mdmpy 0.0.15.18

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MDM Py

This package is a Python implementation of Marginal Distribution Models (MDMs), which can be used in Discrete Choice Modelling.

Documentation

Documentation is kindly hosted by Read The Docs.

Install

This package is uploaded to PyPI. Hence,

pip install mdmpy

should work.

How to use

Simplest Case

Gradient Descent

In the simplest case, we will use the Multinomial Logit (MNL) model, which is used as a default. Assuming numpy, scipy and pandas are installed, we generate choice data assuming a random utility model:

from string import ascii_uppercase as letters
import pandas as pd
import scipy.stats as stats
import numpy as np

NUM_INDIV   = 57
NUM_CHOICES = 3
NUM_ATTR    = 4

np.random.seed(2019)
X = np.random.random((NUM_ATTR, NUM_INDIV * NUM_CHOICES))
true_beta = np.random.random(NUM_ATTR)
V = np.dot(true_beta.T, X)
V = np.reshape(V, (NUM_INDIV,NUM_CHOICES))
eps = stats.gumbel_r.rvs(size=NUM_INDIV * NUM_CHOICES)
eps = np.reshape(eps, (NUM_INDIV, NUM_CHOICES))
U = V + eps
highest_util = np.argmax(U, 1)

df = pd.DataFrame(X.T)
df['choice'] = [1 if idx == x else 0 for idx in highest_util for x in range(NUM_CHOICES)]
df['individual'] = [indiv for indiv in range(NUM_INDIV) for _ in range(NUM_CHOICES)]
df['altvar'] = [altlvl for _ in range(NUM_INDIV) for altlvl in letters[:NUM_CHOICES]]

With this package, we will assume that df is the dataframe which is simply given to us. Instead of having the code itself find out how many individuals, choices and coefficients or attributes there are, we will simply feed them into the class. To perform a gradient descent with this class, we will use the grad_desc method, using the df from above as input,

import mdmpy

# In a typical case one would load df before this line
mdm = mdmpy.MDM(df, 4, 3, [0, 1, 2, 3])
np.random.seed(4)
init_beta = np.random.random(4)
grad_beta = mdm.grad_desc(init_beta)
print(grad_beta)
# expected output [0.30238122 0.07955214 0.86779824 0.50951981]

Solver

The MDM class acts as a wrapper and adds the necessary pyomo variables and sets to model the problem, but requires a solver. IPOPT, an interior point solver, is recommended. If you have such a solver, it can be called. Assuming IPOPT is being used:

import mdmpy

ipopt_exec_path = /path/to/ipopt # Replace with proper path
mdm = mdmpy.MDM(df, 4, 3, [0, 1, 2, 3])
mdm.model_init()
mdm.model_solve("ipopt",ipopt_exec_path)
print([mdm.m.beta[idx].value for idx in mdm.m.beta])
# expected output [0.30238834989235025, 0.07953888508425154, 0.8678050334295714, 0.5095096796373667]