[SciPy-User] optimize.leastsq - Value Error: The truth value of an array with more than one element is ambiguous
Sebastian Walter
sebastian.walter at gmail.com
Sun May 16 12:22:53 EDT 2010
you are using integer arrays...
this should work:
import numpy as np
from scipy.optimize import leastsq
X = np.array([1,2,3,4,5,6,7,8,9],dtype=float)
Y = np.array([-1, 0, 1, 2,7,8,9,10,11],dtype=float)
def func(p, x, y):
a = p
sum = np.array([0,0,0,0,0,0,0,0,0],dtype=float)
for i in range(0, len(x)):
if x[i] > 4:
sum[i] = x[i] + a
else:
sum[i] = x[i] - a
return sum - y
r = leastsq(func,1,args=(X,Y))
print r[0]
regards,
Sebastian
On Sun, May 16, 2010 at 3:58 AM, Zhe Wang <3njoywind at gmail.com> wrote:
> v is calender years e.g
> (1+p)**1984
> May be you could help me with this simple example:
> I have a function defined like this(a0 is a parameter):
> ------------------------------------
> def f(x):
> if x > 4:
> return x + a0
> else:
> return x - a0
> ------------------------------------
> I generated the data when I let a0=2 :
> --------------------------------------
> X = np.array([1,2,3,4,5,6,7,8,9])
> Y = np.array([-1, 0, 1, 2,7,8,9,10,11])
> --------------------------------------
> How can I use leastsq() to fit f(x)? I have wrote some code like this:
> --------------------------------------
> def func(p, x, y):
> a = p
> sum = np.array([0,0,0,0,0,0,0,0,0])
> for i in range(0, len(x)):
> if x[i] > 4:
> sum[i] = x[i] + a
> else:
> sum[i] = x[i] - a
> return sum - y
> r = leastsq(func,1,args=(X,Y))
> print r[0]
> --------------------------------------
> the output is 1.0000000149, much different like 2.
> I doubt whether leastsq() is suitable for this kind of problem. Maybe I
> should try another way?
> Zhe
> On Sun, May 16, 2010 at 12:34 AM, <josef.pktd at gmail.com> wrote:
>>
>> On Sat, May 15, 2010 at 11:06 AM, Zhe Wang <3njoywind at gmail.com> wrote:
>> > Josef:
>> > Thanks,
>> > my example is just for test, it is not always have a fixed start, e.g
>> >>>> T = np.array([1995,1996,1997,1998,1999])
>> >>>> E = np.array([14,12,11,15,12])
>> >>>> T-E
>> > array([1981, 1984, 1986, 1983, 1987])
>> > so, if I define the function:
>> > def func(x):
>> > #....
>> > v = np.arange(...)
>> > Y = np.cumsum((a*(1+p)**v)*I(v))
>> > return Y
>> > when I call leastsq(func, [1,0]), v should change as the element of T
>> > change, e.g.
>> > when t(one element of T) is 1995，v = np.arange(1981, 1995)
>> > when t is 1996, v = np.arange(1984, 1996)
>>
>> are you sure v is supposed to be calender years and not number of
>> years accumulated? i.e
>>
>> (1+p)**1984
>> or
>> (1+p)**(1996-1984)
>>
>> The most efficient would be to use the formula for the sum of a finite
>> geometric series, which would also avoid the sum.
>>
>> Josef
>>
>>
>> > ......
>> > this troubles me so much.
>> > Zhe Wang
>> >
>> >
>> > On Sat, May 15, 2010 at 9:08 PM, <josef.pktd at gmail.com> wrote:
>> >>
>> >> 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
>> >> >> >
>> >> >> > _______________________________________________
>> >> >> > SciPy-User mailing list
>> >> >> > SciPy-User at scipy.org
>> >> >> > http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >> >
>> >> >> >
>> >> >> _______________________________________________
>> >> >> SciPy-User mailing list
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>> >> >
>> >> >
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