在 Python 或 SQL 中使用 Excel 作为求解器
Using Excel like solver in Python or SQL
这是我在 Excel 中做的一个简单计算。我想知道它是否可以完成 python 或任何其他语言。
Loan amount 7692
Period : 12 months
Rate of interest 18 Per Annum
The formula in the B2 cell is =A1*18/100/12
The formula in the A2 cells is =A1+B2-C2
C 栏是借款人每月可能需要偿还的暂定金额。 C2 旁边的所有其他单元格仅指向 200 的第一部分。使用下图中所示的求解器后，我在 C 列中得到正确的部分 705.20。
我想知道这个计算是否可以使用任何脚本语言完成，例如 python（或 SQL）
这是最终版本的样子...
我试过类似的方法，但它没有退出循环并打印所有组合。
loan_amount= 7692
interest = 18
months =12
for rg in range(700, 710):
for i in range(months):
x = loan_amount * interest / 100 / 12
y = loan_amount + x - rg
if x < 0:
print rg, i
exit
else:
loan_amount = y
您的 python 代码有一些问题。一方面，退出命令是 exit()，而不是 exit。这是修改后的版本：
loan_amount= 7692
interest = 18
months = 12
for rg in range(700, 710):
y = loan_amount
for i in range(months):
x = y * interest / 100. / 12.
y = y + x - rg
if y < 0:
print(rg)
exit()
这会打印出 706，这是 705.20 的最接近整数近似值。
如果您想要 python 代码准确打印 705.20，那当然可以。然而，代码会更复杂并且需要花费相当多的精力来编写。电子表格似乎更适合这项工作。
好吧，你可以用数值方法解决它（就像 Excel 那样），你可以通过在某个范围内的某个步骤检查每个数量来用蛮力解决它，或者你可以在一个分析上解决它一张纸。
使用以下表示法
L - initial loan amount = 7692
R - monthly interest rate = 1 + 0.18/12
m - number of months to repay the loan = 12
P - monthly payment to pay the loan in full after m months = unknown
is loan amount after the n-th month. is the initial loan amount (7692). 是m个月后的贷款金额（0）。
第n个月和第(n-1)个月之间的主要关系是：
所以，解析式为：
现在用任何编程语言计算它应该是相当straight-forward。
对于给定的初始参数
顺便说一下，如果您正在模拟真实银行的运作方式，可能很难准确计算到最后一分钱。
您从上述精确解析公式中得到的答案只是近似值。
实际上，所有月度金额（付款和利息）通常都四舍五入到美分。每个月都会有一些舍入误差，这些误差会累积并增长。
除了这些舍入误差之外，不同月份的天数也不同，即使每个月的付款相同，利息通常是针对每个月的每一天计算的，因此每个月都会有所不同。然后还有闰年多一天，也会影响月息
Python 中的一种简单的蛮力方法，带有用于确定所需准确度级别的选项。
"""
Calculate required monthly repayment for a given:
- loan amount, and
- annual interest rate, and
- period of repayments in months
You can nominate the accuracy required by adjusting the value of
ACCURACY_AS_PARTS_OF_CENT. For example:
- .01 = accurate to a dollar
- .1 = accurate to 10 cents
- 1 = accurate to cent
- 100 = accurate to 100th of a cent
"""
# Set constants.
LOAN_AMOUNT = 7692
ANNUAL_INTEREST_PERCENT = 18
REPAY_MONTHS = 12
ACCURACY_AS_PARTS_OF_CENT = 1
loan_amount = int(LOAN_AMOUNT * 100 * ACCURACY_AS_PARTS_OF_CENT)
monthly_interest = float(ANNUAL_INTEREST_PERCENT / 100 / 12)
repay_guess_min = int((LOAN_AMOUNT / REPAY_MONTHS) - 1)
result_found = False
repayment_required = 0
for repay_guess in range(repay_guess_min, loan_amount):
if result_found:
break
loan_balance = loan_amount
for _ in range(REPAY_MONTHS):
interest_to_add = loan_balance * monthly_interest
loan_balance = loan_balance + interest_to_add - repay_guess
if loan_balance <= 0:
repayment_required = repay_guess / 100 / ACCURACY_AS_PARTS_OF_CENT
result_found = True
break
print('Required monthly repayment = $' + str(repayment_required))
代码：
from __future__ import print_function
"""
Formulas: http://mathforum.org/dr.math/faq/faq.interest.html
"""
def annuity_monthly_payment(P, n, q, i, debug = False):
"""
Calculates fixed monthly annuity payment
P - amount of the Principal
n - Number of years
q - the number of times per year that the interest is compounded
i - yearly rate of interest (for example: 0.04 for 4% interest)
"""
if debug:
print('P = %s\t(amount of the Principal)' %P)
print('n = %s\t\t(# of years)' %n)
print('q = %s\t\t(# of periods per year)' %q)
print('i = %s %%\t(Annual interest)' %(i*100))
return P*i/( q*(1 - pow(1 + i/q, -n*q)) )
### Given :
P = 7692
n = 1
q = 12
i = 18/100
print('M = %s' %annuity_monthly_payment(P=P, n=n, q=q, i=i, debug=True))
输出：
P = 7692 (amount of the Principal)
n = 1 (# of years)
q = 12 (# of periods per year)
i = 18.0 % (Annual interest)
M = 705.2025054347173
我决定自己调整一下。所以他是我想出的最后一个版本，现在包括一个“amortization_tab”函数，用于打印摊销 table（PrettyTable 和 CSV 格式）和另一个有用的函数：“rest_amount”，“amount_can_be_payed_in_n_years”，“years_to_pay_off” ”。
我添加了 CSV 格式，因此可以生成初始计算，将其导入 Excel 并在那里继续。
所以现在它或多或少是 annuity loan/mortgage math.
