There are various levels on which to debug a model. One of the simplest is to just print out the values that different variables are taking on.
Because PyMC3
uses Theano
expressions to build the model, and not functions, there is no way to place a print
statement into a likelihood function. Instead, you can use the Theano
Print
operatator. For more information, see: theano Print operator for this before: http://deeplearning.net/software/theano/tutorial/debug_faq.html#how-do-i-print-an-intermediate-value-in-a-function.
Let's build a simple model with just two parameters:
%matplotlib inline
import pymc3 as pm
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
import pandas as pd
import theano.tensor as tt
x = np.random.randn(100)
with pm.Model() as model:
mu = pm.Normal('mu', mu=0, sd=1)
sd = pm.Normal('sd', mu=0, sd=1)
obs = pm.Normal('obs', mu=mu, sd=sd, observed=x)
step = pm.Metropolis()
trace = pm.sample(5000, step)
pm.traceplot(trace);
Multiprocess sampling (2 chains in 2 jobs) CompoundStep >Metropolis: [sd] >Metropolis: [mu] 100%|██████████| 5500/5500 [00:02<00:00, 2358.30it/s]
Hm, looks like something has gone wrong, but what? Let's look at the values getting proposed using the Print
operator:
with pm.Model() as model:
mu = pm.Normal('mu', mu=0, sd=1)
sd = pm.Normal('sd', mu=0, sd=1)
mu_print = tt.printing.Print('mu')(mu)
sd_print = tt.printing.Print('sd')(sd)
obs = pm.Normal('obs', mu=mu_print, sd=sd_print, observed=x)
step = pm.Metropolis()
trace = pm.sample(3, step, tune=0, chains=1, progressbar=False) # Make sure not to draw too many samples
mu __str__ = 0.0 sd __str__ = 0.0
Sequential sampling (1 chains in 1 job) CompoundStep >Metropolis: [sd] >Metropolis: [mu]
sd __str__ = 1.4406091401518182 mu __str__ = 0.0 sd __str__ = 0.0 sd __str__ = 0.0 mu __str__ = -0.7642399016472413 mu __str__ = 0.0 sd __str__ = -0.12899381409884816 mu __str__ = 0.0 sd __str__ = 0.0 sd __str__ = 0.0 mu __str__ = 1.2614994475439572 mu __str__ = 0.0 sd __str__ = 0.7329434719007808 mu __str__ = 0.0 sd __str__ = 0.0 sd __str__ = 0.0 mu __str__ = 0.9331983611718919 mu __str__ = 0.0
In the code above, we set the tune=0, chains=1, progressbar=False
in the pm.sample
, this is done so that the output is cleaner.
Looks like sd
is always 0
which will cause the logp to go to -inf
. Of course, we should not have used a prior that has negative mass for sd
but instead something like a HalfNormal
.
We can also redirect the output to a string buffer and access the proposed values later on (thanks to Lindley Lentati for providing this example):
from io import StringIO
import sys
x = np.random.randn(100)
old_stdout = sys.stdout
mystdout = sys.stdout = StringIO()
with pm.Model() as model:
mu = pm.Normal('mu', mu=0, sd=1)
sd = pm.Normal('sd', mu=0, sd=1)
mu_print = tt.printing.Print('mu')(mu)
sd_print = tt.printing.Print('sd')(sd)
obs = pm.Normal('obs', mu=mu_print, sd=sd_print, observed=x)
step = pm.Metropolis()
trace = pm.sample(5, step, tune=0, chains=1, progressbar=False) # Make sure not to draw too many samples
sys.stdout = old_stdout
output = mystdout.getvalue().split('\n')
mulines = [s for s in output if 'mu' in s]
muvals = [line.split()[-1] for line in mulines]
plt.plot(np.arange(0, len(muvals)), muvals)
plt.xlabel('proposal iteration')
plt.ylabel('mu value');
Sequential sampling (1 chains in 1 job) CompoundStep >Metropolis: [sd] >Metropolis: [mu]
trace['mu']
array([ 0., 0., 0., 0., 0.])
Notice that for each iteration, 3 values were printed and recorded. The printed values are the original value (last sample), the proposed value and the accepted value. Plus the starting value in the very beginning, we recorded in total 1+3*5=16
value above.