在python-cod中简化数据结构和条件语句

Question

我想知道有没有什么方法可以简化下面的代码。如您所见，有许多dict和condition语句用于清除错误的输入数据。注意，行程率值还没有全部输入，现在只是复制和粘贴dicts

编辑

在任何速率中，（x，y）：z。x和y是正确的，z值不是因为它们只是复制/粘贴的

这段代码在您想要复制、粘贴和测试它的情况下起作用

import math


# step 1.4 return trip rates
def trip_rates( population_stratification, analysis_type, low_income, medium_income, high_income ):
  ''' this function returns the proper trip rate tuple to be used based on input 
    data 
    ADPT = Average Daily Person Trips per Household
    pph = person per household
    veh_hh = vehicles per household
    (param_1, param_2): ADPT
  '''
  li = low_income
  mi = medium_income
  hi = high_income
  # table 5 -
  if analysis_type == 1:
    if population_stratification == 1:
      rates = {( li, 1 ):3.6, ( li, 2 ):6.5, ( li, 3 ):9.1, ( li, 4 ):11.5, ( li, 5 ): 13.8,
               ( mi, 1 ):3.9, ( mi, 2 ):7.3, ( mi, 3 ):10.0, ( mi, 4 ):13.1, ( mi, 5 ): 15.9,
               ( hi, 1 ):4.5, ( mi, 2 ):9.2, ( mi, 3 ):12.2, ( mi, 4 ):14.8, ( mi, 5 ): 18.2}
      return rates
    if population_stratification == 2:
      rates = {
               ( li, 1 ):3.1, ( li, 2 ):6.3, ( li, 3 ):9.4, ( li, 4 ):12.5, ( li, 5 ): 14.7,
               ( mi, 1 ):4.8, ( mi, 2 ):7.2, ( mi, 3 ):10.1, ( mi, 4 ):13.3, ( mi, 5 ): 15.5,
               ( hi, 1 ):4.9, ( mi, 2 ):7.7, ( mi, 3 ):12.5, ( mi, 4 ):13.8, ( mi, 5 ): 16.7
              }
      return rates
    if population_stratification == 3: #TODO: input actual rate
      rates = {
               ( li, 1 ):3.6, ( li, 2 ):6.5, ( li, 3 ):9.1, ( li, 4 ):11.5, ( li, 5 ): 13.8,
               ( mi, 1 ):3.9, ( mi, 2 ):7.3, ( mi, 3 ):10.0, ( mi, 4 ):13.1, ( mi, 5 ): 15.9,
               ( hi, 1 ):4.5, ( mi, 2 ):9.2, ( mi, 3 ):12.2, ( mi, 4 ):14.8, ( mi, 5 ): 18.2
              }
      return rates
    if population_stratification == 4: #TODO: input actual rate
      rates = {
               ( li, 1 ):3.1, ( li, 2 ):6.3, ( li, 3 ):9.4, ( li, 4 ):12.5, ( li, 5 ): 14.7,
               ( mi, 1 ):4.8, ( mi, 2 ):7.2, ( mi, 3 ):10.1, ( mi, 4 ):13.3, ( mi, 5 ): 15.5,
               ( hi, 1 ):4.9, ( mi, 2 ):7.7, ( mi, 3 ):12.5, ( mi, 4 ):13.8, ( mi, 5 ): 16.7
              }
      return rates
  #table 6
  elif analysis_type == 2:
    if population_stratification == 1: #TODO: Change rates
      rates = {
               ( 0, 1 ):3.6, ( 0, 2 ):6.5, ( 0, 3 ):9.1, ( 0, 4 ):11.5, ( 0, 5 ): 13.8,
               ( 1, 1 ):3.9, ( 1, 2 ):7.3, ( 1, 3 ):10.0, ( 1, 4 ):13.1, ( 1, 5 ): 15.9,
               ( 2, 1 ):4.5, ( 2, 2 ):9.2, ( 2, 3 ):12.2, ( 2, 4 ):14.8, ( 2, 5 ): 18.2,
               ( 3, 1 ):4.5, ( 3, 2 ):9.2, ( 3, 3 ):12.2, ( 3, 4 ):14.8, ( 3, 5 ): 18.2
              }
      return rates
    if population_stratification == 2: #TODO: Change rates
      rates = {
               ( 0, 1 ):3.6, ( 0, 2 ):6.5, ( 0, 3 ):9.1, ( 0, 4 ):11.5, ( 0, 5 ): 13.8,