的完整库
PS 它已经在 Python v3.5.1 上进行了测试，但它也应该 理论上 可以与另一个 Python 版本一起工作。
from __future__ import print_function
import sys
import math
import io
import csv
import prettytable
# Formulas: http://mathforum.org/dr.math/faq/faq.interest.html
description = {
'P': 'amount of the principal',
'i': 'annual interest rate',
'n': 'number of years',
'q': 'number of times per year that the interest is compounded',
'M': 'fixed monthly payment',
'k': 'number of "payed" payments',
}
def pr_debug(P=None, i=None, n=None, q=None, M=None, k=None, debug=False):
if not debug:
return
columns = ['var','value', 'description']
t = prettytable.PrettyTable(columns)
t.align['var'] = 'l'
t.align['value'] = 'r'
t.align['description'] = 'l'
t.padding_width = 1
t.float_format = '.2'
if P:
t.add_row(['P', P, description['P']])
if i:
t.add_row(['i', i, description['i']])
if n:
t.add_row(['n', n, description['n']])
if q:
t.add_row(['q', q, description['q']])
if M:
t.add_row(['M', M, description['M']])
if k:
t.add_row(['k', k, description['k']])
print(t.get_string() + '\n')
def annuity_monthly_payment(P, n, q, i, debug = False):
"""
Calculates fixed monthly annuity payment
P - amount of the principal
n - number of years
q - number of times per year that the interest is compounded
i - yearly rate of interest (for example: 0.045 for 4.5% interest)
"""
pr_debug(P=P, n=n, q=q, i=i, debug=debug)
i /= 100
return round(P*i/( q*(1 - pow(1 + i/q, -n*q)) ), 2)
def rest_amount(P, M, k, q, i, debug = False):
"""
Calculates rest amount after 'k' payed payments
P - Principal amount
M - fixed amount that have been payed 'k' times
k - # of payments
q - # of periods (12 times per year)
i - yearly interest rate (for example: 0.04 for 4%)
"""
pr_debug(P=P, M=M, k=k, q=q, i=i, debug=debug)
i /= 100
return round((P - M*q/i) * pow(1 + i/q, k) + M*q/i, 2)
def amount_can_be_payed_in_n_years(M, n, q, i, debug = False):
"""
Returns the amount of principal that can be paid off in n years
M - fixed amount that have been payed 'k' times
n - Number of years
q - # of periods (12 times per year)
i - yearly interest rate (for example: 0.04 for 4%)
"""
pr_debug(M=M, n=n, q=q, i=i, debug=debug)
i /= 100
return round( M*(1 - pow(1 + i/q, -n*q) )*q/i, 2)
def years_to_pay_off(P, M, q, i, debug = False):
"""
Returns number of years needed to pay off the loan
P - Principal amount
M - fixed amount that have been payed 'k' times
q - # of periods (12 times per year)
i - yearly interest rate (for example: 0.04 for 4%)
"""
pr_debug(P=P, M=M, q=q, i=i, debug=debug)
i /= 100
return round(-math.log(1 - (P*i/M/q)) / (q*math.log(1 + i/q)), 2)
def amortization_tab(P, n, q, i, M=None, fmt='txt', debug=False):
"""
Generates amortization table
P - Principal amount
M - fixed amount that have been payed 'k' times
q - # of periods (12 times per year)
i - yearly interest rate (for example: 0.04 for 4%)
"""
assert any(fmt in x for x in ['txt', 'csv'])
# calculate monthly payment if it's not given
if not M:
M = annuity_monthly_payment(P=P, n=n, q=q, i=i)
pr_debug(P=P, M=M, n=n, q=q, i=i, debug=debug)
# column headers for the output table
columns=['pmt#','beg_bal','pmt','interest','applied', 'end_bal']
i /= 100
beg_bal = P
term = n*q
if fmt.lower() == 'txt':
t = prettytable.PrettyTable(columns)
t.align = 'r'
t.padding_width = 1
t.float_format = '.2'
elif fmt.lower() == 'csv':
if sys.version_info >= (2,7,0):
out = io.StringIO()
else:
out = io.BytesIO()
t = csv.writer(out, quoting=csv.QUOTE_NONNUMERIC)
t.writerow(columns)
for num in range(1, term+1):
interest = round(beg_bal*i/q , 2)
applied = round(M - interest, 2)
end_bal = round(beg_bal - applied,2)
row = [num, beg_bal, M, interest, applied, end_bal]
if fmt.lower() == 'txt':
t.add_row(row)
elif fmt.lower() == 'csv':
t.writerow(row)
beg_bal = end_bal
if fmt.lower() == 'txt':
return t.get_string()
elif fmt.lower() == 'csv':
return out.getvalue()
############################
P = 7692.0
n = 1
q = 12
i = 18
print(amortization_tab(P, n, q, i, debug=True))
print('#' * 80)
print('#' * 80)
############################
# another example
P = 100000.0
n = 5
q = 12
i = 3.5
k = 36
M = 1200
print(amortization_tab(P, n, q, i, M, fmt='csv', debug=True))
print('*' * 80)
print('Rest amount after %s payments:\t%s' %(k, rest_amount(P=P, M=M, k=k, q=q, i=i)))
输出：
+-----+---------+----------------------------------------------------------+
| var | value | description |
+-----+---------+----------------------------------------------------------+
| P | 7692.00 | amount of the principal |
| i | 18 | annual interest rate |
| n | 1 | number of years |
| q | 12 | number of times per year that the interest is compounded |
| M | 705.20 | fixed monthly payment |
+-----+---------+----------------------------------------------------------+
