我如何应用循环来消除这个复杂的利率计算器的线条？

Question

如何从这段代码中删除行？我觉得我可以使用循环来简化这个过程。
我刚开始学习 python 以使我的数学学习更有趣。我对编码几乎一无所知，所以举个例子会有很大帮助。
该代码必须包含一个选项以投入初始资金。

谢谢！

capital = float(input("Enter initial capital: "))
interest_rate = float(input("Enter interest rate: "))
annual_saving = float(input("Enter annual savings: "))
year = float(input("Enter number of years of saving: "))

# Y stands for Year
Y0 = capital + annual_saving
Y1 = Y0 * (1 + interest_rate)
Y2 = (Y1 + annual_saving) * (1 + interest_rate)
Y3 = (Y2 + annual_saving) * (1 + interest_rate)
Y4 = (Y3 + annual_saving) * (1 + interest_rate)
Y5 = (Y4 + annual_saving) * (1 + interest_rate)
Y6 = (Y5 + annual_saving) * (1 + interest_rate)
Y7 = (Y6 + annual_saving) * (1 + interest_rate)
Y8 = (Y7 + annual_saving) * (1 + interest_rate)
Y9 = (Y8 + annual_saving) * (1 + interest_rate)
Y10 = (Y9 + annual_saving) * (1 + interest_rate)
Y11 = (Y10 + annual_saving) * (1 + interest_rate)

if year == 1:
    print(Y1)
if year == 2:
    print(Y2)
if year == 3:
    print(Y3)
if year == 4:
    print(Y4)
if year == 5:
    print(Y5)
if year == 6:
    print(Y6)
if year == 7:
    print(Y7)
if year == 8:
    print(Y8)
if year == 9:
    print(Y9)
if year == 10:
    print(Y10)

Answer 1

应计利息的公式是 K_interrest = K_start * (1 + (p/100))**n，其中 p 作为您的利率，n 作为年限。这样我们就可以将您的代码放在一个方便的函数中。你的初始资金和年储蓄乘以利率得到你的新资本。这种情况发生 n 次。

def calculate_savings(k, p, a, n):
    for i in range(n):
        k = (k+a) * (1 + (p/100))
    print(f"{k:.2f}", "$")  # will print your result with 2 decimal digits


capital = float(input("Enter initial capital: "))
interest_rate = float(input("Enter interest rate [%]: "))  # input in percent
annual_saving = float(input("Enter annual savings: "))
years_duration = int(input("Enter number of years of saving: "))  # years must be int here

calculate_savings(capital, interest_rate, annual_saving, years_duration)

另一种可能是使用递归函数（调用自身的函数）：

def calculate_savings(k, p, a, n):
    if n <= 0:
        print(f"{k:.2f}", "$")
    else:
        k = (k+a) * (1 + (p/100))
        calculate_savings(k, p, a, n-1)  # calculates k with one year less

通常认为保持代码干燥（不要重复自己）是一种很好的做法。此外，当您的脚本变得更加丰富时，您应该考虑捕获错误的输入。

Answer 2

您可以创建一个函数来进行所有计算并 return 得到正确的结果。函数从输入中获取 4 个变量。

def calculations(capital, interest_rate, annual_saving, year):
    x = (capital + annual_saving) * (1 + interest_rate)
    for i in range(1, year, 1):
        x = (x + annual_saving) * (1 + interest_rate)
    return x

capital = float(input("Enter initial capital: "))
interest_rate = float(input("Enter interest rate: "))
annual_saving = float(input("Enter annual savings: "))
year = float(input("Enter number of years of saving: "))

results = calculations(capital, interest_rate, annual_saving, year)
print(results)

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