How to simplify an array by removing numbers that are close to each other in javascript

Question

I have an array with some numbers

[0, 0.1, 0.2, 0.3, 0.6, 0.8, 1]

how do I remove numbers that are close to each other less than or equal to 0.2 threshold and put the average of those values instead, so the above array becomes

[0.2, 0.8]

0.2 because [0, 0.1, 0.2, 0.3] are close to each other so they become (0 + 0.1 + 0.2 + 0.3) / 4 = 0.2

0.8 because [0.6, 0.8, 1] are close to each other so they become (0.6, 0.8, 1) / 2 = 0.8

Please clarify the actual criteria for removing. Why would the result be `[0.2, 0.8]`? — Unmitigated, May 27 '23 at 19:53
What happens if you have 0,0.1,0.2,0.3,...,0.8,0.9,1? Each neighbor is only 0.1 away from the next? Will they collapse to a single 0.5 value? — derpirscher, May 27 '23 at 19:57
and it doesn't even have to be average if you can further simplify just a value in between those numbers — eguneys, May 27 '23 at 20:00
https://en.wikipedia.org/wiki/Moving_average might be what you are looking for — bas080, May 27 '23 at 20:09

score 1 · Accepted Answer · answered May 27 '23 at 20:10

You could do this by iterating over every element of the array and keeping a temporary close array. If close is empty, or the current element and the last element in close's difference is less than the threshold, then you can add it to close. If not, then average close and add it to the output array. Note that floating point math gets a bit weird here, you can use this rounding trick.

const fixFloat = n => parseFloat(n.toPrecision(12))

function average(arr) {
  let sum = 0;
  for (const el of arr) sum += el;
  return fixFloat(sum / arr.length);
}

function closeRemove(arr, threshold) {
  arr = [...arr].sort()
  const res = [];
  let close = [];
  for (let i = 0; i < arr.length; i++) {
    const diff = fixFloat(arr[i] - close[close.length-1]);
    if (close.length == 0 || diff <= threshold) {
      close.push(arr[i])
    } else {
      res.push(average(close));
      close = [arr[i]];
    }
  }
  res.push(average(close));
  return res;
}

const arr = [0, 0.1, 0.2, 0.3, 0.6, 0.8, 1];
console.log(closeRemove(arr, 0.2));

Note that in your question, the numbers 0, 0.1, 0.2, 0.3 all have a difference of less than 0.2, so they should be averaged together. Their average is 1.5, not 2.

@derpirscher Does OP ask for averages to be rounded to one place after the decimal? I can't see it in the question or any comments. — Michael M., May 27 '23 at 20:25
Not explicitly, but the example seems to point in that direction ... — derpirscher, May 27 '23 at 20:26

How to simplify an array by removing numbers that are close to each other in javascript

1 Answers1