Hi Simon,
Thanks for the pointers! But I'm not sure what to do with the 'time'
variable. (I don't want to make predictions for specific points in
time). I coded as follows (full reproducible example at bottom of
email), but get a warning and error:
N <- 1000 # number of points for smooth to be predicted
# new temperatures and lags for prediction
pred_temp <- seq(min(dat$temp, na.rm = T), max(dat$temp, na.rm = T), length = N)
pred_lag <- rep(c(0, 1,2,3,4,5,6),each=N) ## IS IT CORRECT TO SET
UP LAG LIKE THIS?
# not sure if these covariates are required with type = "terms"
pred_humidity <- rep(median(dat$humidity), N)
pred_rain <- rep(median(dat$rain), N)
pd <- data.frame(temp = pred_temp, lag = pred_lag, humidity =
pred_humidity, rain = pred_rain)
predictions <- predict(mod, pd, type = "terms")
The predict line creates the following warning and error:
Warning in predict.gam(mod, pd, type = "terms") :
not all required variables have been supplied in newdata!
Error in model.frame.default(ff, data = newdata, na.action = na.act) :
object is not a matrix
For ease of reference, I've (re)included the full reproducible example:
library(mgcv)
set.seed(3) # make reproducible example
simdat <- gamSim(1,400)
g <- exp(simdat$f/5)
simdat$y <- rnbinom(g,size=3,mu=g) # negative binomial response var
simdat$time <- 1:400 # create time series
names(simdat) <- c("deaths", "temp", "humidity", "rain", "x3", "f",
"f0", "f1", "f2", "f3", "time")
# lag function based on Wood (book 2017, p.349 and gamair package
documentation p.54
# https://cran.rstudio.com/web/packages/gamair/gamair.pdf)
lagard <- function(x,n.lag=7) {
n <- length(x); X <- matrix(NA,n,n.lag)
for (i in 1:n.lag) X[i:n,i] <- x[i:n-i+1]
X
}
# set up lag, temp, rain and humidity as 7-column matrices
# to create lagged variables
dat <- list(lag=matrix(0:6,nrow(simdat),7,byrow=TRUE),
deaths=simdat$deaths, time = simdat$time)
dat$temp <- lagard(simdat$temp)
dat$rain <- lagard(simdat$rain)
dat$humidity <- lagard(simdat$humidity)
mod <- gam(deaths~s(time, k=70) + te(temp, lag, k=c(12, 4)) +
te(humidity, lag, k=c(12, 4)) + te(rain, lag, k=c(12, 4)), data = dat,
family = nb, method = 'REML', select = TRUE)
# create prediction data
N <- 1000 # number of points for which to predict the smooths
pred_temp <- seq(min(dat$temp, na.rm = T), max(dat$temp, na.rm = T), length = N)
pred_lag <- rep(c(0, 1,2,3,4,5,6),each=N)
pred_humidity <- rep(median(dat$humidity), N)
pred_rain <- rep(median(dat$rain), N)
pd <- data.frame(temp = pred_temp, lag = pred_lag, humidity =
pred_humidity, rain = pred_rain)
# make predictions
predictions <- predict(mod, pd, type = "terms")
On Fri, 22 Jul 2022 at 09:54, Simon Wood <simon.w...@bath.edu> wrote:
>
>
> On 21/07/2022 15:19, jade.shodan--- via R-help wrote:
> > Hello everyone (incl. Simon Wood?),
> >
> > I'm not sure that my original question (see below, including
> > reproducible example) was as clear as it could have been. To clarify,
> > what I would to like to get is:
> >
> > 1) a perspective plot of temperature x lag x relative risk. I know
> > how to use plot.gam and vis.gam but don't know how to get plots on the
> > relative risk scale as opposed to "response" or "link".
>
> - You are on the log scale so I think that all you need to do is to use
> 'predict.gam', with 'type = "terms"' to get the predictions for the
> te(temp, lag) term over the required grid of lags and temperatures.
> Suppose the dataframe of prediction data is 'pd'. Now produce pd0, which
> is identical to pd, except that the temperatures are all set to the
> reference temperature. Use predict.gam to predict te(temp,lag) from pd0.
> Now the exponential of the difference between the first and second
> predictions is the required RR, which you can plot using 'persp',
> 'contour', 'image' or whatever. If you need credible intervals see pages
> 341-343 of my 'GAMs: An intro with R' book (2nd ed).
