Related
I'm trying to learn more about pyhf and my understanding of what the goals are might be limited. I would love to fit my HEP data outside of ROOT, but I could be imposing expectations on pyhf which are not what the authors intended for it's use.
I'd like to write myself a hello-world example, but I might just not know what I'm doing. My misunderstanding could also be gaps in my statistical knowledge.
With that preface, let me explain what I'm trying to explore.
I have some observed set of events for which I calculate some observable and make a binned histogram of that data. I hypothesize that there are two contributing physics processes, which I call signal and background. I generate some Monte Carlo samples for these processes and the theorized total number of events is close to, but not exactly what I observe.
I would like to:
Fit the data to this two process hypothesis
Get from the fit the optimal values for the number of events for each process
Get the uncertainties on these fitted values
If appropriate, calculate an upper limit on the number of signal events.
My starter code is below, where all I'm doing is an ML fit but I'm not sure where to go. I know it's not set up to do what I want, but I'm getting lost in the examples I find on RTD. I'm sure it's me, this is not a criticism of the documentation.
import pyhf
import numpy as np
import matplotlib.pyplot as plt
nbins = 15
# Generate a background and signal MC sample`
MC_signal_events = np.random.normal(5,1.0,200)
MC_background_events = 10*np.random.random(1000)
signal_data = np.histogram(MC_signal_events,bins=nbins)[0]
bkg_data = np.histogram(MC_background_events,bins=nbins)[0]
# Generate an observed dataset with a slightly different
# number of events
signal_events = np.random.normal(5,1.0,180)
background_events = 10*np.random.random(1050)
observed_events = np.array(signal_events.tolist() + background_events.tolist())
observed_sample = np.histogram(observed_events,bins=nbins)[0]
# Plot these samples, if you like
plt.figure(figsize=(12,4))
plt.subplot(1,3,1)
plt.hist(observed_events,bins=nbins,label='Observations')
plt.legend()
plt.subplot(1,3,2)
plt.hist(MC_signal_events,bins=nbins,label='MC signal')
plt.legend()
plt.subplot(1,3,3)
plt.hist(MC_background_events,bins=nbins,label='MC background')
plt.legend()
# Use a very naive estimate of the background
# uncertainties
bkg_uncerts = np.sqrt(bkg_data)
print("Defining the PDF.......")
pdf = pyhf.simplemodels.hepdata_like(signal_data=signal_data.tolist(), \
bkg_data=bkg_data.tolist(), \
bkg_uncerts=bkg_uncerts.tolist())
print("Fit.......")
data = pyhf.tensorlib.astensor(observed_sample.tolist() + pdf.config.auxdata)
bestfit_pars, twice_nll = pyhf.infer.mle.fit(data, pdf, return_fitted_val=True)
print(bestfit_pars)
print(twice_nll)
plt.show()
Note: this answer is based on pyhf v0.5.2.
Alright, so it looks like you've managed to figure most of the big pieces for sure. However, there's two different ways to do this depending on how you prefer to set things up. In both cases, I assume you want an unconstrained fit and you want to...
fit your signal+background model to observed data
fit your background model to observed data
First, let's discuss uncertainties briefly. At the moment, we default to numpy for the tensor background and scipy for the optimizer. See documentation:
numpy backend
scipy optimizer
However, one unfortunate drawback right now with the scipy optimizer is that it cannot return the uncertainties. What you need to do anywhere in your code before the fit (although we generally recommend as early as possible) is to use the minuit optimizer instead:
pyhf.set_backend('numpy', 'minuit')
This will get you the nice features of being able to get the correlation matrix, the uncertainties on the fitted parameters, and the hessian -- amongst other things. We're working to make this consistent for scipy as well, but this is not ready right now.
All optimizations go through our optimizer API which you can currently view through the mixin here in our documentation. Specifically, the signature is
minimize(
objective,
data,
pdf,
init_pars,
par_bounds,
fixed_vals=None,
return_fitted_val=False,
return_result_obj=False,
do_grad=None,
do_stitch=False,
**kwargs)
There are a lot of options here. Let's just focus on the fact that one of the keyword arguments we can pass through is return_uncertainties which will change the bestfit parameters by adding a column for the fitted parameter uncertainty which you want.
1. Signal+Background
In this case, we want to just use the default model
result, twice_nll = pyhf.infer.mle.fit(
data,
pdf,
return_uncertainties=True,
return_fitted_val=True
)
bestfit_pars, errors = result.T
2. Background-Only
In this case, we need to turn off the signal. The way we do this is by setting the parameter of interest (POI) fixed to 0.0. Then we can get the fitted parameters for the background-only model in a similar way, but using fixed_poi_fit instead of an unconstrained fit:
result, twice_nll = pyhf.infer.mle.fixed_poi_fit(
0.0,
data,
pdf,
return_uncertainties=True,
return_fitted_val=True
)
bestfit_pars, errors = result.T
Note that this is quite simply a quick way of doing the following unconstrained fit
bkg_params = pdf.config.suggested_init()
fixed_params = pdf.config.suggested_fixed()
bkg_params[pdf.config.poi_index] = 0.0
fixed_params[pdf.config.poi_index] = True
result, twice_nll = pyhf.infer.mle.fit(
data,
pdf,
init_pars=bkg_params,
fixed_params=fixed_params,
return_uncertainties=True,
return_fitted_val=True
)
bestfit_pars, errors = result.T
Hopefully that clarifies things up more!
Giordon's solution should answer all of your question, but I thought I'd also write out the code to basically address everything we can.
I also take the liberty of changing some of your values a bit so that the signal isn't so strong that the observed CLs value isn't far off to the right of the Brazil band (the results aren't wrong obviously, but it probably makes more sense to be talking about using the discovery test statistic at that point then setting limits. :))
Environment
For this example I'm going to setup a clean Python 3 virtual environment and then install the dependencies (here we're going to be using pyhf v0.5.2)
$ python3 -m venv "${HOME}/.venvs/question"
$ . "${HOME}/.venvs/question/bin/activate"
(question) $ cat requirements.txt
pyhf[minuit,contrib]~=0.5.2
black
(question) $ python -m pip install -r requirements.txt
Code
While we can't easily get the best fit value for both the number of signal events as well as the background events we definitely can do inference to get the best fit value for the signal strength.
The following chunk of code (which is long only because of the visualization) should address all of the points of your question.
# answer.py
import numpy as np
import pyhf
import matplotlib.pyplot as plt
import pyhf.contrib.viz.brazil
# Goals:
# - Fit the model to the observed data
# - Infer the best fit signal strength given the model
# - Get the uncertainties on the best fit signal strength
# - Calculate an 95% CL upper limit on the signal strength
def plot_hist(ax, bins, data, bottom=0, color=None, label=None):
bin_width = bins[1] - bins[0]
bin_leftedges = bins[:-1]
bin_centers = [edge + bin_width / 2.0 for edge in bin_leftedges]
ax.bar(
bin_centers, data, bin_width, bottom=bottom, alpha=0.5, color=color, label=label
)
def plot_data(ax, bins, data, label="Data"):
bin_width = bins[1] - bins[0]
bin_leftedges = bins[:-1]
bin_centers = [edge + bin_width / 2.0 for edge in bin_leftedges]
ax.scatter(bin_centers, data, color="black", label=label)
def invert_interval(test_mus, hypo_tests, test_size=0.05):
# This will be taken care of in v0.5.3
cls_obs = np.array([test[0] for test in hypo_tests]).flatten()
cls_exp = [
np.array([test[1][idx] for test in hypo_tests]).flatten() for idx in range(5)
]
crossing_test_stats = {"exp": [], "obs": None}
for cls_exp_sigma in cls_exp:
crossing_test_stats["exp"].append(
np.interp(
test_size, list(reversed(cls_exp_sigma)), list(reversed(test_mus))
)
)
crossing_test_stats["obs"] = np.interp(
test_size, list(reversed(cls_obs)), list(reversed(test_mus))
)
return crossing_test_stats
def main():
np.random.seed(0)
pyhf.set_backend("numpy", "minuit")
