CODEWITHSUNDEEP
javascriptarrays
AppRoyale
AppRoyale

Reputation: 403

How can I get index of sorted list where element changes?

I have the following array and I am looking to retrieve the index of the original (sorted) array where the element is changing and how often that individual element exists.

ab = [1,1,1,3,3,5,5,5,5,5,6,6]

The desired outcome should be like this:

ac = [0,3,5,10]
ad = [3,2,5,2]

Thank you very much for any suggestion.

Cheers.

Upvotes: 5

Views: 622

Answers (5)

Steve Dutky
Steve Dutky

Reputation: 303

I think this works in R. YMMV
> ab = c(1,1,1,3,3,5,5,5,5,5,6,6)
> i1<-1:length(ab)
> i2<-c(2:length(ab),length(ab))
> i3<-ab[i1]!=ab[i2]
> ac<-c(0,i1[i3])
> ac
[1] 0 3 5 10
> ad<-c(ac[-1],length(ab))-ac
> ad
[1] 3 2 5 2

Upvotes: 0

Djaouad
Djaouad

Reputation: 22776

This code produces similar output to the one you posted:

var ab = [1,1,1,3,3,5,5,5,5,5,6,6];

var ac = Array.from(new Set(ab.map((e) => ab.indexOf(e))));

var ad = [];

for (var i = 0; i < ac.length - 1; i++) {
  ad.push(ac[i + 1] - ac[i]);
}
ad.push(ab.length - ac[ac.length - 1]);

console.log(...ab);
console.log(...ac);
console.log(...ad);

Upvotes: 2

Dmitri Algazin
Dmitri Algazin

Reputation: 3456

so, using https://underscorejs.org/#groupBy you can group by value

_.groupBy([1,1,1,3,3,5,5,5,5,5,6,6]);

or 

_.groupBy([1,1,1,3,3,5,5,5,5,5,6,6], function(num){ return num; })

you will get an object like

{1: [1,1,1], 3: [3,3], 5: [5,5,5,5,5], 6: [6,6]}

so if you take all https://underscorejs.org/#keys and iterate through, value under key is array, take size and append to new array, so you can make ad = [3,2,5,2]

again, iterate through keys and get https://underscorejs.org/#indexOf , you can construct ac = [0,3,5,10]

play around these methods, check examples, and you can do it yourself!

Upvotes: 1

Jonathan K
Jonathan K

Reputation: 550

Try this, should get you what you want

        ab = [1,1,1,3,3,5,5,5,5,5,6,6];

        var items = [];
        var positions = [];
        var count = [];

        ab.map((item, index)=>{

            //check if exist
            let item_index = items.indexOf(item);
            if(item_index == -1) {
                items.push(item);
                positions.push(index);
                count.push(1);
            } else {
                let current_count = count[item_index];
                count[item_index] = ++current_count;
            }
        });

        console.log(positions);
        console.log(count);

Upvotes: 1

Nina Scholz
Nina Scholz

Reputation: 386680

You could iterate the array and check the predecessor. If equal, increment the last count, otherwise add the index and a count of one.

var array = [1, 1, 1, 3, 3, 5, 5, 5, 5, 5, 6, 6],
    { indices, counts } = array.reduce((r, v, i, a) => {
        if (a[i - 1] === v) {
            r.counts[r.counts.length - 1]++;
        } else {
            r.indices.push(i);
            r.counts.push(1);
        }
        return r;
    }, { indices: [], counts: [] });

console.log(...indices);
console.log(...counts);

Upvotes: 5

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