Json 如何将数据帧转换为嵌套列表?
我有一个数据框,我想将其转换为具有自定义嵌套级别的嵌套列表。我就是这样做的,但我相信有更好的方法:Json 如何将数据帧转换为嵌套列表?,json,r,list,dataframe,Json,R,List,Dataframe,我有一个数据框,我想将其转换为具有自定义嵌套级别的嵌套列表。我就是这样做的,但我相信有更好的方法: data <- data.frame(city=c("A", "A", "B", "B"), street=c("a", "b", "a", "b"), tenant=c("Smith","Jones","Smith","Jones"), income=c(100,200,300,400)) nested_data <- lapply(levels(data$city), funct
data <- data.frame(city=c("A", "A", "B", "B"), street=c("a", "b", "a", "b"), tenant=c("Smith","Jones","Smith","Jones"), income=c(100,200,300,400))
nested_data <- lapply(levels(data$city), function(city){
data_city <- subset(data[data$city == city, ], select=-city)
list(city = city, street_values=lapply(levels(data_city$street), function(street){
data_city_street <- subset(data_city[data_city$street == street, ], select=-street)
tenant_values <- apply(data_city_street, 1, function(income_tenant){
income_tenant <- as.list(income_tenant)
list(tenant=income_tenant$tenant, income=income_tenant$income)
})
names(tenant_values) <- NULL
list(street=street, tenant_values=tenant_values)
}))
})
有更好的方法吗?使用
split
可以让您获得最大的收益,最后一步是rapply
:
nestedList <- rapply(lapply(split(data[-1], data[1]),
function(x) split(x[-1], x[1])),
f = function(x) as.character(unlist(x)),
how = "replace")
结构:
> str(nestedList)
List of 2
$ A:List of 2
..$ a:List of 2
.. ..$ tenant: chr "Smith"
.. ..$ income: chr "100"
..$ b:List of 2
.. ..$ tenant: chr "Jones"
.. ..$ income: chr "200"
$ B:List of 2
..$ a:List of 2
.. ..$ tenant: chr "Smith"
.. ..$ income: chr "300"
..$ b:List of 2
.. ..$ tenant: chr "Jones"
.. ..$ income: chr "400"
结构与您要查找的不完全匹配,但这可能有助于您开始使用另一种方法。我通过将函数更改为:
nestedList <- rapply(lapply(split(df[-1], df[1]),
function(x) split(x[-1], x[1])),
f = function(x) as.data.frame(as.list(split(x,x))), how = "replace")
nestedList那么,你是在问一个关于R的JSON
输出的问题,还是如何创建一个R
对象,它是R的定义中的“嵌套列表”,例如fooI刚才显示了JSON
输出,因为它比R list格式更容易理解,但我想从R数据帧转到R list
> str(nestedList)
List of 2
$ A:List of 2
..$ a:List of 2
.. ..$ tenant: chr "Smith"
.. ..$ income: chr "100"
..$ b:List of 2
.. ..$ tenant: chr "Jones"
.. ..$ income: chr "200"
$ B:List of 2
..$ a:List of 2
.. ..$ tenant: chr "Smith"
.. ..$ income: chr "300"
..$ b:List of 2
.. ..$ tenant: chr "Jones"
.. ..$ income: chr "400"
nestedList <- rapply(lapply(split(df[-1], df[1]),
function(x) split(x[-1], x[1])),
f = function(x) as.data.frame(as.list(split(x,x))), how = "replace")