!
答案 0 :(得分：4)
将plm拟合值合并回原始数据集需要一些中间步骤 - plm删除任何缺少数据的行，据我所知，plm对象执行不包含索引信息。数据的顺序不保留（参见Millo Giovanni在this thread中的评论：＆＃34;输入顺序并不总是保留＆＃34;）。
简而言之：
plm对象中获取拟合值。它是单个向量，但条目名称。名称对应于索引中的位置。index()功能获取索引。它可以返回个人和时间指数。请注意，如果删除了缺少数据的行，则索引可能包含的行数多于拟合值。 （也可以直接从原始数据生成索引，但我没有看到保证在plm返回的内容中保留数据的原始顺序。）下面提供了示例代码。有点长，但我试图评论。代码没有优化，我的目的是明确列出步骤。另外，我使用data.table而不是data.frame s。
library(data.table); library(plm)
### Generate dummy data. This way we know the "true" coefficients
set.seed(100)
n <- 500 # Run with more data if you want to get closer to the "true" coefficients
DT <- data.table(CJ(id = c("a","b","c","d","e"), time = c(1:(n / 5))))
DT[, x1 := rnorm(n)]
DT[, x2 := rnorm(n)]
DT[, y := x1 + 2 * x2 + rnorm(n) / 10]
setkey(DT, id, time)
# # Make it an unbalanced panel & put in some NAs
DT <- DT[!(id == "a" & time == 4)]
DT[.("a", 3), x2 := as.numeric(NA)]
DT[.("d", 2), x2 := as.numeric(NA)]
str(DT)
### Run the model -- both individual and time effects; "within" model
summary(PLM <- plm(data = DT, id = c("id", "time"), formula = y ~ x1 + x2, model = "within", effect = "twoways", na.action = "na.omit"))
### Merge the fitted values back into the data.table DT
# Note that PLM$model$y is shorter than the data, i.e. the row(s) with NA have been dropped
cat("\nRows omitted (due to NA): ", nrow(DT) - length(PLM$model$y))
# Since the objects returned by plm() do not contain the index, need to generate it from the data
# The object returned by plm(), i.e. PLM$model$y, has names that point to the place in the index
# Note: The index can also be done as INDEX <- DT[, j = .(id, time)], but use the longer way with index() in case plm does not preserve the order
INDEX <- data.table(index(x = pdata.frame(x = DT, index = c("id", "time")), which = NULL)) # which = NULL extracts both the individual and time indexes
INDEX[, id := as.character(id)]
INDEX[, time := as.integer(time)] # it is returned as a factor, convert back to integer to match the variable type in DT
# Generate the fitted values as the difference between the y values and the residuals
if (all(names(PLM$residuals) == names(PLM$model$y))) { # this should not be needed, but just in case...
FIT <- data.table(
index = as.integer(names(PLM$model$y)), # this index corresponds to the position in the INDEX, from where we get the "id" and "time" below
fit.plm = as.numeric(PLM$model$y) - as.numeric(PLM$residuals)
)
}
FIT[, id := INDEX[index]$id]
FIT[, time := INDEX[index]$time]
# Now FIT has both the id and time variables, can match it back into the original dataset (i.e. we have the missing data accounted for)
DT <- merge(x = DT, y = FIT[, j = .(id, time, fit.plm)], by = c("id", "time"), all = TRUE) # Need all = TRUE, or some data from DT will be dropped!
答案 1 :(得分：0)
残差是模型与公式LHS值的偏差，您没有向我们展示。 'plm'包中有一个fitted.panelmodel函数，但它似乎期望fitted函数默认不返回plm函数，也没有记录这样做，也不是我认为让它咳嗽的方式。
library(plm)
data("Produc", package = "plm")
zz <- plm(log(gsp) ~ log(pcap) + log(pc) + log(emp) + unemp,
data = Produc, index = c("state","year"))
summary(zz) # the example on the plm page:
> str(fitted(zz))
NULL
> names(zz$model)
[1] "log(gsp)" "log(pcap)" "log(pc)" "log(emp)" "unemp"
> Produc[ , c("Yvar", "Fitted")] <- cbind( zz$model[ ,"log(gsp)", drop=FALSE], zz$residuals)
> str(Produc)
'data.frame': 816 obs. of 12 variables:
$ state : Factor w/ 48 levels "ALABAMA","ARIZONA",..: 1 1 1 1 1 1 1 1 1 1 ...
