我想把一个R转换成一个data.frame对象,以便用它来准备d3.js的数据可视化。我发现了很多关于如何将JSON放入R的问题,但很少有关于如何将数据从R写入JSON的问题。
一个特别的问题是,JSON文件需要使用因子进行嵌套,即data.frame的列。我认为从嵌套列表编写代码可能是一种解决方案,但我已经无法从data.frame创建嵌套列表:(
我准备了一个例子:
这表示我的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,我想将其转换为类似于以下有效的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,我将不胜感激!谢谢
发布于 2012-10-11 00:15:33
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)
发布于 2012-10-10 20:37:48
结合使用split
和subset
可能会得到您想要的结果。例如
library(RJSONIO)
list1<-split(subset(MyData,select=c(-Location)),Mydata$Location)
list2<-lapply(list1,function(x){split(subset(x,select=c(-Station)),x$Station,drop=TRUE)})
list3<-lapply(list2,function(x){lapply(x,function(y){split(subset(y,select=c(-Size,-ID)),y$Size,drop=TRUE)})})
jsonOut<-toJSON(list(MyData=list3))
jsonOut1<-gsub('([^\n]*?): \\{\n "Percentage"','\\{"name":\\1,"Percentage"',jsonOut)
jsonOut2<-gsub('"([^"]*?)": \\{','"name":"\\1","children":\\{',jsonOut1)
cat(jsonOut2)
{
"name":"MyData","children":{
"name":"Alpha","children":{
"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":"Beta","children":{
"name":"Meta","children":{
{"name": "Big","Percentage": 0.89
},
{"name":"Medium","Percentage": 0.71
},
{"name":"small","Percentage": 0.59
}
},
"name":"Theta","children":{
{"name": "Big","Percentage": 0.09
},
{"name":"Medium","Percentage": 0.33
},
{"name":"small","Percentage": 0.79
}
}
}
}
}
发布于 2016-05-05 01:46:33
我借鉴了user1609452的回答,回答了关于非常规文件层次结构的问题。如果您有一个列,其中一些数据具有子项,而另一些数据没有子项,请使用以下内容:
makeList<-function(x){
if(ncol(x)>2){
listSplit<-split(x[-1],x[1],drop=T)
lapply(names(listSplit),function(y){
if(as.character(listSplit[[y]][1,1]) > 0){
list(name=y,children=makeList(listSplit[[y]]))
} else {
list(name=y,size=listSplit[[y]][1,2])
}
})
}else{
lapply(seq(nrow(x[1])),function(y){list(name=x[,1][y],size=x[,2][y])})
}
}
基本上,我们检查当前行是否有更多的子行,或者是否只需要向其追加大小。
https://stackoverflow.com/questions/12818864
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