Regex and Array: Most efficient approach

Question

I want to get all words with <=3 letters from an array.

Is iterating the array and checking every entry an efficient approach?

arr = ["cat", "apple", "window", "dog"];

for(i=0; i<arr.length; i++){
    if(arr[i].match(/^[a-z]{3}$/)){
        console.log(arr[i])
    }
}

//returns: "cat" and "dog"

Or is there already a built-in function that does the job for me, so I don't have to explicitly define a for-loop and and if-statement.

It's more about working with regex and arrays. What kind of regex will be executed is less important in my example. — Evgenij Reznik, Feb 17 '16 at 22:59
Maybe you could use `RegRxp.test(string)`, see [*regex.test V.S. string.match to know if a string matches a regular expression*](http://stackoverflow.com/questions/10940137/regex-test-v-s-string-match-to-know-if-a-string-matches-a-regular-expression). — Wiktor Stribiżew, Feb 17 '16 at 23:00
So you want to use a regular expression for the sake of using aregular expression? OK. — Pointy, Feb 17 '16 at 23:01
@Pointy: I'm looking for the most efficient way to process an array with regex. What kind of regular expression it will be (finding <=3 words; finding all words starting with an a, etc.) is not important. This was just an example. — Evgenij Reznik, Feb 17 '16 at 23:03
@user1170330 no, you're not. your question clearly states you want all elements in an array with some kind of length restriction. This has nothing to do with needing regexp; just use `filter`, it's baked into JS and has been for years. — Mike 'Pomax' Kamermans, Feb 17 '16 at 23:04
How about editing the question to say "What's the most cpu-efficient way to process an array with regex?" Then you'll avoid having to explain that you really don't care about the length of the array elements. — Jonathan M, Feb 17 '16 at 23:06
The questions is unclear as it's written. The part about the length is largely irrelevant (as per your comments). — Evan Trimboli, Feb 17 '16 at 23:10

Shafizadeh · Accepted Answer · 2016-02-17T23:37:05.473

3

No need to regex, try this:

var arr = ["cat", "apple", "window", "dog"];
var len = arr.length;

for(len; len<=0; len--){
    if(arr[len].length <= 3){
        console.log(arr[len]);
    }
}

Edit: If you don't want to explicitly define a for-loop and and if statement then try using .filter() and .match():

var arr = ["cat", "apple", "window", "dog"];
var AcceptedItems = arr.filter(function(item) {
    return item.match(/^[a-z]{1,3}$/);
});

edited Feb 17 '16 at 23:37

answered Feb 17 '16 at 23:01

Shafizadeh

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Sadly, the question is posed so poorly that this answer does not hit the intended target, even though it solves the problem posed in the wording of the question. One must dive into the comments to figure out what the OP really wants. – Jonathan M Feb 17 '16 at 23:10
@JonathanM I read the comments and edited my answer ... thank you – Shafizadeh Feb 17 '16 at 23:14
@Shafizadeh: Your 2nd answer is exactly what I wanted. Thank you! – Evgenij Reznik Feb 17 '16 at 23:17
@user1170330 Glad to help `:-)` – Shafizadeh Feb 17 '16 at 23:18

score 2 · Answer 2 · answered Feb 17 '16 at 23:05

2

I assume you want to return words that are even like '1abc' since it has only 3 letters.

The below will work, using filter function of an array.

arr = ["cat", "apple", "window", "dog", "1acb", "1abcd"];

function getLessThanThree(anArray){
    return anArray.filter(function(v){
        return !( /[a-zA-Z]{3}./.exec(v) );
    })
}

console.log(getLessThanThree(arr)); // logs [ 'cat', 'dog', '1acb' ]

Test it here

answered Feb 17 '16 at 23:05

Gabs00

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+1 for answering the OP with a regular expression, without a for loop, and using a regex which can be arbitrarily changed from a length check to any other regex validation method that he may need to implement. – OnlineCop Feb 17 '16 at 23:11

Matías Fidemraizer · Answer 3 · 2016-02-18T07:00:41.950

1

If you want to get items matching something in array you need to iterate it.

Anyway, since ECMA-Script 5 you would refactor your code as follows:

var results = arr.filter(function(item) { return item.lengh <= 3; });

Or in ECMA-Script 6 and above:

var results = arr.filter(item => item.length <= 3);

In terms of productivity, you're going to be faster.

edited Feb 18 '16 at 07:00

answered Feb 17 '16 at 23:03

Matías Fidemraizer

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*I want to get all words with `<=3` letters from an array.* The `<=3 letters` requirement means you need `item.test(/^[a-z]{1,3}$/i);`. Your `/^[a-z]{3}$/` matches exactly 3 lowercase letter words. – Wiktor Stribiżew Feb 17 '16 at 23:11
See what OP says. OP isn't asking for a better regexps but a faster way to find resutls.. – Matías Fidemraizer Feb 17 '16 at 23:11
No, someone baffles OP with unnecessary comments that a regex is not necessary here. – Wiktor Stribiżew Feb 17 '16 at 23:13

score 0 · Answer 4 · answered Feb 17 '16 at 23:03

You're in JavaScript. It can already do this, no regexp required.

var arr = ["cat", "apple", "window", "dog"];
var filtered = arr.filter(function(e) { return e.length <= 3; });

done. Use ES6 for even simpler syntax:

var filtered = arr.filter(e => e.length <= 3);

score 0 · Answer 5 · answered Feb 17 '16 at 23:06

0

Lots of ways to skin this cat, I like:

var arr = ["cat", "apple", "window", "dog"];
var upToThreeLetters = function(word) {
  return word.length <= 3;
}
var filteredArr = arr.filter(upToThreeLetters);

answered Feb 17 '16 at 23:06

linuxdan

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Regex and Array: Most efficient approach

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