How to specify size for bernoulli distribution with pymc3?











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In trying to make my way through Bayesian Methods for Hackers, which is in pymc, I came across this code:



first_coin_flips = pm.Bernoulli("first_flips", 0.5, size=N)


I've tried to translate this to pymc3 with the following, but it just returns a numpy array, rather than a tensor (?):



first_coin_flips = pm.Bernoulli("first_flips", 0.5).random(size=50)


The reason the size matters is that it's used later on in a deterministic variable. Here's the entirety of the code that I have so far:



import pymc3 as pm
import matplotlib.pyplot as plt
import numpy as np
import mpld3
import theano.tensor as tt

model = pm.Model()
with model:
N = 100
p = pm.Uniform("cheating_freq", 0, 1)
true_answers = pm.Bernoulli("truths", p)
print(true_answers)
first_coin_flips = pm.Bernoulli("first_flips", 0.5)
second_coin_flips = pm.Bernoulli("second_flips", 0.5)
# print(first_coin_flips.value)

# Create model variables
def calc_p(true_answers, first_coin_flips, second_coin_flips):
observed = first_coin_flips * true_answers + (1-first_coin_flips) * second_coin_flips
# NOTE: Where I think the size param matters, since we're dividing by it
return observed.sum() / float(N)

calced_p = pm.Deterministic("observed", calc_p(true_answers, first_coin_flips, second_coin_flips))
step = pm.Metropolis(model.free_RVs)
trace = pm.sample(1000, tune=500, step=step)
pm.traceplot(trace)

html = mpld3.fig_to_html(plt.gcf())
with open("output.html", 'w') as f:
f.write(html)
f.close()


And the output:



Output



The coin flips and uniform cheating_freq output look correct, but the observed doesn't look like anything to me, and I think it's because I'm not translating that size param correctly.










share|improve this question


















  • 1




    Could you include a link to the original code you're trying to replicate? Also, the notebooks are already all converted for PyMC3. E.g., each chapter folder has a notebook for PyMC2, PyMC3, and TF Probability, which is what PyMC4 will use.
    – merv
    Nov 7 at 18:31










  • Oh wow, didn't realize that there were already translations for pymc3. That answers my question, I'll create an answer for it. Thanks!
    – Marcus Buffett
    Nov 7 at 19:23















up vote
3
down vote

favorite












In trying to make my way through Bayesian Methods for Hackers, which is in pymc, I came across this code:



first_coin_flips = pm.Bernoulli("first_flips", 0.5, size=N)


I've tried to translate this to pymc3 with the following, but it just returns a numpy array, rather than a tensor (?):



first_coin_flips = pm.Bernoulli("first_flips", 0.5).random(size=50)


The reason the size matters is that it's used later on in a deterministic variable. Here's the entirety of the code that I have so far:



import pymc3 as pm
import matplotlib.pyplot as plt
import numpy as np
import mpld3
import theano.tensor as tt

model = pm.Model()
with model:
N = 100
p = pm.Uniform("cheating_freq", 0, 1)
true_answers = pm.Bernoulli("truths", p)
print(true_answers)
first_coin_flips = pm.Bernoulli("first_flips", 0.5)
second_coin_flips = pm.Bernoulli("second_flips", 0.5)
# print(first_coin_flips.value)

# Create model variables
def calc_p(true_answers, first_coin_flips, second_coin_flips):
observed = first_coin_flips * true_answers + (1-first_coin_flips) * second_coin_flips
# NOTE: Where I think the size param matters, since we're dividing by it
return observed.sum() / float(N)

calced_p = pm.Deterministic("observed", calc_p(true_answers, first_coin_flips, second_coin_flips))
step = pm.Metropolis(model.free_RVs)
trace = pm.sample(1000, tune=500, step=step)
pm.traceplot(trace)

html = mpld3.fig_to_html(plt.gcf())
with open("output.html", 'w') as f:
f.write(html)
f.close()


And the output:



Output



The coin flips and uniform cheating_freq output look correct, but the observed doesn't look like anything to me, and I think it's because I'm not translating that size param correctly.










share|improve this question


















  • 1




    Could you include a link to the original code you're trying to replicate? Also, the notebooks are already all converted for PyMC3. E.g., each chapter folder has a notebook for PyMC2, PyMC3, and TF Probability, which is what PyMC4 will use.
    – merv
    Nov 7 at 18:31










  • Oh wow, didn't realize that there were already translations for pymc3. That answers my question, I'll create an answer for it. Thanks!
    – Marcus Buffett
    Nov 7 at 19:23













up vote
3
down vote

favorite









up vote
3
down vote

favorite











In trying to make my way through Bayesian Methods for Hackers, which is in pymc, I came across this code:



first_coin_flips = pm.Bernoulli("first_flips", 0.5, size=N)


I've tried to translate this to pymc3 with the following, but it just returns a numpy array, rather than a tensor (?):



first_coin_flips = pm.Bernoulli("first_flips", 0.5).random(size=50)


The reason the size matters is that it's used later on in a deterministic variable. Here's the entirety of the code that I have so far:



import pymc3 as pm
import matplotlib.pyplot as plt
import numpy as np
import mpld3
import theano.tensor as tt

model = pm.Model()
with model:
N = 100
p = pm.Uniform("cheating_freq", 0, 1)
true_answers = pm.Bernoulli("truths", p)
print(true_answers)
first_coin_flips = pm.Bernoulli("first_flips", 0.5)
second_coin_flips = pm.Bernoulli("second_flips", 0.5)
# print(first_coin_flips.value)

# Create model variables
def calc_p(true_answers, first_coin_flips, second_coin_flips):
observed = first_coin_flips * true_answers + (1-first_coin_flips) * second_coin_flips
# NOTE: Where I think the size param matters, since we're dividing by it
return observed.sum() / float(N)

calced_p = pm.Deterministic("observed", calc_p(true_answers, first_coin_flips, second_coin_flips))
step = pm.Metropolis(model.free_RVs)
trace = pm.sample(1000, tune=500, step=step)
pm.traceplot(trace)

html = mpld3.fig_to_html(plt.gcf())
with open("output.html", 'w') as f:
f.write(html)
f.close()


And the output:



Output



The coin flips and uniform cheating_freq output look correct, but the observed doesn't look like anything to me, and I think it's because I'm not translating that size param correctly.










share|improve this question













In trying to make my way through Bayesian Methods for Hackers, which is in pymc, I came across this code:



first_coin_flips = pm.Bernoulli("first_flips", 0.5, size=N)


I've tried to translate this to pymc3 with the following, but it just returns a numpy array, rather than a tensor (?):



first_coin_flips = pm.Bernoulli("first_flips", 0.5).random(size=50)


The reason the size matters is that it's used later on in a deterministic variable. Here's the entirety of the code that I have so far:



import pymc3 as pm
import matplotlib.pyplot as plt
import numpy as np
import mpld3
import theano.tensor as tt

model = pm.Model()
with model:
N = 100
p = pm.Uniform("cheating_freq", 0, 1)
true_answers = pm.Bernoulli("truths", p)
print(true_answers)
first_coin_flips = pm.Bernoulli("first_flips", 0.5)
second_coin_flips = pm.Bernoulli("second_flips", 0.5)
# print(first_coin_flips.value)

# Create model variables
def calc_p(true_answers, first_coin_flips, second_coin_flips):
observed = first_coin_flips * true_answers + (1-first_coin_flips) * second_coin_flips
# NOTE: Where I think the size param matters, since we're dividing by it
return observed.sum() / float(N)

calced_p = pm.Deterministic("observed", calc_p(true_answers, first_coin_flips, second_coin_flips))
step = pm.Metropolis(model.free_RVs)
trace = pm.sample(1000, tune=500, step=step)
pm.traceplot(trace)

html = mpld3.fig_to_html(plt.gcf())
with open("output.html", 'w') as f:
f.write(html)
f.close()


And the output:



Output



The coin flips and uniform cheating_freq output look correct, but the observed doesn't look like anything to me, and I think it's because I'm not translating that size param correctly.







python statistics pymc3






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asked Nov 7 at 17:00









Marcus Buffett

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  • 1




    Could you include a link to the original code you're trying to replicate? Also, the notebooks are already all converted for PyMC3. E.g., each chapter folder has a notebook for PyMC2, PyMC3, and TF Probability, which is what PyMC4 will use.
    – merv
    Nov 7 at 18:31










  • Oh wow, didn't realize that there were already translations for pymc3. That answers my question, I'll create an answer for it. Thanks!
    – Marcus Buffett
    Nov 7 at 19:23














  • 1




    Could you include a link to the original code you're trying to replicate? Also, the notebooks are already all converted for PyMC3. E.g., each chapter folder has a notebook for PyMC2, PyMC3, and TF Probability, which is what PyMC4 will use.
    – merv
    Nov 7 at 18:31










  • Oh wow, didn't realize that there were already translations for pymc3. That answers my question, I'll create an answer for it. Thanks!
    – Marcus Buffett
    Nov 7 at 19:23








1




1




Could you include a link to the original code you're trying to replicate? Also, the notebooks are already all converted for PyMC3. E.g., each chapter folder has a notebook for PyMC2, PyMC3, and TF Probability, which is what PyMC4 will use.
– merv
Nov 7 at 18:31




Could you include a link to the original code you're trying to replicate? Also, the notebooks are already all converted for PyMC3. E.g., each chapter folder has a notebook for PyMC2, PyMC3, and TF Probability, which is what PyMC4 will use.
– merv
Nov 7 at 18:31












Oh wow, didn't realize that there were already translations for pymc3. That answers my question, I'll create an answer for it. Thanks!
– Marcus Buffett
Nov 7 at 19:23




Oh wow, didn't realize that there were already translations for pymc3. That answers my question, I'll create an answer for it. Thanks!
– Marcus Buffett
Nov 7 at 19:23












1 Answer
1






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1
down vote



accepted










The pymc3 way to specify the size of a Bernoulli distribution is by using the shape parameter, like:



first_coin_flips = pm.Bernoulli("first_flips", 0.5, shape=N)





share|improve this answer





















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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    The pymc3 way to specify the size of a Bernoulli distribution is by using the shape parameter, like:



    first_coin_flips = pm.Bernoulli("first_flips", 0.5, shape=N)





    share|improve this answer

























      up vote
      1
      down vote



      accepted










      The pymc3 way to specify the size of a Bernoulli distribution is by using the shape parameter, like:



      first_coin_flips = pm.Bernoulli("first_flips", 0.5, shape=N)





      share|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        The pymc3 way to specify the size of a Bernoulli distribution is by using the shape parameter, like:



        first_coin_flips = pm.Bernoulli("first_flips", 0.5, shape=N)





        share|improve this answer












        The pymc3 way to specify the size of a Bernoulli distribution is by using the shape parameter, like:



        first_coin_flips = pm.Bernoulli("first_flips", 0.5, shape=N)






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 7 at 19:25









        Marcus Buffett

        592422




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