Multi Dimensional Mean

Modeling
1

Hello,

I’m having trouble implementing the sum-to-zero constraint of a dichotomous predicted variable with multiple nominal predictors model. I’m attempting to create something of a fusion of the models Kruschke implements in chapters 20 and 21 of Doing Bayesian Data Analysis.

What’s specifically giving me trouble is trying to take the mean of an N dimensional array, and I can’t seem to find a function or implementation that can do it without awkward looping or summing up the entire N dimensional array and diving by the number of elements.

I’ve attached my model code below. Thank you for the assistance!

data {
    int<lower=1> nrows;

    int<lower=1> x1_ncats;
    int<lower=1, upper=x1_ncats> x1[nrows];

    int<lower=1> x2_ncats;
    int<lower=1, upper=x2_ncats> x2[nrows];

    int<lower=1> x3_ncats;
    int<lower=1, upper=x3_ncats> x3[nrows];

    int<lower=0> y[nrows];
    int<lower=1> N[nrows];
}

parameters {
    real a0;
    vector[x1_ncats] a1;
    vector[x2_ncats] a2;
    vector[x3_ncats] a3;
    
    real sigma;
}

model {
    a0 ~ normal(0, 2);
    a1 ~ normal(0, sigma);
    a2 ~ normal(0, sigma);
    a3 ~ normal(0, sigma);

    y ~ binomial_logit(N, a0 + a1[x1] + a2[x2] + a3[x3]);
}

generated quantities {
    real m[x1_ncats, x2_ncats, x3_ncats];
    
    real b0;
    vector[x1_ncats] b1;
    vector[x2_ncats] b2;
    vector[x3_ncats] b3;
    
    for (k1 in 1:x1_ncats)
        for (k2 in 1:x2_ncats)
            for (k3 in 1:x3_ncats)
                m[k1, k2, k3] = a0 + a1[k1] + a2[k2] + a3[k3];

    b0 = mean(m);
        
    for (k1 in 1:x1_ncats) b1[k1] = mean(m[k1, 1:x2_ncats, 1:x3_ncats]) - b0;
    
    for (k2 in 1:x2_ncats) b2[k2] = mean(m[1:x1_ncats, k2, 1:x3_ncats]) - b0;
        
    for (k3 in 1:x3_ncats) b3[k3] = mean(m[1:x1_ncats, 1:x2_ncats, k3]) - b0;
}
2

I’m assuming this code:

b0 = mean(m);

Doesn’t work but you’d like it to?

I think the closest you’ll get is b0 = mean(to_array_1d(m)). to_array_1d will collapse the multidimensional array down to one thing.

Is that what you’re looking for?

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