+------+---------+--------+----------+---------+---------+
| pmt# | beg_bal | pmt | interest | applied | end_bal |
+------+---------+--------+----------+---------+---------+
| 1 | 7692.00 | 705.20 | 115.38 | 589.82 | 7102.18 |
| 2 | 7102.18 | 705.20 | 106.53 | 598.67 | 6503.51 |
| 3 | 6503.51 | 705.20 | 97.55 | 607.65 | 5895.86 |
| 4 | 5895.86 | 705.20 | 88.44 | 616.76 | 5279.10 |
| 5 | 5279.10 | 705.20 | 79.19 | 626.01 | 4653.09 |
| 6 | 4653.09 | 705.20 | 69.80 | 635.40 | 4017.69 |
| 7 | 4017.69 | 705.20 | 60.27 | 644.93 | 3372.76 |
| 8 | 3372.76 | 705.20 | 50.59 | 654.61 | 2718.15 |
| 9 | 2718.15 | 705.20 | 40.77 | 664.43 | 2053.72 |
| 10 | 2053.72 | 705.20 | 30.81 | 674.39 | 1379.33 |
| 11 | 1379.33 | 705.20 | 20.69 | 684.51 | 694.82 |
| 12 | 694.82 | 705.20 | 10.42 | 694.78 | 0.04 |
+------+---------+--------+----------+---------+---------+
################################################################################
################################################################################
+-----+-----------+----------------------------------------------------------+
| var | value | description |
+-----+-----------+----------------------------------------------------------+
| P | 100000.00 | amount of the principal |
| i | 3.50 | annual interest rate |
| n | 5 | number of years |
| q | 12 | number of times per year that the interest is compounded |
| M | 1200 | fixed monthly payment |
+-----+-----------+----------------------------------------------------------+
"pmt#","beg_bal","pmt","interest","applied","end_bal"
1,100000.0,1200,291.67,908.33,99091.67
2,99091.67,1200,289.02,910.98,98180.69
3,98180.69,1200,286.36,913.64,97267.05
4,97267.05,1200,283.7,916.3,96350.75
5,96350.75,1200,281.02,918.98,95431.77
6,95431.77,1200,278.34,921.66,94510.11
7,94510.11,1200,275.65,924.35,93585.76
8,93585.76,1200,272.96,927.04,92658.72
9,92658.72,1200,270.25,929.75,91728.97
10,91728.97,1200,267.54,932.46,90796.51
11,90796.51,1200,264.82,935.18,89861.33
12,89861.33,1200,262.1,937.9,88923.43
13,88923.43,1200,259.36,940.64,87982.79
14,87982.79,1200,256.62,943.38,87039.41
15,87039.41,1200,253.86,946.14,86093.27
16,86093.27,1200,251.11,948.89,85144.38
17,85144.38,1200,248.34,951.66,84192.72
18,84192.72,1200,245.56,954.44,83238.28
19,83238.28,1200,242.78,957.22,82281.06
20,82281.06,1200,239.99,960.01,81321.05
21,81321.05,1200,237.19,962.81,80358.24
22,80358.24,1200,234.38,965.62,79392.62
23,79392.62,1200,231.56,968.44,78424.18
24,78424.18,1200,228.74,971.26,77452.92
25,77452.92,1200,225.9,974.1,76478.82
26,76478.82,1200,223.06,976.94,75501.88
27,75501.88,1200,220.21,979.79,74522.09
28,74522.09,1200,217.36,982.64,73539.45
29,73539.45,1200,214.49,985.51,72553.94
30,72553.94,1200,211.62,988.38,71565.56
31,71565.56,1200,208.73,991.27,70574.29
32,70574.29,1200,205.84,994.16,69580.13
33,69580.13,1200,202.94,997.06,68583.07
34,68583.07,1200,200.03,999.97,67583.1
35,67583.1,1200,197.12,1002.88,66580.22
36,66580.22,1200,194.19,1005.81,65574.41
37,65574.41,1200,191.26,1008.74,64565.67
38,64565.67,1200,188.32,1011.68,63553.99
39,63553.99,1200,185.37,1014.63,62539.36
40,62539.36,1200,182.41,1017.59,61521.77
41,61521.77,1200,179.44,1020.56,60501.21
42,60501.21,1200,176.46,1023.54,59477.67
43,59477.67,1200,173.48,1026.52,58451.15
44,58451.15,1200,170.48,1029.52,57421.63
45,57421.63,1200,167.48,1032.52,56389.11
46,56389.11,1200,164.47,1035.53,55353.58
47,55353.58,1200,161.45,1038.55,54315.03
48,54315.03,1200,158.42,1041.58,53273.45
49,53273.45,1200,155.38,1044.62,52228.83
50,52228.83,1200,152.33,1047.67,51181.16
51,51181.16,1200,149.28,1050.72,50130.44
52,50130.44,1200,146.21,1053.79,49076.65
53,49076.65,1200,143.14,1056.86,48019.79
54,48019.79,1200,140.06,1059.94,46959.85
55,46959.85,1200,136.97,1063.03,45896.82
56,45896.82,1200,133.87,1066.13,44830.69
57,44830.69,1200,130.76,1069.24,43761.45
58,43761.45,1200,127.64,1072.36,42689.09
59,42689.09,1200,124.51,1075.49,41613.6
60,41613.6,1200,121.37,1078.63,40534.97
********************************************************************************
Rest amount after 36 payments: 65574.41
由于title/tag也提到了SQL，我将post一个SQL解决方案：
create table loan (
amount decimal(10,2),
repay_months int,
yearly_interest_rate decimal(4, 4)
);
insert into loan values (7692, 12, 0.18);
select amount * yearly_interest_rate/12 /
(1 - pow(1 + yearly_interest_rate/12, -repay_months))
as monthly_payment
from loan;
结果：
monthly_payment
-----------------
705.2025054347173
如果您想获得整个分期还款时间表，那么一个想法是首先创建一个 table，其中包含连续的月份编号（1、2、...），足以支付最长的贷款pay-off 您将拥有以下数据的持续时间：
create table months (month int);
insert into months -- one year of months
values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),(12);
-- now some record multiplication to avoid long literal lists:
