Javascript 查找数组中项目之间的距离

Question

I've this set of array, each with the same length and 2 of this are sorted

I start with this vars: var sky = [0,3,4,5,6,7,8,9,10,11,12,14,16,17]; var ter = [];

I've to analyze every sky item in that way: for each sky items I've to find the "distance" between the sky item and the sky+1, then from sky item and sky+1+1 etc etc... so

var ter = 3-0
var ter = 4-0
var ter = 5-0
var ter = 6-0
var ter = 7-0
var ter = 8-0
var ter = 9-0
var ter = 10-0
var ter = 11-0
var ter = 12-0
var ter = 14-0
var ter = 16-0
var ter = 17-0

So the second cycle for the sky array have todo the same but start with the second items on sky array so will be

var ter = 4-3
var ter = 5-3
var ter = 6-3
var ter = 7-3
var ter = 8-3
var ter = 9-3
var ter = 10-3
var ter = 11-3
var ter = 12-3
var ter = 14-3
var ter = 16-3
var ter = 17-3

I don't know how to calculate the ter var, and at this point maybe the best is to have it in array, like that

ter = [[3,4,5,6,7,8,9,10,11,12,14,16,17],[1,2,3,4,5,6,7,8,9,11,13,14], and so on];

so in the next phase I can refer to the ter array

For now I've only the start and is not complete because is only a start to try to find a good point, but I don't know why event to start don't works, lol. Ps I don't need the last one the 17 in this case, because I don't have nothing over the last items on sky array

for (j = 0; j < sky.length; j++) {
  if (j !== 0 || j !== sky.length){
    ter.push(sky[j]-sky[0]);
  }
}

console.log(ter);

Any quick idea?

Answer 1

I think a recursion is more efficient than iteration in this case.

JavaScript (ES2015)

let dist = (a,r = []) => {
  if(r.length <= a.length-2) {
    let t = [];
    let b = a[r.length];
    a.forEach(e => t.push(e - b));
    r.push(t.filter(e => e > 0));
    return dist(a,r);
  } else  return r;
}


let sky = [0,3,4,5,6,7,8,9,10,11,12,14,16,17];
let ter = [];

console.log(dist(sky,ter));

Output

[[3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 17], [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 14], [1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 13], [1, 2, 3, 4, 5, 6, 7, 9, 11, 12], [1, 2, 3, 4, 5, 6, 8, 10, 11], [1, 2, 3, 4, 5, 7, 9, 10], [1, 2, 3, 4, 6, 8, 9], [1, 2, 3, 5, 7, 8], [1, 2, 4, 6, 7], [1, 3, 5, 6], [2, 4, 5], [2, 3], [1]]

JS Bin: http://jsbin.com/muhade/edit?js,console

---

If you're not familiar with ES2015, here's the same code in ES5:-

JavaScript (ES5)

var dist = function (a, r) {
    r = r || [];
    if (r.length <= a.length - 2) {
        var t = [];
        var b = a[r.length];
        a.forEach(function (e) { return t.push(e - b); });
        r.push(t.filter(function (e) { return e > 0; }));
        return dist(a, r);
    }
    else return r;
};

var sky = [0, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 17];
var ter = [];

console.log(dist(sky, ter));

Answer 2

I guess you might do as follows if i have understood correctly;

 var sky = [0,3,4,5,6,7,8,9,10,11,12,14,16,17], ters = sky.map((e,i,a) => a.slice(i+1).map(f => [e,f])) .reduce((p,c) => p.concat(c)); console.log(JSON.stringify(ters)); 

Or may be like;

 var sky = [0,3,4,5,6,7,8,9,10,11,12,14,16,17], ters = sky.map((e,i,a) => a.slice(i+1).map(f => fe)); console.log(JSON.stringify(ters)); 

Answer 3

First, reduce your problem to one simple function :

/**
 * @param {Number[]} sky - The original sky array
 * @param {Number} cycle - First cycle is 1, second is 2, so on..
 * @param {Number} itemIndex - Index of item to take distance to
 */
function distance(sky, cycle, itemIndex) {
  cycle = cycle - 1; // the first cycle is actually 0
  if (cycle < 0) return Number.NaN;
  if (cycle > sky.length - 2) return Number.NaN;
  if (itemIndex > sky.length - cycle - 1) return Number.NaN;

  return sky[itemIndex + cycle + 1] - sky[cycle];
}

Now, for instance, you can check:

var sky = [0,3,4,5,6,7,8,9,10,11,12,14,16,17];
console.log(distance(sky, 1, 7)); // 10
console.log(distance(sky, 2, 7)); // 8

Then, to create ter :

var sky = [0,3,4,5,6,7,8,9,10,11,12,14,16,17];

var ter = [];
for (var cycle = 0 ; cycle < sky.length - 1 ; cycle++) {
  var innerTer = [];
  for (var itemIndex = 0; itemIndex <= sky.length - cycle - 1 ; itemIndex++)
    innerTer.push(distance(sky, cycle, itemIndex));
  ter.push(innerTer);
}

console.log(ter);

JSFIDDLE DEMO

Hope this helps.

Answer 4

You could use two nested for loops and push the difference to new array.

 var sky = [0, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 17], i, j, ter = []; for (i = 0; i < sky.length - 1; i++) { ter[i] = []; for (j = i + 1; j < sky.length; j++) { ter[i].push(sky[j] - sky[i]); } } console.log(ter); 

 .as-console-wrapper { max-height: 100% !important; top: 0; } 

Answer 5

A simple way would be:

const getDistance = (arr, numA, numB) => numB - (arr.indexOf(numA)-1)

Then:

const items = [1,2,3,4,5]
const distance = getDistance(items, 2, 4)
console.log(distance) // 2