How to normalize an image using Octave? - algorithm

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.

Related

Trying to put together a teaching-example with pyhf

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!

Error in setting max features parameter in Isolation Forest algorithm using sklearn

I'm trying to train a dataset with 357 features using Isolation Forest sklearn implementation. I can successfully train and get results when the max features variable is set to 1.0 (the default value).
However when max features is set to 2, it gives the following error:
ValueError: Number of features of the model must match the input.
Model n_features is 2 and input n_features is 357
It also gives the same error when the feature count is 1 (int) and not 1.0 (float).
How I understood was that when the feature count is 2 (int), two features should be considered in creating each tree. Is this wrong? How can I change the max features parameter?
The code is as follows:
from sklearn.ensemble.iforest import IsolationForest
def isolation_forest_imp(dataset):
estimators = 10
samples = 100
features = 2
contamination = 0.1
bootstrap = False
random_state = None
verbosity = 0
estimator = IsolationForest(n_estimators=estimators, max_samples=samples, contamination=contamination,
max_features=features,
bootstrap=boostrap, random_state=random_state, verbose=verbosity)
model = estimator.fit(dataset)
In the documentation it states:
max_features : int or float, optional (default=1.0)
The number of features to draw from X to train each base estimator.
- If int, then draw `max_features` features.
- If float, then draw `max_features * X.shape[1]` features.
So, 2 should mean take two features and 1.0 should mean take all of the features, 0.5 take half and so on, from what I understand.
I think this could be a bug, since, taking a look in IsolationForest's fit:
# Isolation Forest inherits from BaseBagging
# and when _fit is called, BaseBagging takes care of the features correctly
super(IsolationForest, self)._fit(X, y, max_samples,
max_depth=max_depth,
sample_weight=sample_weight)
# however, when after _fit the decision_function is called using X - the whole sample - not taking into account the max_features
self.threshold_ = -sp.stats.scoreatpercentile(
-self.decision_function(X), 100. * (1. - self.contamination))
then:
# when the decision function _validate_X_predict is called, with X unmodified,
# it calls the base estimator's (dt) _validate_X_predict with the whole X
X = self.estimators_[0]._validate_X_predict(X, check_input=True)
...
# from tree.py:
def _validate_X_predict(self, X, check_input):
"""Validate X whenever one tries to predict, apply, predict_proba"""
if self.tree_ is None:
raise NotFittedError("Estimator not fitted, "
"call `fit` before exploiting the model.")
if check_input:
X = check_array(X, dtype=DTYPE, accept_sparse="csr")
if issparse(X) and (X.indices.dtype != np.intc or
X.indptr.dtype != np.intc):
raise ValueError("No support for np.int64 index based "
"sparse matrices")
# so, this check fails because X is the original X, not with the max_features applied
n_features = X.shape[1]
if self.n_features_ != n_features:
raise ValueError("Number of features of the model must "
"match the input. Model n_features is %s and "
"input n_features is %s "
% (self.n_features_, n_features))
return X
So, I am not sure on how you can handle this. Maybe figure out the percentage that leads to just the two features you need - even though I am not sure it'll work as expected.
Note: I am using scikit-learn v.0.18
Edit: as #Vivek Kumar commented this is an issue and upgrading to 0.20 should do the trick.

pymc3 Multi-category Bayesian network - how to sample?

I have set up a Bayes net with 3 states per node as below, and can read logp's for particular states from it (as in the code).
Next I would like to sample from it. In the code below, sampling runs but I don't see distributions over the three states in the outputs; rather, I see a mean and variance as if they were continuous nodes. How do I get the posterior over the three states?
import numpy as np
import pymc3 as mc
import pylab, math
model = mc.Model()
with model:
rain = mc.Categorical('rain', p = np.array([0.5, 0. ,0.5]))
sprinkler = mc.Categorical('sprinkler', p=np.array([0.33,0.33,0.34]))
CPT = mc.math.constant(np.array([ [ [.1,.2,.7], [.2,.2,.6], [.3,.3,.4] ],\
[ [.8,.1,.1], [.3,.4,.3], [.1,.1,.8] ],\
[ [.6,.2,.2], [.4,.4,.2], [.2,.2,.6] ] ]))
p_wetgrass = CPT[rain, sprinkler]
wetgrass = mc.Categorical('wetgrass', p_wetgrass)
#brute force search (not working)
for val_rain in range(0,3):
for val_sprinkler in range(0,3):
for val_wetgrass in range(0,3):
lik = model.logp(rain=val_rain, sprinkler=val_sprinkler, wetgrass=val_wetgrass )
print([val_rain, val_sprinkler, val_wetgrass, lik])
#sampling (runs but don't understand output)
if 1:
niter = 10000 # 10000
tune = 5000 # 5000
print("SAMPLING:")
#trace = mc.sample(20000, step=[mc.BinaryGibbsMetropolis([rain, sprinkler])], tune=tune, random_seed=124)
trace = mc.sample(20000, tune=tune, random_seed=124)
print("trace summary")
mc.summary(trace)
answering own question: the trace does contain the discrete values but the mc.summary(trace) function is set up to compute continuous mean and variance stats. To make a histogram of the discrete states, use h = hist(trace.get_values(sprinkler)) :-)

