Question

我写了一些代码来计算IRR，效果很好...

import scipy.optimize as optimize
import datetime


def npv(cf, rate=0.1):
    if len(cf) >= 2:
        first_date = min([x[0] for x in cf])
        dcf = [x[1] * (1 /
                       ((1 + rate) ** ((x[0] - first_date).days / 365))) for x in cf]
        return sum(dcf)
    elif len(cf) == 1:
        return cf[0][1]
    else:
        return 0


def irr(cf):
    f = lambda x: npv(cf, rate=x)
    r = optimize.newton(f, 0, maxiter=70)
    return r

...但是当我尝试此现金流

cf=[(datetime.datetime(2018, 1, 10), -51089.94),
    (datetime.datetime(2022, 10, 6), 4941.0)]

我收到此错误：

File "/Users/maxim/Dropbox/Python/FinProject/fintrack/main/models.py", line 503, in getIRR
    return irr(cf)
  File "/Users/maxim/Dropbox/Python/FinProject/fintrack/main/models.py", line 37, in irr
    r = optimize.newton(f, 0, maxiter=70)
  File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site-packages/scipy/optimize/zeros.py", line 204, in newton
    q1 = func(p1, *args)
  File "/Users/maxim/Dropbox/Python/FinProject/fintrack/main/models.py", line 36, in <lambda>
    f = lambda x: npv(cf, rate=x)
  File "/Users/maxim/Dropbox/Python/FinProject/fintrack/main/models.py", line 27, in npv
    ((1 + rate) ** ((x[0] - first_date).days / 365))) for x in cf]
  File "/Users/maxim/Dropbox/Python/FinProject/fintrack/main/models.py", line 27, in <listcomp>
    ((1 + rate) ** ((x[0] - first_date).days / 365))) for x in cf]
OverflowError: complex exponentiation
[30/Nov/2018 21:28:36] "GET /inv/19/ HTTP/1.1" 500 299065

但是我知道正确答案是-38.912 ..％，我通过Excel得到了这个结果。怎么了使用其他数据，我得到的结果与Excel IRR函数的结果相同...是否应该使用其他函数查找参数？

P.S .：这是迭代的参数和结果列表（财务中的OMG复数％-））：

rate= 0.0 result= -46148.94
rate= 0.0001 result= -46151.281226688276
rate= -1.9711435988300456 result= (-54972.27265283515-4141.40178622848j)
rate= (8.450859228811169-3.967580022971747j) result= (-51089.96465099011+0.07614432462298902j)
rate= (46.548868303534285-96.82120737804672j) result= (-51089.93999938349-9.638432563347345e-07j)
rate= (63880696.05472335+4880892.707757121j) result= (-51089.94-1.7420169038104924e-34j)
rate= (-1.3534185437764045e+18-2.52038641964956e+18j) result= (-51089.94-5.721141417411886e-85j)
rate= (-7.391799165398238e+56+3.969311207511089e+56j) result= (-51089.94+9.185514002355334e-269j)
rate= (3.5446051170119047e+145+6.600895665730368e+145j) result= ERROR!!!

Answer 1

似乎newton无法正确处理（不确定原因）。但是，您可以使用https://github.com/cran/Rpoppler/blob/819bc04dd113f3ec12d8f7d8eecfab79569478d6/configure#L1830，它可以为您带来预期的结果，并且只需要对代码进行很小的修改即可：

import scipy.optimize as optimize
import datetime


def npv(cf, rate=0.1):
    if len(cf) >= 2:
        first_date = min([x[0] for x in cf])
        dcf = [x[1] * (1 /
                       ((1 + rate) ** ((x[0] - first_date).days / 365))) for x in cf]
        return sum(dcf)
    elif len(cf) == 1:
        return cf[0][1]
    else:
        return 0


def irr(cf):
    f = lambda x: npv(cf, rate=x)
    r = optimize.root(f, [0])
    return r

cf = [(datetime.datetime(2018, 1, 10), -51089.94),
      (datetime.datetime(2022, 10, 6), 4941.0)]

print(irr(cf))

这将打印：

    fjac: array([[-1.]])
     fun: array([-2.91038305e-11])
 message: 'The solution converged.'
    nfev: 18
     qtf: array([2.37701897e-06])
       r: array([396401.23327105])
  status: 1
 success: True
       x: array([-0.38912302])

如您所见，x包含预期的-0.38912。

Answer 2

工作代码为：

import scipy.optimize as optimize
import datetime


def npv(cf, rate=0.1):
    if len(cf) >= 2:
        first_date = min([x[0] for x in cf])
        dcf = [x[1] * (1 /
                       ((1 + rate) ** ((x[0] - first_date).days / 365))) for x in cf]
        return sum(dcf)
    elif len(cf) == 1:
        return cf[0][1]
    else:
        return 0


def irr(cf):
    f = lambda x: npv(cf, rate=x)
    r = optimize.root(f, [0])
    return r

非常感谢克莱布！

用Python计算IRR

2 个答案: