Ragged list or data frame to JSON

Question

我想在R中创建一个与D3树结构flare.json对应的不齐列表。我的数据是在data.frame：

path <- data.frame(P1=c("direct","direct","organic","direct"), 
P2=c("direct","direct","end","end"), 
P3=c("direct","organic","",""), 
P4=c("end","end","",""), size=c(5,12,23,45)) 

path 
     P1  P2  P3 P4 size 
1 direct direct direct end 5 
2 direct direct organic end 12 
3 organic end    23 
4 direct end    45 

，但它也可能是一个列表，或者如果有必要重塑：

path <- list() 
path[[1]] <- list(name=c("direct","direct","direct","end"),size=5) 
path[[2]] <- list(name=c("direct","direct","organic","end"), size=12) 
path[[3]] <- list(name=c("organic", "end"), size=23) 
path[[4]] <- list(name=c("direct", "end"), size=45) 

所需的输出是：

rl <- list() 
rl <- list(name="root", children=list()) 
rl$children[1] <- list(list(name="direct", children=list())) 
rl$children[[1]]$children[1] <- list(list(name="direct", children=list())) 
rl$children[[1]]$children[[1]]$children[1] <- list(list(name="direct", children=list())) 
rl$children[[1]]$children[[1]]$children[[1]]$children[1] <- list(list(name="end", size=5)) 

rl$children[[1]]$children[[1]]$children[2] <- list(list(name="organic", children=list())) 
rl$children[[1]]$children[[1]]$children[[2]]$children[1] <- list(list(name="end", size=12)) 

rl$children[[1]]$children[2] <- list(list(name="end", size=23)) 

rl$children[2] = list(list(name="organic", children=list())) 
rl$children[[2]]$children[1] <- list(list(name="end", size=45)) 

所以，当我打印到JSON它是：

require(RJSONIO) 
cat(toJSON(rl, pretty=T)) 

{ 
"name" : "root", 
"children" : [ 
    { 
     "name" : "direct", 
     "children" : [ 
      { 
       "name" : "direct", 
       "children" : [ 
        { 
         "name" : "direct", 
         "children" : [ 
          { 
           "name" : "end", 
           "size" : 5 
          } 
         ] 
        }, 
        { 
         "name" : "organic", 
         "children" : [ 
          { 
           "name" : "end", 
           "size" : 12 
          } 
         ] 
        } 
       ] 
      }, 
      { 
       "name" : "end", 
       "size" : 23 
      } 
     ] 
    }, 
    { 
     "name" : "organic", 
     "children" : [ 
      { 
       "name" : "end", 
       "size" : 45 
      } 
     ] 
    } 
] 
} 

我是我的头很难绕过在R中创建此列表结构所需的递归步骤。在JS我可以非常容易地在节点周围移动，并且在每个节点确定是添加新节点还是继续向下移动根据需要使用推，例如：​​或new = {"name": node, "size": size};，如在此example中那样。我试图split的data.frame在本example：

makeList<-function(x){ 
    if(ncol(x)>2){ 
     listSplit<-split(x,x[1],drop=T) 
     lapply(names(listSplit),function(y){list(name=y,children=makeList(listSplit[[y]]))}) 
    } else { 
     lapply(seq(nrow(x[1])),function(y){list(name=x[,1][y],size=x[,2][y])}) 
    } 
} 

jsonOut<-toJSON(list(name="root",children=makeList(path))) 

，但它给我一个错误

Error: evaluation nested too deeply: infinite recursion/options(expressions=)? 
Error during wrapup: evaluation nested too deeply: infinite recursion/options(expressions=)? 

Answer 1

linked Q&A中给出的函数基本上是您所需要的，但是由于后面的列中某些行的空值，所以它在数据集上失败。相反的，直到你用完列只是一味地重复递归，你需要检查你的“结束”值，并用它来切换到制作树叶：

makeList<-function(x){ 
    listSplit<-split(x[-1],x[1], drop=TRUE); 
    lapply(names(listSplit),function(y){ 
     if (y == "end") { 
      l <- list(); 
      rows = listSplit[[y]]; 
      for(i in 1:nrow(rows)) { 
       l <- c(l, list(name=y, size=rows[i,"size"])); 
      } 
      l; 

     } 
     else { 
      list(name=y,children=makeList(listSplit[[y]])) 
     } 
    }); 
} 

Answer 2

我相信这是你想要的东西，虽然它有一定的局限性。特别是，假设你的网络中的每一个分支是唯一的（即不能有在您的数据帧两行相等比其他尺寸的每一列）：

df.split <- function(p.df) { 
    p.lst.tmp <- unname(split(p.df, p.df[, 1])) 
    p.lst <- lapply(
    p.lst.tmp, 
    function(x) { 
     if(ncol(x) == 2L && nrow(x) == 1L) { 
     return(list(name=x[1, 1], size=unname(x[, 2]))) 
     } else if (isTRUE(is.na(unname(x[ ,2])))) { 
     return(list(name=x[1, 1], size=unname(x[, ncol(x)]))) 
     } 
     list(name=x[1, 1], children=df.split(x[, -1, drop=F])) 
    } 
) 
    p.lst 
} 
all.equal(rl, df.split(path)[[1]]) 
# [1] TRUE 

虽然注意到你有有机大小切换，所以我不得不修复你的rl得到这个结果（rl它有45，但你的path为23）。另外，我修改了path data.frame略：

path <- data.frame(
    root=rep("root", 4), 
    P1=c("direct","direct","organic","direct"), 
    P2=c("direct","direct","end","end"), 
    P3=c("direct","organic",NA,NA), 
    P4=c("end","end",NA,NA), 
    size=c(5,12,23,45), 
    stringsAsFactors=F 
) 

警告：我还没有和其他结构测试这一点，所以有可能会打，你需要调试角落的情况。