javascript-获取元素更改的排序列表的索引

您可以迭代数组并检查前置任务。如果相等,则增加最后一个计数,否则添加索引和一个计数。

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);

此代码生成与您发布的代码类似的输出:

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);

试试这个,你会得到你想要的

        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);

所以,使用 https://underscorejs.org/#groupBy 可以按值分组

_.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; })

你会得到一个像

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

所以如果你把所有的 https://underscorejs.org/#keys 然后遍历,key下的值是array,获取size并追加到新数组中,这样就可以使ad=[3,2,5,2]

再次,遍历键并获取 https://underscorejs.org/#indexOf ,可以构造ac=[0,3,5,10]

玩这些方法,检查例子,你可以自己做!