How to find the greatest common divisor with rounding?

Question

Mathematics

The user taps a short rhythmic pattern of notes of only two durations: xor 2x. About like Morse code, only musically, in tempo. The result is an array of note durations in milliseconds, for example:
[145,160,112,170,281,292,258,305,257]

schedule

How can these durations be classified into notes of multiple durations? How to replace with "1" for short xand "2" for long 2x, so that for the example above it turns out
[1,1,1,1,2,2,2,2,2]

Naive way doesn't work

const a = [145,160,112,170,281,292,258,305,257];
const min = Math.min(...a); // 112
a.map(n => Math.round(n / min)) // [ 1, 1, 1, 2, 3, 3, 2, 3, 2 ]

This question is only about binary classification x || 2x.
Far later, in a different way, it may be interesting to be able to parse a “melody” without the shortest notes, consisting only of multiples 2xand 3x. Or even more difficult - polyrhythm with rational fractions: 4/4 then at the same pace 6/4 or 5/4. But it's not now.
upd . implementation of the solution Vladimir Olokhtonov on JS

spoiler

const AB = data => {
  
  const A = [Math.min(...data)];
  const B = [Math.max(...data)];
  
  const mean = arr => arr.reduce((a,b) => a + b, 0) / arr.length;

  data.forEach(n => {
    if (Math.abs(mean(A) - n) <= Math.abs(mean(B) - n))
      A.push(n);
    else B.push(n);
  });
  
  return [A.slice(1), B.slice(1)];
}
 
AB([145,160,112,170,281,292,258,305,257])

/*

*/

Answer 1

You can use one of the clustering algorithms in scikit-learn , such as KMeans:

>>> import numpy as np
>>> from sklearn.cluster import KMeans
>>> notes = [145, 160, 112, 170, 281, 292, 258, 305, 257]
>>> notes = np.array(notes).reshape(-1, 1)
>>> KMeans(n_clusters=2).fit_predict(notes)
array([0, 0, 0, 0, 1, 1, 1, 1, 1], dtype=int32)

But the order in which numbers are assigned to the clusters is not guaranteed. Clustering simply allows sampling into N parts, let's assign an arbitrary serial number to each cluster. May be so:

>>> KMeans(n_clusters=2).fit_predict(notes)
array([1, 1, 1, 1, 0, 0, 0, 0, 0], dtype=int32)

Answer 2

It seems to me that cluster analysis is needed here.