observable_range = [0.0, 10.0]
bin_width = 0.5
_bins = np.arange(observable_range[0], observable_range[1] + bin_width, bin_width)
n_bkg = 2000
n_signal = int(np.sqrt(n_bkg))
# Generate simulation
bkg_simulation = 10 * np.random.random(n_bkg)
signal_simulation = np.random.normal(5, 1.0, n_signal)
bkg_sample, _ = np.histogram(bkg_simulation, bins=_bins)
signal_sample, _ = np.histogram(signal_simulation, bins=_bins)
# Generate observations
signal_events = np.random.normal(5, 1.0, int(n_signal * 0.8))
bkg_events = 10 * np.random.random(int(n_bkg + np.sqrt(n_bkg)))
observed_events = np.array(signal_events.tolist() + bkg_events.tolist())
observed_sample, _ = np.histogram(observed_events, bins=_bins)
# Visualize the simulation and observations
fig, ax = plt.subplots()
fig.set_size_inches(7, 5)
plot_hist(ax, _bins, bkg_sample, label="Background")
plot_hist(ax, _bins, signal_sample, bottom=bkg_sample, label="Signal")
plot_data(ax, _bins, observed_sample)
ax.legend(loc="best")
ax.set_ylim(top=np.max(observed_sample) * 1.4)
ax.set_xlabel("Observable")
ax.set_ylabel("Count")
fig.savefig("components.png")
# Build the model
bkg_uncerts = np.sqrt(bkg_sample)
model = pyhf.simplemodels.hepdata_like(
signal_data=signal_sample.tolist(),
bkg_data=bkg_sample.tolist(),
bkg_uncerts=bkg_uncerts.tolist(),
)
data = pyhf.tensorlib.astensor(observed_sample.tolist() + model.config.auxdata)
# Perform inference
fit_result = pyhf.infer.mle.fit(data, model, return_uncertainties=True)
bestfit_pars, par_uncerts = fit_result.T
print(
f"best fit parameters:\
\n * signal strength: {bestfit_pars[0]} +/- {par_uncerts[0]}\
\n * nuisance parameters: {bestfit_pars[1:]}\
\n * nuisance parameter uncertainties: {par_uncerts[1:]}"
)
# Perform hypothesis test scan
_start = 0.0
_stop = 5
_step = 0.1
poi_tests = np.arange(_start, _stop + _step, _step)
print("\nPerforming hypothesis tests\n")
hypo_tests = [
pyhf.infer.hypotest(
mu_test,
data,
model,
return_expected_set=True,
return_test_statistics=True,
qtilde=True,
)
for mu_test in poi_tests
]
# Upper limits on signal strength
results = invert_interval(poi_tests, hypo_tests)
print(f"Observed Limit on µ: {results['obs']:.2f}")
print("-----")
for idx, n_sigma in enumerate(np.arange(-2, 3)):
print(
"Expected {}Limit on µ: {:.3f}".format(
" " if n_sigma == 0 else "({} σ) ".format(n_sigma),
results["exp"][idx],
)
)
# Visualize the "Brazil band"
fig, ax = plt.subplots()
fig.set_size_inches(7, 5)
ax.set_title("Hypothesis Tests")
ax.set_ylabel(r"$\mathrm{CL}_{s}$")
ax.set_xlabel(r"$\mu$")
pyhf.contrib.viz.brazil.plot_results(ax, poi_tests, hypo_tests)
fig.savefig("brazil_band.png")
if __name__ == "__main__":
main()
which when run gives
(question) $ python answer.py
best fit parameters:
* signal strength: 1.5884737977889158 +/- 0.7803435235862329
* nuisance parameters: [0.99020988 1.06040191 0.90488207 1.03531383 1.09093327 1.00942088
1.07789316 1.01125627 1.06202964 0.95780043 0.94990993 1.04893286
1.0560711 0.9758487 0.93692481 1.04683181 1.05785515 0.92381263
0.93812855 0.96751869]
* nuisance parameter uncertainties: [0.06966439 0.07632218 0.0611428 0.07230328 0.07872258 0.06899675
0.07472849 0.07403246 0.07613661 0.08606657 0.08002775 0.08655314
0.07564512 0.07308117 0.06743479 0.07383134 0.07460864 0.06632003
0.06683251 0.06270965]
Performing hypothesis tests
/home/stackoverflow/.venvs/question/lib/python3.7/site-packages/pyhf/infer/calculators.py:229: RuntimeWarning: invalid value encountered in double_scalars
teststat = (qmu - qmu_A) / (2 * self.sqrtqmuA_v)
Observed Limit on µ: 2.89
-----
Expected (-2 σ) Limit on µ: 0.829
Expected (-1 σ) Limit on µ: 1.110
Expected Limit on µ: 1.542
Expected (1 σ) Limit on µ: 2.147
Expected (2 σ) Limit on µ: 2.882
Let us know if you have any further questions!
