Comparing numerical values of an array against eachother javascript

Question

let arrayOfNumbers = [1, 2, 3, 4, 5]

What would be the best way to compare the numbers against each other ? For instance, comparing 1 to 2 then 2 to 3 then 3 to four, and so on ?

function t(a) {
  let t = 0
  for (let i = 0; i < a.length; i++) {
    if (a[t] > a[t + 1]) {
      console.log('down')
    } else if (a[t] < a[t + 1]) {
      console.log('up')
     } else if (a[t] === a[t + 1]) {
       console.log('no change')
     }
     t++
   }
}

arrayOfNumbers [i] === arrayOfNumbers [i + 1] arrayOfNumbers [i + 1] === arrayOfNumbers [i + 2] arrayOfNumbers [i + n] === arrayOfNumbers [i + n + 1] you can use a for-loop — Ele
– Ele, Commented Jun 29, 2022 at 19:12
Compare for what? Sorting? If so, the best way is to use Array.sort: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… — capchuck
– capchuck, Commented Jun 29, 2022 at 19:12
Comparing them to see if they're trending in an upwards or downwards trajectory . — bobby_turks
– bobby_turks, Commented Jun 29, 2022 at 19:19
I used a for loop because the array size may change depending on the data source. — bobby_turks
– bobby_turks, Commented Jun 29, 2022 at 19:20

Nina Scholz · Accepted Answer · 2022-06-29 19:31:07Z

1

You could start from index one and check the previous value.

function t(a) {
  for (let i = 1; i < a.length; i++) {
    if (a[i - 1] > a[i]) console.log('down');
    else if (a[i - 1] < a[i]) console.log('up');
    else console.log('no change');
  }
}

t([0, 1, 3, 2, 4, 4, 2]);

answered Jun 29, 2022 at 19:31

Nina Scholz

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Comments

IT goldman · Accepted Answer · 2022-06-29 19:58:30Z

If you want to pick up trend then you can do a linear regression.

function linearRegression(y, x) {
  var lr = {};
  var n = y.length;
  var sum_x = 0;
  var sum_y = 0;
  var sum_xy = 0;
  var sum_xx = 0;
  var sum_yy = 0;

  for (var i = 0; i < y.length; i++) {

    sum_x += x[i];
    sum_y += y[i];
    sum_xy += (x[i] * y[i]);
    sum_xx += (x[i] * x[i]);
    sum_yy += (y[i] * y[i]);
  }

  lr['slope'] = (n * sum_xy - sum_x * sum_y) / (n * sum_xx - sum_x * sum_x);
  lr['intercept'] = (sum_y - lr.slope * sum_x) / n;
  lr['r2'] = Math.pow((n * sum_xy - sum_x * sum_y) / Math.sqrt((n * sum_xx - sum_x * sum_x) * (n * sum_yy - sum_y * sum_y)), 2);

  return lr;
}


function find_next(arr) {
  var y = [];

  for (var i = 0; i < arr.length; i++) {
    y.push(i);
  }

  var reg = linearRegression(y, arr)
  // y = a*x + b
  // x = (y-b)/a
  var next = (arr.length - reg.intercept) / reg.slope
  return next
}

console.log(find_next([1, 2, 3, 4, 5]))
console.log(find_next([0, 1, 3, 2, 4, 4, 2]))

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