如何获取元素更改的排序列表的索引？

Question

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.

Answer 1

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);

Answer 2

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);

Answer 3

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);

Answer 4

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!

Answer 5

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