After some research, I was able to predict the future value using the LSTM code below. I have also attached the Dmd1ahr.csv file in the github link that I am using.
https://github.com/ukeshchawal/hello-world/blob/master/Dmd1ahr.csv
As you all can see below, 90 data points are training sets and 91st to 100th are future value prediction.
However some of the questions that I still have are:
In order to predict these values I had to originally take more than hundred data sets (here, I have taken 500 data sets) which is not exactly what my primary goal is. Is there a way that given 500 data sets, it will predict the rest 10 or 20 out of sample data points? If yes, will you please write me a sample code where you can just take 500 data points from Dmd1ahr.csv file attached below and it will predict some future values (say 501 to 520) based on those 500 points?
The prediction are way off compared to the one who have in your blogs (definitely indicates for parameter tuning - I tried changing epochs, LSTM layers, Activation, optimizer). What other parameter tuning I can do to make it more robust?
Thank you'll in advance.
import numpy as np
import matplotlib.pyplot as plt
import pandas
# By twaking the architecture it could be made more robust
np.random.seed(7)
numOfSamples = 500
lengthTrain = 90
lengthValidation = 100
look_back = 1 # Can be set higher, in my experiments it made performance worse though
transientTime = 90 # Time to "burn in" time series
series = pandas.read_csv('Dmd1ahr.csv')
def generateTrainData(series, i, look_back):
return series[i:look_back+i+1]
trainX = np.stack([generateTrainData(series, i, look_back) for i in range(lengthTrain)])
testX = np.stack([generateTrainData(series, lengthTrain + i, look_back) for i in range(lengthValidation)])
trainX = trainX.reshape((lengthTrain,look_back+1,1))
testX = testX.reshape((lengthValidation, look_back + 1, 1))
trainY = trainX[:,1:,:]
trainX = trainX[:,:-1,:]
testY = testX[:,1:,:]
testX = testX[:,:-1,:]
############### Build Model ###############
import keras
from keras.models import Model
from keras import layers
from keras import regularizers
inputs = layers.Input(batch_shape=(1,look_back,1), name="main_input")
inputsAux = layers.Input(batch_shape=(1,look_back,1), name="aux_input")
# this layer makes the actual prediction, i.e. decides if and how much it goes up or down
x = layers.recurrent.LSTM(300,return_sequences=True, stateful=True)(inputs)
x = layers.recurrent.LSTM(200,return_sequences=True, stateful=True)(inputs)
x = layers.recurrent.LSTM(100,return_sequences=True, stateful=True)(inputs)
x = layers.recurrent.LSTM(50,return_sequences=True, stateful=True)(inputs)
x = layers.wrappers.TimeDistributed(layers.Dense(1, activation="linear",
kernel_regularizer=regularizers.l2(0.005),
activity_regularizer=regularizers.l1(0.005)))(x)
# auxillary input, the current input will be feed directly to the output
# this way the prediction from the step before will be used as a "base", and the Network just have to
# learn if it goes a little up or down
auxX = layers.wrappers.TimeDistributed(layers.Dense(1,
kernel_initializer=keras.initializers.Constant(value=1),
bias_initializer='zeros',
input_shape=(1,1), activation="linear", trainable=False
))(inputsAux)
outputs = layers.add([x, auxX], name="main_output")
model = Model(inputs=[inputs, inputsAux], outputs=outputs)
model.compile(optimizer='adam',
loss='mean_squared_error',
metrics=['mean_squared_error'])
#model.summary()
#model.fit({"main_input": trainX, "aux_input": trainX[look_back-1,look_back,:]},{"main_output": trainY}, epochs=4, batch_size=1, shuffle=False)
model.fit({"main_input": trainX, "aux_input": trainX[:,look_back-1,:].reshape(lengthTrain,1,1)},{"main_output": trainY}, epochs=100, batch_size=1, shuffle=False)
############### make predictions ###############
burnedInPredictions = np.zeros(transientTime)
testPredictions = np.zeros(len(testX))
# burn series in, here use first transitionTime number of samples from test data
for i in range(transientTime):
prediction = model.predict([np.array(testX[i, :, 0].reshape(1, look_back, 1)), np.array(testX[i, look_back - 1, 0].reshape(1, 1, 1))])
testPredictions[i] = prediction[0,0,0]
burnedInPredictions[:] = testPredictions[:transientTime]
# prediction, now dont use any previous data whatsoever anymore, network just has to run on its own output
for i in range(transientTime, len(testX)):
prediction = model.predict([prediction, prediction])
testPredictions[i] = prediction[0,0,0]
# for plotting reasons
testPredictions[:np.size(burnedInPredictions)-1] = np.nan
############### plot results ###############
#import matplotlib.pyplot as plt
plt.plot(testX[:, 0, 0])
plt.show()
plt.plot(burnedInPredictions, label = "training")
plt.plot(testPredictions, label = "prediction")
plt.legend()
plt.show()
pymc.__version__ = '3.0'
theano.__version__ = '0.6.0.dev-RELEASE'
I'm trying to use PyMC3 with a complex likelihood function:
First question: Is this possible?
