Question

假设固定利率债券与下面的示例代码中显示的时间表。

我可以使用businessDaysBetween函数获取期间之间的天数。

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这样做？

以下是预期结果：

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

以下是代码：

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

Answer 1

那是DayCounter个实例的用途。时间取决于您选择的日计数约定（例如，您似乎使用30/360）。

调用

day_count.yearFraction(date1, date2)

将返回date1和date2之间的时间。

如何获取计划的“时间”值

1 个答案: