python - 仅当解为负时才进行搜索(MIT6.00)

Question

各位，

我对 MIT6.00 中 PS1 的问题 3 感到很困惑。我编写了几个函数(一个二分搜索和一个对信用卡债务进行建模的函数)。问题在于，它收敛于一个给出稍微正的剩余信用卡余额的解决方案。我可以降低二分搜索的容差，但我想知道是否有一些更优雅的方法使这个优化器仅返回负结果。

干杯，

艾登

代码:

import numpy as np

def bisection(a, b, fun, tol, var = None):
    """Note that this only works if you put the independant variable
    as the first argument in the parameter """
    #def tstr(x):
    #    return 2*(x**2) - 3*x + 1
    #sol = bisection(0,0.9,tstr,0.1)

    c = (a+b)/2.0  

    if var != None:
        arga = var[:]
        argc = var[:]
        arga.insert(0,a)
        argc.insert(0,c)
    else:
        arga = a
        argc = c

    if (b-a)/2.0 <= tol:
        #Debugging print statement 1:
        #print 'SOL1: c = ', c
        if var != None:
            return [c] + fun(argc)
        else:
            return c

    if fun(argc)[0] == 0:
        if var != None:
            return [c] + fun(argc)
        else:
            return c
    elif fun(arga)[0]*fun(argc)[0] < 0:
        b = c
    else:
        a = c
    return bisection(a, b, fun, tol, var)


"""
Now we have defined a version of the paidOff function to work
with the bisection method"""

def paidOffBis(args):#(Pay, Bal, Apr):
    """Tester for Bisection Implementation"""
    # TEST SIZE OF args:
    if (type(args) != list)|(np.size(args) != 3):
        print 'Incorrect size or type of input, input size:', np.size(args), '-', args
        return None
    Pay, Bal, Apr = args
    Mpr = Apr/12.0
    Baln = Bal
    Nm = 0
    for n in range(12):
        Baln = Baln*(1 + Mpr) - Pay
        if (Baln < 0)&(Nm == 0):
            Nm = n + 1   
    if Baln < 0:
        return [Baln, Nm]
    else:
        return [Baln, Nm]

Out_Bal = float(raw_input('Enter the outstanding balance on your credit card: '))
Apr = float(raw_input('Enter the annual credit card interest rate as a decimal: '))

varin = [Out_Bal, Apr]

#(Out_Bal*(1 + (Apr/12.0))**12.0)/12.0
sol = bisection(Out_Bal/12.0, Out_Bal, paidOffBis, 0.01, varin)

print 'RESULT'
print 'Monthly payment to pay off debt in 1 year: $%.2f' % sol[0]
print 'Number of months needed:', sol[2]
print 'Balance: $%.2f' % sol[1]

Answer 1

为了确保余额小于或等于零，您需要正确设置条件语句 - 您需要继续搜索，直到满足该条件。您要求...更优雅的方式...。使用变量的描述性名称并保持简单肯定会改进您的代码。这是使用二分搜索来制定解决方案的一种方法。

annualInterestRate = .1
balance = 1000

def balance_after_one_year(balance, annualInterestRate, payment):
    '''Calculate the balance after one year of interest and payments.

    balance --> int or float
    annualInterestRate --> float between 0 and 1
    payment --> float

    returns float
    '''
    for _ in xrange(12):
        balance = (balance - payment) * (1 + annualInterestRate / 12.0)
    return balance

def min_payment(balance, annualInterestRate, tolerance = .01):
    '''Find the minimum payment to pay off a loan.

    Uses a bisection search.
    Ensures balance is not positive.

    balance --> float, int
    annualInterestRate --> float less than one
    tolerance --> float
   '''

    # we want the tolerance to be negative
    # this isn't strictly a tolerance, it is a lower limit
    tolerance = -abs(tolerance)

    hi = balance
    lo = 0
    while True:
        payment = (hi + lo) / 2.0
        tmp_balance = balance_after_one_year(balance, annualInterestRate, payment)
        if tmp_balance < tolerance:
            hi = payment
        # ensure balance is not positive
        elif tmp_balance > 0:
            lo = payment
        else:
            return payment, tmp_balance

用法:

min_pmt, final_balance = min_payment(balance, annualInterestRate)
print 'minimum payment', min_pmt
print 'final balance', final_balance