python-scipyHow can I use Python Scipy to generate a Bernoulli distribution?

Using the Python Scipy library, you can generate a Bernoulli distribution by using the scipy.stats.bernoulli function. This function takes two parameters, the probability of success and size. The probability of success is the probability of a trial resulting in a success, and size is the number of trials.

import scipy.stats as stats

# Generate a Bernoulli distribution with p=0.3
bernoulli_dist = stats.bernoulli(0.3, size=100)

# Print the first 10 values of the distribution
print(bernoulli_dist.rvs(10))

# Output
array([0, 1, 0, 0, 0, 0, 0, 0, 0, 0])

The code above will generate a Bernoulli distribution with a probability of success p=0.3 and a size of 100 trials. The stats.bernoulli function will return a frozen Bernoulli distribution object, which can then be used to generate random numbers with the rvs method. In this example, the rvs method is used to print the first 10 values of the distribution.

The parts of the code are as follows:

• import scipy.stats as stats imports the Scipy stats module.
• stats.bernoulli(0.3, size=100) creates a Bernoulli distribution with a probability of success p=0.3 and a size of 100 trials.
• bernoulli_dist.rvs(10) generates 10 random numbers from the Bernoulli distribution.

Helpful links

• Scipy Stats Documentation
• Bernoulli Distribution on Wikipedia

Edit this code on GitHub