如何将数据框转换为嵌套列表？

我有一个数据帧，我想转换为嵌套自定义级别的嵌套列表。这就是我要做的事，但我敢肯定有一个更好的方法：如何将数据框转换为嵌套列表？

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) 
    })) 
})

在JSON的输出是这样的：

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" 
      } 
     ] 
     } 
    ] 
    } 
]

有没有更好的方式来做到这一点？

来源

2013-03-04 nachocab

所以，你问一个关于'与R输出JSON'问题，或者如何创建一个'R'对象，它是R的定义中的“嵌套列表”，例如'foo <-list（bar = NA，snafu =“hello，Dave”）; foo [[bar]] < - list（a = 1，b = 2）'？ – 2013-03-04 16:59:38

我刚才展示了'JSON'输出，因为它比R列表格式更容易理解，但我想从R数据框转到R列表 – nachocab 2013-03-04 17:30:02

怎么样使用split让你最的方式，并rapply最后一步：

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

下面是输出：

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"

，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"

该结构并不完全符合您的要求，但这可能有帮助让你开始使用另一种方法。

来源

2013-03-04 17:43:40 A5C1D2H2I1M1N2O1R2T1

，我发现我的问题的解决，通过改变功能：

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")

来源

2016-01-11 13:33:28 Eli

如何将数据框转换为嵌套列表？

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