Getting started with PYMC for linear regression

Question

Thought I'd start off following this example: http://www.databozo.com/2014/01/17/Exploring_PyMC3.html

But when I follow the example precisely using pymc 2.3 I get an exit and told that the API has changed UserWarning: The MCMC() syntax is deprecated. Please pass in nodes explicitly via M = MCMC(input). 'The MCMC() syntax is deprecated. Please pass in nodes explicitly via M = MCMC(input).') but I don't have a good idea how to change the example to provide exactly what to the model function and how to do with the 'with' clause?

The code in question is:

%pylab inline
import scipy
import numpy as np
x = np.array(range(0,50))
y = np.random.uniform(low=0.0, high=40.0, size=200)
y = map((lambda a: a[0] + a[1]), zip(x,y))

import matplotlib.pyplot as plt
plt.scatter(x,y)

Above sample data generator works fine

import pymc as pm
import numpy as np

trace = None
with pm.Model() as model:         <<<<<<indicated as the error line
    alpha = pm.Normal('alpha', mu=0, sd=20)
    beta = pm.Normal('beta', mu=0, sd=20)
    sigma = pm.Uniform('sigma', lower=0, upper=20)

    y_est = alpha + beta * x

    likelihood = pm.Normal('y', mu=y_est, sd=sigma, observed=y)

    start = pm.find_MAP()
    step = pm.NUTS(state=start)
    trace = pm.sample(2000, step, start=start, progressbar=False)

    pm.traceplot(trace);

dartdog · Accepted Answer · 2014-02-12 21:53:21Z

1

Package author @fonnesbeck informed me that I needed the development version 3 from Github and not the pypi version 2.3. Installed that, via github and I'm good to go. Open source is great!

answered Feb 12, 2014 at 21:53

dartdog

10.9k22 gold badges75 silver badges122 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Getting started with PYMC for linear regression

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related