Remove items from array with splice in for loop

Question

I want to implement a kind of jQuery live search. But before sending the input to the server I'd like to remove all items in my array which have 3 or less characters (because in the german language, those words usually can be ignored in terms of searching) So ["this", "is", "a", "test"] becomes ["this", "test"]

$(document).ready(function() {
var timer, searchInput;
$('#searchFAQ').keyup(function() {
    clearTimeout(timer);
    timer = setTimeout(function() {
        searchInput = $('#searchFAQ').val().match(/\w+/g);
        if(searchInput) {
            for (var elem in searchInput) {
                if (searchInput[elem].length < 4) {
                    //remove those entries
                    searchInput.splice(elem, 1);
                }
            }
            $('#output').text(searchInput);
            //ajax call here
        }
    }, 500);
});
});

Now my problem is that not all items get removed in my for loop. If I for example typ "this is a test" "is" gets removed, "a" stays. JSFIDDLE

I think the problem is the for loop because the indexes of the array change if I remove an item with splice, so it goes on with the "wrong" index.

Perhaps anybody could help me out?

Solution 1

Solution 1

You can loop backwards, with something like the following:

var searchInput, i;

searchInput = ["this", "is", "a", "test"];
i = searchInput.length;
while (i--) {
    if (searchInput[i].length < 4) {
        searchInput.splice(i, 1);
    }
}

DEMO: http://jsfiddle.net/KXMeR/

This is because iterating incrementally through the array, when you splice it, the array is modified in place, so the items are "shifted" and you end up skipping the iteration of some. Looping backwards (with a while or even a for loop) fixes this because you're not looping in the direction you're splicing.

---

Solution 2

At the same time, it's usually faster to generate a new array instead of modifying one in place. Here's an example:

var searchInput, newSearchInput, i, j, cur;

searchInput = ["this", "is", "a", "test"];
newSearchInput = [];
for (i = 0, j = searchInput.length; i < j; i++) {
    cur = searchInput[i];
    if (cur.length > 3) {
        newSearchInput.push(cur);
    }
}

where newSearchInput will only contain valid length items, and you still have the original items in searchInput.

DEMO: http://jsfiddle.net/RYAx2/

---

Solution 3

In addition to the second solution above, a similar, newer Array.prototype method is available to handle that better: filter. Here's an example:

var searchInput, newSearchInput;

searchInput = ["this", "is", "a", "test"];
newSearchInput = searchInput.filter(function (value, index, array) {
    return (value.length > 3);
});

DEMO: http://jsfiddle.net/qky7D/

---

References:

• Array.prototype.filter - https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter
• Array.prototype.filter browser support - http://kangax.github.io/es5-compat-table/#Array.prototype.filter

Solution 2

var myArr = [0,1,2,3,4,5,6];

Problem statement:

myArr.splice(2,1);

  \\ [0, 1, 3, 4, 5, 6];

now 3 moves at second position and length is reduced by 1 which creates problem.

Solution: A simple solution would be iterating in reverse direction when splicing.

var i = myArr.length;
while (i--) {
    // do your stuff
}

Solution 3

If you have the lodash library installed, they have a sweet gem you might want to consider.

The function is _.forEachRight (iterates over elements of a collection from right to left)

Here is an example.

var searchInput = ["this", "is", "a", "test"];

_.forEachRight(searchInput, function(value, key) {

    if (value.length < 4) {
        searchInput.splice(key, 1);
    }
});

Solution 4

You can also use $.grep function to filter an array:

var timer, searchInput;
$('#searchFAQ').keyup(function () {
    clearTimeout(timer);
    timer = setTimeout(function () {
        searchInput = $('#searchFAQ').val().split(/\s+/g); // match is okay too
        searchInput = $.grep(searchInput, function(el) {
            return el.length >= 4;
        });
        console.log(searchInput);
    }, 500);
});

http://jsfiddle.net/dfsq/4Wdp9/

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