Here's my attempt using Thomas Wiecki's post as a guide:
import numpy as np
import theano as th
import pymc as pm
import scipy as sp
# Actual data I'm trying to fit
x = np.array([52.08, 58.44, 60.0, 65.0, 65.10, 66.0, 70.0, 87.5, 110.0, 126.0])
y = np.array([0.522, 0.659, 0.462, 0.720, 0.609, 0.696, 0.667, 0.870, 0.889, 0.919])
yerr = np.array([0.104, 0.071, 0.138, 0.035, 0.102, 0.096, 0.136, 0.031, 0.024, 0.035])
th.config.compute_test_value = 'off'
a = th.tensor.dscalar('a')
with pm.Model() as model:
# Priors
alpha = pm.Normal('alpha', mu=0.3, sd=5)
sig_alpha = pm.Normal('sig_alpha', mu=0.03, sd=5)
t_double = pm.Normal('t_double', mu=4, sd=20)
t_delay = pm.Normal('t_delay', mu=21, sd=20)
nu = pm.Uniform('nu', lower=0, upper=20)
# Some functions needed for calculation of the y estimator
def T(eqd):
doses = np.array([52.08, 58.44, 60.0, 65.0, 65.10,
66.0, 70.0, 87.5, 110.0, 126.0])
tmt_times = np.array([29,29,43,29,36,48,22,11,7,8])
return np.interp(eqd, doses, tmt_times)
def TCP(a):
time = T(x)
BCP = pm.exp(-1E7*pm.exp(-alpha*x*1.2 + 0.69315/t_delay(time-t_double)))
return pm.prod(BCP)
def normpdf(a, alpha, sig_alpha):
return 1./(sig_alpha*pm.sqrt(2.*np.pi))*pm.exp(-pm.sqr(a-alpha)/(2*pm.sqr(sig_alpha)))
def normcdf(a, alpha, sig_alpha):
return 1./2.*(1+pm.erf((a-alpha)/(sig_alpha*pm.sqrt(2))))
def integrand(a):
return normpdf(a,alpha,sig_alpha)/(1.-normcdf(0,alpha,sig_alpha))*TCP(a)
func = th.function([a,alpha,sig_alpha,t_double,t_delay], integrand(a))
y_est = sp.integrate.quad(func(a, alpha, sig_alpha,t_double,t_delay), 0, np.inf)[0]
likelihood = pm.T('TCP', mu=y_est, nu=nu, observed=y_tcp)
start = pm.find_MAP()
step = pm.NUTS(state=start)
trace = pm.sample(2000, step, start=start, progressbar=True)
which produces the following message regarding the expression for y_est:
TypeError: ('Bad input argument to theano function with name ":42" at index 0(0-based)', 'Expected an array-like object, but found a Variable: maybe you are trying to call a function on a (possibly shared) variable instead of a numeric array?')
I've overcome various other hurdles to get this far, and this is where I'm stuck. So, provided the answer to my first question is 'yes', then am I on the right track? Any guidance would be helpful!
N.B. Here is a similar question I found, and another.
Disclaimer: I'm very new at this. My only previous experience is successfully reproducing the linear regression example in Thomas' post. I've also successfully run the Theano test suite, so I know it works.
Yes, its possible to make something with a complex or arbitrary likelihood. Though that doesn't seem like what you're doing here. It looks like you have a complex transformation of one variable into another, the integration step.
Your particular exception is that integrate.quad is expecting a numpy array, not a pymc Variable. If you want to do quad within pymc, you'll have to make a custom theano Op (with derivative) for it.
