[R-sig-ME] Repeated Observations Linear Mixed Model with Outside-Group Spatial Correlation
Ben Bolker
bbolker at gmail.com
Tue Mar 21 22:12:37 CET 2017
Your approach seems about right.
- What precisely does "unsuccessful" mean? warnings, errors,
ridiculous answers?
- Is this your whole data set or a subset?
- centering and scaling predictors is always worth a shot to fix
numeric problems
- INLA is more powerful than lme for fitting spatial correlations,
although it's a *steep* learning curve ...
On Tue, Mar 21, 2017 at 5:08 PM, Michael Hyland
<mhyland at u.northwestern.edu> wrote:
> Hi,
>
> I'm new to the listserv.
>
> A shortened version of my dataset is below. I am developing a model to
> forecast monthly ridership at Bikeshare stations. I want to predict 'Cnts'
> as a function of 'Population' - 'Temperature'. The dataset is unbalanced
> (unequal number of observations for each station) and most of covariates do
> not vary over time, but a few do.
>
> I have successfully used lmer() and lme() in R to capture the dependency
> between the error terms for repeated observations from a given station
> ('id').
>
> model_spatial = lme(log(counts) ~ log(Population)
> +Drive +Med_Income + Buff2 +Rain + Temperature
> , data = Data, random = ~1|id, method = "ML" )
>
> model_All = lmer(log(counts) ~ log(Population)
> +Drive +Med_Income + Buff2 +Rain + Temperature
> + (1|id)
> , data = Data )
>
> However, a Moran's I test of the residuals suggests that the residuals are
> spatially correlated.
> station.dists <- as.matrix(dist(cbind(Data$longitude, Data$latitude)))
> station.dists.inv <- 1/station.dists
> station.dists.inv[is.infinite(station.dists.inv)] <- 0 #Distance value is
> inf for repeated observations from the same station
> Data$resid_all = resid(model_spatial)
> Moran.I(Data$resid_all, station.dists.inv)
>
>
> Hence, I need to develop a model in R that accounts for spatial correlation
> across stations, while simultaneously capturing correlations between
> observations from a single station. I've tried the following updates to
> the lme() model, but have been unsuccessful.
> model_spatial.gau <- update(model_spatial, correlation = corGaus(form = ~
> latitude +longitude ), method = "ML")
> model_spatial.gau <- update(model_spatial, correlation = corGaus(form = ~
> latitude +longitude| id ), method = "ML")
>
> Is there a way to formulate the correlation matrix in lme() or lmer() such
> that the correlation between repeated obvservations of a given station *and*
> the spatial autocorrelation between stations is accounted for?
>
>
> year month id Cnts latitude longitude Population Drive Med_Income Buff2 Rain
> Temperature
> 2015 8 2 2597 41.87264 -87.62398 4256 0.3418054 76857 127 0.07 71.8
> 2015 9 2 2772 41.87264 -87.62398 4256 0.3418054 76857 128 0.15 69
> 2015 10 2 684 41.87264 -87.62398 4256 0.3418054 76857 128 0.07 54.7
> 2015 11 2 394 41.87264 -87.62398 4256 0.3418054 76857 128 0.15 44.6
> 2015 12 2 148 41.87264 -87.62398 4256 0.3418054 76857 129 0.16 39
> 2016 1 2 44 41.87264 -87.62398 4256 0.3418054 76857 129 0.03 24.7
> 2015 5 3 2303 41.86723 -87.61536 16735 0.4312349 75227 90 0.15 60.4
> 2015 6 3 3323 41.86723 -87.61536 16735 0.4312349 75227 98 0.24 67.4
> 2015 7 3 5920 41.86723 -87.61536 16735 0.4312349 75227 98 0.09 72.3
> 2015 8 3 4405 41.86723 -87.61536 16735 0.4312349 75227 98 0.07 71.8
> 2015 9 3 3638 41.86723 -87.61536 16735 0.4312349 75227 99 0.15 69
> 2015 10 3 2061 41.86723 -87.61536 16735 0.4312349 75227 99 0.07 54.7
> 2015 11 3 1074 41.86723 -87.61536 16735 0.4312349 75227 99 0.15 44.6
> 2015 12 3 374 41.86723 -87.61536 16735 0.4312349 75227 100 0.16 39
> 2016 1 3 188 41.86723 -87.61536 16735 0.4312349 75227 100 0.03 24.7
> 2016 2 3 474 41.86723 -87.61536 16735 0.4312349 75227 100 0.04 30.4
> 2015 6 4 2968 41.85627 -87.61335 16735 0.4312349 75227 68 0.24 67.4
> 2015 7 4 4266 41.85627 -87.61335 16735 0.4312349 75227 68 0.09 72.3
> 2015 8 4 3442 41.85627 -87.61335 16735 0.4312349 75227 68 0.07 71.8
> 2015 9 4 2552 41.85627 -87.61335 16735 0.4312349 75227 69 0.15 69
> 2015 10 4 1301 41.85627 -87.61335 16735 0.4312349 75227 69 0.07 54.7
>
>
> Thanks,
> Mike Hyland
>
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>
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