insert into months -- multiply to cover 2 years
select month + 12 from months;
insert into months -- multiply to cover 4 years
select month + 24 from months;
insert into months -- multiply to cover 8 years
select month + 48 from months;
insert into months -- multiply to cover 16 years
select month + 96 from months;
insert into months -- multiply to cover 32 years
select month + 192 from months;
-- OK, we have now months from 1 to 384 (= 32 years)
然后使用以下查询，其中包含上述查询 sub-select:
select month,
monthly_payment * (1 - pow(1 + monthly_interest_rate, month-repay_months))
/ monthly_interest_rate
as loan_balance,
monthly_payment * (1 - pow(1 + monthly_interest_rate, month-1-repay_months))
as interest,
monthly_payment
from months,
(
select amount,
repay_months,
yearly_interest_rate,
yearly_interest_rate/12 as monthly_interest_rate,
amount * yearly_interest_rate/12 /
(1 - pow(1 + yearly_interest_rate/12, -repay_months))
as monthly_payment
from loan
) as loanX
where month <= repay_months
order by 1;
这会产生以下输出：
+-------+--------------------+---------------------+-------------------+
| month | loan_balance | interest | monthly_payment |
+-------+--------------------+---------------------+-------------------+
| 1 | 7102.177494565289 | 115.38 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 2 | 6503.507651549055 | 106.53266241847933 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 3 | 5895.8577608875785 | 97.55261477323582 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 4 | 5279.093121866177 | 88.43786641331367 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 5 | 4653.077013259458 | 79.18639682799265 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 6 | 4017.6706630236345 | 69.79615519889187 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 7 | 3372.7332175342744 | 60.265059945354515 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 8 | 2718.12171036258 | 50.590998263014114 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 9 | 2053.6910305833035 | 40.7718256554387 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 10 | 1379.2938906073389 | 30.805365458749552 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 11 | 694.7807935317352 | 20.689408359110082 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 12 | 0 | 10.421711902976027 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
这里是 SQL fiddle.
所使用的公式在this Wikipedia article中提供和导出。
import numpy as np
for pmt in np.linspace(200, 800, 20):
loan = 7692.00
for n in range(1, 13):
new_balance = loan + ((loan*(1+(0.18/12)))-loan) - pmt
loan = new_balance
print(round(pmt, 2), '->', round(loan,2))
第一列显示 12 个月的等额付款，右列显示 12 个月后的余额。看看余额如何在 705.26 附近趋近于零？这表明零就在附近。
200.0 -> 6588.45
231.58 -> 6176.62
263.16 -> 5764.8
294.74 -> 5352.97
326.32 -> 4941.14
357.89 -> 4529.31
389.47 -> 4117.49
421.05 -> 3705.66
452.63 -> 3293.83
484.21 -> 2882.0
515.79 -> 2470.18
547.37 -> 2058.35
578.95 -> 1646.52
610.53 -> 1234.69
642.11 -> 822.86
673.68 -> 411.04
705.26 -> -0.79
736.84 -> -412.62
768.42 -> -824.45
800.0 -> -1236.27
我有一个类似的 。可能值得一试。
我认为这些 tabular/vector/matrix 类型分析非常适合 numpy 和 pandas。您通常可以编写更紧凑且易于阅读的代码。看看你是否同意。
import numpy as np
import pandas as pd
def mpmt(amt, i, nper):
"""
Calculate the monthly payments on a loan/mortgage
"""
i = i/12 # convert to monthly interest
i1 = i + 1 # used multiple times in formula below
return amt*i1**nper*i/(i1**nper-1)
def ipmt(amt, i, per, nper):
"""
Calculate interest paid in a specific period, per, of a loan/mortgage
"""
i = i/12 # convert to monthly interest
i1 = i + 1 # used multiple times in formula below
return (amt*i*(i1**(nper+1)-i1**per))/(i1*(i1**nper-1))
def amorttable(amt, i, nper):
"""
Create an amortization table for a loan/mortgage
"""
monthlypmt = mpmt(amt, i, nper)
# the following calculations are vectorized
df = pd.DataFrame({'month':np.arange(1, nper+1)})
df['intpaid'] = ipmt(amt, i, df['month'], nper)
df['prinpaid'] = monthlypmt - df['intpaid']
df['balance'] = amt
df['balance'] -= np.cumsum(df['prinpaid'])
return df
print(amorttable(7692, .18, 12).round(2))
结果如下：
month intpaid prinpaid balance
0 1 115.38 589.82 7102.18
1 2 106.53 598.67 6503.51
2 3 97.55 607.65 5895.86
3 4 88.44 616.76 5279.09
4 5 79.19 626.02 4653.08
5 6 69.80 635.41 4017.67
6 7 60.27 644.94 3372.73
7 8 50.59 654.61 2718.12
8 9 40.77 664.43 2053.69
9 10 30.81 674.40 1379.29
10 11 20.69 684.51 694.78
11 12 10.42 694.78 -0.00
这是我在 Excel 中做的一个简单计算。我想知道它是否可以完成 python 或任何其他语言。
Loan amount 7692
Period : 12 months
Rate of interest 18 Per Annum
The formula in the B2 cell is =A1*18/100/12
The formula in the A2 cells is =A1+B2-C2
C 栏是借款人每月可能需要偿还的暂定金额。 C2 旁边的所有其他单元格仅指向 200 的第一部分。使用下图中所示的求解器后，我在 C 列中得到正确的部分 705.20。
我想知道这个计算是否可以使用任何脚本语言完成，例如 python（或 SQL）
这是最终版本的样子...
