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mooc-discrete-choice's Issues

The last request for help

Hoping I get help from anyone in charge of this platform I repost my issue and any help is appreciated and the platform is really meant to bring change in using biogeme software.

Any help is greatly appreciated.
I have difficulty estimating mixed logit model based on the following
information and need help.I tried to go through the biogeme documentation
and unable to see discussion related with unlabelled format modeling using
mixed logit model.
Unlabelled format with three alternatives(1,2,3)represented with
Concept,third alternative is the status quo;Attributes are water
supply(levels 1,2,3),Water quality(two levels:1=yes,0=otherwise),and
prices(5 levels:3,5,10,15 and 20).I want to apply mixed logit with error
components and Quasi-Monte Carlo Simulation,code given below.Any help
please.

Assume you have a pandas DataFrame df with your data

database = db.Database("myDatabase", df)

Use the names of the variables in your data

chosen_alternative = database.variables["Chosen"]
#water_supply = database.variables["Wsup"]
#water_quality = database.variables["Wqua"]
#price_increase = database.variables["Price"]

Define parameters to be estimated (betas)

#beta1 = ex.Beta('beta1', 0, None, None, 0)
#beta2 = ex.Beta('beta2', 0, None, None, 0)
#beta3 = ex.Beta('beta3', 0, None, None, 0)
#beta3_stddev = ex.Beta('beta3_stddev', 0.5, None, None, 0)

Define the alternative specific constants

ASC1 = ex.Beta('ASC1', 1, None, None, 0)
ASC2 = ex.Beta('ASC2', 1, None, None, 0)

Define parameters to be estimated

beta1_mean = ex.Beta('beta1_mean', 0, -1000, 1000, 0) # Mean of the
distribution for beta1
beta1_stddev = ex.Beta('beta1_stddev', 0, None, None, 0) # Standard
deviation of the distribution for beta1
beta1 = beta1_mean + beta1_stddev * ex.bioDraws('beta1', 'UNIFORM')

beta2_mean = ex.Beta('beta2_mean', 0, None, None, 0) # Parameter for
water_quality
beta2_stddev = ex.Beta('beta2_stddev', 0, None, None, 0) # Standard
deviation of the distribution for beta2
beta2 = beta2_mean + beta2_stddev * ex.bioDraws('beta2', 'NORMAL')

beta3_mean = ex.Beta('beta3_mean', 0, None, None, 1) # Parameter for
price_increase
beta3_stddev = ex.Beta('beta3_stddev', 0, None, None, 0) # Standard
deviation of the distribution for beta3
beta3 = beta3_mean + beta3_stddev * ex.bioDraws('beta3', 'UNIFORM')

Define utility functions

V0 = beta1 * water_supply + beta2 * water_quality + beta3 * price_increase
V1 =ASC1 + beta1 * water_supply + beta2 * water_quality + beta3 *
price_increase
V2 =ASC1 + beta1 * water_supply + beta2 * water_quality + beta3 *
price_increase

Associate utility functions with the numbering of alternatives

V = {1: V0, 2: V1,3:V2}

Define the availability of alternatives

You should adjust these based on the specific structure of your data

Define the availability of alternatives

#av = {1: database.variables['Concept'] == 1,

2: database.variables['Concept'] == 2,

3: database.variables['Concept'] == 3}

av = {1:1,2:1,3:1}
num_draws = 10 # Number of Halton draws
draws = ex.bioDraws('draws', 'HALTON')

Estimate the parameters with a maximum of 1000 iterations

#results = biogeme.estimate(maxiter=1000)

Define the model. The choice model is a logit, with availability

conditions
prob = models.logit(V, av, chosen_alternative)

Define the log likelihood function

logprob = ex.MonteCarlo(ex.log(prob))

Create the Biogeme object

biogeme = bio.BIOGEME(database, logprob)

Estimate the parameters

#results = biogeme.estimate()

Get the estimated parameters

#parameters = results.getEstimatedParameters()

Print the parameters

#print(parameters)

Estimate the parameters

results = biogeme.estimate()

Get the estimated parameters

parameters = results.getEstimatedParameters()

Print the parameters

print(parameters)
โ€ฆ

Mixed logit model

while trying to estimate mixed logit model I encountered the following error.Any help is greatly appreciated.
BiogemeError: The norm of the gradient is inf: g=[-1.79769313e+308 -1.79769313e+308 -1.79769313e+308 -1.79769313e+308
-1.79769313e+308]

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