Compare all values of an Array of Objects with an Object

Question

I have the following data:

const myArr = [{
  id: 0,
  company: "microsoft",
  location: "berlin"
}, {
  id: 1,
  company: "google",
  location: "london"
}, {
  id: 2,
  company: "twitter",
  location: "berlin"
}];

let myObj = {
  company: ["google", "twitter"],
  location: ["london"]
}

and given that the myObj.company entries are changing (irrelevant how) I am trying to create a function that filters results and only returns Objects that satisfy the location and company criteria.

In the example above, what we need returned is :

{ 
  id: 1,
  company: "google",
  location: "london"
}

If myObj was

let myObj = {
  company: ["google", "twitter"],
  location: []
}

then the returned result should be

{ 
  id: 1,
  company: "google",
  location: "london"
},
{ 
  id: 2,
  company: "twitter",
  location: "berlin"
}

What is the problem with your iteration? Please add the code you've tried to the question too. — Teemu
– Teemu, Commented Feb 10, 2017 at 16:57
Thanks for all the amazing answers, I wasn't expecting such a feedback. — George Katsanos
– George Katsanos, Commented Feb 10, 2017 at 22:56

Pranav C Balan · Accepted Answer · 2017-02-11 05:06:45Z

2

Use Array#filter method with Array#includes method(for old browser support use Array#indexOf method)

myArr.filter(o => (myObj.company.length == 0 || myObj.company.includes(o.company)) && (myObj.location.length == 0 || myObj.location.includes(o.location)))

const myArr = [{
  id: 0,
  company: "microsoft",
  location: "berlin"
}, {
  id: 1,
  company: "google",
  location: "london"
}, {
  id: 2,
  company: "twitter",
  location: "berlin"
}];

let myObj = {
  company: ["google", "twitter"],
  location: ["london"]
}

console.log(
  myArr.filter(o => (myObj.company.length == 0 || myObj.company.includes(o.company)) && (myObj.location.length == 0 || myObj.location.includes(o.location)))
)

myObj = {
  company: ["google", "twitter"],
  location: []
}

console.log(
  myArr.filter(o => (myObj.company.length == 0 || myObj.company.includes(o.company)) && (myObj.location.length == 0 || myObj.location.includes(o.location)))
)

UPDATE : In case there an n number of properties collection then you need to make some variation, where you can use Object.keys and Array#every methods.

var keys = Object.keys(myObj);

myArr.filter(o => keys.every(k => myObj[k].length == 0 || myObj[k].includes(o[k])))

const myArr = [{
  id: 0,
  company: "microsoft",
  location: "berlin"
}, {
  id: 1,
  company: "google",
  location: "london"
}, {
  id: 2,
  company: "twitter",
  location: "berlin"
}];

let myObj = {
  company: ["google", "twitter"],
  location: ["london"]
}

var keys = Object.keys(myObj);
console.log(
  myArr.filter(o => keys.every(k => myObj[k].length == 0 || myObj[k].includes(o[k])))
)

myObj = {
  company: ["google", "twitter"],
  location: []
}
keys = Object.keys(myObj);
console.log(
  myArr.filter(o => keys.every(k => myObj[k].length == 0 || myObj[k].includes(o[k])))
)

edited Feb 11, 2017 at 5:06

answered Feb 10, 2017 at 17:00

Pranav C Balan

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1 Comment

George Katsanos Over a year ago

thank you. I'm upvoting because 1) didn't know includes ! 2) simplicity and brevity as well as readability.. comment : you are testing every array separately (location, company) so if I have 2-3 different tables with other data it's not going to be so DRY

George Katsanos · Accepted Answer · 2017-02-11 10:54:21Z

1

Use Array.prototype.filter, Array.prototype.every and Object.keys to get the desired result like this: (note that obj could have any number of keys, it flexible that way)

const myArr = [{
  id: 0,
  company: "microsoft",
  location: "berlin"
}, {
  id: 1,
  company: "google",
  location: "london"
}, {
  id: 2,
  company: "twitter",
  location: "berlin"
}];

let myObj = {
  company: ["google", "twitter"],
  location: ["london"]
}

function find(arr, obj) {
  // get only the keys from obj that their corresponding array is not empty
  var keys = Object.keys(obj).filter(k => obj[k].length !== 0);

  // return a filtered array of objects that ...
  return arr.filter(o => {
    // ... met the creteria (for every key k in obj, the current object o must have its value of the key k includded in the array obj[k])
    return keys.every(k => {
      return obj[k].indexOf(o[k]) != -1;   
    });
  });
}

console.log(find(myArr, myObj));

edited Feb 11, 2017 at 10:54

George Katsanos

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answered Feb 10, 2017 at 17:03

ibrahim mahrir

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1 Comment

George Katsanos Over a year ago

Thank you, selecting as correct because: 1/ comments 2/ avoiding unnecessary loops 3/ readable

Nina Scholz · Accepted Answer · 2017-02-11 09:03:04Z

1

You could filter with iterating the keys of criteria and check the content for equality of if the length of the array is not zero.

const filterBy = (array, criteria) => array.filter(o => 
    Object.keys(criteria).every(k =>
        criteria[k].some(c => c === o[k]) || !criteria[k].length)
);

const myArr = [{ id: 0, company: "microsoft", location: "berlin" }, { id: 1, company: "google", location: "london" }, { id: 2, company: "twitter", location: "berlin" }];

console.log(filterBy(myArr, { company: ["google", "twitter"], location: ["london"] }));
console.log(filterBy(myArr, { company: ["google", "twitter"], location: [] }));

.as-console-wrapper { max-height: 100% !important; top: 0; }

With Array#includes instead of Array#some

const filterBy = (array, criteria) => array.filter(o => 
    Object.keys(criteria).every(k => criteria[k].includes(o[k]) || !criteria[k].length)
);

const myArr = [{ id: 0, company: "microsoft", location: "berlin" }, { id: 1, company: "google", location: "london" }, { id: 2, company: "twitter", location: "berlin" }];

console.log(filterBy(myArr, { company: ["google", "twitter"], location: ["london"] }));
console.log(filterBy(myArr, { company: ["google", "twitter"], location: [] }));

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Feb 11, 2017 at 9:03

answered Feb 10, 2017 at 19:24

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

1 Comment

George Katsanos Over a year ago

+1 because of the console-wrapper hack :) as well as the very scalable solution, amazing! it does take me a while to read through all those nested array functions though, and in the end we're using, filter, keys, every and some so I'm left wondering if there's a more verbose but also readable way of doing it

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