TypeScript 按任意类型过滤元组类型

Question

如何通过提供的元组中的任意类型过滤提供的元组类型来生成新的元组类型？

type Journey = ["don't", 'stop', 'believing'];

type ExcludeFromTuple<T extends unknown[], E> = ????;

type DepressingJourney = ExcludeFromTuple<Journey, "don't">; // type should be ['stop', 'believing']

请注意，该解决方案不需要事先确保类型 E 存在于类型 T 中，如果存在，只需将其删除即可。

虽然这里的示例很简单，但我有一个更复杂的用例，我希望能够通过我正在编写的库的使用者定义的任意类型进行过滤。

尽管 TypeScript 原生支持 exclude type，但它仅适用于联合类型，我一直无法找到元组的等效项。

像 ExcludeFromTuple 这样的类型对于生成其他实用程序类型非常有用。

type RemoveStringsFromTuple<T extends unknown[]> = ExcludeFromTuple<T, string>;
type RemoveNumbersFromTuple<T extends unknown[]> = ExcludeFromTuple<T, number>;
type RemoveNeversFromTuple<T extends unknown[]> = ExcludeFromTuple<T, never>;
type RemoveUndefinedsFromTuple<T extends unknown[]> = ExcludeFromTuple<T, undefined>;

我觉得该类型需要利用 TypeScript 2.8 的组合 conditional types， TypeScript 3.1 的 mapped types on tuples，以及某种递归类型的魔法，但我一直无法弄清楚，也找不到任何人弄清楚。

Answer 1

TS 4.1+ 更新：

随着 TS4.1 中引入的 variadic tuple types introduced in TS 4.0, and recursive conditional types，您现在可以将 ExcludeFromTuple 更简单地写为：

type ExcludeFromTuple<T extends readonly any[], E> =
    T extends [infer F, ...infer R] ? [F] extends [E] ? ExcludeFromTuple<R, E> :
    [F, ...ExcludeFromTuple<R, E>] : []

您可以验证这是否按预期工作：

type DepressingJourney = ExcludeFromTuple<Journey, "don't">;
// type should be ['stop', 'believing']

type SlicedPi = ExcludeFromTuple<[3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9], 1 | 9>
// type SlicedPi = [3, 4, 5, 2, 6, 5, 3, 5, 8, 7]

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TS-4.1 之前的答案：

哎呀，这才是真正需要的东西 recursive conditional types which are not supported in TypeScript yet. If you want to use them you do so at your own risk。通常我宁愿写一个应该递归的类型，然后将它展开到一个固定的深度。所以我写 type F<X> = ...F0<X>...; type F0<X> = ...F1<X>...;.

而不是 type F<X> = ...F<X>...

为了写这篇文章，我想对元组使用基本的“列表处理”类型，即 Cons<H, T> 将类型 H 添加到元组 T 上； Head<T> 获取元组 T 的第一个元素，Tail<T> 获取删除第一个元素的元组 T。您可以这样定义：

type Cons<H, T> = T extends readonly any[] ? ((h: H, ...t: T) => void) extends ((...r: infer R) => void) ? R : never : never;
type Tail<T extends readonly any[]> = ((...t: T) => void) extends ((h: any, ...r: infer R) => void) ? R : never;
type Head<T extends readonly any[]> = T[0];

那么递归类型看起来像这样：

/* type ExcludeFromTupleRecursive<T extends readonly any[], E> = 
     T["length"] extends 0 ? [] : 
     ExcludeFromTupleRecursive<Tail<T>, E> extends infer X ? 
     Head<T> extends E ? X : Cons<Head<T>, X> : never; */

想法是：取出元组的尾部 T 并对其执行 ExcludeFromTupleRecursive。这就是递归。然后，对于结果，当且仅当它不匹配 E.

时，你应该在元组的头部添加前缀

但这是非法循环，所以我这样展开它：

type ExcludeFromTuple<T extends readonly any[], E> = T["length"] extends 0 ? [] : X0<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type X0<T extends readonly any[], E> = T["length"] extends 0 ? [] : X1<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type X1<T extends readonly any[], E> = T["length"] extends 0 ? [] : X2<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type X2<T extends readonly any[], E> = T["length"] extends 0 ? [] : X3<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type X3<T extends readonly any[], E> = T["length"] extends 0 ? [] : X4<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type X4<T extends readonly any[], E> = T["length"] extends 0 ? [] : X5<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type X5<T extends readonly any[], E> = T["length"] extends 0 ? [] : X6<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type X6<T extends readonly any[], E> = T["length"] extends 0 ? [] : X7<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type X7<T extends readonly any[], E> = T["length"] extends 0 ? [] : X8<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type X8<T extends readonly any[], E> = T["length"] extends 0 ? [] : X9<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type X9<T extends readonly any[], E> = T["length"] extends 0 ? [] : XA<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type XA<T extends readonly any[], E> = T["length"] extends 0 ? [] : XB<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type XB<T extends readonly any[], E> = T["length"] extends 0 ? [] : XC<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type XC<T extends readonly any[], E> = T["length"] extends 0 ? [] : XD<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type XD<T extends readonly any[], E> = T["length"] extends 0 ? [] : XE<Tail<T>, E> extends infer X ? Head<T> extends E ? X : Cons<Head<T>, X> : never;
type XE<T extends readonly any[], E> = T; // bail out

玩得开心吗？让我们看看它是否有效：

type DepressingJourney = ExcludeFromTuple<Journey, "don't">;
// type should be ['stop', 'believing']

type SlicedPi = ExcludeFromTuple<[3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9], 1 | 9>
// type SlicedPi = [3, 4, 5, 2, 6, 5, 3, 5, 8, 7]

我觉得不错。

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TypeScript 按任意类型过滤元组类型

TypeScript filter tuple type by an arbitrary type

types

typescript

typescript-generics

typescript-typings