Javascript: filter array of objects by array of strings

Question

I wonder if there is a more elegant way of doing this. Suppose i have an array of objects like this:

a = [
  {
    "id": "kpi02",
    "value": 10
  },
  {
    "id": "kpi02",
    "value": 30
  },
  {
    "id": "kpi02",
    "value": 11
  },
  {
    "id": "kpi02",
    "value": 33
  },
  {
    "id": "kpi03",
    "value": 1
  },
  {
    "id": "kpi03",
    "value": 0.5
  },
  {
    "id": "kpi04",
    "value": 0.5
  }
]

Now i want to filter on the id property, to return all objects with a match in another array

var kpis = ["kpi03", "kpi02"];

I came up with this solution:

var b = [];
for (j in kpis) {
 for (i in a) { 
    if (a[i].id == kpis[j]) {
    b.push(a[i]);
    }
 }
}

Coming from R, this seems a bit complicated, is there any way to do that with the filter prototype? Like this but with an array of strings to compare with instead of a single string:

 var b = a.filter( function(item){return (item.id == "kpi03");} );

Thanks a lot!

Oleksandr T. · Accepted Answer · 2015-01-25 15:41:38Z

22

You can use indexOf in filter, like this

var res = a.filter(function (el) {
  return kpis.indexOf(el.id) >= 0; 
});

Example

answered Jan 25, 2015 at 15:41

Oleksandr T.

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1 Comment

Schw4rz Over a year ago

Thank you for the quick answer and the example! I knew there must be a more elegant solution :)

itaydafna · Accepted Answer · 2018-08-06 08:20:00Z

14

Another nice alternative is using .filter with .includes:

var result = a.filter(item => kpis.includes(item.id))

answered Aug 6, 2018 at 8:20

itaydafna

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Comments

Amit Chhatbar · Accepted Answer · 2020-07-17 10:38:35Z

3

If you want them in the order of the array of string then

    var result = [];
kpis.map((item) => {
    const matchedObject = a.find(
        (option) => option.id === item
    );
    result.push(matchedObject);
});

answered Jul 17, 2020 at 10:38

Amit Chhatbar

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Comments

Amit Joki · Accepted Answer · 2015-01-25 15:41:52Z

2

Just make use of Array.indexOf

var b = a.filter(function(item){return kpids.indexOf(item.id) > -1 });

Array.indexOf returns the index of the argument passed in the array on which indexOf is being called on. It returns -1 if there isn't the element which we are looking for.

So, we make sure that it index is greater than -1

answered Jan 25, 2015 at 15:41

Amit Joki

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