Javascript Question posted in
A very good W3school tutorial can be found here.

How to remove multiply dictionary keys from array of dictionaries in JavaScript?

I have

let Remove = ["Last", "Gender"];
let Contact = [{"First": "Bob", "Last": "Jim", "Gender": "M"}, 
                {"First": "Amy", "Last": "Christen", "Gender": "F"}];

How I remove multiple keys from the dictionary with the keys I want to remove stored in an array?

To get

[{"First": "Bob"}, {"First": "Amy"}]

TS Playground

Tags:

3

Answers


  1. 0

    You could map the objects and reduce the copy by removing keys.

    const remove = ["Last", "Gender"];
const contacts = [
  {"First": "Bob", "Last": "Jim", "Gender": "M"}, 
  {"First": "Amy", "Last": "Christen", "Gender": "F"}
];

const removed = contacts.map(contact => {
  return remove.reduce((copy, keyToRemove) => {
    delete copy[keyToRemove];
    return copy;
  }, { ...contact });
});

console.log(removed);
    .as-console-wrapper { top: 0; max-height: 100% !important; }

    Here is a TS implementation. You can use a Record<string, string> type to represent your object.

    const remove: string[] = ["Last", "Gender"];

const contacts = [
  {"First": "Bob", "Last": "Jim", "Gender": "M"}, 
  {"First": "Amy", "Last": "Christen", "Gender": "F"}
];

const removed = prune(contacts, remove);

console.log(removed);

function prune<TObj extends Record<string, string>>(
        records: TObj[], keysToRemove: string[]): TObj[] {
    return records.map(contact => {
        return keysToRemove.reduce((copy: TObj, keyToRemove: string) => {
            delete copy[keyToRemove]; // Remove they key from the copy
            return copy;
        }, { ...contact });
    });
}
    Login or Signup to reply.
  2. 0

    Another possible solution using forEach and for cycles:

    let Remove = ["Last", "Gender"];

let Contact = [{"First": "Bob", "Last": "Jim", "Gender": "M"}, 
                {"First": "Amy", "Last": "Christen", "Gender": "F"}];

//Want [{"First": "Bob"}, {"First": "Amy"}]      
Contact.forEach((oneContact) => {
  for (let i=0; i<Remove.length; i++) {
    delete oneContact[Remove[i]];
  }
})
console.log(Contact);
    .as-console-wrapper { top: 0; max-height: 100% !important; }
    Login or Signup to reply.
  3. 0

    You can use .map() on your array to map each object to a new object without the keys specified in remove. For each object, you can grab the entries which you can then filter to remove the entries that have a key within your remove array. Once you’ve removed the entries you don’t need, you can rebuild a new object by using Object.fromEntries():

    const res = contact.map(obj => Object.fromEntries(Object.entries(obj).filter(([key]) => !remove.includes(key))));

    const remove = ["Last", "Gender"];
const contact = [{"First": "Bob", "Last": "Jim", "Gender": "M"}, {"First": "Amy", "Last": "Christen", "Gender": "F"}];

const res = contact.map(obj => Object.fromEntries(
  Object.entries(obj)
    .filter(([key]) => !remove.includes(key))
));
console.log(res);

    TS Playground Link

    Login or Signup to reply.
Please signup or login to give your own answer.

Back To Top
Search