I do have an array of objects and I want to create a filter function to return an new array

var items = [{
        "row": 0,
        "column": 0,
        "items": [2, 3, 4, 5, 6, 7]
    }, {
        "row": 0,
        "column": 1,
        "items": [8, 9, 10, 11, 12, 13]
            ....



        {
            "row": 2,
            "column": 2,
            "items": [50, 51, 52, 53, 54, 55]
        }]

    var newArray = function (items, row) {
        //filter items and return new array

        return filtered
    }

newArray should contain all values from 'items' that have the same row value.

I do have an array of objects and I want to create a filter function to return an new array

var items = [{
        "row": 0,
        "column": 0,
        "items": [2, 3, 4, 5, 6, 7]
    }, {
        "row": 0,
        "column": 1,
        "items": [8, 9, 10, 11, 12, 13]
            ....



        {
            "row": 2,
            "column": 2,
            "items": [50, 51, 52, 53, 54, 55]
        }]

    var newArray = function (items, row) {
        //filter items and return new array

        return filtered
    }

newArray should contain all values from 'items' that have the same row value.

• javascript
• arrays
• underscore.js

Share Improve this question Follow edited Dec 5, 2013 at 13:48 Dhaval Marthak 17.4k66 gold badges4848 silver badges6969 bronze badges asked Dec 5, 2013 at 13:45 lunacafulunacafu 36688 silver badges2121 bronze badges 7

• Hey, wele to StackOverflow. It would help if you explained what you have tried so far and why it didn't work for you. – Tibos Commented Dec 5, 2013 at 13:46
• Tried to group them with var rowGrps = _.groupBy(items, function(p){ return p.row; }); but didn't feel right – lunacafu Commented Dec 5, 2013 at 13:48
• Could you demonstrate the result you're expecting to be created from this? – Jonathan Lonowski Commented Dec 5, 2013 at 13:50
• 1 @lunacafu: But that's just what you want? How else should the expected output look like? Btw, you could shorten that to _.groupBy(items, "row") – Bergi Commented Dec 5, 2013 at 13:50
• Question is confusing because the top-level array is called items, but also each item contains a property named items – broofa Commented Dec 5, 2013 at 13:55

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3 Answers 3

Sorted by: Reset to default 3

If I understand the question correctly, the result would be given by

function filter(items, row) {
    return _.chain(items)    // initiate method chaining for convenience
        .where({row: row})   // filter out objects from rows we don't care about
        .pluck("items")      // get the "items" arrays from the filtered objects
        .flatten()           // concatenate them into a single array
        .value();            // unwrap the result to return it
}

Calling filter(items, 0) in the example given would return

[2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

which is the concatenated aggregate of items arrays inside objects with row equal to 0.

Maybe you're looking for this:

_.reduce(items, function(res, item) {
    var key = item.row;
    if (!(key in res)) res[key] = [];
    [].push.apply(res[key], item.items);
    return res;
}, {})

In underscore, to get all items where the row value matches the index of the item in the array:

var items = [...];
var filteredItems = _.filter(items, function(item, i) {
  return item.row == i;
});

Or with the native Array.prototype.map method. E.g.

var items = [...];
var filteredItems = items.filter(function(item, i) {
  return item.row == i;
});