Question

在贝叶斯方法黑客的this notebook中，他们从python函数中创建了一个确定性变量：

# from code line 9 in the notebook
@pm.deterministic
def lambda_(tau=tau, lambda_1=lambda_1, lambda_2=lambda_2):
    out = np.zeros(n_count_data)
    out[:tau] = lambda_1  # lambda before tau is lambda1
    out[tau:] = lambda_2  # lambda after (and including) tau is lambda2
    return out

我试图几乎完全重新创建这个实验，但显然@pm.deterministic不是pymc3中的东西。知道如何在pymc3中做到这一点吗？

Answer 1

此模型在＆＃34; Probabilistic Programming and Bayesian Methods for Hackers＆＃34;的PyMC3端口中翻译。如

with pm.Model() as model:
    alpha = 1.0/count_data.mean()  # Recall count_data is the
                                   # variable that holds our txt counts
    lambda_1 = pm.Exponential("lambda_1", alpha)
    lambda_2 = pm.Exponential("lambda_2", alpha)

    tau = pm.DiscreteUniform("tau", lower=0, upper=n_count_data - 1)

    # These two lines do what the deterministic function did above
    idx = np.arange(n_count_data) # Index
    lambda_ = pm.math.switch(tau > idx, lambda_1, lambda_2)

    observation = pm.Poisson("obs", lambda_, observed=count_data)
    trace = pm.sample()

请注意，我们只是使用pm.math.switch（使用theano.tensor.switch的别名）来计算lambda_。还有pm.Deterministic，但这里不需要它。

Pymc3 python函数确定性

1 个答案: