Group by on JSON array using Underscore JS

Question

I have a JSON array object as follows:

var orders = [{
    orderId: 1,
    firstName: 'John',
    lastName: 'Smith',
    address: {
        street: '123 Main Street',
        city: 'New York',
        zip: 10001
    }
}, {
    orderId: 2,
    firstName: 'John',
    lastName: 'Smith',
    address: {
        street: '456 Main Street',
        city: 'New York',
        zip: 10001
    }
}, {
    orderId: 3,
    firstName: 'John',
    lastName: 'Smith',
    address: {
        street: '123 Main Street',
        city: 'New York',
        zip: 10001
    }
}, {
    orderId: 4,
    firstName: 'John',
    lastName: 'Smith',
    address: {
        street: '123 Main Street',
        city: 'New York',
        zip: 10002
    }
}];

I am trying to use underscore.js to create a new array object, grouped by the address to meet the use case of displaying all orders that have been shipped to 123 Main Street, New York, 1001.

Is underscore.js the right approach to do this? If so, how should I do so? Any hints will be helpful.

What would be the output of this? Object pointing to arrays? What would you key the objects by? Arrays of Arrays? — Ruan Mendes
– Ruan Mendes, Commented Dec 1, 2014 at 23:02
@JuanMendes - the output would be an array of arrays. I'm trying the suggestion you made about _.groupBy now. — orchard
– orchard, Commented Dec 1, 2014 at 23:14

Bergi · Accepted Answer · 2014-12-02 14:03:36Z

3

See _.groupBy

console.log(_.groupBy(orders, function(obj){
    return obj.address.street + '|' + obj.address.city + '|' + obj.address.zip;
}));

See http://jsfiddle.net/mendesjuan/gc47ruyL/1/

This example assumes you cannot have a | in an address, you may need a better separator, or use JSON.stringify:

console.log(_.groupBy(orders, function(obj){
    return JSON.stringify([obj.address.street, obj.address.city, obj.address.zip]);
}));

edited Dec 2, 2014 at 14:03

Bergi

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answered Dec 1, 2014 at 23:09

Ruan Mendes

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8 Comments

orchard Over a year ago

thanks @JuanMendes - I think groupBy will solve my problem. The JSON I posted above was an over simplified version of what I have so I am trying to put the pieces together. I will update here either way to inform if it works or not. Thanks again - everyone - for the help.

Bergi Over a year ago

return JSON.stringify(…) always works for escaping if you don't want to choose a specific separator.

Bergi Over a year ago

@JuanMendes: Actually I meant to stringify each string on its own, or maybe an array of them, to get a concatenation like in your answer. I didn't even think of the possibility to JSON.stringify(obj.address), which would indeed suffer from the mentioned problem.

orchard Over a year ago

@JuanMendes - thanks for suggesting to use _.groupBy - it was exactly what I needed. I am a new user here and am amazed by the help I was provided. I will make every effort to contribute back.

Ruan Mendes Over a year ago

@orchard Same thing happened to me 5 years ago, I've been addicted since stackoverflow.com/questions/1868333/…

|

amit_g · Accepted Answer · 2014-12-01 23:13:51Z

2

If you already have underscore in your page, you could use

var filteredOrders = _.filter(orders, function(order){
  return (order.address.street === '123 Main Street') &&
          (order.address.city === 'New York') &&
          (order.address.zip === 10001);
});

Fiddle

edited Dec 1, 2014 at 23:13

answered Dec 1, 2014 at 22:49

amit_g

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9 Comments

orchard Over a year ago

Thanks @amit_g - that is helpful, but this would work for this specific example where there is only 1 address. The JSON example I posted was representative of an array which could contain multiple users with different addresses. I need to be able to compare the address object across all users, so that I can find all users who share the same address.

amit_g Over a year ago

@orchard, please update the question with a sample to demonstrate this requirement.

Ruan Mendes Over a year ago

@JordanGray Actually it cannot be easily extended, it's easy if you know what you're searching for, otherwise, some heavy work is needed to find duplicates

Jordan Gray Over a year ago

@JuanMendes Good point; I neglected to consider excluding duplicates. Still, I feel that Amit deserves some credit for answering what was asked. :)

orchard Over a year ago

@JordanGray - the first thing I tried was to mark the answer as helpful :) but I am a new user and found that I don't have the required reputation (15 points) to do so ... I completely concur Amit's response is very helpful indeed.

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