Jupyter notebook answering the question: which way of splitting people up in three groups by something to do with their birth date produces the most evenly-sized groups?

birthday-thirds.ipynb

{
  "metadata": {
    "language_info": {
      "codemirror_mode": {
        "name": "python",
        "version": 3
      },
      "file_extension": ".py",
      "mimetype": "text/x-python",
      "name": "python",
      "nbconvert_exporter": "python",
      "pygments_lexer": "ipython3",
      "version": "3.8"
    },
    "kernelspec": {
      "name": "python",
      "display_name": "Python (Pyodide)",
      "language": "python"
    }
  },
  "nbformat_minor": 4,
  "nbformat": 4,
  "cells": [
    {
      "cell_type": "markdown",
      "source": "Data from https://www.ons.gov.uk/peoplepopulationandcommunity/birthsdeathsandmarriages/livebirths/articles/howpopularisyourbirthday/2015-12-18",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "import csv\n\nwith open('data.csv') as f:\n    d = csv.DictReader(f)\n    rows = list(d)\n\nfrom datetime import datetime\n\nrows = [(datetime.strptime('2000 '+r['date'],'%Y %d-%b'), float(r['average'])) for r in rows]",
      "metadata": {
        "trusted": true
      },
      "execution_count": 38,
      "outputs": []
    },
    {
      "cell_type": "code",
      "source": "months = [31,28,31,30,31,30,31,31,30,31,30,31]",
      "metadata": {
        "trusted": true
      },
      "execution_count": 39,
      "outputs": []
    },
    {
      "cell_type": "code",
      "source": "from collections import defaultdict\n\nmethods = [\n    (\"Month number mod 3\", lambda d: d.month % 3),\n    (\"Day number mod 3\",   lambda d: d.day % 3),\n    (\"Contiguous thirds of the year\", lambda d: (d.month -1)//3),\n    (\"Day of the year mod 3\", lambda d: (sum(months[:d.month-1]) + d.day) % 3),\n]\n\nresults = []\n\nfor desc,method in methods:\n    f = defaultdict(lambda: 0)\n\n    for d,n in rows:\n        m = method(d)\n        f[m] += n\n\n    t = sum(f.values())\n\n    p = [n/t for n in f.values()]\n    results.append((desc, max(p) - min(p)))\n\nresults.sort(key=lambda x:x[1])\n\nfor desc, p in results:\n    print(f\"{p}\\t{desc}\")",
      "metadata": {
        "trusted": true
      },
      "execution_count": 40,
      "outputs": [
        {
          "name": "stdout",
          "text": "8.054183179989627e-05\tDay of the year mod 3\n0.006354592308398355\tMonth number mod 3\n0.014516064139928453\tContiguous thirds of the year\n0.02048407072431757\tDay number mod 3\n",
          "output_type": "stream"
        }
      ]
    }
  ]
}