Ragged list or data frame to JSON

Question

我想在R中創建一個與D3樹結構flare.json對應的不齊列表。我的數據是在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 

，但它也可能是一個列表，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) 

所需的輸出是：

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，當我打印到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 
      } 
     ] 
    } 
] 
} 

我是我的頭很難繞過在R中創建此列表結構所需的遞歸步驟。在JS我可以非常容易地在節點周圍移動，並且在每個節點確定是添加新節點還是繼續向下移動根據需要使用推，例如：​​或new = {"name": node, "size": size};，如在此example中那樣。我試圖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))) 

，但它給我一個錯誤

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中給出的函數基本上是您所需要的，但是由於後面的列中某些行的空值，所以它在數據集上失敗。相反的，直到你用完列只是一味地重複遞歸，你需要檢查你的「結束」值，並用它來切換到製作樹葉：

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

我相信這是你想要的東西，雖然它有一定的侷限性。特別是，假設你的網絡中的每一個分支是唯一的（即不能有在您的數據幀兩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 

雖然注意到你有有機大小切換，所以我不得不修復你的rl得到這個結果（rl它有45，但你的path爲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：我還沒有和其他結構測試這一點，所以有可能會打，你需要調試角落的情況。