Get array difference between multiple arrays of different sizes

Question

Been going around in circles with this for a while.

For example let's say I have different times slots stored as this

let timechunks1 = [["09:00", "10:00", "11:00"], ["09:00", "10:00", "11:00"], ["09:00", "10:00", "11:00", "12:00"]] 
let timechunks2 = [["09:00", "10:00", "11:00"], ["09:00", "10:00", "11:00"], ["09:00", "10:00"]]


 for (let i=0; i< timechunks.length; i++) {
 differences.push(_.difference(timechunks[0], timechunks[i]))      
}

Would need to extract 12:00 since it doesn't appear in others. Closest I got was with underscore _difference but alas it's not doing what I expected.

For timechunks1 I'm looking to return "12:00" For timechunks2 I'm looking for a way to return "11:00"

Added another example

[
["09:00", "10:00", "11:00" ,"13:00"],
["09:00", "10:00", "11:00", "13:00"], 
["09:00", "10:00", "11:00", "12:00"]
]

for which it would return ["13:00", "12:00"]

Julian Gonggrijp · Accepted Answer · 2020-07-31 18:36:37Z

2

In set-theoretic terms, what you are looking for is the symmetric difference. You can find this set by taking the union and subtracting the intersection:

function symmetricDifference(...sets) {
    return _.difference(_.union(...sets), _.intersection(...sets));
}

answered Jul 31, 2020 at 18:36

Julian Gonggrijp

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2 Comments

C.Astraea Over a year ago

Thanks ! Learned something new :) Seeing the venn diagrams helps visualize it better ) If I knew how it's called would have found that Lodash even has a function that does this called _.xor() but doesn't go all the way lodash.com/docs/4.17.15#xor

Julian Gonggrijp Over a year ago

What do you mean by "doesn't go all the way"?

Scott Sauyet · Accepted Answer · 2020-07-31 13:13:30Z

1

You can build this on implementations of intersection, difference, and uniq. Here's how I might do it with Ramda's versions:

const notCommon = (xs, common = xs .reduce (intersection)) => 
  uniq (xs .flatMap (x => difference (x, common)))


const timechunks1 = [["09:00", "10:00", "11:00"], ["09:00", "10:00", "11:00"], ["09:00", "10:00", "11:00", "12:00"]] 
const timechunks2 = [["09:00", "10:00", "11:00"], ["09:00", "10:00", "11:00"], ["09:00", "10:00"]]
const timechunks3 = [["09:00", "10:00", "11:00" ,"13:00"], ["09:00", "10:00", "11:00", "13:00"], ["09:00", "10:00", "11:00", "12:00"]]

console .log (notCommon (timechunks1))
console .log (notCommon (timechunks2))
console .log (notCommon (timechunks3))

<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.27.1/ramda.js"></script>
<script> const {intersection, uniq, difference} = R                  </script>

We find the elements in common to all the sets by using intersection Here we reduce over Ramda's binary version. Other versions might let you do something like common = intersection (xs) or common = intersection (...xs), depending on their exact API. Then we flatMap a difference function over the timechunks, returning the values not found in common. And finally we take only the uniq elements of that list.

answered Jul 31, 2020 at 13:13

Scott Sauyet

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3 Comments

C.Astraea Over a year ago

Thanks ! Quite powerful data manipulation will look more into ramda! This covers all edge cases. I will implement this right away. In the meanwhile I had looked at the problem the other way around and only return duplicates that appear less times than the timechunks array length which I knew beforehand represents the timeslots for each available agent.

Julian Gonggrijp Over a year ago

For what it's worth, this is equivalent to my answer.

Scott Sauyet Over a year ago

@Julian: yes it is. Interesting, I've never thought about symmetric difference extending beyond two inputs.

Vasanth Gopal · Accepted Answer · 2020-07-31 10:17:47Z

0

Here is an immutable way of doing it

function findTimeSlot(timechunks){
    const slots = _.map(_.zip(...timechunks), _.compact);
    const maxChunk = _.max(_.map(slots, slot => slot.length));

    return _.unique(_.find(slots, (slot) => slot.length < maxChunk));
}

returns

["11:00"] for findTimeSlot(timechunks2)

And

["12:00"] for findTimeSlot(timechunks1)

Let know if it works.

answered Jul 31, 2020 at 10:17

Vasanth Gopal

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Comments

nip · Accepted Answer · 2020-07-31 10:24:20Z

0

function findUniqueTimeSlots(timechunks) {
  const hm = {};
  timechunks.flat(1).forEach(el => {
    if (hm.hasOwnProperty(el)) {
      hm[el] += 1;
    } else {
      hm[el] = 1;
    }
  });
  return Object.keys(hm).filter(key => hm[key] === 1);
}

answered Jul 31, 2020 at 10:24

nip

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Comments

hgb123 · Accepted Answer · 2020-07-31 10:46:14Z

0

You could achieve that by following these steps:

find the common timechunk, _.intersection(...timechunks)
iterate the timechunks and find the difference of each from the common, _.difference(timechunk, commons)

Below snippet could help you

const timechunks1 = [
  ["09:00", "10:00", "11:00"],
  ["09:00", "10:00", "11:00"],
  ["09:00", "10:00", "11:00", "12:00"],
]
const timechunks2 = [
  ["09:00", "10:00", "11:00"],
  ["09:00", "10:00", "11:00"],
  ["09:00", "10:00"],
]
const timechunks3 = [
  ["09:00", "10:00", "11:00", "13:00"],
  ["09:00", "10:00", "11:00", "13:00"],
  ["09:00", "10:00", "11:00", "12:00"],
]

const getDifferences = (timechunks) =>
  _.chain(timechunks)
    .map((timechunk) => _.difference(timechunk, _.intersection(...timechunks)))
    .flatten()
    .uniq()
    .value()

console.log(getDifferences(timechunks1))
console.log(getDifferences(timechunks2))
console.log(getDifferences(timechunks3))

<script src="https://cdn.jsdelivr.net/npm/[email protected]/lodash.min.js"></script>

answered Jul 31, 2020 at 10:46

hgb123

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1 Comment

Scott Sauyet Over a year ago

Note that you call intersection repeatedly on the same values.

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