iPython Notebooks

Drag promedio.ipynb

{
 "metadata": {
  "name": ""
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import numpy as np"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Despu\u00e9s de importar Numpy extraemos los datos del archivo que hemos exportado de STAR-CCM+:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "data = np.genfromtxt('Drag report.csv',\n",
      "                     dtype=float,\n",
      "                     delimiter=',',\n",
      "                     names=True)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 6
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "data.dtype.names"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 7,
       "text": [
        "('Iteration_Iteration', 'Drag_Force_Coefficient')"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Nos quedamos con las 3000 \u00faltimas iteraciones."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "drag = data[data.dtype.names[1]][-3000:]"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 12
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Obtenemos el valor m\u00e1ximo, el m\u00ednimo y la media:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "np.max(drag)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 13,
       "text": [
        "0.37309515224949907"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "np.min(drag)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 14,
       "text": [
        "0.31619330461407386"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "np.mean(drag)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 15,
       "text": [
        "0.34351105987568487"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Otra forma de afrontar el problema es obtener el \u00faltimo m\u00e1ximo y m\u00ednimo local. Para ello necesitamos importar un paquete de Scipy (ver http://stackoverflow.com/questions/4624970/finding-local-maxima-minima-with-numpy-in-a-1d-numpy-array)."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from scipy.signal import argrelextrema"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 16
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "lmax = argrelextrema(drag, np.greater)\n",
      "lmin = argrelextrema(drag, np.less)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 17
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Lo que obtenemos son \u00edndices. Pero veamos c\u00f3mo nos los presenta."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "print(lmin)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(array([ 117,  244,  371,  497,  624,  751,  878, 1004, 1131, 1258, 1384,\n",
        "       1511, 1638, 1765, 1891, 2018, 2145, 2271, 2398, 2525, 2651, 2778,\n",
        "       2905], dtype=int64),)\n"
       ]
      }
     ],
     "prompt_number": 25
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Como podemos observar, el array est\u00e1 dentro de una tupla. Por tanto, si queremos obtener el \u00faltimos m\u00ednimos y m\u00e1ximos locales deberemos escribir `lmin[0][-1]` y `lmax[0][-1]`."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "np.mean(drag[lmin[0][-1]:lmax[0][-1]])"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 27,
       "text": [
        "0.34290937463108467"
       ]
      }
     ],
     "prompt_number": 27
    }
   ],
   "metadata": {}
  }
 ]
}