Using underscore js reduce

Question

How does underscorejs reduce work?

It's simple to get _.reduce([1, 2, 3], function(memo, num){ return memo + num; }, 0); (the result is 6).

But how do the other optional parameters work? In the docs it says:

Memo is the initial state of the reduction, and each successive step of it should be returned by iterator. The iterator is passed four arguments: the memo, then the value and index (or key) of the iteration, and finally a reference to the entire list."

But I don't understand. I tried to use reduce for the following problem, and I couldn't figure it out:

var input = [{"score": 2, "name": "Jon", "venue": "A"}, {"score": 3, "name": "Jeff", "venue":"A"}, {"score": 4, "name": "Jon", "venue":"B"}, {"score": 4, "name": "Jeff", "venue":"B"}];

var output = [{"score": 6, "name":"Jon", "venue": ["A", "B"]}, {"score": 7, "name":"Jeff", "venue": ["A", "B"]}];

How can I get as output using _reduce for input? And it will really helpful how it works inside reduce.

Please show what you tried so that we can see specifically what you're misunderstanding. The memo is simply the initial value at first, then the previous return value for the subsequent iterations. The rest of the arguments are the typical args you'd have for methods like .map() or .forEach() — user2437417
– user2437417, Commented Aug 1, 2013 at 22:08

Alberto · Accepted Answer · 2013-08-01 22:22:15Z

10

Reduce takes a list of values, and reduces it to a single value. What you are trying is not just reduce. You are trying to first group (by name) and then reduce each group. A possible implementation would be something like this, where you first group, and then map each group to a reduce operation that accumulates the score and appends the venue.

var input = [
    {"score": 2, "name": "Jon", "venue": "A"}, 
    {"score": 3, "name": "Jeff", "venue":"A"}, 
    {"score": 4, "name": "Jon", "venue":"B"}, 
    {"score": 4, "name": "Jeff", "venue":"B"}];

var output = _.map(_.groupBy(input, "name"), function(group, name) {
    return _.reduce(group, function(memo, elem) { 
            memo.score += elem.score;
            memo.venue.push(elem.venue);
            return memo;
        },
        { name: name, score: 0, venue: [] });
});

answered Aug 1, 2013 at 22:22

Alberto

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Serabe · Accepted Answer · 2013-08-01 22:25:19Z

Instead of reduce, try using plain and simple each in this way:

_.each admits a third parameter for a context. So, for example:

var input = ["Alice", "Bob", "Charlie"];
var context = {};

_.each(input, function(o,i) { this[i] = o; }, context);

console.log(context) //=> {1: "Alice", 2: "Bob", 3: "Charlie"}

Ruby has a new method call each_with_object similar to reduce for that exact pattern, where the memo is not necessarily the value return (in fact, my function is not returning anything!); but underscore simplifies its API design simply admitting a context for the function as the third parameter.

You can use reduce, but you would need to return the memo each time, making it a bit less elegant IMHO.

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