               ( 1, 1 ):3.9, ( 1, 2 ):7.3, ( 1, 3 ):10.0, ( 1, 4 ):13.1, ( 1, 5 ): 15.9,
               ( 2, 1 ):4.5, ( 2, 2 ):9.2, ( 2, 3 ):12.2, ( 2, 4 ):14.8, ( 2, 5 ): 18.2,
               ( 3, 1 ):4.5, ( 3, 2 ):9.2, ( 3, 3 ):12.2, ( 3, 4 ):14.8, ( 3, 5 ): 18.2
              }
      return rates
    if population_stratification == 3: #TODO: Change rates
      rates = {
               ( 0, 1 ):3.6, ( 0, 2 ):6.5, ( 0, 3 ):9.1, ( 0, 4 ):11.5, ( 0, 5 ): 13.8,
               ( 1, 1 ):3.9, ( 1, 2 ):7.3, ( 1, 3 ):10.0, ( 1, 4 ):13.1, ( 1, 5 ): 15.9,
               ( 2, 1 ):4.5, ( 2, 2 ):9.2, ( 2, 3 ):12.2, ( 2, 4 ):14.8, ( 2, 5 ): 18.2,
               ( 3, 1 ):4.5, ( 3, 2 ):9.2, ( 3, 3 ):12.2, ( 3, 4 ):14.8, ( 3, 5 ): 18.2
              }
      return rates
    if population_stratification == 4: #TODO: Change rates
      rates = {
               ( 0, 1 ):3.6, ( 0, 2 ):6.5, ( 0, 3 ):9.1, ( 0, 4 ):11.5, ( 0, 5 ): 13.8,
               ( 1, 1 ):3.9, ( 1, 2 ):7.3, ( 1, 3 ):10.0, ( 1, 4 ):13.1, ( 1, 5 ): 15.9,
               ( 2, 1 ):4.5, ( 2, 2 ):9.2, ( 2, 3 ):12.2, ( 2, 4 ):14.8, ( 2, 5 ): 18.2,
               ( 3, 1 ):4.5, ( 3, 2 ):9.2, ( 3, 3 ):12.2, ( 3, 4 ):14.8, ( 3, 5 ): 18.2
              }
      return rates
  # table 7
  elif analysis_type == 3:
    if population_stratification == 1: #TODO: input actual rate
      rates = {
               ( li, 0 ):3.6, ( li, 1 ):6.5, ( li, 2 ):9.1, ( li, 3 ):11.5,
               ( mi, 0 ):3.9, ( mi, 1 ):7.3, ( mi, 2 ):10.0, ( mi, 3 ):13.1,
               ( hi, 0 ):4.5, ( mi, 1 ):9.2, ( mi, 2 ):12.2, ( mi, 3 ):14.8
              }
      return rates
    if population_stratification == 2: #TODO: input actual rate
      rates = {
               ( li, 0 ):3.6, ( li, 1 ):6.5, ( li, 2 ):9.1, ( li, 3 ):11.5,
               ( mi, 0 ):3.9, ( mi, 1 ):7.3, ( mi, 2 ):10.0, ( mi, 3 ):13.1,
               ( hi, 0 ):4.5, ( mi, 1 ):9.2, ( mi, 2 ):12.2, ( mi, 3 ):14.8
              }
      return rates
    if population_stratification == 3: #TODO: input actual rate
      rates = {
               ( li, 0 ):3.6, ( li, 1 ):6.5, ( li, 2 ):9.1, ( li, 3 ):11.5,
               ( mi, 0 ):3.9, ( mi, 1 ):7.3, ( mi, 2 ):10.0, ( mi, 3 ):13.1,
               ( hi, 0 ):4.5, ( mi, 1 ):9.2, ( mi, 2 ):12.2, ( mi, 3 ):14.8
              }
      return rates
    if population_stratification == 4: #TODO: input actual rate
      rates = {
               ( li, 0 ):3.6, ( li, 1 ):6.5, ( li, 2 ):9.1, ( li, 3 ):11.5,
               ( mi, 0 ):3.9, ( mi, 1 ):7.3, ( mi, 2 ):10.0, ( mi, 3 ):13.1,
               ( hi, 0 ):4.5, ( mi, 1 ):9.2, ( mi, 2 ):12.2, ( mi, 3 ):14.8
              }
      return rates

def interpolate( population_stratification, analysis_type, low_income, medium_income, high_income, x, y ):
  #get rates dict
  rates = trip_rates( population_stratification, analysis_type, low_income, medium_income, high_income )