我试过类似的方法，但它没有退出循环并打印所有组合。
loan_amount= 7692
interest = 18
months =12
for rg in range(700, 710):
for i in range(months):
x = loan_amount * interest / 100 / 12
y = loan_amount + x - rg
if x < 0:
print rg, i
exit
else:
loan_amount = y
您的 python 代码有一些问题。一方面，退出命令是 exit()，而不是 exit。这是修改后的版本：
loan_amount= 7692
interest = 18
months = 12
for rg in range(700, 710):
y = loan_amount
for i in range(months):
x = y * interest / 100. / 12.
y = y + x - rg
if y < 0:
print(rg)
exit()
这会打印出 706，这是 705.20 的最接近整数近似值。
如果您想要 python 代码准确打印 705.20，那当然可以。然而，代码会更复杂并且需要花费相当多的精力来编写。电子表格似乎更适合这项工作。
好吧，你可以用数值方法解决它（就像 Excel 那样），你可以通过在某个范围内的某个步骤检查每个数量来用蛮力解决它，或者你可以在一个分析上解决它一张纸。
使用以下表示法
L - initial loan amount = 7692
R - monthly interest rate = 1 + 0.18/12
m - number of months to repay the loan = 12
P - monthly payment to pay the loan in full after m months = unknown
n-th month. m个月后的贷款金额（0）。
第n个月和第(n-1)个月之间的主要关系是：
所以，解析式为：
现在用任何编程语言计算它应该是相当straight-forward。
对于给定的初始参数
顺便说一下，如果您正在模拟真实银行的运作方式，可能很难准确计算到最后一分钱。
您从上述精确解析公式中得到的答案只是近似值。
实际上，所有月度金额（付款和利息）通常都四舍五入到美分。每个月都会有一些舍入误差，这些误差会累积并增长。
除了这些舍入误差之外，不同月份的天数也不同，即使每个月的付款相同，利息通常是针对每个月的每一天计算的，因此每个月都会有所不同。然后还有闰年多一天，也会影响月息
Python 中的一种简单的蛮力方法，带有用于确定所需准确度级别的选项。
"""
Calculate required monthly repayment for a given:
- loan amount, and
- annual interest rate, and
- period of repayments in months
You can nominate the accuracy required by adjusting the value of
ACCURACY_AS_PARTS_OF_CENT. For example:
- .01 = accurate to a dollar
- .1 = accurate to 10 cents
- 1 = accurate to cent
- 100 = accurate to 100th of a cent
"""
# Set constants.
LOAN_AMOUNT = 7692
ANNUAL_INTEREST_PERCENT = 18
REPAY_MONTHS = 12
ACCURACY_AS_PARTS_OF_CENT = 1
loan_amount = int(LOAN_AMOUNT * 100 * ACCURACY_AS_PARTS_OF_CENT)
monthly_interest = float(ANNUAL_INTEREST_PERCENT / 100 / 12)
repay_guess_min = int((LOAN_AMOUNT / REPAY_MONTHS) - 1)
result_found = False
repayment_required = 0
for repay_guess in range(repay_guess_min, loan_amount):
if result_found:
break
loan_balance = loan_amount
for _ in range(REPAY_MONTHS):
interest_to_add = loan_balance * monthly_interest
loan_balance = loan_balance + interest_to_add - repay_guess
if loan_balance <= 0:
repayment_required = repay_guess / 100 / ACCURACY_AS_PARTS_OF_CENT
result_found = True
break
print('Required monthly repayment = $' + str(repayment_required))
代码：
from __future__ import print_function
"""
Formulas: http://mathforum.org/dr.math/faq/faq.interest.html
"""
def annuity_monthly_payment(P, n, q, i, debug = False):
"""
Calculates fixed monthly annuity payment
P - amount of the Principal
n - Number of years
q - the number of times per year that the interest is compounded
i - yearly rate of interest (for example: 0.04 for 4% interest)
"""
if debug:
print('P = %s\t(amount of the Principal)' %P)
print('n = %s\t\t(# of years)' %n)
print('q = %s\t\t(# of periods per year)' %q)
print('i = %s %%\t(Annual interest)' %(i*100))
return P*i/( q*(1 - pow(1 + i/q, -n*q)) )
### Given :
P = 7692
n = 1
q = 12
i = 18/100
print('M = %s' %annuity_monthly_payment(P=P, n=n, q=q, i=i, debug=True))
输出：
P = 7692 (amount of the Principal)
n = 1 (# of years)
q = 12 (# of periods per year)
i = 18.0 % (Annual interest)
M = 705.2025054347173
我决定自己调整一下。所以他是我想出的最后一个版本，现在包括一个“amortization_tab”函数，用于打印摊销 table（PrettyTable 和 CSV 格式）和另一个有用的函数：“rest_amount”，“amount_can_be_payed_in_n_years”，“years_to_pay_off” ”。 我添加了 CSV 格式，因此可以生成初始计算，将其导入 Excel 并在那里继续。 所以现在它或多或少是 annuity loan/mortgage math.
