Python 分期付款和利息产出
Python Amortization Payment and Interest Output
我有一个正在为学校做的项目,我很确定我做对了,但我不确定如何检查输出。我只需要知道我是否正在正确地处理这个问题。任务是让它显示每月付款和总利息。
这是我的代码
principal = float(input("Input Loan Principal: "))
# Principal amount borrowed input by user
loan_term = float(input("Input Loan Term In Years: "))
# Loan Term in years input by user
loan_interest = float(input("Input Loan Interest Rate: "))
# Loan interest in percentage input by user
monthly_rate = loan_interest / 100
# Loan interest percentage to decimal
loan_term_months = loan_term * 12
# Loan term from years to months
balance = principal
# Separate variable for calculations
math1 = (1.0 + monthly_rate) ** loan_term
math2 = (monthly_rate/(math1 - 1.0))
# Calculations
monthly_payment = (monthly_rate + math2) * balance
# Final Calculation for monthly payment
interest = (monthly_payment * loan_term) - principal
# Final Calculation for interest paid
final_monthly_payment = str(round(monthly_payment, 2))
final_interest = str(round(interest, 2))
# Rounding to two decimal points
print("Monthly payment: ", final_monthly_payment)
print("Effective Interest Paid: ", final_interest)
# Final print
提前感谢您的帮助
这是使用 scipy fsolve 检查答案的奇特方法:
import numpy as np
from scipy.optimize import fsolve
fv = 0
pv = 200000
rate = 0.075 / 12
nper = 15 * 12
def f(pmt):
return fv + pv*(1 + rate)**nper + pmt*(1 + rate*0) / rate*((1 + rate)**nper - 1)
pmt_amt = fsolve(f, [100], xtol=0.000001)
print(pmt_amt)
正在打印 result
returns:
[-1854.02472001]
上面的主要概念是让scipy求解公式中的变量pmt
...就是支付金额。解释是需要支付 1,854.02 美元才能在 15 年内以 7.5% 的利率还清 200,000 美元的贷款。
要查看每笔付款有多少本金和利息,numpy 可以再次帮助您解决 np.ppmt
和 np.ipmt
问题。示例:
import numpy as np
fv = 0
pv = 200000
rate = 0.075 / 12
nper = 15 * 12
for per in range(nper):
principal = -np.ppmt(rate, per, nper, pv)
interest = -np.ipmt(rate, per, nper, pv)
print(principal, interest, principal + interest)
Returns:
600.27301367 1253.75170634 1854.02472001
604.024720005 1250.0 1854.02472001
607.799874505 1246.2248455 1854.02472001
611.598623721 1242.42609628 1854.02472001
615.421115119 1238.60360489 1854.02472001
619.267497089 1234.75722292 1854.02472001
623.137918946 1230.88680106 1854.02472001
627.032530939 1226.99218907 1854.02472001
630.951484257 1223.07323575 1854.02472001
...
1797.15714344 56.8675765608 1854.02472001
1808.38937559 45.6353444143 1854.02472001
1819.69180919 34.3329108169 1854.02472001
1831.064883 22.9598370094 1854.02472001
我有一个正在为学校做的项目,我很确定我做对了,但我不确定如何检查输出。我只需要知道我是否正在正确地处理这个问题。任务是让它显示每月付款和总利息。 这是我的代码
principal = float(input("Input Loan Principal: "))
# Principal amount borrowed input by user
loan_term = float(input("Input Loan Term In Years: "))
# Loan Term in years input by user
loan_interest = float(input("Input Loan Interest Rate: "))
# Loan interest in percentage input by user
monthly_rate = loan_interest / 100
# Loan interest percentage to decimal
loan_term_months = loan_term * 12
# Loan term from years to months
balance = principal
# Separate variable for calculations
math1 = (1.0 + monthly_rate) ** loan_term
math2 = (monthly_rate/(math1 - 1.0))
# Calculations
monthly_payment = (monthly_rate + math2) * balance
# Final Calculation for monthly payment
interest = (monthly_payment * loan_term) - principal
# Final Calculation for interest paid
final_monthly_payment = str(round(monthly_payment, 2))
final_interest = str(round(interest, 2))
# Rounding to two decimal points
print("Monthly payment: ", final_monthly_payment)
print("Effective Interest Paid: ", final_interest)
# Final print
提前感谢您的帮助
这是使用 scipy fsolve 检查答案的奇特方法:
import numpy as np
from scipy.optimize import fsolve
fv = 0
pv = 200000
rate = 0.075 / 12
nper = 15 * 12
def f(pmt):
return fv + pv*(1 + rate)**nper + pmt*(1 + rate*0) / rate*((1 + rate)**nper - 1)
pmt_amt = fsolve(f, [100], xtol=0.000001)
print(pmt_amt)
正在打印 result
returns:
[-1854.02472001]
上面的主要概念是让scipy求解公式中的变量pmt
...就是支付金额。解释是需要支付 1,854.02 美元才能在 15 年内以 7.5% 的利率还清 200,000 美元的贷款。
要查看每笔付款有多少本金和利息,numpy 可以再次帮助您解决 np.ppmt
和 np.ipmt
问题。示例:
import numpy as np
fv = 0
pv = 200000
rate = 0.075 / 12
nper = 15 * 12
for per in range(nper):
principal = -np.ppmt(rate, per, nper, pv)
interest = -np.ipmt(rate, per, nper, pv)
print(principal, interest, principal + interest)
Returns:
600.27301367 1253.75170634 1854.02472001
604.024720005 1250.0 1854.02472001
607.799874505 1246.2248455 1854.02472001
611.598623721 1242.42609628 1854.02472001
615.421115119 1238.60360489 1854.02472001
619.267497089 1234.75722292 1854.02472001
623.137918946 1230.88680106 1854.02472001
627.032530939 1226.99218907 1854.02472001
630.951484257 1223.07323575 1854.02472001
...
1797.15714344 56.8675765608 1854.02472001
1808.38937559 45.6353444143 1854.02472001
1819.69180919 34.3329108169 1854.02472001
1831.064883 22.9598370094 1854.02472001