Question

如何使用data.frame（MyData）创建一个JSON对象，如下所示。我尝试了其他方法，但这个问题看起来很独特。

ID   Station   Size 
1      Zeta    Big  
2      Zeta Medium  
3      Zeta  small  
4      Yota    Big  
5      Yota Medium  
6      Yota  small

预期结果

{
    "name": "bubble",
    "children": [{
        "name": "Zeta",
        "children": [{
            "name": "big"
        }, {
            "name": "Medium"
        }, {
            "name": "small"
        }]
    }, {
        "name": "Yota",
        "children": [{
            "name": "big"
        }, {
            "name": "Medium"
        }, {
            "name": "small"
        }]
    }]
}

这是我搜索过但无法改变的内容

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])})
  }
}


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

Answer 1

数据

我正在使用的数据是

df <- data.frame(Station = c(rep("Zeta", 3), rep("Yota", 3)), Size = rep(c("Big","Medium","Small"),2), stringsAsFactors = F)

方式

向后工作，我们可以看到我们追求的结构

t <- "{\"name\": \"bubble\",\"children\": [{\"name\": \"Zeta\",\"children\": [{\"name\": \"big\"}, {\"name\": \"Medium\" }, {\"name\": \"small\"}]}, {\"name\": \"Yota\",\"children\": [{\"name\": \"big\"}, {\"name\": \"Medium\"}, {\"name\": \"small\"}]}]}\"" library(jsonlite) l <- fromJSON(t) str(l) #List of 2 #$ name : chr "bubble" #$ children:'data.frame': 2 obs. of 2 variables: # ..$ name : chr [1:2] "Zeta" "Yota" # ..$ children:List of 2 # .. ..$ :'data.frame': 3 obs. of 1 variable: # .. .. ..$ name: chr [1:3] "big" "Medium" "small" # .. ..$ :'data.frame': 3 obs. of 1 variable: # .. .. ..$ name: chr [1:3] "big" "Medium" "small"

要重建这个，我们需要

## first element lst <- list(name = "bubble") ## second element l_child1 <- l$children$children[[1]] l_child2 <- l$children$children[[2]] l_child <- list(data.frame(name=l_child1), data.frame(name=l_child2)) n <- c("Zeta", "Yota") df_child <- data.frame(name = n, stringsAsFactors = F) df_child$children <- l_child lst <- list(name = "bubble", children = df_child) toJSON(lst, pretty=F) # {"name":["bubble"],"children":[{"name":"Zeta","children":[{"name":"big"},{"name":"Medium"},{"name":"small"}]},{"name":"Yota","children":[{"name":"big"},{"name":"Medium"},{"name":"small"}]}]}

所以我们知道我们需要l_child1, l_child2, l_child, df_child的结构，但是我们如何从原始的df那里得到那些结果？对于一般情况呢？

<强>解决方案

df$Size向我们提供了l_child个列表，每个Station都有一个大小。因此，我们可以使用lapply向我们提供我们的每个“小组”的儿童名单。

n <- unique(df$Station) l_child <- lapply(1:length(n), FUN=function(x){ t <- data.frame(name = (df[df$Station == n[x], "Size"]), stringsAsFactors=F) return(t) })

我们现在可以构建我们的最终列表

df_child <- data.frame(name = n, stringsAsFactors = FALSE) df_child$children <- l_child lst <- list(name = "bubble", children = df_child)

并检查：

> toJSON(lst, pretty=T) { "name": ["bubble"], "children": [ { "name": "Zeta", "children": [ { "name": "Big" }, { "name": "Medium" }, { "name": "Small" } ] }, ... etc...

在R中创建具有多个子节点的Json对象

1 个答案: