Remove all matching elements from array

Question

I have this array:

["onelolone", "twololtwo", "three"]

How can I remove the elements using only 'lol' as value and as result be:

["three"]

• Achtung: Does this answer your question? Remove all elements from array that match specific string
• Hef: You wanna remove the items that don't contain "lol"?
• S. Walker: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

Answer 1

Use Array.prototype.filter, and check that it doesn't include "lol":

const result = ["onelolone","twololtwo","three"].filter(ele => !ele.includes("lol"))
console.log(result)// ["three"]

• Gabriel: Nice exactly what I been looking for. I have a question, if I want to include more values like ("lol","th") and expect = console.log(result)// [ ]. How can that be achieved?
• Aplet123: .filter(ele => !["lol", "th"].some(phrase => ele.includes(phrase))

Answer 2

Array#filter is used to keep elements that satisfy a predicate. But we can build a reject function that removes the elements that satisfy a predicate:

const reject = fn => xs => xs.filter(x => fn(x) === false);

Then we can build a curried function for matching a string:

const matches = x => y => y.includes(x);

const reject_lol = reject(matches("lol"));
reject_lol(["onelolone", "twololtwo", "three"]);
//=> ["three"]

We can tweak matches to support any number of strings:

const matches = (...xs) => y => xs.some(x => y.includes(x));

const reject_abbr = reject(matches("lol", "wtf"));
reject_abbr(["lolone", "twowtf", "three"]);
//=> ["three"]