Unexpected result when filtering one object array against two other object arrays

Question

I'm trying to filter out objects from array "a" that match with objects in array "b" and "c". here is a link to jsfiddle to test the code.

Here is what I currently have:

const a = [{
  "name": "sondre",
  "uq_id": "abc1"
}, {
  "name": "sofie",
  "uq_id": "abc2"
}, {
  "name": "casper",
  "uq_id": "abc3"
}, {
  "name": "odin",
  "uq_id": "abc4"
}];

const b = [{
  "name": "sondre",
  "uq_id": "abc1"
}, {
  "name": "odin",
  "uq_id": "abc4"
}];

const c = [{
  "name": "casper",
  "uq_id": "abc3"
}];

function sort(a, b, c) {
  result = [];
  console.log(result);
  if (b !== null) {
    result = a.filter(function(item) {
      return !b.includes(item.uq_id);
    })
  }
  if (c !== null) {
    result = result.filter(function(item) {
      return !c.includes(item.uq_id);
    })
  }
  console.log(result);
}

sort(a, b, c);

I expect the following output:

[{name="sofie", uq_id="abc2"}]

But for some reason it outputs:

[{name="sondre", uq_id="abc1"},
{name="sofie", uq_id="abc2"},
{name="casper", uq_id="abc3"},
{name="odin", uq_id="abc4"}]

Does anyone know how I can get this to work as I intended it?

*"heres a link to jsfiddle to test the code"* Please do your runnable example **here, on-site** using Stack Snippets (the `[<>]` toolbar button; [here's how to do one](https://meta.stackoverflow.com/questions/358992/)). — T.J. Crowder, Jul 08 '19 at 10:53
Why is the function called `sort`? It doesn't do any sorting. — T.J. Crowder, Jul 08 '19 at 10:53
^ indeed, it does more "filtering". Or rather intersection between three arrays. — VLAZ, Jul 08 '19 at 10:54

T.J. Crowder · Accepted Answer · 2019-07-08T11:03:30.753

If the goal is to filter out entries from a whose name matches an entry on either b or c, you can't use includes unless the entries in a, b, and c refer to the same objects (not just equivalent ones).

Assuming they don't, you can use some to find out whether an array contains a match for a name. You'll want to use && to see that there's no match in either b or c:

const filtered = a.filter(entry => {
    return !b.some(({name}) => entry.name === name) &&
           !c.some(({name}) => entry.name === name);
});

Live Copy:

const a = [{
  "name": "sondre",
  "uq_id": "abc1"
}, {
  "name": "sofie",
  "uq_id": "abc2"
}, {
  "name": "casper",
  "uq_id": "abc3"
}, {
  "name": "odin",
  "uq_id": "abc4"
}];

const b = [{
  "name": "sondre",
  "uq_id": "abc1"
}, {
  "name": "odin",
  "uq_id": "abc4"
}];

const c = [{
  "name": "casper",
  "uq_id": "abc3"
}];

function filter(a, b, c) {
    const filtered = a.filter(entry => {
        return !b.some(({name}) => entry.name === name) &&
               !c.some(({name}) => entry.name === name);
    });
    return filtered;
}

console.log(filter(a, b, c));

That can also be expressed with every, whichever you prefer:

const filtered = a.filter(entry => {
    return b.every(({name}) => entry.name !== name) &&
           c.every(({name}) => entry.name !== name);
});

If b and c are really large (hundreds of thousands of entries, perhaps millions) that could be inefficient enough to justify creating Sets of names first:

const names = new Set([
  ...b.map(({name}) => name),
  ...c.map(({name}) => name)
]);
const filtered = a.filter(entry => {
    return !names.has(entry.name);
});

Or you might just do that for preference or clarity.

Live Copy:

const a = [{
  "name": "sondre",
  "uq_id": "abc1"
}, {
  "name": "sofie",
  "uq_id": "abc2"
}, {
  "name": "casper",
  "uq_id": "abc3"
}, {
  "name": "odin",
  "uq_id": "abc4"
}];

const b = [{
  "name": "sondre",
  "uq_id": "abc1"
}, {
  "name": "odin",
  "uq_id": "abc4"
}];

const c = [{
  "name": "casper",
  "uq_id": "abc3"
}];

function filter(a, b, c) {
    const names = new Set([
      ...b.map(({name}) => name),
      ...c.map(({name}) => name)
    ]);
    const filtered = a.filter(entry => {
        return !names.has(entry.name);
    });
    return filtered;
}

console.log(filter(a, b, c));

Apologies, I had a silly logic error (and typo) that I didn't catch until doing the runnable examples. :-/ — T.J. Crowder, Jul 08 '19 at 11:04
How would this handle B or C being empty, undefined or null? — Sondre Ljovshin, Jul 08 '19 at 11:20
@SondreLjovshin - Empty isn't a problem. If they might be `undefined` or `null`, you'll want to handle that. Probably the easiest way is `if (!b) { b = []; } if (!c) { c = []; }` before you start the filtering. — T.J. Crowder, Jul 08 '19 at 11:31

score 1 · Answer 2 · answered Jul 08 '19 at 10:57

b and c are arrays of objects it only includes objects. You need to use some() on then to compare uq_id

const a = [{
    "name": "sondre",
    "uq_id": "abc1"
}, {
    "name": "sofie",
    "uq_id": "abc2"
}, {
    "name": "casper",
    "uq_id": "abc3"
}, {
    "name": "odin",
    "uq_id": "abc4"
}];

const b = [{
    "name": "sondre",
    "uq_id": "abc1"
}, {
    "name": "odin",
    "uq_id": "abc4"
}];

const c = [{
    "name": "casper",
    "uq_id": "abc3"
}];

sort(a, b, c);



function sort(a, b, c) {
    let result = [];
    if (b !== null) {
        result = a.filter(function(item) {
            return !b.some(x => x.uq_id === item.uq_id);
        })
    }
    if (c !== null) {
        result = result.filter(function(item) {
            return !c.some(x => x.uq_id === item.uq_id);
        })
    }
    console.log(result);
}

Thank you for the answer, this seems to resolve my problem in the most efficient way — Sondre Ljovshin, Jul 08 '19 at 11:07

score 0 · Answer 3 · answered Jul 08 '19 at 11:13

Single filter is enough :

const a = [ { "name": "sondre", "uq_id": "abc1" }, 
            { "name": "sofie" , "uq_id": "abc2" }, 
            { "name": "casper", "uq_id": "abc3" }, 
            { "name": "odin"  , "uq_id": "abc4" } ];

const b = [ { "name": "sondre", "uq_id": "abc1" }, 
            { "name": "odin"  , "uq_id": "abc4" } ];

const c = [ { "name": "casper", "uq_id": "abc3" } ];


const result = a.filter(x => !b.some(y => x.uq_id === y.uq_id) 
                          && !c.some(y => x.uq_id === y.uq_id));

console.log(result);

Unexpected result when filtering one object array against two other object arrays

3 Answers3

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