How to convert a dataframe into a nested list?

Question

I have a dataframe that I want to convert to a nested list with a custom level of nesting. This is how I do it, but I'm sure there is a better way:

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)
    }))
})

The output in JSON looks like:

library(rjson)
write(toJSON(nested_data), "")
[{"city":"A","street_values":[{"street":"a","tenant_values":[{"tenant":"Smith","income":"100"}]},{"street":"b","tenant_values":[{"tenant":"Jones","income":"200"}]}]},{"city":"B","street_values":[{"street":"a","tenant_values":[{"tenant":"Smith","income":"300"}]},{"street":"b","tenant_values":[{"tenant":"Jones","income":"400"}]}]}]

# or prettified:

[
  {
    "city": "A",
    "street_values": [
      {
        "street": "a",
        "tenant_values": [
          {
            "tenant": "Smith",
            "income": "100"
          }
        ]
      },
      {
        "street": "b",
        "tenant_values": [
          {
            "tenant": "Jones",
            "income": "200"
          }
        ]
      }
    ]
  },
  {
    "city": "B",
    "street_values": [
      {
        "street": "a",
        "tenant_values": [
          {
            "tenant": "Smith",
            "income": "300"
          }
        ]
      },
      {
        "street": "b",
        "tenant_values": [
          {
            "tenant": "Jones",
            "income": "400"
          }
        ]
      }
    ]
  }
]

Is there a better way to do this?

So, are you asking a question about JSON output from R, or how to create an R object which is a "nested list" in R's definition, e.g. foo<-list(bar=NA,snafu="hello, Dave"); foo[[bar]] <- list(a=1,b=2) ? — Carl Witthoft
– Carl Witthoft, Commented Mar 4, 2013 at 16:59
I just showed the JSON output because it is easier to understand than the R list format, but I want to go from R dataframe to R list — nachocab
– nachocab, Commented Mar 4, 2013 at 17:30

A5C1D2H2I1M1N2O1R2T1 · Accepted Answer · 2013-03-04 17:43:40Z

What about using split to get you most of the way, and rapply for the last step:

nestedList <- rapply(lapply(split(data[-1], data[1]), 
                            function(x) split(x[-1], x[1])), 
                     f = function(x) as.character(unlist(x)), 
                     how = "replace")

Here's the output:

nestedList
# $A
# $A$a
# $A$a$tenant
# [1] "Smith"
# 
# $A$a$income
# [1] "100"
# 
# 
# $A$b
# $A$b$tenant
# [1] "Jones"
# 
# $A$b$income
# [1] "200"
# 
# 
# 
# $B
# $B$a
# $B$a$tenant
# [1] "Smith"
# 
# $B$a$income
# [1] "300"
# 
# 
# $B$b
# $B$b$tenant
# [1] "Jones"
# 
# $B$b$income
# [1] "400"

And the structure:

> 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"

The structure doesn't exactly match what you're looking for, but this might help get you started with an alternative approach.

Eli · Accepted Answer · 2016-01-11 13:33:28Z

1

I found the solution of my question by changing the function as:

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")

answered Jan 11, 2016 at 13:33

Eli

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