我对此有疑问,我知道代码太长太复杂,但我开始了: 这是我正在使用的数据:
Data:
date = dt.datetime(2018, 6, 26)
maturity_dtime = DatetimeIndex(['2020-04-07', '2020-08-07', '2020-12-07', '2023-12-07',
'2027-12-07', '2032-12-07', '2040-02-07'],
dtype='datetime64[ns]', freq=None)
curve = ['act/365','Lineal','Anual',
[datetime.datetime(2018, 6, 27, 0, 0), 4.105922851627142e-05],
[datetime.datetime(2018, 7, 26, 0, 0), 0.001200502096981415],
[datetime.datetime(2018, 9, 26, 0, 0), 0.0034882824213942065],
[datetime.datetime(2018, 12, 26, 0, 0), 0.006427227712844319],
[datetime.datetime(2019, 3, 26, 0, 0), 0.008915157135919838],
[datetime.datetime(2019, 6, 26, 0, 0), 0.011097508773927123],
[datetime.datetime(2020, 6, 26, 0, 0), 0.0171882727144943]]
然后我有这个功能:
def day_count(start_date, end_date, basis):
if basis == 'act/365':
days = (end_date - start_date).days
else:
print('fail')
return days
def year_fraction(start_date, end_date, basis):
if basis == "act/365":
yf = day_count(start_date, end_date,basis) / 360.0
else:
print('fail')
return yf
def interpol_curva(date,maturity_date,curve):
base=curve[0]
interpol=curve[1]
#compo_fg=curve[2]
nrows=int(len(curve))
if maturity_date > curve[nrows-1][0]: #Here is the mistake
maturity_date=curve[nrows-1][0]
if maturity_date<curve[3][0]:
maturity_date=curve[3][0]
r1=3
while maturity_date>curve[r1][0] and r1<nrows-1:
r1=r1+1
r1=r1-1
if r1==2:
r1=3
if r1>=nrows-1:
r1=nrows-2
r2=r1+1
#t1=year_fraction_2(date, curve[r1][0], base)
#t2=year_fraction_2(date, curve[r2][0], base)
#tt=year_fraction_2(date, matDate, base)
if base=='act/360' or base=='act/365':
yf1=(maturity_date-curve[r1][0]).days
yf2=(curve[r2][0]-maturity_date).days
yftt=(curve[r2][0]-curve[r1][0]).days
else:
print("fail")
if interpol=='Lineal':
return (curve[r1][1]*yf2+curve[r2][1]*yf1)/yftt
def Discount_Factor_2(value_date,maturity_date,curve):
basis=curve[0]
Composition=curve[2]
yf = year_fraction(value_date, maturity_date, basis)
r=interpol_curva(value_date,maturity_date,curve)
if Composition == "Anual":
df = 1 / (1 + r * yf)
else:
print("fail")
return df
然后我尝试运行函数 Discount_Factor_2,但出现此错误:
Discount_Factor_2(date,maturity_dtime,curve)
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
错误在行 if maturity_date > curve[nrows-1][0]
我想知道是否有办法修复它。稍后我想通过更改创建可变曲线的参数来最小化该函数的结果。
非常感谢,如果不清楚,请见谅。感谢您抽出宝贵时间。
编辑:添加完整的回溯:
Discount_Factor_2(value_date, maturity_dtime, curve)
Traceback (most recent call last):
File "<ipython-input-30-9cbb6735a3e3>", line 1, in <module>
Discount_Factor_2(value_date, maturity_dtime, curve)
File "<ipython-input-20-181d050d0cd4>", line 251, in Discount_Factor_2
r=interpol_curva(value_date,maturity_date,curve)
File "<ipython-input-20-181d050d0cd4>", line 178, in interpol_curva
if maturity_date > curve[nrows-1][0]:
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
最佳答案
问题出在您的 if
语句中 -
if maturity_date > curve[nrows-1][0]
想想这是干什么的——
假设 maturity_dtime
是一个看起来像这样的 pd.Series
-
2020-04-07
2020-08-07
2023-12-07
现在,如果您执行 maturity_date > curve[nrows-1][0]
,这将迭代检查 maturity_date
中的每个元素是否大于 curve[ nrows-1][0]
因此,这将产生另一个 pandas.Series
,它可能看起来像这样 -
True
False
True
python 中的 if
语句,作为一个简单的例子,需要一个 bool
值,而你只是通过提供一堆 bool 值来混淆它。因此,您需要在最后使用 .all()
或 .any()
(这些是通常要做的事情),具体取决于您想要什么
关于python - 如何修复错误 : The truth value of an array with more than one element is ambiguous,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53279238/