Question

编辑：以下是根据我收到的反馈/答案的工作代码。

这个问题源于我之前在使用麻省理工学院开放课件学习Python / CS时提出的问题。 - See my previous question here -

我使用以下代码列出月付款和其他内容。但是，在循环结束时，我需要给出已经支付的总金额的累计金额。

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,)

问题在于我无法弄清楚如何保持支付金额的总计。我尝试添加totalPaid = round(interestPaid + principalPaid, 2)，但这仅仅导致了一个月的总数，我似乎无法让它保持每个月的价值，然后在最后将它们全部添加到打印出来。

我也知道结果数量应为1131.12

我已经找到了许多通过列表知道每个值时这样做的例子，但我似乎无法正确地推断出这一点。

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，初始化变量以累积值：

total_paid = 0

然后，在循环体中，添加适当的量。您可以使用+=运算符添加到现有变量，例如

total_paid += 1

是total_paid = total_paid + 1的缩写。您不希望每次迭代都为total_paid赋予一个新值，而是希望添加到其现有值。

我不确定问题的具体细节，但这是循环时累积值的一般形式。

Answer 2

您是否始终支付最低金额？只需使用minPayment而不是再次计算出数学。保持运行总计，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，因此您只需检查月份== 12，但这里根本没有必要。

Answer 3

您实际上必须将totalPaid初始化为0然后

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

循环内部。你的问题是你没有累积总数，你只是在每次迭代时设置一个新的。

Answer 4

听起来你很亲密。问题是你每次都要覆盖总数。尝试这样的事情：

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

Answer 5

这 answer 对我有用：

首先创建派生值：

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

然后遍历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']

<头>

索引日期	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

计算for循环期间的运行总计 - Python

5 个答案: