在 f# 中加入基于相同标签的 2 个记录列表

Question

我有 2 个具有相同标签的记录列表 id1。我需要一种加入他们的方式。

type A = { id1: int; name: string }

type B = { id1: int; id2: Option<int> }

let a1 = { id1 =  1; name = "nafis" }
let a2 = { id1 =  2; name = "habib" }

let b1 = { id1 = 1; id2 = Some(5) }
let b2 = { id1 = 1; id2 = None }
let b3 = { id1 = 2; id2 = None }

let a = [a1; a2]
let b = [b1; b2; b3]

printfn "%A" a =>  [({id1 = 1;name = "nafis";}, {id1 = 2;name = "habib";})]
printfn "%A" b =>  
[({id1 = 1; id2 = Some 5;}, {id1 = 1; id2 = None;}, {id1 = 2;id2 = None;})]

如何加入这两个基于id1的列表？

我想要这样的输出=>

[({id1 = 1;name = "nafis"; id2 = [Some 5; None];}, {id1 = 2;name = "habib"; id2 =[None];})]

某种形式的教程或博客 link 会有所帮助。

Answer 1

MSDN: Query expressions enable you to query a data source and put the data in a desired form. Query expressions provide support for LINQ in F#.

正如评论中正确提到的那样，您需要有第三个类型 C 来承载连接结果，然后使用良好的旧 LINQ 来连接列表：

type A = { id1: int; name: string }
type B = { id1: int; id2: Option<int> }
type C = { id1: int; name: string; id2: Option<int> }

let a1 = { id1 =  1; name = "nafis" }
let a2 = { id1 =  2; name = "habib" }

let b1 = { id1 = 1; id2 = Some(5) }
let b2 = { id1 = 1; id2 = None }
let b3 = { id1 = 2; id2 = None }

let a = [a1; a2]
let b = [b1; b2; b3]

let result (a:A list) (b: B list) = query {
    for list1 in a do
    join list2 in b on
        (list1.id1 = list2.id1)
    select {id1 = list1.id1; name = list1.name; id2= list2.id2}
}

let c = result a b |> List.ofSeq

结果：

val c : C list = [{id1 = 1;
               name = "nafis";
               id2 = Some 5;}; {id1 = 1;
                                name = "nafis";
                                id2 = None;}; {id1 = 2;
                                               name = "habib";
                                               id2 = None;}]

在 f# 中加入基于相同标签的 2 个记录列表

Join 2 list of recoreds based on same label in f#

f#

list-manipulation