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假设固定利率债券的时间表如下面的示例代码所示。

我可以通过使用该businessDaysBetween功能获得男高音之间的天数。

现在我想要“时间价值”。有没有办法在不创建新功能的情况下做到这一点?

这是预期的结果:

May 14th, 2012          .5
November 14th, 2012     .5
May 14th, 2013          .5
November 14th, 2013     .5
May 14th, 2014          .5
November 14th, 2014     .5
May 14th, 2015          .5
November 16th, 2015     .505556
May 16th, 2016          .5
November 14th, 2016     .49444

这是代码:

from QuantLib import *
import pandas as pd


effective_date = Date(14, 11, 2011)
termination_date = Date(14, 11, 2016)
tenor = Period(Semiannual)
calendar = UnitedStates()
business_convention = ModifiedFollowing
termination_business_convention = Following
date_generation = DateGeneration.Forward
end_of_month = False
day_count = Thirty360()
schedule = Schedule(effective_date,
                    termination_date,
                    tenor,
                    calendar,
                    business_convention,
                    termination_business_convention,
                    date_generation,
                    end_of_month)
t = []
for i, d in  enumerate(schedule):
    tmp = i+1, d,
    t.append(tmp)
df = pd.DataFrame(t,columns = ['tenorNo','tenorDate'])

nbDays = []
for x in df['tenorNo'] :
    if  x == 1:
        tmp = 0
    else:
        tmp = calendar.businessDaysBetween(df['tenorDate'][x-2],df['tenorDate'][x-1])
        nbDays.append(tmp)

    df['nbDays'] = nbDays

print df
    tenorNo            tenorDate  nbDays
0         1  November 14th, 2011       0
1         2       May 14th, 2012     125
2         3  November 14th, 2012     127
3         4       May 14th, 2013     124
4         5  November 14th, 2013     127
5         6       May 14th, 2014     124
6         7  November 14th, 2014     127
7         8       May 14th, 2015     124
8         9  November 16th, 2015     127
9        10       May 16th, 2016     125
10       11  November 14th, 2016     125
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1 回答 1

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这就是DayCounter实例的用途。时间取决于您选择的计日惯例(例如,您似乎使用的是 30/360)。

打电话

day_count.yearFraction(date1, date2)

将返回 和 之间的date1时间date2

于 2017-03-04T20:43:39.713 回答