I have this problem: I have a cohort of individuals grouped in 5 age groups. Initially all of them are susceptible and then they develop disease and finally they have cancers. I have information about the age group distribution of the susceptible and then the cancer carrier. Between the susceptible state and the cancer they pass through 7 stages , with same transition rate.
I'm trying to create a model that simulate each transition as a binomial extraction and fit the data I have.
I tried something but in the moment of analysing the traces , nothing work
You can see the code
Where am I getting wrong?
Thanks for any help
from pylab import *
from pymc import *
from pymc.Matplot import plot as plt
#susceptible_data = array([647,1814,8838,9949,1920])
susceptible_data = array([130,398,1415,1303,206])
infected_data_100000 = array([0,197,302,776,927])
infected_data = array([0,7,38,90,17])
prior_values=np.zeros(len(infected_data))
for i in range(0,len(infected_data)):
prior_values[i]=infected_data[i]/susceptible_data[i]
# stochastic priors
beta1 = Uniform('beta1', 0., 1.)
lambda_0_temp=susceptible_data[0]
lambda_0_1=pymc.Binomial("lambda_0_1",lambda_0_temp,pow(beta1,1))
lambda_0_2=pymc.Binomial("lambda_0_2",lambda_0_1.value,pow(beta1,1))
lambda_0_3=pymc.Binomial("lambda_0_3",lambda_0_2.value,pow(beta1,1))
lambda_0_4=pymc.Binomial("lambda_0_4",lambda_0_3.value,pow(beta1,1))
lambda_0_5=pymc.Binomial("lambda_0_5",lambda_0_4.value,pow(beta1,1))
lambda_0_6=pymc.Binomial("lambda_0_6",lambda_0_5.value,pow(beta1,1))
lambda_0_7=pymc.Binomial("lambda_0_7",n=lambda_0_6.value,p=pow(beta1,1),value=infected_data[0],observed=True)
lambda_1_temp=susceptible_data[1]
lambda_1_1=pymc.Binomial("lambda_1_1",lambda_1_temp,pow(beta1,1))
lambda_1_2=pymc.Binomial("lambda_1_2",lambda_1_1.value,pow(beta1,1))
lambda_1_3=pymc.Binomial("lambda_1_3",lambda_1_2.value,pow(beta1,1))
lambda_1_4=pymc.Binomial("lambda_1_4",lambda_1_3.value,pow(beta1,1))
lambda_1_5=pymc.Binomial("lambda_1_5",lambda_1_4.value,pow(beta1,1))
lambda_1_6=pymc.Binomial("lambda_1_6",lambda_1_5.value,pow(beta1,1))
lambda_1_7=pymc.Binomial("lambda_1_7",n=lambda_1_6.value,p=pow(beta1,1),value=infected_data[1],observed=True)
lambda_2_temp=susceptible_data[2]
lambda_2_1=pymc.Binomial("lambda_2_1",lambda_2_temp,pow(beta1,1))
lambda_2_2=pymc.Binomial("lambda_2_2",lambda_2_1.value,pow(beta1,1))
lambda_2_3=pymc.Binomial("lambda_2_3",lambda_2_2.value,pow(beta1,1))
lambda_2_4=pymc.Binomial("lambda_2_4",lambda_2_3.value,pow(beta1,1))
lambda_2_5=pymc.Binomial("lambda_2_5",lambda_2_4.value,pow(beta1,1))
lambda_2_6=pymc.Binomial("lambda_2_6",lambda_2_5.value,pow(beta1,1))
lambda_2_7=pymc.Binomial("lambda_2_7",n=lambda_2_6.value,p=pow(beta1,1),value=infected_data[2],observed=True)
lambda_3_temp=susceptible_data[3]
lambda_3_1=pymc.Binomial("lambda_3_1",lambda_3_temp,pow(beta1,1))
lambda_3_2=pymc.Binomial("lambda_3_2",lambda_3_1.value,pow(beta1,1))
lambda_3_3=pymc.Binomial("lambda_3_3",lambda_3_2.value,pow(beta1,1))
lambda_3_4=pymc.Binomial("lambda_3_4",lambda_3_3.value,pow(beta1,1))
lambda_3_5=pymc.Binomial("lambda_3_5",lambda_3_4.value,pow(beta1,1))
lambda_3_6=pymc.Binomial("lambda_3_6",lambda_3_5.value,pow(beta1,1))
lambda_3_7=pymc.Binomial("lambda_3_7",n=lambda_3_6.value,p=pow(beta1,1),value=infected_data[3],observed=True)
lambda_4_temp=susceptible_data[4]
lambda_4_1=pymc.Binomial("lambda_4_1",lambda_4_temp,pow(beta1,1))
lambda_4_2=pymc.Binomial("lambda_4_2",lambda_4_1.value,pow(beta1,1))
