如何在R中复制带符号链接的文件夹/目录？

I have an R function that uses a directory with relative symbolic links (e.g. ../../data) as a template. I know the relative links will be valid since the function puts the template in a specific part of a known file structure. The file.copy function can copy directories with recrusive = TRUE, but it turns the symbolic links into copies of the directories they point to. The linux system command cp -r path1 path2 will copy the links correctly, but I want to use R functions when possible.

我有一个R函数,它使用一个带有相对符号链接的目录(例如../../data)作为模板。我知道相对链接是有效的,因为函数将模板放在已知文件结构的特定部分。 file.copy函数可以使用recrusive = TRUE复制目录,但它会将符号链接转换为它们指向的目录的副本。 linux系统命令cp -r path1 path2将正确复制链接,但我希望尽可能使用R函数。

How do I copy a directory in R that contains relative symbolic links that point outside the directory?

如何复制R中包含指向目录外部的相对符号链接的目录?

I know I could write my own function that recursivly lists files (list.dirs), finds symbolic links (Sys.readlink), and recreates them (file.link), while copying all other files, but I wanted to know if that functionality already exists.

我知道我可以编写自己的函数,递归列出文件(list.dirs),找到符号链接(Sys.readlink),并重新创建它们(file.link),同时复制所有其他文件,但我想知道是否有这个功能已经存在。

1 个解决方案

#1

I have come up with a solution, so I figured I would post it in case others might need to do this and there is actually not a more conventional way to do it. I am still interested in hearing other answers!

我想出了一个解决方案,所以我想我会发布它,以防其他人可能需要这样做,实际上没有更传统的方法来做到这一点。我仍然有兴趣听到其他答案!

I came up with the following function:

我想出了以下功能:

#===================================================================================================
#' Copy folders with links
#' 
#' Copies folders like \link{\code{file.copy}} except it replicates links correctly on unix-like 
#' systems.
#' 
#' @param from (\code{character}) The path to the folder to be copied
#' @param to (\code{character}) Where to copy the folder to. 
#' 
copy_folder_with_links <- function(from, to) {
  target <- file.path(to, basename(from))
  if (file.exists(target)) stop(paste0("Target folder ", target, " already exists."))
  # Get list of all files/folders to copy ----------------------------------------------------------
  path <- data.frame(target = list.files(from, recursive = TRUE, all.files = TRUE, include.dirs = TRUE))
  path$from  <- file.path(from, path$target)
  path$to  <- file.path(to, basename(from), path$target)
  # Get type of file/folders -----------------------------------------------------------------------
  path$type <- factor("file", levels = c("file", "folder", "link"))
  path$type[file.info(path$from)$isdir] <- "folder"
  path$type[Sys.readlink(path$from) != ""] <- "link"
  # Remove all files that are descendants of links -------------------------------------------------
  is_child <- function(query, refs) {
    sapply(refs, function(x) grepl(paste0("^", x), query) & query != x)
  }
  path <- path[!sapply(path$from, function(x) any(is_child(x, path$from) & path$type == "link")), ]
  # Make copy --------------------------------------------------------------------------------------
  invisible(lapply(path$to[path$type == "folder"], dir.create, recursive = TRUE))
  invisible(file.copy(from = path$from[path$type == "file"], to = path$to[path$type == "file"]))
  invisible(file.symlink(Sys.readlink(path$from[path$type == "link"]), path$to[path$type == "link"]))
}

It works like file.copy (with less options), but correctly copies links. It works by

它的工作方式类似于file.copy(选项较少),但可以正确复制链接。它的工作原理

Getting a list of all files/folders to be copied using list.files(from, recursive = TRUE, all.files = TRUE, include.dirs = TRUE)

使用list.files获取要复制的所有文件/文件夹的列表(from,recursive = TRUE,all.files = TRUE,include.dirs = TRUE)

Determining whether each is a file, folder, or link

确定每个文件是文件,文件夹还是链接

Removing descendants of linked folders from the list (since links are followed by list.files)

从列表中删除链接文件夹的后代(因为链接后面是list.files)

Copying the folders, files, and links (in that order)

复制文件夹,文件和链接(按此顺序)

#1