I am having trouble sampling from a Dirichlet/Multinomial distribution with pymc3.
I tried to create a simple test-case to recreate a Beta/Binomial using Dirichlet/Multinomial with n=2, but I can't get it to work.
Below I have some code that works for Binomial but fails for Multinomial.
One of the obvious differences is that the Multinomial model is more constrained:
i.e. to start, rating is set to 10 in the Binomial model, and [10,10] in the Multinomial.
The pymc3 Dirichlet code does say "Only the first k-1 elements of x are expected" but only arrays of shape 2 seem to work in my code.
The output shows that num_friends and rating are being sampled in the Binomial case, but not in the Multinomial case. friends_ratings is being sampled in both. Thanks!
Oh, also Dirichlet('d', np.array([1,1])) crashes with "Floating point error 8". It only appears to fail when two integers of value 1 are passed in. np.array([1.,1.]) works.
import pymc as pm
import numpy as np
print "TEST BINOMIAL"
with pm.Model() as model:
friends_ratings = pm.Beta('friends_ratings', alpha=1, beta=2)
num_friends = pm.DiscreteUniform('num_friends', lower=0, upper=100)
rating = pm.Binomial('rating', n=num_friends, p=friends_ratings)
step = pm.Metropolis([num_friends, friends_ratings, rating])
start = {"friends_ratings":.5, "num_friends":20, 'rating':10}
tr = pm.sample(5, step, start=start, progressbar=False)
print "friends", [tr[i]['num_friends'] for i in range(len(tr))]
print "friends_ratings", [tr[i]['friends_ratings'] for i in range(len(tr))]
print "rating", [tr[i]['rating'] for i in range(len(tr))]
print "TEST DIRICHLET"
with pm.Model() as model:
friends_ratings = pm.Dirichlet('friends_ratings', np.array([1.,1.]), shape=2)
num_friends = pm.DiscreteUniform('num_friends', lower=0, upper=100)
rating = pm.Multinomial('rating', n=num_friends, p=friends_ratings, shape=2)
step = pm.Metropolis([num_friends, friends_ratings, rating])
start = {'friends_ratings': np.array([0.5,0.5]), 'num_friends': 20, 'rating': [10,10]}
tr = pm.sample(5, step, start=start, progressbar=False)
print "friends", [tr[i]['num_friends'] for i in range(len(tr))]
print "friends_ratings", [tr[i]['friends_ratings'] for i in range(len(tr))]
print "rating", [tr[i]['rating'] for i in range(len(tr))]
Output:
TEST BINOMIAL
friends [22.0, 24.0, 24.0, 23.0, 23.0]
friends_ratings [0.5, 0.5, 0.41, 0.41, 0.41]
ratingf [10.0, 11.0, 11.0, 11.0, 11.0]
TEST DIRICHLET
friends [20.0, 20.0, 20.0, 20.0, 20.0]
friends_ratings [array([ 0.51369621, 1.490608 ]), ... ]
rating [array([ 10., 10.]), array([ 10., 10.]), ... ]
PyMC3 does not automatically normalize the Dirichlet. So far you have to do this explicitly using simplextransform. See here for an example.
There is an issue of making this transform automatic though: https://github.com/pymc-devs/pymc3/issues/315
EDIT (9/14/2015): PyMC3 now automatically transforms the dirichlet distribution (as any other distribution). So you don't need to specify that manually anymore.
Related
I'm trying to learn how to use lightgbm distributed.
I wrote a simple hello world kind of code where I use iris dataset with 150 rows, split it into train (100 rows) and test(50 rows). Then training the train test set are further split into two parts. Each part is fed into two machines with appropriate rank.
The problem I see is that lgb.train hangs.
