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简化python代码中的数据结构和条件语句

conditional data-structures python

dassouki · 技术社区 · 15 年前

我想知道是否有任何方法可以简化下面的代码。正如您所看到的,有许多DICT和条件语句被用来清除错误的输入数据。请注意,跳闸率值尚未全部输入,目前仅复制和粘贴DICT

编辑

如果您想复制、粘贴和测试此代码,则此代码可以正常工作

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 )

2 回复 | 直到 15 年前

BenMorel Manish Pradhan 11 年前

那么,从哪里开始呢?这只是第一个观察结果:

这里有很多数据,代码和数据似乎是相互混合的。

数据和代码应该分开。数据是一个外部源,您可以修改或读取它。您可能需要修改代码,以便将数据从良好的可编辑表示快速解析为对算法有用的表示。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,并且您错过了很多“hi”键?)。

如果您需要更好的抽象,如矩阵和数据数组,请查看 numpy

你数过你的尺寸了吗?这里有一个具有X维的多维矩阵: 分析类型、人口分层、收入水平、指数

编辑2

好的,我将介绍一些较小的python风格:测试集合或范围中的值。

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

你可以用

if analysis_type in (1, 2, 4):

或者甚至使用可读的名称,如建议的那样(立方,…)。

而不是:

if x < high_income and x >= medium_income:

您可以使用链接条件;Python是为数不多的编程语言之一,其中条件链生成nautral if语句:

if medium_income <= x < high_income:

当然,比小代码量更重要的是代码设计和重构。编辑2只能给你一些润色。

你应该学会厌恶重复的代码。

analysis_type

你明白你的职能吗 trip_rates 可以这样实施吗?可能吗?

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

Kathy Van Stone 15 年前

除了kaizer关于数据和代码的建议外,下面是一些简单的清理:

if y >= 5:
      y_1 = 5
      y_2 = 5
    else:
      y_1 = int( math.floor( y ) )
      y_2 = int( math.ceil( y ) )

可以写成

min(5, int(math.floor(y))

或

int(math.floor(min(5, y))

甚至做了一个功能:

def limitedInt(v, maxV):
   return min(5, int(math.floor(y))

我也建议你这样做,而不是说 analysis_type == 1 你说什么 analysis_type = CUBIC

你可能会找到这本书 Refactoring Refactoring Workbook the website 也可以使用,但是如果不知道书中描述的“代码气味”,它就没有那么大的帮助。