是否有类似make.path.relative(base.path, target.path)
的内容?
我想将完整路径转换为给定基本路径的相对路径(如项目目录)。
答案 0 :(得分:1)
确定。我自己写了这个函数:
make.path.relative<-function(base.path, target.path)
{
base.s<-strsplit(base.path,'/',fixed=TRUE)[[1]]
target.s<-strsplit(target.path,'/',fixed=TRUE)[[1]]
idx<-1
maxidx<-min(length(target.s),length(base.s))
while(idx<=maxidx)
{
if (base.s[[idx]]!=target.s[[idx]])
break
idx<-idx+1
}
dotscount<-length(base.s)-idx+1
ans1<-paste0(paste(rep('..',times=dotscount),collapse='/'))
if (idx<=length(target.s))
ans2<-paste(target.s[idx:length(target.s)],collapse='/')
else
ans2<-''
ans<-character(0)
if (ans1!='')
ans[[length(ans)+1]]<-ans1
if (ans2!='')
ans[[length(ans)+1]]<-ans2
ans<-paste(ans,collapse='/')
return(ans)
}
您必须首先清理路径以确保它们使用相同的斜杠约定。您可以使用我对Function to concatenate paths?
的回答中的path.cat函数示例:
> make.path.relative('C:/home/adam', 'C:/home/adam/tmp/R')
[1] "tmp/R"
> make.path.relative('/home/adam/tmp', '/home/adam/Documents/R')
[1] "../Documents/R"
> make.path.relative('/home/adam/Documents/R/Project', '/home/adam/minetest')
[1] "../../../minetest"
答案 1 :(得分:1)
类似,但更短:
make.path.relative = function(base, target) {
common = sub('^([^|]*)[^|]*(?:\\|\\1[^|]*)$', '^\\1/?', paste0(base, '|', target))
paste0(gsub('[^/]+/?', '../', sub(common, '', base)),
sub(common, '', target))
}
make.path.relative('C:/home/adam', 'C:/home/adam/tmp/R')
#[1] "tmp/R"
make.path.relative('/home/adam/tmp', '/home/adam/Documents/R')
#[1] "../Documents/R"
make.path.relative('/home/adam/Documents/R/Project', '/home/adam/minetest')
#[1] "../../../minetest"
Voodoo正则表达式from here。