我有一个名为foo的脚本。R包含另一个脚本other。R,在同一个目录下:

#!/usr/bin/env Rscript
message("Hello")
source("other.R")

但我想让R找到另一个。R,不管当前工作目录是什么。

换句话说,就是foo。R需要知道自己的路径。我该怎么做呢?


当前回答

只是在上面的答案的基础上,作为安全检查,您可以添加一个包装器,当sys.frame(1)失败时(如果interactive() == TRUE可能会失败),或者源脚本不在主脚本所期望的位置时,它会要求用户找到文件。

fun_path = tryCatch(expr = 
                      {file.path(dirname(sys.frame(1)$ofile), "foo.R")},
                    error = function(e){'foo.R'}
                    )
if(!file.exists(fun_path))
{
  msg = 'Please select "foo.R"'
  # ask user to find data
  if(Sys.info()[['sysname']] == 'Windows'){#choose.files is only available on Windows
    message('\n\n',msg,'\n\n')
    Sys.sleep(0.5)#goes too fast for the user to see the message on some computers
    fun_path  = choose.files(
      default = file.path(gsub('\\\\', '/', Sys.getenv('USERPROFILE')),#user
                          'Documents'),
      caption = msg
    )
  }else{
    message('\n\n',msg,'\n\n')
    Sys.sleep(0.5)#goes too fast for the user to see the message on some computers
    fun_path = file.choose(new=F)
  }
}
#source the function
source(file = fun_path, 
       encoding = 'UTF-8')

其他回答

I would use a variant of @steamer25 's approach. The point is that I prefer to obtain the last sourced script even when my session was started through Rscript. The following snippet, when included on a file, will provided a variable thisScript containing the normalized path of the script. I confess the (ab)use of source'ing, so sometimes I invoke Rscript and the script provided in the --file argument sources another script that sources another one... Someday I will invest in making my messy code turns into a package.

thisScript <- (function() {
  lastScriptSourced <- tail(unlist(lapply(sys.frames(), function(env) env$ofile)), 1)

  if (is.null(lastScriptSourced)) {
    # No script sourced, checking invocation through Rscript
    cmdArgs <- commandArgs(trailingOnly = FALSE)
    needle <- "--file="
    match <- grep(needle, cmdArgs)
    if (length(match) > 0) {
      return(normalizePath(sub(needle, "", cmdArgs[match]), winslash=.Platform$file.sep, mustWork=TRUE))
    }
  } else {
    # 'source'd via R console
    return(normalizePath(lastScriptSourced, winslash=.Platform$file.sep, mustWork=TRUE))
  }
})()

这对我很有用

library(rstudioapi)    
rstudioapi::getActiveDocumentContext()$path
frame_files <- lapply(sys.frames(), function(x) x$ofile)
frame_files <- Filter(Negate(is.null), frame_files)
PATH <- dirname(frame_files[[length(frame_files)]])

不要问我它是如何工作的,因为我已经忘记了:/

这对我很有用。只是从命令行参数中greps它,去掉不需要的文本,执行dirname,最后从它获得完整的路径:

args <- commandArgs(trailingOnly = F)  
scriptPath <- normalizePath(dirname(sub("^--file=", "", args[grep("^--file=", args)])))

解决方案在2016年出现。非常感谢作者,Sahil Seth!

CRAN和github上的funr包提供了sys.script()函数,该函数获取当前脚本的完整路径。它甚至引用了一个类似的SO帖子。

因此,解是:

myscript。接待员:

#!/usr/bin/env Rscript
f  <-  funr::sys.script()
show(f)

然后执行命令:

user@somewhere:/home$ Rscript myscript.R

将在命令行输出,例如:

"/home/path/to/myscript.R"

到控制台。