Here is the code:
import argparse
import logging
import lightgbm as lgb
import pandas as pd
from sklearn import datasets
import socket
print('lightgbm', lgb.__version__)
HOST = socket.gethostname()
ip_address = socket.gethostbyname(HOST)
print("IP=", ip_address)
# looks like lightgbm operates only with ip addresses
IPS = ['10.121.22.166', '10.121.22.83']
assert ip_address in IPS
logger = logging.getLogger(__name__)
pd.set_option('display.max_rows', 4)
pd.set_option('display.max_columns', 100)
pd.set_option('display.width', 10000)
pd.set_option('max_colwidth', 100)
pd.set_option('precision', 5)
def read_train_data(rank):
iris = datasets.load_iris()
iris_df = pd.DataFrame(iris.data, columns=iris.feature_names)
partition = rank
assert partition < 2
separate = 100
train_df = iris_df.iloc[:separate]
test_df = iris_df.iloc[separate:]
separate_train = 60
separate_test = 30
if partition == 0:
train_df = train_df.iloc[:separate_train]
test_df = test_df.iloc[:separate_test]
else:
train_df = train_df.iloc[separate_train:]
test_df = test_df.iloc[separate_test:]
def get_lgb_dataset(df):
target_column = df.columns[-1]
columns = df.columns[:-1]
assert target_column not in columns
print('Target column', target_column)
x = df[columns]
y = df[target_column]
print(x)
ds = lgb.Dataset(free_raw_data=False, data=x, label=y, params={
"enable_bundle": False
})
ds.construct()
return ds
dtrain = get_lgb_dataset(train_df)
dtest = get_lgb_dataset(test_df)
return dtrain, dtest
def train(args):
port0 = 56456
rank = IPS.index(ip_address)
print("Rank=", rank, HOST)
print("RR", rank)
dtrain, dtest = read_train_data(rank=rank)
params = {'boosting_type': 'gbdt',
'class_weight': None,
'colsample_bytree': 1.0,
'importance_type': 'split',
'learning_rate': 0.1,
'max_depth': 2,
'min_child_samples': 20,
'min_child_weight': 0.001,
'min_split_gain': 0.0,
'n_estimators': 1,
'num_leaves': 31,
'objective': 'regression',
'metric': 'rmse',
'random_state': None,
'reg_alpha': 0.0,
'reg_lambda': 0.0,
'silent': False,
'subsample': 1.0,
'subsample_for_bin': 200000,
'subsample_freq': 0,
'tree_learner': 'data_parallel',
'num_threads': 48,
'machines': ','.join([f'{machine}:{port0}' for i, machine in enumerate(IPS)]),
'local_listen_port': port0,
'time_out': 120,
'num_machines': len(IPS)
}
print(params)
logging.info("starting to train lgb at node with rank %d", rank)
evals_result = {}
if args.scikit == 1:
print("Using scikit learn")
bst = lgb.sklearn.LGBMRegressor(**params)
bst.fit(
dtrain.data,
dtrain.label,
eval_set=[(dtest.data, dtest.label)],
)
else:
print("Using regular LGB")
bst = lgb.train(params,
dtrain,
valid_sets=[dtest],
evals_result=evals_result)
print(evals_result)
logging.info("finish xgboost training at node with rank %d", rank)
return bst
def main(args):
logging.info("starting the train job")
model = train(args)
pd.set_option('display.max_rows', 500)
print("OUT", model.__class__)
try:
print(model.trees_to_dataframe())
except:
print(model.booster_.trees_to_dataframe())
if __name__ == '__main__':
parser = argparse.ArgumentParser()
parser.add_argument(
'--scikit',
help='scikit',
default=0,
type=int,
)
main(parser.parse_args())
I can run it with the scikit fit interface by running: python simple_distributed_lgb_test.py --scikit 1
On the two machines. It produces a reasonable result.
However, when I use -- scikit 0 (which uses lgb.train), then fitting just hangs on both nodes. Last messages before it hangs:
[LightGBM] [Info] Total Bins 22
[LightGBM] [Info] Number of data points in the train set: 40, number of used features: 2
[LightGBM] [Warning] Found whitespace in feature_names, replace with underlines
[LightGBM] [Info] Start training from score 0.873750
Is that a bug or an expected behavior? dask.py in lightgbm does use scikit learn fit interface.
I use an overnight master version 3.2.1.99. 5b7a6f3e7150aeb704d1dd2b852d246af3e913a3 tag to be exact from Jul 12.
UPDATE 1
I'm trying to dig into the code. So far I see few things:
scikit.train interface appears to have an extra syncronization step before fitting first tree. lgb.train doesn't have it. Dunno yet where it comes from. (I see some Network::Allreduce operations)
It appears that scikit.train has workers syncronized - each worker knows the correct sizes of the blocks to send and receive during reducescatter operations. For example one the first allreduce worker1 sends 208 blocks and receives 368 blocks of data (in Linkers::SendRecv), while worker2 is reversed - sends 368 and receives 208. So allreduce completes fine. ()
On the contrary, lgb.train has workers not syncronized - each worker has numbers for send and receive blocks during reducescatter at the first DataParallelTreeLearner::FindBestSplits encounter. But they don't match. Worker1 sends 208 abd wants to receive 400. Worker2 sends 192 and wants to receive 176. So, the worker that wants to receive more just hangs. The other worker eventually hangs too.
