Sorting F#值为'的记录排序顺序；尚未按顺序分配给类型

我试图按照每条记录中的第二个元素对一系列记录进行排序。问题在于，它们不是一个值，而是一个类型。我有一个函数，它根据它的类型返回值

这就是我所拥有的：

type Suit = Spades | Clubs | Hearts | Diamonds
type Rank = Ace | Two | Three | Four | Five | Six | Seven | Eight | Nine | Ten | Jack | Queen | King
type Card = { suit: Suit; rank: Rank}

type Hand = Card seq

let cardValue(card:Card) = 
    if (card.rank = Ace) then 1
    elif (card.rank = Two) then 2
    elif (card.rank = Three) then 3
    elif (card.rank = Four) then 4
    elif (card.rank = Five) then 5
    elif (card.rank = Six) then 6
    elif (card.rank = Seven) then 7
    elif (card.rank = Eight) then 8
    elif (card.rank = Nine) then 9
    elif (card.rank = Ten) then 10
    elif (card.rank = Jack) then 10
    elif (card.rank = Queen) then 10
    elif (card.rank = King) then 10
    else 0

let sortHandByValue(hand:Hand) = 
......missing code here......

我想做的是按照等级作为一个值来排序

例如，现在的手是：{{红心；三个}；{黑桃；杰克}；{钻石；两个}

它将对手进行排序，结果是：{{钻石；两个}；{红心；三个}；{黑桃；杰克}

我尝试过使用hand |>Seq.sort |>Seq.groupBy id |>Seq.map snd，但这并没有按值排序，只按字母顺序排序

我不能更改任何类型，但我可以更改其他所有类型。

---

任何想法都将不胜感激，谢谢

您需要稍微清理一下代码，因为它显示记录，我将卡片类型重写为元组，并将匹配函数也重写为元组。然后，您只需通过管道进入

Seq.sortBy

或

Seq.sortByDescending

（按升序排序）：

type套装=黑桃|梅花|红心|钻石
类型等级=王牌|二|三|四|五|六|七|八|九|十|杰克|女王|国王
类型卡=西装*等级
手动类型=卡片顺序
let cardValue（卡：卡）=
配牌
|_u，Ace->1
|两个->两个
|(三)->三
|四个->四个
|五，五->五
|(六)->六
|七点，七点->七点
|八点，八点->八点
|_uu9->9
|十，十，杰克，王后，国王->十
let hand=seq[（红桃，三颗）；（黑桃，杰克）；（钻石，两颗）]
hand |>Seq.sortBy cardValue
//val it:seq=
//顺序[（钻石，二）；（红桃，三）；（黑桃，杰克）]

有记录的版本： 我把这个和你原来的很接近

/// Version with Records
type Suit = Spades | Clubs | Hearts | Diamonds
type Rank = Ace | Two | Three | Four | Five | Six | Seven | Eight | Nine | Ten | Jack | Queen | King
type Card = { suit: Suit; rank: Rank}
type Hand = seq<Card>

let cardValue(card:Card) = 
    if (card.rank = Ace) then 1
    elif (card.rank = Two) then 2
    elif (card.rank = Three) then 3
    elif (card.rank = Four) then 4
    elif (card.rank = Five) then 5
    elif (card.rank = Six) then 6
    elif (card.rank = Seven) then 7
    elif (card.rank = Eight) then 8
    elif (card.rank = Nine) then 9
    elif (card.rank = Ten) then 10
    elif (card.rank = Jack) then 10
    elif (card.rank = Queen) then 10
    elif (card.rank = King) then 10
    else 0

let hand = seq [{suit=Hearts; rank=Three}; {suit=Spades;rank=Jack}; {suit=Diamonds;rank= Two}]

hand |> Seq.sortBy cardValue 

///有记录的版本
类型套装=黑桃|梅花|红心|钻石
类型等级=王牌|二|三|四|五|六|七|八|九|十|杰克|女王|国王
类型卡片={suit:suit；rank:rank}
手动类型=顺序
let cardValue（卡：卡）=
如果（card.rank=Ace），则1
elif（card.rank=2）则为2
elif（card.rank=3）则为3
elif（card.rank=4）则为4
elif（card.rank=5）则为5
elif（card.rank=6）则为6
elif（card.rank=7）则为7
elif（card.rank=8）则为8
elif（card.rank=9）则为9
elif（card.rank=10）则为10
elif（card.rank=Jack）然后10
elif（card.rank=Queen）然后10
elif（card.rank=King）然后是10
其他0
let hand=seq[{suit=Hearts；rank=Three}；{suit=Spades；rank=Jack}；{suit=Diamonds；rank=Two}]
hand |>Seq.sortBy cardValue

valit:seq=seq[{suit=Diamonds； 等级=二；}；{suit=心； 等级=三；}；{suit=黑桃； 秩=杰克；}]

---

您的序列不是元组而是记录。

hand |>Seq.sortBy cardValue

。？我还认为

Rank*Suit

更容易理解…我感谢您的帮助和解释，但不幸的是我无法更改类型卡。因此，当我尝试将卡片与值进行匹配时，它现在抱怨的不是类型“card”，而是“a*”b“，这不是真正的问题。您只需要最后一行：-）让我添加一个记录版本。你应该去掉元组标签，把问题重新表述一下。可以了，抱歉给你带来了困惑。我做F#的时间不长，我的术语也不是最好的哈哈。嗯，这并没有真正改变解决方案的逻辑，顺便说一句。我真的很感谢你的帮助和建议，现在我知道了这一点，我应该可以做剩下的事情了！

/// Version with Records
type Suit = Spades | Clubs | Hearts | Diamonds
type Rank = Ace | Two | Three | Four | Five | Six | Seven | Eight | Nine | Ten | Jack | Queen | King
type Card = { suit: Suit; rank: Rank}
type Hand = seq<Card>

let cardValue(card:Card) = 
    if (card.rank = Ace) then 1
    elif (card.rank = Two) then 2
    elif (card.rank = Three) then 3
    elif (card.rank = Four) then 4
    elif (card.rank = Five) then 5
    elif (card.rank = Six) then 6
    elif (card.rank = Seven) then 7
    elif (card.rank = Eight) then 8
    elif (card.rank = Nine) then 9
    elif (card.rank = Ten) then 10
    elif (card.rank = Jack) then 10
    elif (card.rank = Queen) then 10
    elif (card.rank = King) then 10
    else 0

let hand = seq [{suit=Hearts; rank=Three}; {suit=Spades;rank=Jack}; {suit=Diamonds;rank= Two}]

hand |> Seq.sortBy cardValue