lambda_4_3=pymc.Binomial("lambda_4_3",lambda_4_2.value,pow(beta1,1))
lambda_4_4=pymc.Binomial("lambda_4_4",lambda_4_3.value,pow(beta1,1))
lambda_4_5=pymc.Binomial("lambda_4_5",lambda_4_4.value,pow(beta1,1))
lambda_4_6=pymc.Binomial("lambda_4_6",lambda_4_5.value,pow(beta1,1))
lambda_4_7=pymc.Binomial("lambda_4_7",n=lambda_4_6.value,p=pow(beta1,1),value=infected_data[4],observed=True)
model=pymc.Model([lambda_0_7,lambda_1_7,lambda_2_7,lambda_3_7,lambda_4_7,beta1])
mcmc =pymc.MCMC(model)
mcmc.sample(iter=100000, burn=50000, thin=10, verbose=1)
lambda_0_samples=mcmc.trace('lambda_0_7')[:]
lambda_1_samples=mcmc.trace('lambda_1_7')[:]
lambda_2_samples=mcmc.trace('lambda_2_7')[:]
lambda_3_samples=mcmc.trace('lambda_3_7')[:]
lambda_4_samples=mcmc.trace('lambda_4_7')[:]
beta1_samples=mcmc.trace('beta1')[:]
What you have implemented above only associates data with the 7th distribution in each set; the others are seemingly-redundant hierarchies on the binomial probability. I would think you want data informing each stage. I'm not sure there is information to inform what the values of p should be at each stage, based on what is provided.
In their paper describing Viola-Jones object detection framework (Robust Real-Time Face Detection by Viola and Jones), it is said:
All example sub-windows used for training were variance normalized to minimize the effect of different lighting conditions.
My question is "How to implement image normalization in Octave?"
I'm NOT looking for the specific implementation that Viola & Jones used but a similar one that produces almost the same output. I've been following a lot of haar-training tutorials(trying to detect a hand) but not yet able to output a good detector(xml).
I've tried contacting the authors, but still no response yet.
I already answered how to to it in general guidelines in this thread.
Here is how to do method 1 (normalizing to standard normal deviation) in octave (Demonstrating for a random matrix A, of course can be applied to any matrix, which is how the picture is represented):
>>A = rand(5,5)
A =
0.078558 0.856690 0.077673 0.038482 0.125593
0.272183 0.091885 0.495691 0.313981 0.198931
0.287203 0.779104 0.301254 0.118286 0.252514
0.508187 0.893055 0.797877 0.668184 0.402121
0.319055 0.245784 0.324384 0.519099 0.352954
>>s = std(A(:))
s = 0.25628
>>u = mean(A(:))
u = 0.37275
>>A_norn = (A - u) / s
A_norn =
-1.147939 1.888350 -1.151395 -1.304320 -0.964411
-0.392411 -1.095939 0.479722 -0.229316 -0.678241
-0.333804 1.585607 -0.278976 -0.992922 -0.469159
0.528481 2.030247 1.658861 1.152795 0.114610
-0.209517 -0.495419 -0.188723 0.571062 -0.077241
In the above you use:
To get the standard deviation of the matrix: s = std(A(:))
To get the mean value of the matrix: u = mean(A(:))
And then following the formula A'[i][j] = (A[i][j] - u)/s with the
vectorized version: A_norm = (A - u) / s
Normalizing it with vector normalization is also simple:
>>abs = sqrt((A(:))' * (A(:)))
abs = 2.2472
>>A_norm = A / abs
A_norm =
0.034959 0.381229 0.034565 0.017124 0.055889
0.121122 0.040889 0.220583 0.139722 0.088525
0.127806 0.346703 0.134059 0.052637 0.112369
0.226144 0.397411 0.355057 0.297343 0.178945
0.141980 0.109375 0.144351 0.231000 0.157065
In the abvove:
abs is the absolute value of the vector (its length), which is calculated with vectorized multiplications (A(:)' * A(:) is actually sum(A[i][j]^2))
Then we use it to normalize the vector so it will be of length 1.