问题
I am trying to fit line profiles as detected with a spectrograph on a CCD. For ease of consideration, I have included a demonstration that, if solved, is very similar to the one I actually want to solve.
I've looked at this: https://stats.stackexchange.com/questions/46626/fitting-model-for-two-normal-distributions-in-pymc and various other questions and answers, but they are doing something fundamentally different than what I want to do.
import pymc as mc
import numpy as np
import pylab as pl
def GaussFunc(x, amplitude, centroid, sigma):
return amplitude * np.exp(-0.5 * ((x - centroid) / sigma)**2)
wavelength = np.arange(5000, 5050, 0.02)
# Profile 1
centroid_one = 5025.0
sigma_one = 2.2
height_one = 0.8
profile1 = GaussFunc(wavelength, height_one, centroid_one, sigma_one, )
# Profile 2
centroid_two = 5027.0
sigma_two = 1.2
height_two = 0.5
profile2 = GaussFunc(wavelength, height_two, centroid_two, sigma_two, )
# Measured values
noise = np.random.normal(0.0, 0.02, len(wavelength))
combined = profile1 + profile2 + noise
# If you want to plot what this looks like
pl.plot(wavelength, combined, label="Measured")
pl.plot(wavelength, profile1, color='red', linestyle='dashed', label="1")
pl.plot(wavelength, profile2, color='green', linestyle='dashed', label="2")
pl.title("Feature One and Two")
pl.legend()
My question: Can PyMC (or some variant) give me the mean, amplitude, and sigma for the two components used above?
Please note that the functions that I will actually fit on my real problem are NOT Gaussians -- so please provide the example using a generic function (like GaussFunc in my example), and not a "built-in" pymc.Normal() type function.
Also, I understand model selection is another issue: so with the current noise, 1 component (profile) might be all that is statistically justified. But I'd like to see what the best solution for 1, 2, 3, etc. components would be.
I'm also not wed to the idea of using PyMC -- if scikit-learn, astroML, or some other package seems perfect, please let me know!
EDIT:
I failed a number of ways, but one of the things that I think was on the right track was the following:
sigma_mc_one = mc.Uniform('sig', 0.01, 6.5)
height_mc_one = mc.Uniform('height', 0.1, 2.5)
centroid_mc_one = mc.Uniform('cen', 5015., 5040.)
But I could not construct a mc.model that worked.
回答1:
Not the most concise PyMC code, but I made that decision to help the reader. This should run, and give (really) accurate results.
I made the decision to use Uniform priors, with liberal ranges, because I really have no idea what we are modelling. But probably one has an idea about the centroid locations, and can use a better priors there.
### Suggested one runs the above code first.
### Unknowns we are interested in
est_centroid_one = mc.Uniform("est_centroid_one", 5000, 5050 )
est_centroid_two = mc.Uniform("est_centroid_two", 5000, 5050 )
est_sigma_one = mc.Uniform( "est_sigma_one", 0, 5 )
est_sigma_two = mc.Uniform( "est_sigma_two", 0, 5 )
est_height_one = mc.Uniform( "est_height_one", 0, 5 )
est_height_two = mc.Uniform( "est_height_two", 0, 5 )
#std deviation of the noise, converted to precision by tau = 1/sigma**2
precision= 1./mc.Uniform("std", 0, 1)**2
#Set up the model's relationships.
@mc.deterministic( trace = False)
def est_profile_1(x = wavelength, centroid = est_centroid_one, sigma = est_sigma_one, height= est_height_one):
return GaussFunc( x, height, centroid, sigma )
@mc.deterministic( trace = False)
def est_profile_2(x = wavelength, centroid = est_centroid_two, sigma = est_sigma_two, height= est_height_two):
return GaussFunc( x, height, centroid, sigma )
@mc.deterministic( trace = False )
def mean( profile_1 = est_profile_1, profile_2 = est_profile_2 ):
return profile_1 + profile_2
observations = mc.Normal("obs", mean, precision, value = combined, observed = True)
model = mc.Model([est_centroid_one,
est_centroid_two,
est_height_one,
est_height_two,
est_sigma_one,
est_sigma_two,
precision])
#always a good idea to MAP it prior to MCMC, so as to start with good initial values
map_ = mc.MAP( model )
map_.fit()
mcmc = mc.MCMC( model )
mcmc.sample( 50000,40000 ) #try running for longer if not happy with convergence.
Important
Keep in mind the algorithm is agnostic to labeling, so the results might show profile1
with all the characteristics from profile2
and vice versa.
来源:https://stackoverflow.com/questions/15184293/fit-two-normal-distributions-histograms-with-mcmc-using-pymc