如何创建多种类型的数据框与RTypeProvider

Question

这似乎是RTypeProvider只能处理相同类型的namedParams。是这样吗？

例如，

open RDotNet 
open RProvider 

type foo = { 
    Which: string 
    Qty: float option 
    }  

let someFoos = [{Which = "that"; Qty = Some 4.0}; {Which = "other"; Qty = Some 2.0}] 

let thingForR = 
    namedParams [ 
     "which", someFoos |> List.map (fun x -> x.Which); 
     "qty", someFoos |> List.map (fun x -> x.Qty); 
     ] 
    |> R.data_frame 

，因为我如果我扭转了thingForR让利的顺序得到的x.Qty一个错误说

This expression was expected to have type 
    string 
but here has type 
    float option 

不工作，那么我得到相反的错误：

let thingForR = 
    namedParams [ 
     "qty", someFoos |> List.map (fun x -> x.Qty); 
     "which", someFoos |> List.map (fun x -> x.Which); 
     ] 
    |> R.data_frame 

在这里，错误x.Which是

This expression was expected to have type 
    float option 
but here has type 
    string 

namedParams中的字典能不能有不同的类型吗？如果是这样，你如何在F＃中创建一个不同类型的数据帧并将它们传递给R？

Answer 1

您需要框中字典里面的值。这样他们都只是对象。所以：

let thingForR = 
    namedParams [ 
     "which", box (someFoos |> List.map (fun x -> x.Which)); 
     "qty", box (someFoos |> List.map (fun x -> x.Qty) |> List.map (Option.toNullable >> float)); 
     ] 
    |> R.data_frame 

给我：

val thingForR :
SymbolicExpression = which qty
1 that 4
2 other 2

请参阅您刚才的问题上float option转换的Option list到float list。如有必要，还需要string option。

你可以通过Deedle（如果不是选项值）：

let someFoos' = [{Which = "that"; Qty = 4.0}; {Which = "other"; Qty = 2.0}] 
let df' = someFoos' |> Frame.ofRecords 
df' |> R.data_frame