[SciPy-User] How to fit a curve/function?

josef.pktd at gmail.com josef.pktd at gmail.com
Thu Jun 9 12:07:47 EDT 2011 

On Thu, Jun 9, 2011 at 11:52 AM, Johannes Radinger <JRadinger at gmx.at> wrote:
> Hello again...
>
> i try no to fit a curve using integrals as conditions.
> the scipy manual says that integrations to infinite are possible with Inf,
>
> I tried following but I fail (saying inf is not defined):
>
> cond2 = 5.0/10/2 - integrate.quad(pdf,35000,Inf,args=(s1,s2))[0]
>
> what causes the problem? Do I use quad/Inf in a wrong way?

numpy.inf    inf doesn't exist in python itself

Josef

>
> The error is:
> NameError: global name 'Inf' is not defined
>
> /Johannes
>
> -------- Original-Nachricht --------
>> Datum: Wed, 8 Jun 2011 11:37:52 -0400
>> Von: josef.pktd at gmail.com
>> An: SciPy Users List <scipy-user at scipy.org>
>> Betreff: Re: [SciPy-User] How to fit a curve/function?
>
>> On Wed, Jun 8, 2011 at 11:37 AM,  <josef.pktd at gmail.com> wrote:
>> > On Wed, Jun 8, 2011 at 10:56 AM, Johannes Radinger <JRadinger at gmx.at>
>> wrote:
>> >>
>> >> -------- Original-Nachricht --------
>> >>> Datum: Wed, 8 Jun 2011 10:33:45 -0400
>> >>> Von: josef.pktd at gmail.com
>> >>> An: SciPy Users List <scipy-user at scipy.org>
>> >>> Betreff: Re: [SciPy-User] How to fit a curve/function?
>> >>
>> >>> On Wed, Jun 8, 2011 at 10:27 AM, Johannes Radinger <JRadinger at gmx.at>
>> >>> wrote:
>> >>> >
>> >>> > -------- Original-Nachricht --------
>> >>> >> Datum: Wed, 8 Jun 2011 10:12:58 -0400
>> >>> >> Von: josef.pktd at gmail.com
>> >>> >> An: SciPy Users List <scipy-user at scipy.org>
>> >>> >> Betreff: Re: [SciPy-User] How to fit a curve/function?
>> >>> >
>> >>> >> On Wed, Jun 8, 2011 at 9:41 AM, Johannes Radinger
>> <JRadinger at gmx.at>
>> >>> >> wrote:
>> >>> >> >
>> >>> >> > -------- Original-Nachricht --------
>> >>> >> >> Datum: Wed, 8 Jun 2011 07:10:38 -0400
>> >>> >> >> Von: josef.pktd at gmail.com
>> >>> >> >> An: SciPy Users List <scipy-user at scipy.org>
>> >>> >> >> Betreff: Re: [SciPy-User] How to fit a curve/function?
>> >>> >> >
>> >>> >> >> On Wed, Jun 8, 2011 at 6:52 AM, Johannes Radinger
>> <JRadinger at gmx.at>
>> >>> >> >> wrote:
>> >>> >> >> > Hello,
>> >>> >> >> >
>> >>> >> >> > I've got following function describing any kind of animal
>> >>> dispersal
>> >>> >> >> kernel:
>> >>> >> >> >
>> >>> >> >> > def pdf(x,s1,s2):
>> >>> >> >> >    return
>> >>> >> >>
>> >>> >>
>> >>>
>> (p/(math.sqrt(2*math.pi*s1**2))*numpy.exp(-((x-0)**(2)/(2*s1**(2)))))+((1-p)/(s2*math.sqrt(2*math.pi))*numpy.exp(-((x-0)**(2)/(2*s2**(2)))))
>> >>> >> >> >
>> >>> >> >> > On the other hand I've got data from literature with which I
>> want
>> >>> to
>> >>> >> fit
>> >>> >> >> the function so that I get s1, s2 and x.
>> >>> >> >> > Ususally the data in the literature are as follows:
>> >>> >> >> >
>> >>> >> >> > Example 1: 50% of the animals are between -270m and +270m and
>> 90%
>> >>> >>  are
>> >>> >> >> between -500m and + 500m
>> >>> >> >> >
>> >>> >> >> > Example 2: 84% is between - 5000 m and +5000m, and 73% are
>> between
>> >>> >> >> -1000m and +1000m
>> >>> >> >> >
>> >>> >> >> > So far as I understand an integration of the function is
>> needed to
>> >>> >> solve
>> >>> >> >> for s1 and s2 as all the literature data give percentage (area
>> under
>> >>> >> the
>> >>> >> >> curve) Can that be used to fit the curve or can that create
