Question

这个问题/解决方案让我想到了另一个相关的问题here - 我们将不胜感激！

根据初步反馈更新了以下当前代码

我是Python的新手（这是我的第二个程序）。我目前正在使用麻省理工学院的开放课件来使用Python Academic Earth videos来介绍CS，我正在研究问题集1 Viewable Here。我已经创建了这个程序，在12个月内成功地重新创建了“测试用例1”（不包括“结果”部分......仍在努力）但我的问题是，以下（我的）代码是否尽可能高效？当我没有必要时，我觉得我正在重复自己。：

原始代码：

balance = float(raw_input("Outstanding Balance: "))
interestRate = float(raw_input("Interest Rate: "))
minPayRate = float(raw_input("Minimum Monthly Payment Rate: "))
interestPaid = round((interestRate/12.0)*balance, 2)
minPayment = round(minPayRate*balance, 2)
principalPaid = round(minPayment-interestPaid, 2)
remainingBalance = round(balance-principalPaid, 2)
month = 1

while month < 12 :    
    if month > 1 :
        balance = remainingBalance
    interestPaid = round((interestRate/12.0)*balance, 2)
    minPayment = round(minPayRate*balance, 2)
    principalPaid = round(minPayment-interestPaid, 2)
    remainingBalance = round(balance-principalPaid , 2)   
    month = month+1

    print 'Month: ' + str(month)
    print 'Minimum monthly payment: ' + str(minPayment)
    print 'Principle paid: ' + str(principalPaid)
    print 'Remaining balance: ' + str(remainingBalance)

当前代码

balance = float(raw_input("Outstanding Balance: "))
interestRate = float(raw_input("Interest Rate: "))
minPayRate = float(raw_input("Minimum Monthly Payment Rate: "))

for month in xrange(1, 12+1):
    interestPaid = round(interestRate / 12.0 * balance, 2)
    minPayment = round(minPayRate * balance, 2)
    principalPaid = round(minPayment - interestPaid, 2)
    remainingBalance = round(balance - principalPaid, 2)

    print 'Month: %d' % (month,)
    print 'Minimum monthly payment: %.2f' % (minPayment,)
    print 'Principle paid: %.2f' % (principalPaid,)
    print 'Remaining balance: %.2f' % (remainingBalance,)

    balance = remainingBalance

如果您在此新代码中看到任何其他内容，请与我们联系！

非常感谢那些帮助我实现这一目标的人。

Answer 1

这与任何可能的效率问题无关*，但如果您正在进行金融算术，则应查看decimal模块。否则，你最终会遇到奇怪的rounding and representation errors。

*更准确地说：它会降低效率，但会提高代码的正确性。

Answer 2

print "x: " + str(x)

应替换为：

print "x:", x

这是打印的特例。

将循环更改为：

for month in xrange(1, 12+1):

放弃第一个循环的检查，只需将余额设置为remainingBalance作为结束。

因为您手动增加月份，所以每次都会打印错误的值。

如果你的效率和执行效率一样，那你就会担心too soon。如果你的意思是复制代码，你会在循环之前不必要地复制数学。结合以上内容：

balance = float(raw_input("Outstanding Balance: "))
interestRate = float(raw_input("Interest Rate: "))
minPayRate = float(raw_input("Minimum Monthly Payment Rate: "))

for month in xrange(1, 12+1):    
  interestPaid = round(interestRate / 12.0 * balance, 2)
  minPayment = round(minPayRate * balance, 2)
  principalPaid = round(minPayment - interestPaid, 2)
  remainingBalance = round(balance - principalPaid, 2)   

  print 'Month:', month
  print 'Minimum monthly payment:', minPayment
  print 'Principle paid:', principalPaid
  print 'Remaining balance:', remainingBalance

  balance = remainingBalance

Answer 3

您应该使用string interpolation或formatting而不是传递给str()并添加。

print 'Month: %d' % (month,)

Answer 4

对于（速度）优化，您可能希望阅读An Optimization Anecdote

Python初学者。这个Python代码是否尽可能高效？

4 个答案: