javascript - What is the most efficient inverse of lodash _.intersection() - Stack Overflow

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Using lodash, I want to find all the elements between two arrays that are different, basically the opposite or inverse of _.intersection[with|by]. Of course I reviewed _.difference[with|by] but it's not an inverse version of _.intersection as it only pares the first array with the second, rather than paring them against each other. The closest I was able to get is very cludgy which has motivated me to ask here if I'm missing a more efficient and/or elegant option. I'm only interested in lodash based solutions.

Here is the closest I could e up with to get a unique array of values that are different between both arrays. I'm interested in values with id properties that aren't in both arrays and values with matching id properties but different v properties.

const n = [{id: 0, v: 9.7}, {id: 1, v: 1.7}, {id: 3, v: 2.6}, {id: 4, v: 1.89}]
const o = [{id: 1, v: 1.7}, {id: 3, v: 3.6}, {id: 7, v: 0.89}, {id: 4, v: 1.89}]

_.uniqBy(_.concat(
    _.differenceWith(n, o, _.isEqual), _.differenceWith(o, n, _.isEqual)), 'id')

That code will yield:

[{id: 0, v: 9.7}, {id: 3, v: 2.6}, {id: 7, v: 0.89}]

Using lodash, I want to find all the elements between two arrays that are different, basically the opposite or inverse of _.intersection[with|by]. Of course I reviewed _.difference[with|by] but it's not an inverse version of _.intersection as it only pares the first array with the second, rather than paring them against each other. The closest I was able to get is very cludgy which has motivated me to ask here if I'm missing a more efficient and/or elegant option. I'm only interested in lodash based solutions.

Here is the closest I could e up with to get a unique array of values that are different between both arrays. I'm interested in values with id properties that aren't in both arrays and values with matching id properties but different v properties.

const n = [{id: 0, v: 9.7}, {id: 1, v: 1.7}, {id: 3, v: 2.6}, {id: 4, v: 1.89}]
const o = [{id: 1, v: 1.7}, {id: 3, v: 3.6}, {id: 7, v: 0.89}, {id: 4, v: 1.89}]

_.uniqBy(_.concat(
    _.differenceWith(n, o, _.isEqual), _.differenceWith(o, n, _.isEqual)), 'id')

That code will yield:

[{id: 0, v: 9.7}, {id: 3, v: 2.6}, {id: 7, v: 0.89}]

• javascript
• lodash

Share Improve this question Follow asked Mar 31, 2020 at 4:54 scubastevescubasteve 2,86844 gold badges4141 silver badges5151 bronze badges 7

• how did you decided to keep between these two {id: 3, v: 3.6} and {id: 3, v: 2.6}? – Mechanic Commented Mar 31, 2020 at 6:31
• @CaptainMhmdrz_A the v property is different – scubasteve Commented Mar 31, 2020 at 14:47
• I mean how did you decided to keep 2.6 not 3.6 in the output – Mechanic Commented Mar 31, 2020 at 14:49
• Ah, I see. Actually I don't care, I'm only interested in the fact they are different. – scubasteve Commented Mar 31, 2020 at 17:46
• Is it the coronavirus that's responsible for lack of activity or is my question boring or irrelevant? – scubasteve Commented Apr 1, 2020 at 18:58

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2 Answers 2

Sorted by: Reset to default 3

_.xorWith is the closest I could think of, but it doesn't work with the example you provided because two objects have an id of 3.

const n = [{id: 0, v: 9.7}, {id: 1, v: 1.7}, {id: 3, v: 2.6}, {id: 4, v: 1.89}]
const o = [{id: 1, v: 1.7}, {id: 3, v: 3.6}, {id: 7, v: 0.89}, {id: 4, v: 1.89}]
_.xorWith(n, o, _.isEqual)

output would be:

[
  { id: 0, v: 9.7 },
  { id: 3, v: 2.6 },
  { id: 3, v: 3.6 },
  { id: 7, v: 0.89 }
]

I found this question after looking for the same thing, but I since came to realize that for my use case - identifying records that don't match, and then why they don't match - it isn't useful to have a list of all differences like that.

That list can't tell me anything I can use, so I would still need to do a further operation to see which array/s are the problem.

So for me, it makes more sense to first check for equality, then to use _.difference twice, because then the result of it yields usable information, e.g.

if (!_.isEqual(a, b) {
  const idsFromBThatAreNotInA = _.difference(a, b);
  const idsFromAThatAreNotInB = _.difference(b, a);
}

I don't know what the use case of the OP is, so I don't know if this is directly relevant, but maybe it can help others.

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