Data-Science-from-Scratch_Chapter5_Statistics_Joel-Grus.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "8fe10b4e",
   "metadata": {},
   "source": [
    "# Statistics"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "77505265",
   "metadata": {},
   "source": [
    "## Describing Data"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "933a1458",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<Figure size 432x288 with 1 Axes>"
      ]
     },
     "metadata": {
      "needs_background": "light"
     },
     "output_type": "display_data"
    }
   ],
   "source": [
    "# How to describe data\n",
    "\n",
    "num_friends = [100, 49, 41, 40, 25]\n",
    "\n",
    "# In the first approach, we put the friends count in a histogram using Counter and plt.bar()\n",
    "\n",
    "from matplotlib import pyplot as plt\n",
    "from collections import Counter\n",
    "\n",
    "friend_counts = Counter(num_friends)\n",
    "xs = range(101)                              # biggest value is 100\n",
    "ys = [friend_counts[x] for x in xs]          # height is only # (number) of friends\n",
    "plt.bar(xs, ys)\n",
    "plt.axis([0, 101, 0, 25])\n",
    "plt.title(\"Friends count histogram\")\n",
    "plt.xlabel(\"# of friends\")\n",
    "plt.ylabel(\"# of people\")\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "584f8c9a",
   "metadata": {},
   "outputs": [],
   "source": [
    "# Unfortunately, it is hard to include this graph in discussions. So we better get to generate some statistics. \n",
    "# The simplest statistic is the number of points in data:\n",
    "num_points = len(num_friends)    # 5\n",
    "\n",
    "# Maximum and minimum values\n",
    "largest_value = max(num_friends)      # 100\n",
    "smallest_value = min(num_friends)     # 25\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "36b70e04",
   "metadata": {},
   "source": [
    "## Tendências Centrais"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "5a08f0be",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "51.0\n"
     ]
    }
   ],
   "source": [
    "# Usually we want to have a notion of where our data is centered. The mean will be more used because it is the \n",
    "# sum of the data divided by its count:\n",
    "\n",
    "def mean(x):\n",
    "    return sum(x)/len(x)\n",
    "\n",
    "print(mean(num_friends))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "2249505d",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "41\n"
     ]
    }
   ],
   "source": [
    "# Sometimes we may get interested by the median, which is the biggest value of the middle (if the number of \n",
    "# points is odd) or the mean of the two values in the middle (if the number of points is even).\n",
    "\n",
    "# Different from the mean, the mean doesn't depends on eachvalue of its data. E.g. if you increase the biggest \n",
    "# point (or decrease the smaller point) the points in the middle remain intact, hence the median is mantained.\n",
    "\n",
    "# The median function is a little more complicated, mainly because the \"even\" part:\n",
    "\n",
    "def median(v):\n",
    "    \"\"\"finds the middle value of v\"\"\"\n",
    "    n = len(v)\n",
    "    sorted_v = sorted(v)\n",
    "    midpoint = n//2\n",
    "    \n",
    "    if n % 2 == 1:\n",
    "        # if is odd, returns the value in the middle\n",
    "        return sorted_v[midpoint]\n",
    "    else:\n",
    "        # if is even, return the mean of the middle values\n",
    "        lo = midpoint-1\n",
    "        hi = midpoint\n",
    "        return (sorted_v[lo] + sorted_v[hi]) / 2\n",
    "    \n",
    "print(median(num_friends))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "1ff0137c",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "25\n",
      "40\n",
      "49\n",
      "100\n"
     ]
    }
   ],
   "source": [
    "# Clearly the mean is easier to compute and varies in a smoother way while the data changes.\n",
    "# At the same time, the mean is sensible to discrepant values in our data. If the discrepant values have the \n",
    "# possibility to be bad data (or, in another way, non-representative of any phenomenon we are trying to \n",
    "# understand), then the mean can lead us to a mistake. For example, it is said that around the 80's, the degree \n",
    "# with the biggest average initial salary at the North Carolina University was geography, mainly due to the NBA\n",
    "# star (and a discrepant value) Michael Jordan.\n",
    "\n",
    "# A generalization of the mean is the quantile, which represents the value under where a certain percentage of \n",
    "# the data are (the median represents the value under where 50% of the data are).\n",
    "def quantile(x, p):\n",
    "    \"\"\"returns the p-ith percentual in x\"\"\"\n",
    "    p_index = int(p*len(x))\n",
    "    return sorted(x)[p_index]\n",
    "\n",
    "print(quantile(num_friends, 0.10))\n",
    "print(quantile(num_friends, 0.25))\n",
    "print(quantile(num_friends, 0.75))\n",
    "print(quantile(num_friends, 0.90))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "id": "4c29d9f0",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[100, 49, 41, 40, 25]\n"
     ]
    }
   ],
   "source": [
    "# Less commonly, you maybe want to take a look at the mode\n",
    "from collections import Counter\n",
    "def mode(x):\n",
    "    \"\"\"returns a list, it can have more than one mode\"\"\"\n",
    "    counts = Counter(x)\n",
    "    max_count = max(counts.values())\n",
    "    return [x_i for x_i, count in counts.items() if count == max_count]\n",
    "\n",
    "print(mode(num_friends))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "edc31c2e",
   "metadata": {},
   "source": [
    "## Dispersion"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "id": "73fecf17",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "75\n"
     ]
    }
   ],
   "source": [
    "# Dispersion refers to the measure of how spread are our data. Tipically, are statistics in which values near zero\n",
    "# are not spread and for bigger values (whatever it means) means a lot spread. E.g. one simple measure is the \n",
    "# amplitude, which is the difference between the biggest and the smallest element:\n",
    "\n",
    "# \"amplitude\" already has a meaning in Python, so we are going to use a different name\n",
    "def data_range(x):\n",
    "    return max(x) - min(x)\n",
    "\n",
    "print(data_range(num_friends))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "bfbc852f",
   "metadata": {},
   "outputs": [],
   "source": []
  }
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