TypeScript Type-safe Omit Function

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;
}

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

Simplest way:

export const omit = <T extends object, K extends keyof T>(obj: T, ...keys: K[]): Omit<T, K> => {
  keys.forEach((key) => delete obj[key])
  return obj
}

As a pure function:

export const omit = <T extends object, K extends keyof T>(obj: T, ...keys: K[]): Omit<T, K> => {
  const _ = { ...obj }
  keys.forEach((key) => delete _[key])
  return _
}

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];
    };
}