如何写与来自R的孩子的json

Question

我想将R data.frame转换为JSON对象,以便使用它来使用d3.js准备数据可视化.我发现了许多问题,询问如何将JSON引入R,但很少有关于如何将数据从R写入JSON的问题.

一个特殊的问题是JSON文件需要使用因子嵌套,即data.frame的列.我认为从嵌套列表编写可能是一个解决方案,但我已经无法从data.frame创建嵌套列表:(

我有预制红外线的例子:

这代表我的data.frame(称为"MyData").

ID  Location Station   Size Percentage
1     Alpha    Zeta    Big       0.63
2     Alpha    Zeta Medium       0.43
3     Alpha    Zeta  small       0.47
4     Alpha    Yota    Big       0.85
5     Alpha    Yota Medium       0.19
6     Alpha    Yota  small       0.89
7      Beta   Theta    Big       0.09
8      Beta   Theta Medium       0.33
9      Beta   Theta  small       0.79
10     Beta    Meta    Big       0.89
11     Beta    Meta Medium       0.71
12     Beta    Meta  small       0.59

现在,我想把它变成类似这种有效的json格式,包括子节点:

   {
 "name":"MyData",
 "children":[
   {
     "name":"Alpha",
     "children":[
        {
           "name":"Zeta",
           "children":[
              {
                 "name":"Big",
                 "Percentage":0.63
              },
              {
                 "name":"Medium",
                 "Percentage":0.43
              },
              {
                 "name":"Small",
                 "Percentage":0.47
              }
           ]
        },
        {
           "name":"Yota",
           "children":[
              {
                 "name":"Big",
                 "Percentage":0.85
              },
              {
                 "name":"Medium",
                 "Percentage":0.19
              },
              {
                 "name":"Small",
                 "Percentage":0.89
              }
           ]
        }
    ]   
},
    {
     "name":"Zeta",
     "children":[
        {
           "name":"Big",
           "Percentage":0.63
        },
        {
           "name":"Medium",
           "Percentage":0.43
        },
        {
           "name":"Small",
           "Percentage":0.47
        }
     ]
  },
  {
     "name":"Yota",
     "children":[
        {
           "name":"Big",
           "Percentage":0.85
        },
        {
           "name":"Medium",
           "Percentage":0.19
        },
        {
           "name":"Small",
           "Percentage":0.89
        }
     ]
  }
  ]
 }

如果有人能帮助我,我将非常感激!谢谢

Answer 1

这是一种更简洁的递归方法:

require(RJSONIO)

makeList<-function(x){
  if(ncol(x)>2){
    listSplit<-split(x[-1],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],Percentage=x[,2][y])})
  }
}


jsonOut<-toJSON(list(name="MyData",children=makeList(MyData[-1])))
cat(jsonOut)