简而言之，你无法得到你想要的.这是因为在 R 中，数字形式的日期必须包含一天.

The short answer is that you can't get exactly what you want. This is because in R, dates with numeric form must include a day.

为了在 R 中成功导入 Stata 日期，您首先可以转换相应的Stata 中从每月到日期时间的变量:

For successfully importing a Stata date in R, you first can convert the respective variable in Stata from a monthly to a date-time one:

clear
set obs 1

generate date = monthly("2000-Jan", "YM")

display %tmCCYY-Mon date
2000-Jan

display date
480

replace date = dofm(date)

display %tdCCYY-Mon date
2000-Jan

display date
14610

replace date = cofd(date) + tc(00:00:35)

display %tc date
01jan2000 00:01:40

display %15.0f date
1262304100352

然后在 R 中您可以执行以下操作:

Then in R you can do the following:

statadatetime <-  1262304100352

rdatetime <- as.POSIXct(statadatetime/1000, origin = "1960-01-01")
rdatetime
[1] "2000-01-01 02:01:40 EET"

typeof(rdatetime)
[1] "double"

rdate <- as.Date(rdatetime)
rdate
[1] "2000-01-01"

typeof(rdate)
[1] "double"

您可以通过以下方式获得所需的年-(缩写)月形式:

You can get the Year-(abbreviated) Month form you want with the following:

rdate = format(rdate,"%Y-%b")
[1] "2000-Jan"

typeof(rdate)
[1] "character"

然而，正如你所看到的，这将改变 rdate 持有的类型日期.

However, as you can see, this will change the type of rdate holding the date.

尝试将其改回:

rdate <- as.Date(rdate)
Error in charToDate(x) : 
  character string is not in a standard unambiguous format