  # dealing with x parameters
  #when using income levels, x_1 and x_2 are li, mi, or hi
  if analysis_type == 1 or analysis_type == 2 or analsis_type == 4:
    if x < high_income and x >= medium_income:
      x_1 = medium_income
      x_2 = high_income
    elif x < medium_income:
      x_1 = low_income
      x_2 = medium_income
    else:
      x_1 = high_income
      x_2 = high_income
  if analysis_type == 3:
    if x >= 3:
      x_1 = 3
      x_2 = 3
    else:
      x_1 = int( math.floor( x ) )
      x_2 = int( math.ceil( x ) )

  # dealing with y parametrs
  #when using persons per household, max number y = 5
  if analysis_type == 1 or analysis_type == 4:
    if y >= 5:
      y_1 = 5
      y_2 = 5
    else:
      y_1 = int( math.floor( y ) )
      y_2 = int( math.ceil( y ) )
  elif analysis_type == 2 or analysis_type == 3:
    if y >= 5:
      y_1 = 5
      y_2 = 5
    else:
      y_1 = int( math.floor( y ) )
      y_2 = int( math.ceil( y ) )

  # denominator
  z = ( ( x_2 - x_1 ) * ( y_2 - y_1 ) )

  result = ( ( ( rates[( x_1, y_1 )] ) * ( ( x_2 - x ) * ( y_2 - y ) ) / ( z ) ) +
             ( ( rates[( x_2, y_1 )] ) * ( ( x - x_1 ) * ( y_2 - y ) ) / ( z ) ) +
             ( ( rates[( x_1, y_2 )] ) * ( ( x_2 - x ) * ( y - y_1 ) ) / ( z ) ) +
             ( ( rates[( x_2, y_2 )] ) * ( ( x - x_1 ) * ( y - y_1 ) ) / ( z ) ) )

  return result

#test
low_income = 20000 #this is calculated using exchange rates
medium_income = 40000 # this is calculated using exchange rates
high_income = 60000 # this is calculated using exchange rates
population_stratification = 1 #inputed by user
analysis_type = 1 #inputed by user
x = 35234.34 #test income
y = 3.5 # test pph

print interpolate( population_stratification, analysis_type, low_income, medium_income, high_income, x, y )

Answer 1

好吧，从哪里开始？以下是第一次观察：

你有很多数据，似乎代码和数据是混合在一起的。在

数据和代码应该分开。数据是一个外部源，可以修改或读入。你可以调整你的代码来快速地将数据从一个好的可编辑的表示转换成对你的算法有用的表示。我怀疑您的代码会更短、更清晰、更不容易出错（您是否注意到所有的“rates”字典都有多个键，而您错过了很多“hi”键？）。在

如果您需要更好的抽象，如矩阵和数据数组，请查看numpy

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编辑1

你数过你的维度数了吗？这里有一个X维的多维矩阵： 分析类型、人口分层、收入水平、指数

如果我看到右边，这是一个3x4x3x3（=108个条目）“矩阵”或“查找表”。如果这是你的模型建立的数据，好的。但你不能把这些数字放进一个你读到的文件或表格里吗？你的代码几乎是微不足道的。在

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编辑2

好的，我将介绍一些小的python风格：测试集合或范围中的值。在

而不是：

if analysis_type == 1 or analysis_type == 2 or analsis_type == 4:

你可以用

^{pr2}$

甚至使用可读的名字作为（立方，…）建议。在

而不是：

if x < high_income and x >= medium_income:

您可以使用链式条件；Python是为数不多的编程语言之一，在这种语言中，条件链可以生成nautral if语句：

if medium_income <= x < high_income:

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编辑3

当然，比小代码数字更重要的是代码设计和重构。编辑2只能给你一些润色。在

你应该学会厌恶重复的代码。在

另外，在一个函数中有很多分支。这是个好兆头，你应该把它分成多个功能。它还可以减少重复。例如，当一个像analysis_type这样的变量可以完全改变函数的功能时，为什么在一个函数中有两个不同的行为呢？你不应该把整个程序放在一个函数里。也许analysis_type==3更好地用它自己的函数表示（例如）？在

您是否知道您的函数trip_rates基本上执行数组查找，其中数组查找是硬编码的，就像..:return。。如果：return..，那么数组在函数中是完全写出的？如果trip_rates可以这样实现呢？有可能吗？在

data_model = compute_table(low_income, ...)
return data_model[analysis_type][population_stratification]