的完整库PS 它已经在 Python v3.5.1 上进行了测试，但它也应该 理论上 可以与另一个 Python 版本一起工作。
from __future__ import print_function
import sys
import math
import io
import csv
import prettytable
# Formulas: http://mathforum.org/dr.math/faq/faq.interest.html
description = {
'P': 'amount of the principal',
'i': 'annual interest rate',
'n': 'number of years',
'q': 'number of times per year that the interest is compounded',
'M': 'fixed monthly payment',
'k': 'number of "payed" payments',
}
def pr_debug(P=None, i=None, n=None, q=None, M=None, k=None, debug=False):
if not debug:
return
columns = ['var','value', 'description']
t = prettytable.PrettyTable(columns)
t.align['var'] = 'l'
t.align['value'] = 'r'
t.align['description'] = 'l'
t.padding_width = 1
t.float_format = '.2'
if P:
t.add_row(['P', P, description['P']])
if i:
t.add_row(['i', i, description['i']])
if n:
t.add_row(['n', n, description['n']])
if q:
t.add_row(['q', q, description['q']])
if M:
t.add_row(['M', M, description['M']])
if k:
t.add_row(['k', k, description['k']])
print(t.get_string() + '\n')
def annuity_monthly_payment(P, n, q, i, debug = False):
"""
Calculates fixed monthly annuity payment
P - amount of the principal
n - number of years
q - number of times per year that the interest is compounded
i - yearly rate of interest (for example: 0.045 for 4.5% interest)
"""
pr_debug(P=P, n=n, q=q, i=i, debug=debug)
i /= 100
return round(P*i/( q*(1 - pow(1 + i/q, -n*q)) ), 2)
def rest_amount(P, M, k, q, i, debug = False):
"""
Calculates rest amount after 'k' payed payments
P - Principal amount
M - fixed amount that have been payed 'k' times
k - # of payments
q - # of periods (12 times per year)
i - yearly interest rate (for example: 0.04 for 4%)
"""
pr_debug(P=P, M=M, k=k, q=q, i=i, debug=debug)
i /= 100
return round((P - M*q/i) * pow(1 + i/q, k) + M*q/i, 2)
def amount_can_be_payed_in_n_years(M, n, q, i, debug = False):
"""
Returns the amount of principal that can be paid off in n years
M - fixed amount that have been payed 'k' times
n - Number of years
q - # of periods (12 times per year)
i - yearly interest rate (for example: 0.04 for 4%)
"""
pr_debug(M=M, n=n, q=q, i=i, debug=debug)
i /= 100
return round( M*(1 - pow(1 + i/q, -n*q) )*q/i, 2)
def years_to_pay_off(P, M, q, i, debug = False):
"""
Returns number of years needed to pay off the loan
P - Principal amount
M - fixed amount that have been payed 'k' times
q - # of periods (12 times per year)
i - yearly interest rate (for example: 0.04 for 4%)
"""
pr_debug(P=P, M=M, q=q, i=i, debug=debug)
i /= 100
return round(-math.log(1 - (P*i/M/q)) / (q*math.log(1 + i/q)), 2)
def amortization_tab(P, n, q, i, M=None, fmt='txt', debug=False):
"""
Generates amortization table
P - Principal amount
M - fixed amount that have been payed 'k' times
q - # of periods (12 times per year)
i - yearly interest rate (for example: 0.04 for 4%)
"""
assert any(fmt in x for x in ['txt', 'csv'])
# calculate monthly payment if it's not given
if not M:
M = annuity_monthly_payment(P=P, n=n, q=q, i=i)
pr_debug(P=P, M=M, n=n, q=q, i=i, debug=debug)
# column headers for the output table
columns=['pmt#','beg_bal','pmt','interest','applied', 'end_bal']
i /= 100
beg_bal = P
term = n*q
if fmt.lower() == 'txt':
t = prettytable.PrettyTable(columns)
t.align = 'r'
t.padding_width = 1
t.float_format = '.2'
elif fmt.lower() == 'csv':
if sys.version_info >= (2,7,0):
out = io.StringIO()
else:
out = io.BytesIO()
t = csv.writer(out, quoting=csv.QUOTE_NONNUMERIC)
t.writerow(columns)
for num in range(1, term+1):
interest = round(beg_bal*i/q , 2)
applied = round(M - interest, 2)
end_bal = round(beg_bal - applied,2)
row = [num, beg_bal, M, interest, applied, end_bal]
if fmt.lower() == 'txt':
t.add_row(row)
elif fmt.lower() == 'csv':
t.writerow(row)
beg_bal = end_bal
if fmt.lower() == 'txt':
return t.get_string()
elif fmt.lower() == 'csv':
return out.getvalue()
############################
P = 7692.0
n = 1
q = 12
i = 18
print(amortization_tab(P, n, q, i, debug=True))
print('#' * 80)
print('#' * 80)
############################
# another example
P = 100000.0
n = 5
q = 12
i = 3.5
k = 36
M = 1200
print(amortization_tab(P, n, q, i, M, fmt='csv', debug=True))
print('*' * 80)
print('Rest amount after %s payments:\t%s' %(k, rest_amount(P=P, M=M, k=k, q=q, i=i)))
输出：
+-----+---------+----------------------------------------------------------+
| var | value | description |
+-----+---------+----------------------------------------------------------+
| P | 7692.00 | amount of the principal |
| i | 18 | annual interest rate |
| n | 1 | number of years |
| q | 12 | number of times per year that the interest is compounded |