>> ranges
>> >>> for
>> >>> >> s1
>> >>> >> >> and s2.
>> >>> >> >>
>> >>> >> >> I don't see a way around integration.
>> >>> >> >>
>> >>> >> >> If you have exactly 2 probabilities, then you can you a solver
>> like
>> >>> >> >> scipy.optimize.fsolve to match the probabilites
>> >>> >> >> eg.
>> >>> >> >> 0.5 = integral pdf from -270 to 270
>> >>> >> >> 0.9 = integral pdf from -500 to 500
>> >>> >> >>
>> >>> >> >> If you have more than 2 probabilities, then using optimization
>> of a
>> >>> >> >> weighted function of the moment conditions would be better.
>> >>> >> >>
>> >>> >> >> Josef
>> >>> >> >
>> >>> >> >
>> >>> >> >
>> >>> >> > Hello again
>> >>> >> >
>> >>> >> > I tried following, but without success so far. What do I have to
>> do
>> >>> >> excactly...
>> >>> >> >
>> >>> >> > import numpy
>> >>> >> > from scipy import stats
>> >>> >> > from scipy import integrate
>> >>> >> > from scipy.optimize import fsolve
>> >>> >> > import math
>> >>> >> >
>> >>> >> > p=0.3
>> >>> >> >
>> >>> >> > def pdf(x,s1,s2):
>> >>> >> >    return
>> >>> >>
>> >>>
>> (p/(math.sqrt(2*math.pi*s1**2))*numpy.exp(-((x-0)**(2)/(2*s1**(2)))))+((1-p)/(s2*math.sqrt(2*math.pi))*numpy.exp(-((x-0)**(2)/(2*s2**(2)))))
>> >>> >> >
>> >>> >> > def equ(s1,s2):
>> >>> >> >    0.5==integrate.quad(pdf,-270,270,args=(s1,s2))
>> >>> >> >    0.9==integrate.quad(pdf,-500,500,args=(s1,s2))
>> >>> >> >
>> >>> >> > result=fsolve(equ, 1,500)
>> >>> >> >
>> >>> >> > print result
>> >>> >>
>> >>> >> equ needs to return the deviation of the equations (I changed some
>> >>> >> details for s1 just to try it)
>> >>> >>
>> >>> >> import numpy
>> >>> >> from scipy import stats
>> >>> >> from scipy import integrate
>> >>> >> from scipy.optimize import fsolve
>> >>> >> import math
>> >>> >>
>> >>> >> p=0.3
>> >>> >>
>> >>> >> def pdf(x,s1,s2):
>> >>> >>     return
>> >>> >>
>> >>>
>> (p/(math.sqrt(2*math.pi*s1**2))*numpy.exp(-((x-0)**(2)/(2*s1**(2)))))+((1-p)/(math.sqrt(2*math.pi*s2**2))*numpy.exp(-((x-0)**(2)/(2*s2**(2)))))
>> >>> >>
>> >>> >> def equ(arg):
>> >>> >>     s1,s2 = numpy.abs(arg)
>> >>> >>     cond1 = 0.5 - integrate.quad(pdf,-270,270,args=(s1,s2))[0]
>> >>> >>     cond2 = 0.9 - integrate.quad(pdf,-500,500,args=(s1,s2))[0]
>> >>> >>     return [cond1, cond2]
>> >>> >>
>> >>> >> result=fsolve(equ, [200., 1200])
>> >>
>> >> thank you for your last reply...seems that the parameters of the two
>> normals are nearly identical... anyway just two small addtional questions:
>> >>
>> >> 1)in fsolve(equ, [200., 1200]) the 200 and 1200 are kind of start
>> values so far as I understand...how should these be choosen? what is
>> recommended?
>> >
>> > There is no general solution for choosing starting values, in your
>> > case it should be possible to
>> >
>> >>>> q = np.array([0.5, 0.9])
>> >>>> cr = x/stats.norm.ppf(0.5 + q/2.)
>> >>>> x = [270, 500]
>> >>>> q = np.array([0.5, 0.9])
>> >>>> x = [270, 500]
>> >>>> cr = x/stats.norm.ppf(0.5 + q/2.)
>> >>>> stats.norm.cdf(500, scale=cr[1]) - stats.norm.cdf(-500, scale=cr[1])
>> > 0.89999999999999991
>> -------
>> I forgot to remove the typos
>> >>>> stats.norm.cdf(q[0], scale=cr[1]) - stats.norm.cdf(-q[0],
>> scale=cr[0])
>> > 0.0011545021185267457
>> >>>> stats.norm.cdf(q[0], scale=cr[0]) - stats.norm.cdf(-q[0],
>> scale=cr[0])
>> > 0.000996601515122153
>> ---------
>> >>>> stats.norm.cdf(x[0], scale=cr[0]) - stats.norm.cdf(-x[0],
