计算 for 循环期间的运行总计 - Python

Question

编辑：下面是我的工作代码，基于我收到的反馈/答案。

这个问题源于我之前使用麻省理工学院的开放课件学习 Python/CS 时提出的问题。 --在这里查看我之前的问题< /a>--

我正在使用以下代码来列出每月付款和其他事项的列表。然而，在循环结束时，我需要给出月份已支付总额的运行总计。

原始代码

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

    if month in xrange(12, 12+1):
        print 'RESULTS'
        print 'Total amount paid: '
        print 'Remaining balance: %.2f' % (remainingBalance,)

问题是我无法弄清楚如何保持支付金额的运行总额。我尝试添加 totalPaid = round(interestPaid +principalPaid, 2) 但这只会导致一个月的总计，我似乎无法让它保留每个月的该值，然后将它们全部添加放在最后打印出来。

我还知道结果金额应该是 1131.12

我已经找到了许多在知道每个值时通过列表执行此操作的示例，但我似乎无法正确推断。

固定代码

balance = float(raw_input("Outstanding Balance: "))
interestRate = float(raw_input("Interest Rate: "))
minPayRate = float(raw_input("Minimum Monthly Payment Rate: "))
totalPaid = 0

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)
    totalPaid += round(minPayment, 2)

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

    balance = remainingBalance

    if month in xrange(12, 12+1):
        print 'RESULTS'
        print 'Total amount paid: %.2f' % (totalPaid,)
        print 'Remaining balance: %.2f' % (remainingBalance,)

Answer 1

在循环之前，初始化一个变量来累加值：

total_paid = 0

然后，在循环体中，向其添加适当的金额。您可以使用+= 运算符添加到现有变量，例如

total_paid += 1

total_paid =total_paid + 1 的缩写形式。您不想每次迭代都为 total_paid 提供新值，而是希望添加到其现有值。

我不确定你的问题的具体情况，但这是循环时累积值的一般形式。

Answer 2

你总是支付最低还款额吗？只需使用 minPayment 而不是再次计算数学。保留运行总计，然后在循环后将其打印出来。

balance = float(raw_input("Outstanding Balance: "))
interestRate = float(raw_input("Interest Rate: "))
minPayRate = float(raw_input("Minimum Monthly Payment Rate: "))
paid = 0

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)
    paid += minPayment

    print  # Make the output easier to read.
    print 'Month: %d' % (month,)
    print 'Minimum monthly payment: %.2f' % (minPayment,)
    print 'Principle paid: %.2f' % (principalPaid,)
    print 'Remaining balance: %.2f' % (remainingBalance,)

    balance = remainingBalance

print
print 'RESULTS'
print 'Total amount paid:', paid
print 'Remaining balance: %.2f' % (remainingBalance,)

另请注意，范围只有一个值，因此您只需检查 Month == 12，但这里根本没有必要。

Answer 3

这个答案对我有用：

首先，创建派生值：

df.loc[0, 'C'] = df.loc[0, 'D']

然后迭代剩余的行并填充计算值：

for i in range(1, len(df)):
    df.loc[i, 'C'] = df.loc[i-1, 'C'] * df.loc[i, 'A'] + df.loc[i, 'B']

Index_Date	A	B	C	D
2015/01/31	10	10	10	10
2015/02/01	2	3	23	22
2015/02/02	10	60	290	280

Answer 4

您实际上必须将totalPaid初始化为0，然后

totalPaid = round(interestPaid + principalPaid, 2) + totalPaid

在循环内。你的问题是你没有累积总数，你只是在每次迭代中设置一个新的。

Answer 5

听起来你很接近。问题是你每次都会覆盖总数。尝试这样的事情：

totalPaid = totalPaid + round(interestPaid + principalPaid, 2)