| M | 705.20 | fixed monthly payment |
+-----+---------+----------------------------------------------------------+
+------+---------+--------+----------+---------+---------+
| pmt# | beg_bal | pmt | interest | applied | end_bal |
+------+---------+--------+----------+---------+---------+
| 1 | 7692.00 | 705.20 | 115.38 | 589.82 | 7102.18 |
| 2 | 7102.18 | 705.20 | 106.53 | 598.67 | 6503.51 |
| 3 | 6503.51 | 705.20 | 97.55 | 607.65 | 5895.86 |
| 4 | 5895.86 | 705.20 | 88.44 | 616.76 | 5279.10 |
| 5 | 5279.10 | 705.20 | 79.19 | 626.01 | 4653.09 |
| 6 | 4653.09 | 705.20 | 69.80 | 635.40 | 4017.69 |
| 7 | 4017.69 | 705.20 | 60.27 | 644.93 | 3372.76 |
| 8 | 3372.76 | 705.20 | 50.59 | 654.61 | 2718.15 |
| 9 | 2718.15 | 705.20 | 40.77 | 664.43 | 2053.72 |
| 10 | 2053.72 | 705.20 | 30.81 | 674.39 | 1379.33 |
| 11 | 1379.33 | 705.20 | 20.69 | 684.51 | 694.82 |
| 12 | 694.82 | 705.20 | 10.42 | 694.78 | 0.04 |
+------+---------+--------+----------+---------+---------+
################################################################################
################################################################################
+-----+-----------+----------------------------------------------------------+
| var | value | description |
+-----+-----------+----------------------------------------------------------+
| P | 100000.00 | amount of the principal |
| i | 3.50 | annual interest rate |
| n | 5 | number of years |
| q | 12 | number of times per year that the interest is compounded |
| M | 1200 | fixed monthly payment |
+-----+-----------+----------------------------------------------------------+
"pmt#","beg_bal","pmt","interest","applied","end_bal"
1,100000.0,1200,291.67,908.33,99091.67
2,99091.67,1200,289.02,910.98,98180.69
3,98180.69,1200,286.36,913.64,97267.05
4,97267.05,1200,283.7,916.3,96350.75
5,96350.75,1200,281.02,918.98,95431.77
6,95431.77,1200,278.34,921.66,94510.11
7,94510.11,1200,275.65,924.35,93585.76
8,93585.76,1200,272.96,927.04,92658.72
9,92658.72,1200,270.25,929.75,91728.97
10,91728.97,1200,267.54,932.46,90796.51
11,90796.51,1200,264.82,935.18,89861.33
12,89861.33,1200,262.1,937.9,88923.43
13,88923.43,1200,259.36,940.64,87982.79
14,87982.79,1200,256.62,943.38,87039.41
15,87039.41,1200,253.86,946.14,86093.27
16,86093.27,1200,251.11,948.89,85144.38
17,85144.38,1200,248.34,951.66,84192.72
18,84192.72,1200,245.56,954.44,83238.28
19,83238.28,1200,242.78,957.22,82281.06
20,82281.06,1200,239.99,960.01,81321.05
21,81321.05,1200,237.19,962.81,80358.24
22,80358.24,1200,234.38,965.62,79392.62
23,79392.62,1200,231.56,968.44,78424.18
24,78424.18,1200,228.74,971.26,77452.92
25,77452.92,1200,225.9,974.1,76478.82
26,76478.82,1200,223.06,976.94,75501.88
27,75501.88,1200,220.21,979.79,74522.09
28,74522.09,1200,217.36,982.64,73539.45
29,73539.45,1200,214.49,985.51,72553.94
30,72553.94,1200,211.62,988.38,71565.56
31,71565.56,1200,208.73,991.27,70574.29
32,70574.29,1200,205.84,994.16,69580.13
33,69580.13,1200,202.94,997.06,68583.07
34,68583.07,1200,200.03,999.97,67583.1
35,67583.1,1200,197.12,1002.88,66580.22
36,66580.22,1200,194.19,1005.81,65574.41
37,65574.41,1200,191.26,1008.74,64565.67
38,64565.67,1200,188.32,1011.68,63553.99
39,63553.99,1200,185.37,1014.63,62539.36
40,62539.36,1200,182.41,1017.59,61521.77
41,61521.77,1200,179.44,1020.56,60501.21
42,60501.21,1200,176.46,1023.54,59477.67
43,59477.67,1200,173.48,1026.52,58451.15
44,58451.15,1200,170.48,1029.52,57421.63
45,57421.63,1200,167.48,1032.52,56389.11
46,56389.11,1200,164.47,1035.53,55353.58
47,55353.58,1200,161.45,1038.55,54315.03
48,54315.03,1200,158.42,1041.58,53273.45
49,53273.45,1200,155.38,1044.62,52228.83
50,52228.83,1200,152.33,1047.67,51181.16
51,51181.16,1200,149.28,1050.72,50130.44
52,50130.44,1200,146.21,1053.79,49076.65
53,49076.65,1200,143.14,1056.86,48019.79
54,48019.79,1200,140.06,1059.94,46959.85
55,46959.85,1200,136.97,1063.03,45896.82
56,45896.82,1200,133.87,1066.13,44830.69
57,44830.69,1200,130.76,1069.24,43761.45
58,43761.45,1200,127.64,1072.36,42689.09
59,42689.09,1200,124.51,1075.49,41613.6
60,41613.6,1200,121.37,1078.63,40534.97
********************************************************************************
Rest amount after 36 payments: 65574.41
由于title/tag也提到了SQL，我将post一个SQL解决方案：
create table loan (
amount decimal(10,2),
repay_months int,
yearly_interest_rate decimal(4, 4)
);
insert into loan values (7692, 12, 0.18);
select amount * yearly_interest_rate/12 /
(1 - pow(1 + yearly_interest_rate/12, -repay_months))
as monthly_payment
from loan;
结果：
monthly_payment
-----------------
705.2025054347173
如果您想获得整个分期还款时间表，那么一个想法是首先创建一个 table，其中包含连续的月份编号（1、2、...），足以支付最长的贷款pay-off 您将拥有以下数据的持续时间：
