{
 "metadata": {
  "name": "9932_03_01"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Chapter 3, example 1\n",
      "====================\n",
      "\n",
      "We illustrate the notion of array computation with a simple example consisting of finding, among a large list of locations, the closest one from a position of interest."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "## Pure Python version\n",
      "\n",
      "The pure Python version loops through all positions, computes the distance from the position of interest, and returns the index of the closest location."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "def closest(position, positions):\n",
      "    x0, y0 = position\n",
      "    dbest, ibest = None, None\n",
      "    for i, (x, y) in enumerate(positions):\n",
      "        d = (x - x0) ** 2 + (y - y0) ** 2\n",
      "        if dbest is None or d < dbest:\n",
      "            dbest, ibest = d, i\n",
      "    return ibest\n",
      "import random"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "We generate a list of random positions."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "positions = [(random.random(), random.random()) for _ in xrange(10000000)]"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Now we evaluate the time required to compute the closest position from `(0.5, 0.5)`."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%timeit closest((.5, .5), positions)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "1 loops, best of 3: 12.8 s per loop\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "## Vectorized version\n",
      "\n",
      "This version uses NumPy to compute all distances in a vectorized way, without using Python loops. This version is far more efficient than the pure Python version."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "First, we need to activate the pylab mode so that NumPy is loaded and the NumPy objects are available in the current namespace."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%pylab"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
        "For more information, type 'help(pylab)'.\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Generating random positions with NumPy's `rand` function is much faster than using a list."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "positions = rand(10000000, 2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 5
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "`positions` is a NumPy `ndarray` object."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "type(positions)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "pyout",
       "prompt_number": 6,
       "text": [
        "numpy.ndarray"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "It has two dimensions and a shape of `(10000000, 2)`."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "positions.ndim, positions.shape"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "pyout",
       "prompt_number": 7,
       "text": [
        "(2, (10000000, 2))"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "We can easily get the columns of this matrix, which contain here the x and y coordinates of all positions."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "x, y = positions[:,0], positions[:,1]"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 8
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Now we compute the distances in a vectorized fashion, which is much more efficient than with a Python loop."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "distances = (x - .5) ** 2 + (y - .5) ** 2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 9
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%timeit exec In[9]"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "1 loops, best of 3: 341 ms per loop\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%timeit ibest = distances.argmin()"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "10 loops, best of 3: 24.2 ms per loop\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Using NumPy is about 35 times faster than using pure Python code here."
     ]
    }
   ],
   "metadata": {}
  }
 ]
}