有包含xirr和xnpv函数的金融Python库吗？

Question

numpy有irr和npv这两个函数，但我需要xirr和xnpv这两个函数。

这个链接提到xirr和xnpv很快就会推出。
http://www.projectdirigible.com/documentation/spreadsheet-functions.html#coming-soon

有没有哪个Python库里有这两个函数？谢谢。

Answer 1

我创建了一个用于快速计算XIRR（内部收益率）的工具包，叫做PyXIRR

这个工具包不需要其他外部库，运行速度比现有的任何实现都要快。

from datetime import date
from pyxirr import xirr

dates = [date(2020, 1, 1), date(2021, 1, 1), date(2022, 1, 1)]
amounts = [-1000, 1000, 1000]

# feed columnar data
xirr(dates, amounts)

# feed tuples
xirr(zip(dates, amounts))

# feed DataFrame
import pandas as pd
xirr(pd.DataFrame({"dates": dates, "amounts": amounts}))

Answer 2

这里有一种实现这两个功能的方法。

import scipy.optimize

def xnpv(rate, values, dates):
    '''Equivalent of Excel's XNPV function.

    >>> from datetime import date
    >>> dates = [date(2010, 12, 29), date(2012, 1, 25), date(2012, 3, 8)]
    >>> values = [-10000, 20, 10100]
    >>> xnpv(0.1, values, dates)
    -966.4345...
    '''
    if rate <= -1.0:
        return float('inf')
    d0 = dates[0]    # or min(dates)
    return sum([ vi / (1.0 + rate)**((di - d0).days / 365.0) for vi, di in zip(values, dates)])

def xirr(values, dates):
    '''Equivalent of Excel's XIRR function.

    >>> from datetime import date
    >>> dates = [date(2010, 12, 29), date(2012, 1, 25), date(2012, 3, 8)]
    >>> values = [-10000, 20, 10100]
    >>> xirr(values, dates)
    0.0100612...
    '''
    try:
        return scipy.optimize.newton(lambda r: xnpv(r, values, dates), 0.0)
    except RuntimeError:    # Failed to converge?
        return scipy.optimize.brentq(lambda r: xnpv(r, values, dates), -1.0, 1e10)

Answer 3

在网上找到了一些不同的实现方法后，我自己写了一个Python的实现：

def xirr(transactions):
    years = [(ta[0] - transactions[0][0]).days / 365.0 for ta in transactions]
    residual = 1
    step = 0.05
    guess = 0.05
    epsilon = 0.0001
    limit = 10000
    while abs(residual) > epsilon and limit > 0:
        limit -= 1
        residual = 0.0
        for i, ta in enumerate(transactions):
            residual += ta[1] / pow(guess, years[i])
        if abs(residual) > epsilon:
            if residual > 0:
                guess += step
            else:
                guess -= step
                step /= 2.0
    return guess-1

from datetime import date
tas = [ (date(2010, 12, 29), -10000),
    (date(2012, 1, 25), 20),
    (date(2012, 3, 8), 10100)]
print xirr(tas) #0.0100612640381