>> scale=cr[0])
>> > 0.5
>> >>>> sol = fsolve(equ, np.sort(cr))
>> >
>> > there are some numerical problems finding the solution (???)
>> >
>> >>>> equ(sol)
>> > array([-0.05361093,  0.05851309])
>> >>>> from pprint import pprint
>> >>>> pprint(fsolve(equ, np.sort(cr), xtol=1e-10, full_output=1))
>> > (array([ 354.32616549,  354.69918062]),
>> >  {'fjac': array([[-0.7373189 , -0.67554484],
>> >       [ 0.67554484, -0.7373189 ]]),
>> >  'fvec': array([-0.05361093,  0.05851309]),
>> >  'nfev': 36,
>> >  'qtf': array([  1.40019135e-07,  -7.93593929e-02]),
>> >  'r': array([ -5.21390161e-04,  -1.21700831e-03,   3.88274320e-07])},
>> >  5,
>> >  'The iteration is not making good progress, as measured by the \n
>> > improvement from the last ten iterations.')
>> >
>> >>
>> >> 2) How can that be solve if I have I third condition (overfitted) can
>> that be used as well or how does the alternative look like?
>> >
>> > use optimize.leastsq on equ (I never tried this for this case)
>> > use fmin on the sum of squared errors
>> >
>> > if the intervals for the probabilities are non-overlapping (interval
>> > data), then there is an optimal weighting matrix, (but my code for
>> > that in the statsmodels.sandbox is not verified).
>> >
>> > Josef
>> >
>> >
>> >>
>> >> /johannes
>> >>
>> >>> >>
>> >>> >> print result
>> >>> >>
>> >>> >> but in the results I get the parameters are very close to each
>> other
>> >>> >> [-356.5283675   353.82544075]
>> >>> >>
>> >>> >> the pdf looks just like a mixture of 2 normals both with loc=0,
>> then
>> >>> >> maybe the cdf of norm can be used directly
>> >>> >
>> >>> >
>> >>> > Thank you for that hint... First yes these are 2 superimposed
>> normals
>> >>> but for other reasons I want to use the original formula instead of
>> the
>> >>> stats.functions...
>> >>> >
>> >>> > anyway there is still a thing...the locator s1 and s2 are like the
>> scale
>> >>> parameter of stats.norm so the are both + and -. For fsolve above it
>> seems
>> >>> that I get only one parameter (s1 or s2) but for the positive and
>> negative
>> >>> side of the distribution. So in actually there are four parameters
>> -s1,
>> >>> +s1, -s2, +s2. How can I solve that? Maybe I can restrict the fsolve
>> to look
>> >>> for the two values only in the positive range...
>> >>>
>> >>> It doesn't really matter, if the scale only shows up in quadratic
>> >>> terms, or as in my initial change I added a absolute value, so whether
>> >>> it's positive or negative, it's still only one value, and we
>> >>> interprete it as postive scale
>> >>>
>> >>> s1 = sqrt(s1**2)
>> >>>
>> >>> Josef
>> >>>
>> >>> >
>> >>> > any guesses?
>> >>> >
>> >>> > /J
>> >>> >
>> >>> >>
>> >>> >> >>> from scipy import stats
>> >>> >> >>> stats.norm.cdf(270, scale=350) - stats.norm.cdf(-270,
>> scale=350)
>> >>> >> 0.55954705470577526
>> >>> >> >>>
>> >>> >> >>> stats.norm.cdf(270, scale=354) - stats.norm.cdf(-270,
>> scale=354)
>> >>> >> 0.55436474670960978
>> >>> >> >>> stats.norm.cdf(500, scale=354) - stats.norm.cdf(-500,
>> scale=354)
>> >>> >> 0.84217642881921018
>> >>> >>
>> >>> >> Josef
>> >>> >> >
>> >>> >> >
>> >>> >> > /Johannes
>> >>> >> >>
>> >>> >> >> >
>> >>> >> >> > /Johannes
>> >>> >> >> >
>> >>> >> >> > --
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>> >>> >> >> >
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