Is it possible to extract keys from a parameter for type inference in Typescript?

Question

I was trying to extract some properties from an object.
The keys which should be extracted are given as parameter to the function.

For example, for this code, it copies the specific keys from the object and returns new, another object with the same keys.

function pick(source: object, ...keys: string[]) {
    const returnValue = {} 
    for (const key in source) {
        const index = keys.indexOf(key)
        if (index < 0) continue
        keys.splice(index, 1)
        returnValue[ key ] = source[ key ]
    }
    return returnValue
}

When I use this function, TypeScript infers the return type as {} which is an empty object. Is there a way I can get this as { [key: .. in parameter 'keys' ..]: any }?

I tried this but it did not work.

function pick<O>(source: object, ...keys: (keyof O)[]): O {
    const returnValue: O = {} 
    for (const key in source) {
        const index = keys.indexOf(key)
        if (index < 0) continue
        keys.splice(index, 1)
        returnValue[ key ] = source[ key ]
    }
    return returnValue
}

Thank You

artem · Accepted Answer · 2017-11-20 18:25:52Z

2

There is built-in Pick type that was added in TypeScript 2.1 exactly for describing such a function:

function pick<T, K extends keyof T>(source: T, ...keys: K[]): Pick<T, K> {

Just for reference, it's a very simple mapped type defined as

type Pick<T, K extends keyof T> = {[k in K]: T[k]};

However, your implementation does not compile without type casts because TypeScript complains about keys.indexOf(key) - key is now keyof T and it's not compatible with the type of indexOf parameter, which is keys element type - K.

So I made slightly different implementation with a loop that goes over keys array. I do not remove "used" keys from keys - if there are duplicates (which is not likely I suppose), assigning the same value twice will do no harm here:

function pick<T, K extends keyof T>(source: T, ...keys: K[]): Pick<T, K> {
    const returnValue = {} as Pick<T, K>;
    keys.forEach(k => {
        returnValue[k] = source[k];
    });
    return returnValue;
}

answered Nov 20, 2017 at 18:25

artem

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1 Comment

Tarık İNCE Over a year ago

Thank you very much for your answer. It has solved my problem. And also thank you for your explanation.

mickaelw · Accepted Answer · 2017-11-20 10:58:04Z

0

I think you can do a simple mapper between your objects:

interface OriginalObject {
  property1: string
  property2: string
  property3: string
}

interface MappedObject {
   mappedProperty1: string,
   mappedProperty2: string
}

class Mapper {
   static mapOriginalObjectToMappedObject(object: OrginalObject): MappedObject {
      return {
         mappedProperty1: object.property2, 
         mappedProperty2: object.property3
      }
   }

   static mapFooToBaz ...
   static mapBarToFoo ...
}

In your code:

const objectA: OriginalObject = {
  property1: "foo",
  property2: "bar",
  property3: "baz"
}
const mappedObjectA = Mapper.mapOriginalObjectToMappedObject(objectA)

With this way, your code stay simple

answered Nov 20, 2017 at 10:58

mickaelw

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1 Comment

Tarık İNCE Over a year ago

Thank you for your response but the thing is, there are lots of combinations (nearly 240 different possibilities in usage) and that much code would be redundancy I think.

andrea06590 · Accepted Answer · 2017-11-20 18:35:55Z

0

If you want to "parse" two objects using same key, you should use Object.keys

let's assume you have two objects myObject and myOtherObj, if the keys are the same just use only one mapper :

Object.keys(myObject).map((key) => {
   console.log(key) // should be the same for both obj
   // your two obj values
   console.log(myObject[key]);
   console.log(myOtherObj[key]);
});

EDIT : you can also do Object.keys(myOtherObj).map....

answered Nov 20, 2017 at 18:35

andrea06590

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1 Comment

Tarık İNCE Over a year ago

Thank you but it is not actually for parsing/joining or mapping. Just picking some properties, say, "dynamically".

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