gistfile1.txt

{
 "metadata": {
  "name": "hw3-solutions-1-and-2"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "## Week 3, problem 1 -- load in files, calculate averages/stddevs, and plot"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "import csv\nimport math\n\n# function to load x, y values from a CSV file.  Note, should take 'filename' and use 'csv' module.\ndef load_xy(filename):\n    fp = open(filename, 'rb')\n    r = csv.reader(fp)\n    \n    xx = []\n    yy = []\n    for x, y in r:\n        x = float(x)\n        y = float(y)\n        xx.append(x)\n        yy.append(y)\n        \n    return xx, yy\n\ndef average(yy):\n    # could do like this:\n    #return sum(yy) / float(len(yy))\n\n    # or:\n    total = 0.\n    for y in yy:\n        total += y\n    return total / float(len(yy))\n\ndef stddev(yy):\n    avg = average(yy)\n    diffsq = 0.\n    \n    for y in yy:\n        diffsq += (y - avg)**2\n    return math.sqrt(diffsq) / float(len(yy))",
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 11
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "x1, y1 = load_xy('/usr/local/notebooks/quartet/q1.csv')\nx2, y2 = load_xy('/usr/local/notebooks/quartet/q2.csv')\nx3, y3 = load_xy('/usr/local/notebooks/quartet/q3.csv')\nx4, y4 = load_xy('/usr/local/notebooks/quartet/q4.csv')\n\nprint 'q1:', average(y1), stddev(y1)\nprint 'q2:', average(y2), stddev(y2)\nprint 'q3:', average(y3), stddev(y3)\nprint 'q4:', average(y4), stddev(y4)\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "q1: 7.50090909091 0.584034775577\nq2: 7.50090909091 0.584060246164\nq3: 7.5 0.583705744956\nq4: 7.50090909091 0.583750278525\n"
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "plot(x1, y1, 'r.')\nplot(x2, y2, 'b+')\nplot(x3, y3, 'gx')\nplot(x4, y4, 'y*')",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "pyout",
       "prompt_number": 12,
       "text": "[<matplotlib.lines.Line2D at 0x7fb96448c0d0>]"
      },
      {
       "output_type": "display_data",
       "png": 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       "text": "<matplotlib.figure.Figure at 0x7fb96420b510>"
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "## Problem 2 -- invert dictionaries"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "def invert_d(dd):\n    e = {}\n    for k in dd:\n        value = dd[k]\n        e[value] = k\n    return e\n",
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 13
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "# test --\n\nfirst_d = { 'first': 1, 'second': 2 }\ninverted_test = { 1: 'first', 2: 'second' }\n\nassert invert_d(first_d) == inverted_test",
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 14
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "bad_d = { 'one': 1, 'uno': 1 }\n\nprint invert_d(bad_d)",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "{1: 'one'}\n"
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "What's happening with 'bad_d', above, is that the value '1' belongs to two keys, 'one' and 'uno'.  When you invert this so that the value 'uno' belongs to the key 1, and the value 'one' belongs to the key 1, you run into the problem that keys have to be unique -- so either 'uno' or 'one' will be lost.  Which one depends on the order in which the 'for' loop iterates over the dictionary -- the *last* key (of 'one' and 'uno') to be returned from bad_d will be assigned to the inverted dictionary as the value for the key 1."
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "",
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}