TypeScript Type-safe Omit Function

Question

I want to replicate lodash's _.omit function in plain typescript. omit should return an object with certain properties removed specified via parameters after the object parameter which comes first.

Here is my best attempt:

function omit<T extends object, K extends keyof T>(obj: T, ...keys: K[]): {[k in Exclude<keyof T, K>]: T[k]} {
    let ret: any = {};
    let key: keyof T;
    for (key in obj) {
        if (!(keys.includes(key))) {
            ret[key] = obj[key];
        }
    }
    return ret;
}

Which gives me this error:

Argument of type 'keyof T' is not assignable to parameter of type 'K'.
  Type 'string | number | symbol' is not assignable to type 'K'.
    Type 'string' is not assignable to type 'K'.ts(2345)
let key: keyof T

My interpretation of the error is that:

1. Since key is a keyof T and T is an object, key can be a symbol, number or string.

2. Since I use the for in loop, key can only be a string but includes might take a number if I pass in an array, for example? I think. So that means there's a type error here?

Any insights as to why this doesn't work and how to make it work are appreciated!

Solution 1

interface Omit {
    <T extends object, K extends [...(keyof T)[]]>
    (obj: T, ...keys: K): {
        [K2 in Exclude<keyof T, K[number]>]: T[K2]
    }
}

const omit: Omit = (obj, ...keys) => {
    const ret = {} as {
        [K in keyof typeof obj]: (typeof obj)[K]
    };
    let key: keyof typeof obj;
    for (key in obj) {
        if (!(keys.includes(key))) {
            ret[key] = obj[key];
        }
    }
    return ret;
};

For convenience I've pulled most of the typings to an interface.

The problem was that K had been being inferred as a tuple, not as a union of keys. Hence, I changed it's type constraint accordingly:

[...(keyof T)[]] // which can be broke down to:
keyof T // a union of keys of T
(keyof T)[] // an array containing keys of T
[...X] // a tuple that contains X (zero or more arrays like the  described one above)

Then, we need to transform the tuple K to a union (in order to Exclude it from keyof T). It is done with K[number], which is I guess is self-explaining, it's the same as T[keyof T] creating a union of values of T.

Playground

Solution 2

Object.keys or for in returns keys as string and excludes symbols. Numeric keys are also converted to strings.

You need to convert numeric string keys to numbers otherwise it will return the object with string keys.

function omit<T extends Record<string | number, T['']>,
 K extends [...(keyof T)[]]>(
    obj: T,
    ...keys: K
): { [P in Exclude<keyof T, K[number]>]: T[P] } {
    return (Object.keys(obj)
         .map((key) => convertToNumbers(keys, key)) as Array<keyof T>)
        .filter((key) => !keys.includes(key))
        .reduce((agg, key) => ({ ...agg, [key]: obj[key] }), {}) as {
        [P in Exclude<keyof T, K[number]>]: T[P];
    };
}

function convertToNumbers(
    keys: Array<string | number | symbol>,
    value: string | number
): number | string {
    if (!isNaN(Number(value)) && keys.some((v) => v === Number(value))) {
        return Number(value);
    }

    return value;
}


// without converToNumbers omit({1:1,2:'2'}, 1) will return {'1':1, '2':'2'}
// Specifying a numeric string instead of a number will fail in Typescript

To include symbols you can use the code below.

function omit<T, K extends [...(keyof T)[]]>(
    obj: T,
    ...keys: K
): { [P in Exclude<keyof T, K[number]>]: T[P] } {
    return (Object.getOwnPropertySymbols(obj) as Array<keyof T>)
        .concat(Object.keys(obj)
        .map((key) => convertToNumbers(keys, key)) as Array<keyof T>)
        .filter((key) => !keys.includes(key))
        .reduce((agg, key) => ({ ...agg, [key]: obj[key] }), {}) as {
        [P in Exclude<keyof T, K[number]>]: T[P];
    };
}

Solution 3

The accepted answer from Nurbol above is probably the more typed version, but here is what I am doing in my utils-min.

It uses the typescript built-in Omit and is designed to only support string key names. (still need to loosen up the Set to Set, but everything else seems to work nicely)

export function omit<T extends object, K extends Extract<keyof T, string>>(obj: T, ...keys: K[]): Omit<T, K> {
  let ret: any = {};
  const excludeSet: Set<string> = new Set(keys); 
  // TS-NOTE: Set<K> makes the obj[key] type check fail. So, loosing typing here. 

  for (let key in obj) {
    if (!excludeSet.has(key)) {
      ret[key] = obj[key];
    }
  }
  return ret;
}

Comments

[Salami]

Yes I don't think it's a good idea either, because then omit won't work for regular arrays.

[Andreas Linnert]

ESLint complains about T extends object but this can be replaced with T extends Record<string, unknown> and it still works.

[cvss]

Very useful. Just a small correction, in convertToNumbers method, the comparison type should be == instead of === as it will return false("2" === 2).

[Baptiste Arnaud]

Cheers mate. I didn't know I needed this function so much 🥰

[Lonli-Lokli]

Thanks about sample, I rewrote it a little bit to export const copyWithout = <T extends Record<string, unknown>, K extends [...(keyof T)[]]>( source: T, ...toExclude: K ) => { return Object.keys(source).filter(key => !toExclude.includes(key)).reduce<{ [K2 in Exclude<keyof T, K[number]>]: T[K2]; }>((acc, curr) => { return { ...acc, [curr]: source[curr] }; }, {} as { [K in keyof typeof source]: typeof source[K] }); };

[Per Enström]

The resulting type when calling this omit function is just an empty interface {}, not the type I'm expecting. Is this expected, or is there a workaround for that?