[SciPy-User] GIS Raster Calculation: Combination of NumPy Arrays and SciPy Density function
josef.pktd at gmail.com
josef.pktd at gmail.com
Fri Jan 7 07:08:56 EST 2011
On Fri, Jan 7, 2011 at 6:50 AM, Johannes Radinger <JRadinger at gmx.at> wrote:
>
> -------- Original-Nachricht --------
>> Datum: Thu, 6 Jan 2011 11:50:20 -0500
>> Von: josef.pktd at gmail.com
>> An: SciPy Users List <scipy-user at scipy.org>
>> Betreff: Re: [SciPy-User] GIS Raster Calculation: Combination of NumPy Arrays and SciPy Density function
>
>> On Thu, Jan 6, 2011 at 9:59 AM, Johannes Radinger <JRadinger at gmx.at>
>> wrote:
>> > Hello...
>> >
>> > I am working mostly in ArcGIS but as it's Raster Calculator resp. Map
>> Algebra is limited I want to use NumPy Arrays and perform calcultations in
>> SciPy.
>> >
>> > I can get my Rasterfiles in ArcGIS quite easily transformed/exported
>> into a NumPy Array with the function:
>> >
>> > newArray = arcpy.RasterToNumPyArray(inRaster)
>> >
>> > This Raster/Array contains distance values (ranging from around -20000
>> to + 20000 float).
>> >
>> > Some time before you helped with some distance-density functions which I
>> want to apply now. New Values should be assigned to the cells/elements in
>> the array by following function:
>> >
>> > DensityRaster = p * stats.norm.pdf(x, loc=m, scale=s)
>> >
>> > where p = 0.3, m=0 and s=200, the x-value is the distance value from the
>> single cells in newArray. Can I just define x=newArray ? and use an array
>> as input variable?
>>
>> Yes, just try it out. It's fully vectorized.
>> loc=0 is the default, so in this case you wouldn't need to specify loc=m.
>>
>> If your array is very large, there is a shortcut to (maybe) avoid some
>> intermediate arrays.
>
> Oh thank you, that works perfectly. What do you mean with a shortcut?
>
> And just another short question about the stats.norm.cdf function: That is calculating the cumulative probabilty from the smallest value (most negative x) to the point x, but is there also a function which gives the cumulative sum away from 0 to point x in both directions (negativ and positive? Like an integration of the probabilty function from 0 to (-)x?
cdf is the standard definition of the cdf, integration from lower
support limit, attribute .a, for normal it's -np.inf, to point x.
there is also sf(x) = 1-cdf(x)
in general
Prob(0<y<=x) = cdf(x) - cdf(0)
for symmetric centered Prob(0<y<=x) = (1-2*cdf(-x))/2 , but I suspect
for the normal you don't gain anything compared to cdf(x) - 0.5
Josef
>
> thank you
> /j
>>
>> Josef
>>
>>
>> >
>> > thank you
>> > /j
>> >
>> >
>> > -------- Original-Nachricht --------
>> >> Datum: Thu, 6 Jan 2011 07:10:58 -0500
>> >> Von: josef.pktd at gmail.com
>> >> An: SciPy Users List <scipy-user at scipy.org>
>> >> Betreff: Re: [SciPy-User] solving integration, density function
>> >
>> >> On Thu, Jan 6, 2011 at 7:01 AM, Johannes Radinger <JRadinger at gmx.at>
>> >> wrote:
>> >> > Thank you for the simplification of the formula,
>> >> > but I still get a different result in the case
>> >> > when x<s1 (eg. x=2, s1=3).
