我一直在使用
hmisc
包中的一个名为 haven_labelled
的类(有时只是 labelled
)。其目的是从 Stata .dta
数据集中导入列标签。当尝试在数据帧上使用 plm
时,我收到错误:
Error in as.data.frame.default(x[[i]], optional = TRUE) :
cannot coerce class ‘c("pseries", "haven_labelled")’ to a data.frame
课程如下:
> class(actualdataset)
[1] "pdata.frame" "data.frame"
> class(actualdataset$examplevar)
[1] "pseries" "haven_labelled"
因此,我想从该数据库中删除
haven_labelled
类。遗憾的是我无法重现该错误。我认为这与我的 var
属于双重类别(其中包括有 actualdataset
)有关。请参阅以下示例数据集。haven_labelled
有什么建议吗?
编辑:我在想以下事情:
library(data.table)
library(plm)
library(Hmisc)
set.seed(1)
DT <- data.table(panelID = sample(50,50), # Creates a panel ID
Country = c(rep("A",30),rep("B",50), rep("C",20)),
some_NA = sample(0:5, 6),
some_NA_factor = sample(0:5, 6),
Group = c(rep(1,20),rep(2,20),rep(3,20),rep(4,20),rep(5,20)),
Time = rep(seq(as.Date("2010-01-03"), length=20, by="1 month") - 1,5),
norm = round(runif(100)/10,2),
Income = sample(100,100),
Happiness = sample(10,10),
Sex = round(rnorm(10,0.75,0.3),2),
Age = round(rnorm(10,0.75,0.3),2),
Educ = round(rnorm(10,0.75,0.3),2))
DT [, uniqueID := .I] # Creates a unique ID
DT[DT == 0] <- NA # https://stackoverflow.com/questions/11036989/replace-all-0-values-to-na
DT$some_NA_factor <- factor(DT$some_NA_factor)
labels <- data.table::fread("Varcode Variables
panelID a
Country b
Group c
Time d
norm e
Income f
Happiness g
Sex h
Age i
Educ j
uniqueID k
", header = TRUE)
for (i in seq_len(ncol(DT))) {
label(DT[[i]]) <- labels$Variables[match(names(DT)[i], labels$Varcode)]
}
DTp <- plm::pdata.frame(DT, index= c("panelID", "Time"))
result <- plm(Happiness ~ Income, data=DTp, model="within")
> class(DTp)
[1] "pdata.frame" "data.frame"
> class(DTp$Income)
[1] "pseries" "labelled" "integer"
编辑2:答案可以防止应用时出现任何错误
for for (i in seq_len(ncol(DT)) {
if (sapply(DT, function(x) class(x)[1L]) == "haven_labelled") {
attr(DT[,i],"class[1L]") <- "integer"
}
}
。遗憾的是,所有
plm
、coefficients
都为零。 standard errors
和 P-values
是 t-values
。我不确定是什么原因造成的。,根据您的原始数据集更改NA
和
labelled
labelled_ch
for (i in seq_len(ncol(DTp))) {
if (any(class(DTp[,i]) == "labelled")) {
#browser()
ind = which(class(DTp[,i])=="labelled")
attr(DTp[,i],"class")[ind] <- "labelled_ch"
}
}
)。我在当前的 R 版本中总是遇到这样的错误:
avector
感谢您使用
Can't convert `y` <double> to match type of `x` <avector>.
函数的想法。这是我的成果:
class()
这可以与 tidyverse 一起用于数据帧
simplify_classes = function(x, strip=c("avector")) {
# We may handle single variables ...
if ((length(class(x)) > 1) || (class(x) != "data.frame")) {
class(x) = setdiff(class(x), strip)
return(x)
}
# ... or a data.frames of multiple variables
dfn = data.frame(row.names = row.names(x))
for (v in names(x)) {
tmp = x[[v]]
class(tmp) = setdiff(class(tmp), strip)
dfn[[v]] = tmp
}
return(dfn)
}
对于单个变量也是如此
my_data %>% select(!!items) %>% simplify_classes() %>% str()
希望这可以节省您一些时间。