[SciPy-User] optimize.leastsq - Value Error: The truth value of an array with more than one element is ambiguous

[josef.pktd at gmail.com]

On Sat, May 15, 2010 at 8:49 AM, Zhe Wang <3njoywind at gmail.com> wrote:
> Josef:
> Thanks for your reply:)
> Actually I want to fit this equation:
> Y(t) = sigma(v=t-e(t), t)[(a*(1+p)**v)*I(v)]
> I got {t} and {Y(t)} and a, p are parameters. e(t) and I(v) can be
> calculated by e() and I().

Do you always have a fixed start date as in your example

>>> T = np.array([1995,1996,1997,1998,1999])
>>> E = np.array([10,11,12,13,14])
>>> T-E
array([1985, 1985, 1985, 1985, 1985])

so that always v =range(T0, T+1)     with fixed T0=1985

this would make it easier to work forwards than backwards, e.g. something like
v = np.arange(...)
Y = np.cusum((a*(1+p)**v)*I(v))

Josef




> I rewrote my code like this:
> ----------------------------------------------------------------------------------------------
> from scipy.optimize import leastsq
> import numpy as np
> def Iv(t):
>     return 4
> def Yt(x, et):
>     a, pa = x
>     sum = np.array([0,0,0,0,0])
>     for i in range(0, len(et)):
>         for j in range(0, et[i]):
>             v = T[i] - et[i] + j
>             sum[i] += a*(1+pa)**(v)*Iv(v)
>     return sum - Y
> T = np.array([1995,1996,1997,1998,1999])
> Y =
> np.array([639300.36866,664872.383407,691467.278743,719125.969893,747891.008688])
> E = np.array([10,11,12,13,14])
> r = leastsq(Yt, [1,0], args = (E), maxfev=10000000)
> A, Pa = r[0]
> print "A=",A,"Pa=",Pa
> ----------------------------------------------------------------------------------------------
> the output is:
> A= 1.0 Pa = 0.0
> ----------------------------------------------------------------------------------------------
> I don't think it is correct. Hope for your guidence.
> On Sat, May 15, 2010 at 7:02 PM, <josef.pktd at gmail.com> wrote:
>>
>> On Sat, May 15, 2010 at 5:25 AM, Zhe Wang <3njoywind at gmail.com> wrote:
>> > Traceback (most recent call last):
>> >   File "D:\Yt.py", line 31, in <module>
>> >     r = leastsq(residuals, [1,0], args=(Y,T), maxfev=10000000)
>> >   File "D:\Python26\lib\site-packages\scipy\optimize\minpack.py", line
>> > 266,
>> > in leastsq
>> >     m = check_func(func,x0,args,n)[0]
>> >   File "D:\Python26\lib\site-packages\scipy\optimize\minpack.py", line
>> > 12,
>> > in check_func
>> >     res = atleast_1d(thefunc(*((x0[:numinputs],)+args)))
>> >   File "D:\Yt.py", line 26, in residuals
>> >     return y - Yt(x, p)
>> >   File "D:\Yt.py", line 20, in Yt
>> >     for i in range(0, Et(x)):
>> >   File "D:\Yt.py", line 11, in Et
>> >     if t == 1995:
>> > ValueError: The truth value of an array with more than one element is
>> > ambiguous. Use a.any() or a.all()
>> >
>> > ---------------------------------------------------------------------------------------------------------
>> >
>> > When running the following code:
>> >
>> >
>> > --------------------------------------------------------------------------------------------
>> >
>> > from scipy.optimize import leastsq
>> > import numpy as np
>> >
>> > def Iv(t):
>> >     if t == 1995:
>> >         return t + 2
>> >     else:
>> >         return t
>> >
>> > def Et(t):
>> >     if t == 1995:
>> >         return t + 2
>> >     else:
>> >         return t
>> >
>> > def Yt(x, p):
>> >     a, pa = p
>> >     sum = 0
>> >
>> >     for i in range(0, Et(x)):
>> >         v = x - et + i
>> >         sum += a*(1+p)**(v)*Iv(v)
>> >     return sum
>> >
>> > def residuals(p, y, x):
>> >     return y - Yt(x, p)
>> >
>> > T = np.array([1995,1996,1997,1998,1999])
>> > Y =
>> >
>> > np.array([639300.36866,664872.383407,691467.278743,719125.969893,747891.008688])
>> >
>> > r = leastsq(residuals, [1,0], args=(Y,T), maxfev=10000000)
>> > A, Pa = r[0]
>> > print "A=",A,"Pa=",Pa
>> >
>> >
>> > ----------------------------------------------------------------------------------------------
>> >
>> > I know the error occurs when I compare t like: "if t == 1995",but I have
>> > no
>> > idea how to handle it correctly.
>>
>> try the vectorized version of a conditional assignment, e.g.
>> np.where(t == 1995, t, t+2)
>>
>> I didn't read enough of your example, to tell whether your Yt loop can
>> be vectorized with a single sum, but I guess so.
>>
>> optimize leastsq expects an array, so residuals (and Yt) need to
>> return an array not a single value, maybe np.cusum and conditional or
>> data dependent slicing/indexing works
>>
>> Josef
>>
>>
>> >
>> > Any help would be greatly appreciated.
>> >
>> > Zhe Wang
>> >
>> > _______________________________________________
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>> > http://mail.scipy.org/mailman/listinfo/scipy-user
>> >
>> >
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