SPSS: Syntax for Calculating FSCORE from rotated factors

I found the proper syntax to import my centriod factor extraction into SPSS and rotate it. (The semi-bumbling tale is here.
The next issue is this: because of limitations in SPSS on what subcommands can be used when reading a matrix in (only /ROTATE and /EXTRACTION are permitted), I can't get the factor scores. SPSS displays this error:
"Factor scores cannot be computed with matrix input."
I still need to find a way to get the FSCORE of the newly rotated factors of each factor for all cases by running a regression using the newly rotated factors and saving the regression values as a new variable (/SAVE REG(ALL).
Ideas are welcome. Thank you for your expertise!
Assets: Dataset A of 36 cases and 74 variables (the basis of the centriod factor extraction); centriod factor extraction matrix
Here's the SPSS syntax that almost does what I need - except it uses PCA extraction instead of centroid.
FACTOR
/VARIABLES VAR00001 VAR00002 VAR00003 VAR00004 VAR00005 VAR00006 VAR00007 VAR00008 VAR00009 VAR00010 VAR00011 VAR00012 VAR00013 VAR00014 VAR00015 VAR00016 VAR00017 VAR00018 VAR00019 VAR00020 VAR00021 VAR00022 VAR00023 VAR00024 VAR00025 VAR00026 VAR00027 VAR00028 VAR00029 VAR00030 VAR00031 VAR00032 VAR00033 VAR00034 VAR00035 VAR00036 VAR00037 VAR00038 VAR00039 VAR00040 VAR00041 VAR00042 VAR00043 VAR00044 VAR00045 VAR00046 VAR00047 VAR00048 VAR00049 VAR00050 VAR00051 VAR00052 VAR00053 VAR00054 VAR00055 VAR00056 VAR00057 VAR00058 VAR00059 VAR00060 VAR00061 VAR00062 VAR00063 VAR00064 VAR00065 VAR00066 VAR00067 VAR00068 VAR00069 VAR00070 VAR00071 VAR00072 VAR00073 VAR00074
/MISSING LISTWISE
/ANALYSIS VAR00001 VAR00002 VAR00003 VAR00004 VAR00005 VAR00006 VAR00007 VAR00008 VAR00009 VAR00010 VAR00011 VAR00012 VAR00013 VAR00014 VAR00015 VAR00016 VAR00017 VAR00018 VAR00019 VAR00020 VAR00021 VAR00022 VAR00023 VAR00024 VAR00025 VAR00026 VAR00027 VAR00028 VAR00029 VAR00030 VAR00031 VAR00032 VAR00033 VAR00034 VAR00035 VAR00036 VAR00037 VAR00038 VAR00039 VAR00040 VAR00041 VAR00042 VAR00043 VAR00044 VAR00045 VAR00046 VAR00047 VAR00048 VAR00049 VAR00050 VAR00051 VAR00052 VAR00053 VAR00054 VAR00055 VAR00056 VAR00057 VAR00058 VAR00059 VAR00060 VAR00061 VAR00062 VAR00063 VAR00064 VAR00065 VAR00066 VAR00067 VAR00068 VAR00069 VAR00070 VAR00071 VAR00072 VAR00073 VAR00074
/PRINT INITIAL CORRELATION SIG DET INV REPR AIC EXTRACTION ROTATION FSCORE
/FORMAT BLANK(.544)
/CRITERIA FACTORS(6) ITERATE(80)
/EXTRACTION PC <---Here's the rub.
/CRITERIA ITERATE(80) DELTA(0)
/ROTATION OBLIMIN
/SAVE REG(ALL)
/METHOD=CORRELATION.
I referred to the MATRIX command in my reply to your other post (Rotations). You will need to research the appropriate equations for performing this calculation and set up the matrix algebra within a MATRIX END - MATRIX code block. Easy once you have the math right. I'm too busy/lazy to research and write it but this should provide a fertile lead.

Dealing with under flow while calculating GMM parameters using EM

I am currently runnuing training in matlab on a matrix of logspecrum samples I am constantly dealing with underflow problems.I understood that I need to work with log's in order to deal with underflowing.
I am still strugling with uderflow though , when i calculate the mean (mue) bucause it is negetive i cant work with logs so i need the real values that underflow.
These are equasions i am working with:
In MATLAB code i calulate log_tau in oreder avoid underflow but when calulating mue i need exp(log(tau)) which goes to zero.
I am attaching relevent MATLAB code
**in the code i called the variable alpha is tau ...
for i = 1 : 50
log_c = Logsum(log_alpha,1) - log(N);
c = exp(log_c);
mue = DataMat*alpha./(repmat(exp(Logsum(log_alpha,1)),FrameSize,1));
log_abs_mue = log(abs(mue));
log_SigmaSqr = log((DataMat.^2)*alpha) - repmat(Logsum(log_alpha,1),FrameSize,1) - 2*log_abs_mue;
SigmaSqr = exp(log_SigmaSqr);
for j=1:N
rep_DataMat(:,:,j) = repmat(DataMat(:,j),1,M);
log_gamma(j,:) = log_c - 0.5*(FrameSize*log(2*pi)+sum(log_SigmaSqr)) + sum((rep_DataMat(:,:,j) - mue).^2./(2*SigmaSqr));
end
log_alpha = log_gamma - repmat(Logsum(log_gamma,2),1,M);
alpha = exp(log_alpha);
end
c = exp(log_c);
SigmaSqr = exp(log_SigmaSqr);
does any one see how i can avoid this? or what needs to be fixed in code?
What i did was add this line to the MATLAB code:
mue(isnan(mue))=0; %fix 0/0 problem
and this one:
SigmaSqr(SigmaSqr==0)=1;%fix if mue_k = x_k
not sure if this is the best solution but is seems to work...
any have a better idea?

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