CODEWITHSUNDEEP
javascriptnode.js
kentor
kentor

Reputation: 18534

Most performant way to crawl json object array

I've got a list with ~1000 possible file extensions and it's aliases. The structure looks like this:

var iconMap = [
    { icon: 'photoshop', extensions: ['psd'], format: 'svg' },
    { icon: 'photoshop2', extensions: ['psd'], format: 'svg', disabled: true },
    { icon: 'php', extensions: ['php1', 'php2', 'php3', 'php4', 'php5', 'php6', 'phps', 'phpsa', 'phpt', 'phtml', 'phar'], format: 'svg' }]

I want to get the 'icon' property for the according extension. Therefore I want to create a getExtensionIcon(fileExtension) function which should return:

var phpIcon = getExtensionIcon('php') // 'php'
var pharIcon = getExtensionIcon('phar') // 'php'
var php5Icon = getExtensionIcon('php5') // 'php'

My question:

I only got a few ideas which would be very ineffecient, so I am curious how you would realize the getExtensionicon function?

Upvotes: 0

Views: 667

Answers (2)

Jaromanda X
Jaromanda X

Reputation: 1

Here is how I would do it - this "pre-processes" the iconMap once, creating an object that keys extension with a value of icon

var iconMap = [
    { icon: 'photoshop', extensions: ['psd'], format: 'svg' },
    { icon: 'photoshop2', extensions: ['psd'], format: 'svg', disabled: true },
    { icon: 'php', extensions: ['php1', 'php2', 'php3', 'php4', 'php5', 'php6', 'phps', 'phpsa', 'phpt', 'phtml', 'phar'], format: 'svg' }
];
var getExtensionIcon = (src => {
    var extensions = src.filter(item => !item.disabled).reduce((result, item) => {
        item.extensions.forEach(ext => {
            result[ext] = item.icon;
        });
        return result;
    }, {});
    return ext => extensions[ext];
})(iconMap);
console.log('psd is', getExtensionIcon('psd'));
console.log('php1 is', getExtensionIcon('php1'));

Upvotes: 1

Luca Kiebel
Luca Kiebel

Reputation: 10096

If you can use jQuery, it has the $.grep() function which seems like a good fit for this purpose:

function getExtensionIcon(arr, lang){
    var result = $.grep(arr, function(e){ return e.extensions.indexOf(lang) != -1; });
    return result[0].icon;
}

I don't really know how performant this is, but you should at least give it a try ;)

Testing with your Example:

function getExtensionIcon(arr, lang){
    var result = $.grep(arr, function(e){ return e.extensions.indexOf(lang) != -1; });
    return result[0].icon;
}

var iconMap = [
    { icon: 'photoshop', extensions: ['psd'], format: 'svg' },
    { icon: 'photoshop2', extensions: ['psd'], format: 'svg', disabled: true },
    { icon: 'php', extensions: ['php1', 'php2', 'php3', 'php4', 'php5', 'php6', 'phps', 'phpsa', 'phpt', 'phtml', 'phar'], format: 'svg' }];
    
console.log(getExtensionIcon(iconMap, "php6"));
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Upvotes: 0

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