>
> > 2) a plot of relative risk (accumulated across all lags) vs
> > temperature, given a reference temperature. An example of such a plot
> > can be found in figure 2 (bottom) of this paper by Gasparrini et al:
> > https://onlinelibrary.wiley.com/doi/abs/10.1002/sim.3940
>
> - I guess this only makes sense if you have the same temperature at all
> lags. So this time produce a data.frame with each desired temperature
> repeated for each lag: 'pd1'. Again use predict.gam(...,type="terms").
> Then sum the predictions over lags for each temperature, subtract the
> minimum, and take the exponential. Same as above for CIs.
>
> best,
>
> Simon
>
> > I've seen Simon Wood's response to a related issue here:
> > https://stat.ethz.ch/pipermail/r-help/2012-May/314387.html
> > However, I'm not sure how to apply this to time series data with
> > distributed lag, to get the above mentioned figures.
> >
> > Would be really grateful for suggestions!
> >
> > Jade
> >
> > On Tue, 19 Jul 2022 at 16:07, jade.sho...@googlemail.com
> > <jade.sho...@googlemail.com> wrote:
> >> Dear list members,
> >>
> >> Does anyone know how to obtain a relative risk/ risk ratio from a GAM
> >> with a distributed lag model implemented in mgcv? I have a GAM
> >> predicting daily deaths from time series data consisting of daily
> >> temperature, humidity and rainfall. The GAM includes a distributed lag
> >> model because deaths may occur over several days following a high heat
> >> day.
> >>
> >> What I'd like to do is compute (and plot) the relative risk
> >> (accumulated across all lags) for a given temperature vs the
> >> temperature at which the risk is lowest, with corresponding confidence
> >> intervals. I am aware of the predict.gam function but am not sure if
> >> and how it should be used in this case. (Additionally, I'd also like
> >> to plot the relative risk for different lags separately).
> >>
> >> I apologise if this seems trivial to some. (Actually, I hope it is,
> >> because that might mean I get a solution!) I've been looking for
> >> examples on how to do this, but found nothing so far. Suggestions
> >> would be very much appreciated!
> >>
> >> Below is a reproducible example with the GAM:
> >>
> >> library(mgcv)
> >> set.seed(3) # make reproducible example
> >> simdat <- gamSim(1,400) # simulate data
> >> g <- exp(simdat$f/5)
> >> simdat$y <- rnbinom(g,size=3,mu=g) # negative binomial response var
> >> simdat$time <- 1:400 # create time series
> >> names(simdat) <- c("deaths", "temp", "humidity", "rain", "x3", "f",
> >> "f0", "f1", "f2", "f3", "time")
> >>
> >> # lag function based on Simon Wood (book 2017, p.349 and gamair
> >> package documentation p.54
> >> # https://cran.rstudio.com/web/packages/gamair/gamair.pdf)
> >> lagard <- function(x,n.lag=7) {
> >> n <- length(x); X <- matrix(NA,n,n.lag)
> >> for (i in 1:n.lag) X[i:n,i] <- x[i:n-i+1]
> >> X
> >> }
> >>
> >> # set up lag, temp, rain and humidity as 7-column matrices
> >> # to create lagged variables - based on Simon Wood's example
> >> dat <- list(lag=matrix(0:6,nrow(simdat),7,byrow=TRUE),
> >> deaths=simdat$deaths, time = simdat$time)
> >> dat$temp <- lagard(simdat$temp)
> >> dat$rain <- lagard(simdat$rain)
> >> dat$humidity <- lagard(simdat$humidity)
> >>
> >> mod <- gam(deaths~s(time, k=70) + te(temp, lag, k=c(12, 4)) +
> >> te(humidity, lag, k=c(12, 4)) + te(rain, lag, k=c(12, 4)), data = dat,
> >> family = nb, method = 'REML', select = TRUE)
> >>
> >> summary(mod)
> >> plot(mod, scheme = 1)
> >> plot(mod, scheme = 2)
> >>
> >> Thanks for any suggestions you may have,
> >>
> >> Jade
> > ______________________________________________
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
> --
> Simon Wood, School of Mathematics, University of Edinburgh,
> https://www.maths.ed.ac.uk/~swood34/
>
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