>> >> >
>> >> > here a code to try:
>> >> >
>> >> > ********************
>> >> > import math
>> >> > from scipy import stats
>> >> >
>> >> > s1 = 3
>> >> > m = 0
>> >> > p = 1
>> >> > x = 2
>> >> >
>> >> > func = stats.norm.pdf(x, loc=m, scale=(s1))
>> >> > func2 = (1/(s1*math.sqrt(2*math.pi)) * math.exp(-0.5*((x-m)/s1)**2))
>> >> >
>> >> > print func
>> >> > print func2
>> >> > ********************************
>> >>
>> >> use floats, I think you just run into integer division
>> >>
>> >> (x-m)/s1
>> >>
>> >> Josef
>> >>
>> >> >
>> >> > /j
>> >> >
>> >> > -------- Original-Nachricht --------
>> >> >> Datum: Thu, 6 Jan 2011 05:48:25 -0600
>> >> >> Von: Warren Weckesser <warren.weckesser at enthought.com>
>> >> >> An: SciPy Users List <scipy-user at scipy.org>
>> >> >> Betreff: Re: [SciPy-User] solving integration, density function
>> >> >
>> >> >> On Thu, Jan 6, 2011 at 5:27 AM, Johannes Radinger <JRadinger at gmx.at>
>> >> >> wrote:
>> >> >>
>> >> >> > Hey
>> >> >> >
>> >> >> > Last time you helped me a lot with my normal
>> >> >> > probabilty density function. My problem now is
>> >> >> > quite simple, I think it's just a problem with
>> >> >> > the syntax (brackets):
>> >> >> >
>> >> >> > There are two ways to calculate the pdf, with the
>> >> >> > stats-function and with pure mathematically, but
>> >> >> > the give different results and I can't find the
>> >> >> > where I make the mistake:
>> >> >> >
>> >> >> >
>> >> >> > func1 = stats.norm.pdf(x, loc=m, scale=(s1))
>> >> >> > func2 =
>> >> >> >
>> 1/((s1)*(math.sqrt(2*math.pi))))*(math.exp(((-0.5)*((x-m)/(s1)))**2)
>> >> >> >
>> >> >> > Where is the problem
>> >> >> >
>> >> >>
>> >> >>
>> >> >> func2 = 1/(s1*math.sqrt(2*math.pi)) * math.exp(-0.5*((x-m)/s1)**2)
>> >> >>
>> >> >>
>> >> >> Warren
>> >> >>
>> >> >>
>> >> >>
>> >> >> > thank you...
>> >> >> >
>> >> >> > /j
>> >> >> >
>> >> >> > -------- Original-Nachricht --------
>> >> >> > > Datum: Tue, 21 Dec 2010 09:18:15 -0500
>> >> >> > > Von: Skipper Seabold <jsseabold at gmail.com>
>> >> >> > > An: SciPy Users List <scipy-user at scipy.org>
>> >> >> > > Betreff: Re: [SciPy-User] solving integration, density function
>> >> >> >
>> >> >> > > On Tue, Dec 21, 2010 at 7:48 AM, Johannes Radinger
>> >> <JRadinger at gmx.at>
>> >> >> > > wrote:
>> >> >> > > >
>> >> >> > > > -------- Original-Nachricht --------
>> >> >> > > >> Datum: Tue, 21 Dec 2010 13:20:47 +0100
>> >> >> > > >> Von: Gregor Thalhammer <Gregor.Thalhammer at gmail.com>
>> >> >> > > >> An: SciPy Users List <scipy-user at scipy.org>
>> >> >> > > >> Betreff: Re: [SciPy-User] solving integration, density
>> function
>> >> >> > > >
>> >> >> > > >>
>> >> >> > > >> Am 21.12.2010 um 12:06 schrieb Johannes Radinger:
>> >> >> > > >>
>> >> >> > > >> > Hello,
>> >> >> > > >> >
>> >> >> > > >> > I am really new to python and Scipy.
>> >> >> > > >> > I want to solve a integrated function with a python script
>> >> >> > > >> > and I think Scipy should do that :)
>> >> >> > > >> >
>> >> >> > > >> > My task:
>> >> >> > > >> >
>> >> >> > > >> > I do have some variables (s, m, K,) which are now
>> absolutely
>> >> set,
>> >> >> > but
>> >> >> > > in
>> >> >> > > >> future I'll get the values via another process of pyhton.