$ year : int 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 ...
$ pcap : num 15033 15502 15972 16406 16763 ...
$ hwy : num 7326 7526 7765 7908 8026 ...
$ water : num 1656 1721 1765 1742 1735 ...
$ util : num 6051 6255 6442 6756 7002 ...
$ pc : num 35794 37300 38670 40084 42057 ...
$ gsp : int 28418 29375 31303 33430 33749 33604 35764 37463 39964 40979 ...
$ emp : num 1010 1022 1072 1136 1170 ...
$ unemp : num 4.7 5.2 4.7 3.9 5.5 7.7 6.8 7.4 6.3 7.1 ...
$ Yvar :Classes 'pseries', 'pseries', 'integer' atomic [1:816] 10.3 10.3 10.4 10.4 10.4 ...
.. ..- attr(*, "index")='data.frame': 816 obs. of 2 variables:
.. .. ..$ state: Factor w/ 48 levels "ALABAMA","ARIZONA",..: 1 1 1 1 1 1 1 1 1 1 ...
.. .. ..$ year : Factor w/ 17 levels "1970","1971",..: 1 2 3 4 5 6 7 8 9 10 ...
$ Fitted: num -0.04656 -0.03064 -0.01645 -0.00873 -0.02708 ...
答案 2 :(得分：0)
我有一个简化的方法。这里的主要问题是双重的：
1）pdata.frames按名称排序输入，然后按年份排序。这可以通过在运行plm之前先对数据框进行排序来解决。
2）删除公式中包含的NA中包含NA的行。我通过创建第二个公式来处理这个问题，包括我的id和时间变量，然后使用model.frame来提取回归中使用的数据（不包括NA，但现在还包括id和时间）
library(plm)
set.seed(100)
n <- 10 # Run with more data if you want to get closer to the "true" coefficients
DT <- data.frame(id = c("a","c","b","d","e"), time = c(1:(n / 5)),x1 = rnorm(n),x2= rnorm(n),x3=rnorm(n))
DT$Y = DT$x2 + 2 * DT$x3 + rnorm(n) / 10 # make x1 a function of other variables
DT$x3[3]=NA # add an NA to show this works with missing data
DT
# now can add drop.index = F, but note that DT is now sorted by order(id,time)
pdata.frame(DT,index=c('id','time'),drop.index = F)
# order DT to match pdata.frame that will be used for plm
DT=DT[order(DT$id,DT$time),]
# formulas
formulas =Y~x1+x2+x3
formulas_dataframe = Y~x1+x2+x3 +id+time # add id and time for model.frame
# estimate
random <- plm(formulas, data=DT, index=c("id", "time"), model="random",na.action = 'na.omit')
summary(random)
# merge prediction and and model.frame
fitted = data.frame(fitted = random$model[[1]] - random$residuals)
model_data = cbind(as.data.frame(as.matrix(random$model)),fitted) # this isn't really needed but shows that input and model.frame are same
model_data = cbind(model_data,na.omit(model.frame(formulas_dataframe,DT)))
model_data
答案 3 :(得分：0)
在使用predict.out.plm估算第一差异或固定效果模型后，我写了一个函数（plm）来做样本预测。
该函数进一步将预测值添加到原始数据的索引。这是通过使用rownames - plm中保存的attributes(plmobject)$index和rownames
model.matrix来完成的
有关详细信息，请参阅此处发布的功能：
答案 4 :(得分：0)
这篇文章已经有一段时间了，但是我相信现在最简单的方法是：
data-bs-... Fe.model.fitted <- cbind(FE.model$model,
resid=FE.model$residuals,
fitted=plm:::fitted_exp.plm(FE.model))
包未导出函数fitted_exp.plm，但是我们可以使用plm来提取它。