递归对象列表

Question

R是否有一种优雅的方式递归地将任意深度列表（包含列表和向量）转换为路径向量？例如，转换这样的：

list(
    home = list(
    jerry = c("R", "bla"), 
    mary = "xx" 
), 
    tmp = c("foo", "bar"), 
    c("etc") 
) 

要一个对象作为这样的：unlist

c(
    "/home/jerry/R", 
    "/home/jerry/bla", 
    "/home/mary/xx", 
    "/tmp/foo", 
    "/tmp/bar", 
    "/etc" 
) 

Answer 1

的名称不约你想要什么：

> test <- list(
+  home = list(
+   jerry = c("R", "bla"), 
+   mary = "xx" 
+ ), 
+  tmp = c("foo", "bar"), 
+  etc = c() 
+) 
> unlist(test) 
home.jerry1 home.jerry2 home.mary  tmp1  tmp2 
     "R"  "bla"  "xx"  "foo"  "bar" 

把手递归的多个级别，以及：

> test <- list(
+  home = list(
+   jerry = list(a="R", b="bla"), 
+   mary = list(c="xx") 
+ ), 
+  tmp = list(d="foo", e="bar"), 
+  etc = list(nothing=NULL) 
+) 
> unlist(test) 
home.jerry.a home.jerry.b home.mary.c  tmp.d  tmp.e 
     "R"  "bla"   "xx"  "foo"  "bar" 

从那里可以很容易地添加你想要的（具有终值是最后一个路径elemtn）上的最后一点：

> unl <- unlist(test) 
> res <- names(unl) 
> res <- paste(res,unl,sep=".") 
> res 
[1] "home.jerry.a.R" "home.jerry.b.bla" "home.mary.c.xx" "tmp.d.foo"  "tmp.e.bar"  

Answer 2

只要不关你的列表中的元素都有名字，其中包括点（与在您的列表麻烦NULL -valued etc元）外，这应该工作：

ll <- rapply(l, function(X) sapply(X,list), how="replace") #for tip element names 
nms <- names(unlist(ll)) 
gsub(".", "/", nms, fixed=TRUE) 
# [1] "home/jerry/R" "home/jerry/bla" "home/mary/xx" "tmp/foo"  
# [5] "tmp/bar"  

Answer 3

对于一个更通用的方法，包括用点和空元素posdibly名称，使用此：

f <- function(test, parent=character(0)) 
{ 
    if(is.null(test)) return(parent) 

    if(is.list(test)) 
    { 
     result <- character(0) 

     for(i in seq_along(test)) 
     { 
      result <- c(result, Recall(test[[i]], parent=paste0(parent,"/",names(test)[i]))) 
     } 
    } 
    else 
    { 
     result <- paste0(parent,"/",as.character(test)) 
    } 

    result 
} 

f(test) 

#[1] "/home/jerry/R" "/home/jerry/bla" "/home/mary/xx" "/tmp/foo"  "/tmp/bar"  "/etc"