>> >> >> > > >> >
>> >> >> > > >> > s = 400
>> >> >> > > >> > m = 0
>> >> >> > > >> > K = 1
>> >> >> > > >> >
>> >> >> > > >> > And have have following function:
>> >> >> > > >> > (1/((s*K)*sqrt(2*pi)))*exp((-1/2*(((x-m)/s*K))^2) which is
>> the
>> >> >> > > density
>> >> >> > > >> function of the normal distribution a symetrical curve with
>> the
>> >> >> mean
>> >> >> > > (m) of
>> >> >> > > >> 0.
>> >> >> > > >> >
>> >> >> > > >> > The total area under the curve is 1 (100%) which is for an
>> >> >> > > integration
>> >> >> > > >> from -inf to +inf.
>> >> >> > > >> > I want to know x in the case of 99%: meaning that the
>> integral
>> >> >> (-x
>> >> >> > to
>> >> >> > > >> +x) of the function is 0.99. Due to the symetry of the curve
>> you
>> >> >> can
>> >> >> > > also set
>> >> >> > > >> the integral from 0 to +x equal to (0.99/2):
>> >> >> > > >> >
>> >> >> > > >> > 0.99 =
>> >> >> integral((1/((s*K)*sqrt(2*pi)))*exp((-1/2*(((x-m)/s*K))^2)),
>> >> >> > > -x,
>> >> >> > > >> x)
>> >> >> > > >> > resp.
>> >> >> > > >> > (0.99/2) =
>> >> >> > > integral((1/((s*K)*sqrt(2*pi)))*exp((-1/2*(((x-m)/s*K))^2)),
>> >> >> > > >> 0, x)
>> >> >> > > >> >
>> >> >> > > >> > How can I solve that question in Scipy/python
>> >> >> > > >> > so that I get x in the end. I don't know how to write
>> >> >> > > >> > the code...
>> >> >> > > >>
>> >> >> > > >>
>> >> >> > > >> --->
>> >> >> > > >> erf(x[, out])
>> >> >> > > >>
>> >> >> > > >> y=erf(z) returns the error function of complex argument
>> >> defined
>> >> >> > > as
>> >> >> > > >> as 2/sqrt(pi)*integral(exp(-t**2),t=0..z)
>> >> >> > > >> ---
>> >> >> > > >>
>> >> >> > > >> from scipy.special import erf, erfinv
>> >> >> > > >> erfinv(0.99)*sqrt(2)
>> >> >> > > >>
>> >> >> > > >>
>> >> >> > > >> Gregor
>> >> >> > > >>
>> >> >> > > >
>> >> >> > > >
>> >> >> > > > Thank you Gregor,
>> >> >> > > > I only understand a part of your answer... I know that the
>> >> integral
>> >> >> of
>> >> >> > > the density function is a error function and I know that the
>> >> argument
>> >> >> > "from
>> >> >> > > scipy.special import erf, erfinv" is to load the module.
>> >> >> > > >
>> >> >> > > > But how do I write the code including my orignial function so
>> >> that I
>> >> >> > can
>> >> >> > > modify it (I have also another function I want to integrate).
>> how
>> >> do i
>> >> >> > > start? I want to save the whole code to a python-script I can
>> then
>> >> >> load
>> >> >> > e.g.
>> >> >> > > into ArcGIS where I want to use the value of x for further
>> >> >> calculations.
>> >> >> > > >
>> >> >> > >
>> >> >> > > Are you always integrating densities? If so, you don't want to
>> >> use
>> >> >> > > integrals probably, but you could use scipy.stats
>> >> >> > >
>> >> >> > > erfinv(.99)*np.sqrt(2)
>> >> >> > > 2.5758293035489004
>> >> >> > >
>> >> >> > > from scipy import stats
>> >> >> > >
>> >> >> > > stats.norm.ppf(.995)
>> >> >> > > 2.5758293035489004
>> >> >> > >
>> >> >> > > Skipper
>> >> >> > > _______________________________________________
>> >> >> > > SciPy-User mailing list
>> >> >> > > SciPy-User at scipy.org
>> >> >> > > http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >> >
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