[R-sig-ME] spatial autocorrelation as random effect with count data

Ben Bolker bbolker at gmail.com
Wed Jan 10 22:30:57 CET 2018

  It will be hard to make this work with glmer at present.

 If you want to use geostatistical models (i.e., specifying spatial
autocorrelation function explicitly), as far as I know at present your
best options are

  INLA  (possibly via the inlabru package): looks like this implements
the Matern family (which includes exponential and Gaussian
autocorrelation functions as special cases)

  spaMM: spatial models etc. via h-likelihood

  would welcome more informed advice from other people on the list!

On Wed, Jan 10, 2018 at 3:59 PM, Sima Usvyatsov <ghiaco at gmail.com> wrote:
> Hello,
> I am working on a spatially autocorrelated dataset with a negative binomial
> (count) response variable. I have been using the glmmPQL approach (MASS),
> but I seem to have a hard time fitting the fixed effects. I came across the
> mention that one could build the spatial autocorrelation into a random
> effect (https://stat.ethz.ch/pipermail/r-sig-mixed-models/2011q1/015364.html).
> I've done some searching but could not find a straightforward example of
> this practice. I have 20 sampling locations (sampled repeatedly to a 4,000
> point dataset) and I know that there is spatial autocorrelation between
> them (by looking at autocorrelation plots of a naive model). The 20 grid
> points are clustered into 4 strata, and I am interested in the strata
> effects (so would like to keep the strata as fixed).
> How would I go about expressing the spatial autocorrelation in this setup?
> In the future I'd like to explore GAMs for this application, but for now
> I'm stuck with a GLM approach... I would love to be able to use glmer()
> with a random effect that expresses spatial autocorrelation.
> Here's a fake dataset.
> library(MASS)
> df <- data.frame(Loc = as.factor(rep(1:20, each = 5)), Lat = rep(rnorm(20,
> 30, 0.1), each = 5), Lon = rep(rnorm(20, -75, 1), each = 5), x =
> rnegbin(100, 1, 1), Stratum = rep(1:5, each = 20))
> Thank you so much!
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