CODEWITHSUNDEEP
javascript
whytheq
whytheq

Reputation: 35557

Can I make this function more concise

I have this javascript snippet and am wondering if I can calculate amount and users in a single pass of the reduce function?

root.children.forEach(function(v) {

    v.amount = v.children.reduce(function(a, b) {
        console.log(a);
        return {
            amount: a.amount + b.amount
        }
    }, {
        'amount': 0
    }).amount

    v.users = v.children.reduce(function(a, b) {
        console.log(a);
        return {
            users: a.users + b.users
        }
    }, {
        'users': 0
    }).users

})

Upvotes: 1

Views: 81

Answers (3)

Nina Scholz
Nina Scholz

Reputation: 386570

You could a single Array#forEach loop.

var root = {
        children: [
            { children: [
                { amount: 2, users: 3 },
                { amount: 7, users: 5 }
            ]}
        ]
    };

root.children.forEach(function(v) {
    v.amount = 0;
    v.users = 0;

    v.children.forEach(function(a) {
        v.amount += a.amount;
        v.users += a.users;
    });
});
console.log(root);
.as-console-wrapper { max-height: 100% !important; top: 0; }

Upvotes: 1

Cerbrus
Cerbrus

Reputation: 72857

It looks like you can literally combine the two methods into one:

root.children.forEach(function(v) {        
    var result = v.children.reduce(
        function(a, b) {
            return {
                amount: a.amount + b.amount,
                users: a.users + b.users
            };
        },
        { amount: 0, users: 0 }
    ); // ^ Note that I left out the quotes there. In this case, they're optional.

    v.amount = result.amount;
    v.users= result.users;
});

Upvotes: 6

Rajaprabhu Aravindasamy
Rajaprabhu Aravindasamy

Reputation: 67207

Yes you can do that like below,

root.children.forEach(function(v) {
    var obj = v.children.reduce(function(a, b) {
        a.amount += b.amount;
        a.users += a.users;
    }, {'amount': 0, 'users' : 0 });
    v.amount = obj.amount;
    v.users = obj.users;
});

Upvotes: 6

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