不需要泛型函数，只需写例如：

const { omit1, omit2, ...filtered } = source

filtered将包含source除omit1和omit2之外的所有属性。
如果你有字符串数组的键，你可以使用这样的方法：

function omit<T extends object, P extends keyof T>(obj: T, okeys: P[]): Omit<T, P> {
  const ret = { ...obj }
  okeys.forEach((key) => delete ret[key])
  return ret
}

可以使用Object.entries和Object.fromEntries过滤掉对象的属性。它也会使打字变得不那么麻烦。你只需要对最终结果进行强制转换：

function omit2<T extends {}, P extends keyof T>(obj: T, okeys: P[]): Omit<T, P> {
  const b = Object.fromEntries(Object.entries(obj).filter(([k, _]) => !okeys.includes(k as P)))
  return b as Omit<T, P>;
}

Playground链接

export function omit<T extends Object, P extends keyof T>(obj: T, ...omitkeys: P[]): Omit<T, P> {
    return (Object.keys(obj)).reduce((newObj, key) => {
        for (const omitKey of omitkeys) {
            if (omitKey != key) {
                newObj[key as Exclude<keyof T, P>] = obj[key as Exclude<keyof T, P>];
            }
        }
        return newObj;
    }, {} as Omit<T, P>)
}