Calculate Masses.ipynb.json

{
 "metadata": {
  "name": "Calculate Masses"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "In an attempt to simplivy your problem, let's focus on the nitty-gritty of it: the actual calculation. Let's save the user input for another moment (maybe we can convince Eddie and Jeremy to do it for you\u2026)"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "'''\n    Assume you have the following equation and that the user has identified\n    for which molecule they have given you the desired mass. Assume also that\n    you have the mass of a particular reactant or product already entered by\n    the user (ignore that problem, at least for now). For example, suppose\n    you have a structure like masses, in which one mass has been specified.\n\n    How would you go about implementing calculateMasses()?\n    (HINT: figure out which mass you _have_ first!)\n\n    I'll throw in a handy conversion function for you, in case you need it...\n'''\n\n'''\n    Let's imagine the user interface for a second (inputs are prefaced by >>>)\n\n    C3H8 + 5O2 \u2014> 4H2O + 3CO2\n    (1)    (2)    (3)    (4)\n\n    What reactant or product do you have a mass for?\n    >>> 3\n    What is the mass of 4H2O?\n    >>> 23.746kg\n\n    This would generate a masses that looks like:\n\n    masses = [\n        [\n            {'mass': -1,\n             'unit': '?'},\n            {'mass': -1,\n             'unit': '?'}\n        ],\n        [\n            {'mass': 23.746,\n             'unit': 'kg'},\n            {'mass': -1,\n             'unit': '?'}\n        ]\n    ]\n\n    '''\n\ndef convert(mass, oldUnit, newUnit):\n    '''\n        Convert between metric mass units (can be extended, btw...).\n        For example convert(10, 'kg', 'g') would return 1000.0, while\n        convert(10, 'g', 'kg') would return 0.01\n    '''\n    conversionTable = dict()\n    conversionTable['t'] = 1000000.0\n    conversionTable['kg'] = 1000.0\n    conversionTable['g'] = 1.0\n    conversionTable['mg'] = 0.001\n    return mass * (conversionTable[oldUnit]/conversionTable[newUnit])\n\ndef calculateMasses(equation, masses):\n    '''\n        Use the (assume balanced) equation to calculate all of the masses,\n        given one mass filled in for you by the user...\n    '''\n    # magic goes here\n    return calculatedMasses\n\nmasses = [\n    [\n        {'mass': 23.2,\n         'unit': 'kg'},\n        {'mass': -1,\n         'unit': '?'}\n    ],\n    [\n        {'mass': -1,\n         'unit': '?'},\n        {'mass': -1,\n         'unit': '?'}\n    ]\n]\n\nknownMass = 0 # the first in the list of reactants and products\n    \nequation = [\n    [\n        {'coefficient': 1,\n         'formula': {\n            'C': 3,\n            'H': 8\n        }},\n        {'coefficient': 5,\n         'formula': {\n            'O': 2\n        }}\n    ],\n    [\n        {'coefficient': 4,\n         'formula': {\n            'H': 2,\n            'O': 1\n        }},\n        {'coefficient': 3,\n         'formula': {\n            'C': 1,\n            'O': 2\n        }}\n    ]\n]",
     "language": "python",
     "metadata": {},
     "outputs": []
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "But\u2026 what if we modified the printEquation functions from the class' sample code to return strings (whose length we can calculate) rather than just printing them..."
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "def printEquation(equation):\n    for side in equation:\n        for molecule in side:\n            # printOperand(molecule)\n            operandText = generateOperandText(molecule)\n            print(operandText, sep = '', end = '')\n            print(' is',len(operandText), 'long')\n            if molecule != side[-1]:\n                print(' + ', sep = '', end = '')\n        if side != equation[-1]:\n            print(' --> ', sep = '', end = '')\n    print()\n\ndef generateOperandText(operand):\n    operandText = ''\n    if operand['coefficient'] > 1:\n        # operand['coefficient'], '(', sep = '', end = '')\n        operandText = operandText + str(operand['coefficient']) + '('\n    operandText = operandText + generateMoleculeText(operand['formula'])\n    if operand['coefficient'] > 1:\n        # print(')', sep = '', end = '')\n        operandText = operandText + ')'\n    return operandText\n\ndef generateMoleculeText(molecule):\n    # FIXME why does this print in the wrong order?\n    moleculeText = ''\n    for symbol, subscript in molecule.items():\n        if subscript > 1:\n            # print(symbol, subscript, sep = '_', end = '')\n            moleculeText = moleculeText + symbol + '_' + str(subscript)\n        else:\n            # print(symbol, sep = '', end = '')\n            moleculeText = moleculeText + symbol\n    return moleculeText\n\npropane = dict() # of coefficient and formula\npropane['coefficient'] = 1\npropane['formula'] = dict()\npropane['formula']['C'] = 3\npropane['formula']['H'] = 8\n\noxygen = dict()\noxygen['coefficient'] = 5\noxygen['formula'] = dict()\noxygen['formula']['O'] = 2\n\nreactants = [propane, oxygen]\n\nwater = dict()\nwater['coefficient'] = 4\nwater['formula'] = dict()\nwater['formula']['H'] = 2\nwater['formula']['O'] = 1\n\ncarbondioxide = dict()\ncarbondioxide['coefficient'] = 3\ncarbondioxide['formula'] = dict()\ncarbondioxide['formula']['C'] = 1\ncarbondioxide['formula']['O'] = 2\n\nproducts = [water, carbondioxide]\n\nequation = [reactants, products]\n\nprintEquation(equation)",
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}