Is there a way to more efficiently perform the following calculations in RStan?
I have only provided the minimal amount of coded that is needed:
parameters {
real beta_0;
real beta_1;
}
model {
vector [n] p_i = exp(beta_0 + beta_1*x)/[1 + exp(beta_0 + beta_1*x)];
y ~ bernoulli(p_i);
/* Likelihood:
for(i in 1:n){
p_i[i] = exp(beta_0 + beta_1*x[i])/(1 + exp(beta_0 + beta_1*x[i]));
y[i] ~ bernoulli(p_i[i]);
*/}
// Prior:
beta_0 ~ normal(m_beta_0, s_beta_0);
beta_1 ~ normal(m_beta_1, s_beta_1);
}
I obtain the following error message: "Matrix expression elements must be type row_vector and row vector expression elements must be int or real, but found element of type vector". If I use the for loop (which is commented out), the code works fine, but I would like to limit the use of for loops in my code. In the above code, x, is a vector of length n.
Another example:
parameters {
real gamma1;
real gamma2;
real gamma3;
real gamma4;
}
model {
// Likelihood:
real lambda;
real beta;
real phi;
for(i in 1:n){
lambda = exp(gamma1)*x[n_length[i]]^gamma2;
beta = exp(gamma3)*x[n_length[i]]^gamma4;
phi = lambda^(-1/beta);
y[i] ~ weibull(beta, phi);
}
//y ~ weibull(exp(gamma1)*x^gamma2, exp(gamma3)*x^gamma4); //cannot raise a vector to a power
// Prior:
gamma1 ~ normal(m_gamma1, s_gamma1);
gamma2 ~ normal(m_gamma2, s_gamma2);
gamma3 ~ normal(m_gamma3, s_gamma3);
gamma4 ~ normal(m_gamma4, s_gamma4);
}
The above code works, but the commented out likelihood calculation does not work since I "cannot raise a vector to a power" (but you can in R). I would, once again, like to not be forced to use for loops. In the above code, n_length, is a vector of length n.
A final example. If I want to draw 10000 samples from a normal distribution in R, I can simply specify
rnorm(10000, mu, sigma)
But in RStan, I would have to use a for loop, for example
parameters {
real mu;
real sigma;
}
generated quantities {
vector[n] x;
for(i in 1:n) {
x[i] = normal_rng(mu, sigma);
}
}
Is there anything that I can do to speed up my RStan examples?
This line of code:
vector [n] p_i = exp(beta_0 + beta_1*x)/[1 + exp(beta_0 + beta_1*x)];
is not valid syntax in the Stan language because square brackets are only used for indexing. It could instead be
vector [n] p_i = exp(beta_0 + beta_1*x) ./ (1 + exp(beta_0 + beta_1*x));
which utilizes the elementwise division operator, or better yet
vector [n] p_i = inv_logit(beta_0 + beta_1*x);
in which case y ~ bernoulli(p_i);
would work as a likelihood. Better still, just do
y ~ bernoulli_logit(beta_0 + beta_1 * x);
and it will do the transformation for you in a numerically stable fashion. You could also use bernoulli_logit_glm
, which is slightly faster particularly with large datasets.
In Stan 2.19.x, I think you can draw N values from a probability distribution in the generated quantities block. But you are too worried about for
loops. The Stan program is transpiled to C++ where loops are fast and almost all of the functions in the Stan language that accept vector inputs and produce vector outputs actually involve the same loop in C++ as if you had done the loop yourself.