Check if there are a list of words contained in a string

Question

Let's consider the following string text "I just tried the new burgers in burger land!" and the following array of strings ["eat", "burger", "fries"]

What I want is to know if any of the words of text exist in the array, but in a optimized way... not just with two for-loops...

Any ideas?

Actually, hardcoding the result is the most optimized way :P at least for one example, anyway... — Patrick Roberts, Mar 15 '19 at 15:43
[.indexOf](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/indexOf) — j08691, Mar 15 '19 at 15:44

Get Off My Lawn · Accepted Answer · 2019-03-18T15:59:36.670

1

You could do this via regular expression by joining the array to create a string.

RegEx References

Word Boundary – Allows you to perform a "whole words only" search
Non-Capturing Groups – This is a group that is not stored in memory to be referenced later on.
Pipe Symbol – Search for the literal text

Once the RegEx string is created via JavaScript it looks like this (string escaping removed):

\b(?:eat|burger|fries)\b

const str = 'I just tried the new burgers in burger land!'
const arr = ['eat', 'burger', 'fries']

// "\b" This is a word boundary - Allows you to perform a "whole words only" search
// "(?:xxx)" This is a non-capturing group. It's not stored in memory to be referenced later on.
// 'arr.join()' creates the following string: "eat|burger|fries"
console.log(str.match(new RegExp(`\\b(?:${arr.join('|')})\\b`, 'ig')))

If needing to escape regex is a concern, you can do it by escaping special characters. and mapping over the array of words that need to be tested.

const str = 'I just tried the new burgers in burger land!'
const arr = ['eat', 'burger', 'fries', '(']
const esc = s => s.replace(/[.*+?^${}()|[\]\\]/g, '\\$&')

console.log(str.match(new RegExp(`\\b(?:${arr.map(esc).join('|')})\\b`, 'ig')))

edited Mar 18 '19 at 15:59

answered Mar 15 '19 at 16:02

Get Off My Lawn

34,175
38
176
338

I guess you can just drop the `\\b` in join. simply `join('|')` will be enough as you already have `word boundaries` at start and end. – Code Maniac Mar 15 '19 at 16:19
That picks up `burger` and `burgers`... not sure if that is the desired result. – Get Off My Lawn Mar 15 '19 at 16:27
That can be handled with capturing group. `\\b(?:${arr.join('|')})\\b` you can try like this – Code Maniac Mar 15 '19 at 16:32
Anyways nice answer +1 – Code Maniac Mar 15 '19 at 16:34
This works only as long as none of your strings has any special RegExp characters. Consider `arr = ['(']`. – Jordan Running Mar 15 '19 at 16:55
That can be handled by implementing this: https://stackoverflow.com/questions/3446170/escape-string-for-use-in-javascript-regex – Get Off My Lawn Mar 15 '19 at 16:57
Hi @GetOffMyLawn and thanks for your answer! Could you please explain the RegExp part, because I haven't used it before? What does each character do? Thanks in advance! – Sotiris Kaniras Mar 16 '19 at 14:20

Code Maniac · Answer 2 · 2019-03-15T16:25:37.903

Well the most optimized way IMO is to have wordsTest as object instead of array so lookup becomes O(1) operation. and than you can use filter to get any matching value.

const test = "I just tried the new burgers in burger land!";
const wordsTest = {"eat":1, "burger":1, "fries":1}

const wordsInString = (test) => test.split(' ').filter(e => wordsTest[e] )

console.log(wordsInString(test))
console.log(wordsInString('No matching words'))

prasanth · Answer 3 · 2019-03-15T15:49:32.193

0

use Array#Filter and indexOf().

Array#Filter is filter the array based index match in string
use toLowerCase() for match case sensitive string

var a ="I just tried the new burgers in burger land!";
var arr = ["eat", "burger", "fries"];

var res = arr.filter(i=> a.toLowerCase().indexOf(i.toLowerCase()) > -1)
console.log(res)

edited Mar 15 '19 at 15:49

answered Mar 15 '19 at 15:47

prasanth

22,145
4
29
53

1

I like the way you used filter – Saurabh Sarathe Mar 15 '19 at 15:49

Check if there are a list of words contained in a string

3 Answers3