Python 在numpy rate函数中获取特定值集的负速率
我不熟悉Python 在numpy rate函数中获取特定值集的负速率,python,numpy,rate,Python,Numpy,Rate,我不熟悉numpy并使用numpy.rate()计算每月支付贷款的统一利率的APR no_of_month = 24 payment = 8584 loan_amount = 50000 apr = rate(no_of_month,-payment,loan_amount,0.0) *12 apr计算为-22.816,但实际值应为2.0102(在LibreOffice Calc中计算) 对于no\u of u month=23而言,apr为2.00079但对于no\u of u month=
numpy
并使用numpy.rate()
计算每月支付贷款的统一利率的APR
no_of_month = 24
payment = 8584
loan_amount = 50000
apr = rate(no_of_month,-payment,loan_amount,0.0) *12
apr
计算为-22.816
,但实际值应为2.0102
(在LibreOffice Calc
中计算)
对于no\u of u month=23
而言,apr
为2.00079
但对于no\u of u month=24
而言,apr
为-22.816
以下是我目前对为什么会发生这种情况的猜测(他们可能是错的)
- 因为持续时间是12的倍数(又是一整年)
- 对特定范围内的
有一些限制numpy.rate
no_of_month - 1 to 36
payment - 1000 to 1000000
loan_amount - 10000 to 10000000
所有组合都是可能的np.rate返回用x**24解多项式方程的利率。这有24个解决方案,其中一些可能是重复的,一些可能是复杂的。对于该特定数据:
pv = 50000
payment = 8584
mpr= np.rate(24, -payment, pv, 0.0)
np.pv(mpr, 24, payment)
# -49999.999999789325 This represents the 50000 pv
mpr
# -1.901406995298687 # mpr = -190.1% per month!
mpr1 = np.rate(24, -payment, pv, 0.0, guess = .15)
# guess lets you change the starting point for the search
mpr1
# 0.16750654293672343 # mar = 16.8% per month
np.pv(mpr1, 24, payment)
# -49999.99999999999
def apr(mpr, periods = 12):
""" apr is ( 1 + monthly_rate ) ** 12 - 1 """
return (1+mpr)**periods-1
apr(apr)
# -0.7122263079633477 apr = -71.2%
apr(mpr1)
# 5.4137477809069345 apr of 541.4%
i、 e.16.8%和-190.1%都是方程的数学正确解-在金融环境下,190.1%没有多大意义
有两个时期可能更容易理解
loan_amount = 10
payment = 6
n_periods = 2
Solve 10 - 6r -6r**2
r = (6 +-sqrt(36-4*(-6)*10))/(2*-6)
r = -1.8844373105 and 0.8844373105
r = 1/(1+i) where i is the interest rate
i = 1/r - 1
r = -1.8844373105
1/r-1
# -1.5306623862879625
r1 = 0.8844373105
1/r1-1
# 0.13066238627435212
mpr = np.rate(2, -payment, loan_amount, 0.0)
mpr
# 0.13066238629183413
mpr1 = np.rate(2, -payment, loan_amount, 0.0, guess = -1.5)
mpr1
# -1.5306623862918336
在这种情况下,numpy将求解两个速率,它们来自二次方程的两个根
这可能没有帮助,但确实解释了为什么np.rate
(和np.irr
)可以解决意外答案
编辑:
我意识到,如果r=1/(1+利率),r只有一个真正的正解。这是通常最具商业意义的解决方案
import numpy as np
"""
To simplify the analysis let
r = 1 / ( 1 + interest_rate )
p = periodic payments
n = number of periods
then:
pv = loan - p*r - p*r**2 - ... -p*r**n
dpv/dr = -p -2*p*r - ... -p*n*r**(n-1)
If r > 0 and p > 0 dpv/dr is negative
Therefore there is at most one solution to pv == 0 for r > 0.
pv == loan when r == 0
For large r -p*r**n will dominate and pv will be negative.
Therefore there will be one positive real solution to pv == 0
"""
def polynomial_from(nper, loan, pay):
""" Create numpy array to represent the polynomial """
return np.array([-pay]*nper+[loan])
# np.roots returns one root per nper. Filter to real roots only.
def real_roots(poly):
roots_ = np.roots(poly)
return roots_[np.isclose(roots_.imag, 0)].real
# return the real part of the roots- with a zero imaginary part
def feasible_rate(nper, loan, pay):
poly = polynomial_from(nper, loan, pay)
reals = real_roots(poly)
r = reals[reals>0][0] # r is the real root > 0
return 1/r - 1
def apr(int_rate, nperiods = 12):
return ( 1 + int_rate ) ** nperiods - 1
mpr = feasible_rate( 24, 50000, 8584 )
print( 'Monthly rate: {:%}, Annual Rate: {:%}'.format(mpr, apr(mpr)) )
# Monthly rate: 16.750654%, Annual Rate: 541.374778%
回答得好,但你能解释一下提问者是如何达到预期结果的吗?@Trilarion没有数学逻辑来决定哪个答案是“正确的”。通过为猜测选择不同的值,有时会生成不同的答案。如果有人想使用numpy(与电子表格相反),他们可能不希望检查每个结果的可行性<代码>np。根可能有助于识别所有解决方案。一个单独的算法可以在可行范围内从实际解(相对于复杂解)中进行选择。业务客户必须建议适合他们的解决方案。@Trilarion我已经确定了一个最具“商业意识”的解决方案。请参见我答案末尾的“编辑”。@TlsChris使用了
guess
参数和大多数可能的组合,并选择了适用于所有人的值。