Possibly it has something to do with lgb.Dataset. That thing may need to have same bins or something. I tried to force it by forcedbins_filename parameter. But it doesn't seem to help with lgb.train.
UPDATE 2
Success. If I remove the following line from the example:
ds.construct()
Everything works. So I guess we can't use construct on Dataset when using distributed training.
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!
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.
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)) :-)
pymc.__version__ = '3.0'
theano.__version__ = '0.6.0.dev-RELEASE'
I'm trying to use PyMC3 with a complex likelihood function:
First question: Is this possible?
Here's my attempt using Thomas Wiecki's post as a guide:
import numpy as np
import theano as th
import pymc as pm
import scipy as sp
# Actual data I'm trying to fit
x = np.array([52.08, 58.44, 60.0, 65.0, 65.10, 66.0, 70.0, 87.5, 110.0, 126.0])
y = np.array([0.522, 0.659, 0.462, 0.720, 0.609, 0.696, 0.667, 0.870, 0.889, 0.919])
yerr = np.array([0.104, 0.071, 0.138, 0.035, 0.102, 0.096, 0.136, 0.031, 0.024, 0.035])
th.config.compute_test_value = 'off'
a = th.tensor.dscalar('a')
with pm.Model() as model:
# Priors
alpha = pm.Normal('alpha', mu=0.3, sd=5)
sig_alpha = pm.Normal('sig_alpha', mu=0.03, sd=5)
t_double = pm.Normal('t_double', mu=4, sd=20)
t_delay = pm.Normal('t_delay', mu=21, sd=20)
nu = pm.Uniform('nu', lower=0, upper=20)
# Some functions needed for calculation of the y estimator
def T(eqd):
doses = np.array([52.08, 58.44, 60.0, 65.0, 65.10,
66.0, 70.0, 87.5, 110.0, 126.0])
tmt_times = np.array([29,29,43,29,36,48,22,11,7,8])
return np.interp(eqd, doses, tmt_times)
def TCP(a):
time = T(x)
BCP = pm.exp(-1E7*pm.exp(-alpha*x*1.2 + 0.69315/t_delay(time-t_double)))
return pm.prod(BCP)
def normpdf(a, alpha, sig_alpha):
return 1./(sig_alpha*pm.sqrt(2.*np.pi))*pm.exp(-pm.sqr(a-alpha)/(2*pm.sqr(sig_alpha)))
def normcdf(a, alpha, sig_alpha):
return 1./2.*(1+pm.erf((a-alpha)/(sig_alpha*pm.sqrt(2))))
def integrand(a):
return normpdf(a,alpha,sig_alpha)/(1.-normcdf(0,alpha,sig_alpha))*TCP(a)
func = th.function([a,alpha,sig_alpha,t_double,t_delay], integrand(a))
y_est = sp.integrate.quad(func(a, alpha, sig_alpha,t_double,t_delay), 0, np.inf)[0]
likelihood = pm.T('TCP', mu=y_est, nu=nu, observed=y_tcp)
start = pm.find_MAP()
step = pm.NUTS(state=start)
trace = pm.sample(2000, step, start=start, progressbar=True)
which produces the following message regarding the expression for y_est:
TypeError: ('Bad input argument to theano function with name ":42" at index 0(0-based)', 'Expected an array-like object, but found a Variable: maybe you are trying to call a function on a (possibly shared) variable instead of a numeric array?')
I've overcome various other hurdles to get this far, and this is where I'm stuck. So, provided the answer to my first question is 'yes', then am I on the right track? Any guidance would be helpful!
N.B. Here is a similar question I found, and another.
Disclaimer: I'm very new at this. My only previous experience is successfully reproducing the linear regression example in Thomas' post. I've also successfully run the Theano test suite, so I know it works.
Yes, its possible to make something with a complex or arbitrary likelihood. Though that doesn't seem like what you're doing here. It looks like you have a complex transformation of one variable into another, the integration step.
Your particular exception is that integrate.quad is expecting a numpy array, not a pymc Variable. If you want to do quad within pymc, you'll have to make a custom theano Op (with derivative) for it.