Need JS array negative filter based on another array

Question

I need help on filtering a Javascript array to remove names that DO NOT include at least one of a series of letters listed in a second array. I have tried over and over, using combinations of filter(), and also nested FOR loops. EXAMPLE:

name = ["chuck","lina","wanda","denise","reggie","candy"];
letter = ["h","l","z","e"];

The array returned by filter() should be:
nametwo = ["wanda","candy"];

It's simple to explain but I just cannot make it work in code. I've found many examples of filter() on here but none has helped. Thanks!

HI all - thanks for your responses. All good... <3<3<3 you all. :-) — cmitt0
– cmitt0, Commented Feb 5, 2022 at 17:39

cmgchess · Accepted Answer · 2022-02-05 16:31:51Z

1

Basically the same question here

const name = ["chuck","lina","wanda","denise","reggie","candy"];
const letter = ["h","l","z","e"];

const checker = value =>
  !letter.some(element => value.includes(element));

console.log(name.filter(checker));

answered Feb 5, 2022 at 16:31

cmgchess

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Maxim Petrikov · Accepted Answer · 2022-02-05 16:45:59Z

1

For searching letters inside a string, you should use indexOf function.

If you want one line solution you can do it this way:

const names = ["chuck","lina","wanda","denise","reggie","candy"];
const letters = ["h","l","z","e"];
const result = names.filter(name => letters.filter(letter => name.indexOf(letter) !== -1).length === 0);

console.log(result)

And you can do it use two cycles:

const names = ["chuck","lina","wanda","denise","reggie","candy"];
const letters = ["h","l","z","e"];

const result = [];
for (let i = 0; i < names.length; i++) {
    let isContainLetter = false;
    for (let j = 0; j < letters.length; j++) {
      if (names[i].indexOf(letters[j]) !== -1) {
        isContainLetter = true;
        break;
      }
    }
  if (!isContainLetter) result.push(names[i]);
}

console.log(result)

edited Feb 5, 2022 at 16:45

answered Feb 5, 2022 at 16:39

Maxim Petrikov

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1 Comment

Noam Over a year ago

I like this answer because it explains the logic, not just provides code.

Nina Scholz · Accepted Answer · 2022-02-05 17:14:56Z

1

You could filter with every and check if the name does not contains letters.

const
    names = ["chuck", "lina", "wanda", "denise", "reggie", "candy"],
    letters = ["h", "l", "z", "e"],
    result = names.filter(n => letters.every(l => !n.includes(l)));

console.log(result); // ["wanda", "candy"]

answered Feb 5, 2022 at 17:14

Nina Scholz

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1 Comment

cmitt0 Over a year ago

Brilliant, absolutely brilliant. Works like a charm. Wow. Thank you!

Rohìt Jíndal · Accepted Answer · 2022-02-08 16:01:35Z

0

We can use :

Array.every() method to test whether all elements in the array pass the test implemented by the provided function or not. It returns a Boolean value.
Then we can use Array.filter() method which creates a new array with all elements that pass the test implemented by the provided function.

Working Demo :

const name = ["chuck","lina","wanda","denise","reggie","candy"];
const letter = ["h","l","z","e"];

const resultArr = name.filter((word) => letter.every((char) => !word.includes(char)));

console.log(resultArr);

answered Feb 8, 2022 at 16:01

Rohìt Jíndal

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