create table months (month int);
insert into months -- one year of months
values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),(12);
-- now some record multiplication to avoid long literal lists:
insert into months -- multiply to cover 2 years
select month + 12 from months;
insert into months -- multiply to cover 4 years
select month + 24 from months;
insert into months -- multiply to cover 8 years
select month + 48 from months;
insert into months -- multiply to cover 16 years
select month + 96 from months;
insert into months -- multiply to cover 32 years
select month + 192 from months;
-- OK, we have now months from 1 to 384 (= 32 years)
然后使用以下查询，其中包含上述查询 sub-select:
select month,
monthly_payment * (1 - pow(1 + monthly_interest_rate, month-repay_months))
/ monthly_interest_rate
as loan_balance,
monthly_payment * (1 - pow(1 + monthly_interest_rate, month-1-repay_months))
as interest,
monthly_payment
from months,
(
select amount,
repay_months,
yearly_interest_rate,
yearly_interest_rate/12 as monthly_interest_rate,
amount * yearly_interest_rate/12 /
(1 - pow(1 + yearly_interest_rate/12, -repay_months))
as monthly_payment
from loan
) as loanX
where month <= repay_months
order by 1;
这会产生以下输出：
+-------+--------------------+---------------------+-------------------+
| month | loan_balance | interest | monthly_payment |
+-------+--------------------+---------------------+-------------------+
| 1 | 7102.177494565289 | 115.38 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 2 | 6503.507651549055 | 106.53266241847933 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 3 | 5895.8577608875785 | 97.55261477323582 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 4 | 5279.093121866177 | 88.43786641331367 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 5 | 4653.077013259458 | 79.18639682799265 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 6 | 4017.6706630236345 | 69.79615519889187 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 7 | 3372.7332175342744 | 60.265059945354515 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 8 | 2718.12171036258 | 50.590998263014114 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 9 | 2053.6910305833035 | 40.7718256554387 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 10 | 1379.2938906073389 | 30.805365458749552 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 11 | 694.7807935317352 | 20.689408359110082 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
| 12 | 0 | 10.421711902976027 | 705.2025054347173 |
+-------+--------------------+---------------------+-------------------+
这里是 SQL fiddle.
所使用的公式在this Wikipedia article中提供和导出。
import numpy as np
for pmt in np.linspace(200, 800, 20):
loan = 7692.00
for n in range(1, 13):
new_balance = loan + ((loan*(1+(0.18/12)))-loan) - pmt
loan = new_balance
print(round(pmt, 2), '->', round(loan,2))
第一列显示 12 个月的等额付款，右列显示 12 个月后的余额。看看余额如何在 705.26 附近趋近于零？这表明零就在附近。
200.0 -> 6588.45
231.58 -> 6176.62
263.16 -> 5764.8
294.74 -> 5352.97
326.32 -> 4941.14
357.89 -> 4529.31
389.47 -> 4117.49
421.05 -> 3705.66
452.63 -> 3293.83
484.21 -> 2882.0
515.79 -> 2470.18
547.37 -> 2058.35
578.95 -> 1646.52
610.53 -> 1234.69
642.11 -> 822.86
673.68 -> 411.04
705.26 -> -0.79
736.84 -> -412.62
768.42 -> -824.45
800.0 -> -1236.27
我有一个类似的
我认为这些 tabular/vector/matrix 类型分析非常适合 numpy 和 pandas。您通常可以编写更紧凑且易于阅读的代码。看看你是否同意。
import numpy as np
import pandas as pd
def mpmt(amt, i, nper):
"""
Calculate the monthly payments on a loan/mortgage
"""
i = i/12 # convert to monthly interest
i1 = i + 1 # used multiple times in formula below
return amt*i1**nper*i/(i1**nper-1)
def ipmt(amt, i, per, nper):
"""
Calculate interest paid in a specific period, per, of a loan/mortgage
"""
i = i/12 # convert to monthly interest
i1 = i + 1 # used multiple times in formula below
return (amt*i*(i1**(nper+1)-i1**per))/(i1*(i1**nper-1))
def amorttable(amt, i, nper):
"""
Create an amortization table for a loan/mortgage
"""
monthlypmt = mpmt(amt, i, nper)
# the following calculations are vectorized
df = pd.DataFrame({'month':np.arange(1, nper+1)})
df['intpaid'] = ipmt(amt, i, df['month'], nper)
df['prinpaid'] = monthlypmt - df['intpaid']
df['balance'] = amt
df['balance'] -= np.cumsum(df['prinpaid'])
return df
print(amorttable(7692, .18, 12).round(2))
结果如下：
month intpaid prinpaid balance
0 1 115.38 589.82 7102.18
1 2 106.53 598.67 6503.51
2 3 97.55 607.65 5895.86
3 4 88.44 616.76 5279.09
4 5 79.19 626.02 4653.08
5 6 69.80 635.41 4017.67
6 7 60.27 644.94 3372.73
7 8 50.59 654.61 2718.12
8 9 40.77 664.43 2053.69
9 10 30.81 674.40 1379.29
10 11 20.69 684.51 694.78
11 12 10.42 694.78 -0.00