[R] mgcv: relative risk from GAM with distributed lag

[j@de@shod@@ m@iii@g oii googiem@ii@com]

Hi Simon,

Thanks for that example. It's been very useful!  It turned out that I
created the prediction data for lag incorrectly. I've now managed to
get perspective plots on the relative risk scale.

I've now moved on to trying to get a plot of overall RR (y-axis) for a
range of temperature values (x-axis), where RR for each temperature
value is summed over lag periods 0-6 (see bottom plot page 2230 in
Gasparrini's paper: https://doi.org/10.1002/sim.3940). This now seems
straighforward with predict.gam() and type = "terms" in the way you
suggested, but as acknowledged, that doesn't give me credible
intervals. I've just been told that I MUST present results with
confidence/ credible intervals (which is fair enough).

For the last two days I've tried making the plot of overall RR with
CIs, using the lpmatrix, as you suggested (looked it up in your book,
and also an earlier post of yours with an example:
https://stat.ethz.ch/pipermail/r-help/2012-May/314387.html), but I
don't even understand how the book example can be applied to my
problem, and I can't get the example in the online post to work. I
don't really understand what an lpmatrix is, (other than that it seems
to give me a value for each smooth term for each basis function?) or
what the code in the book example does, so I have no idea what I am
doing wroing, or what good results are supposed to look like. (I'm
sorry, my background is in public health rather than statistics).

I am hoping that I can ask for your (or anyone's)  help one final
time. (Once I have the CI's, my modelling will be finished, and I will
be out of everyone's hair with questions about GAMs... at least for a
while!).

Here's my code with reproducible data to get a plot of overall RR
(y-axis) vs temperature (x-axis) where overall RR for each temperature
value is obtained by summing RR over lag periods 0-6. But I have
absolutely no idea how to use predict.gam with type = lpmatrix to get
from this curve to one with CIs.

library(mgcv)

# simulate data
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  Simon Wood's book (2017, p.349)
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
}

# create lagged variables as 7-column matrices
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
m1 <- 40 # number of points to predict across the range of each weather variable
m2 <- 7 # number of lag periods
n <- m1*m2 # total number of prediction points

pred_temp <- seq(min(dat$temp, na.rm = T), max(dat$temp, na.rm = T),
length = m1)
pred_lag <- 0:6
pred_rain <- seq(min(dat$rain, na.rm = T), max(dat$rain, na.rm = T),
length = m1)
pred_humidity <- seq(min(dat$humidity, na.rm = T), max(dat$humidity,
na.rm = T), length = m1)
pred_time <- seq(min(dat$time, na.rm = T), max(dat$time, na.rm = T),
length = m1)

pd <- data.frame(temp=rep(pred_temp,m2),lag=rep(pred_lag,each=m1),rain=rep(pred_rain,m2),humidity=rep(pred_humidity,m2),time=rep(pred_time,m2))

#  prediction data with reference temperature set to median
temperature, all else equal to pd
pred_temp0 <- median(dat$temp, na.rm=T)
pd0 <- data.frame(temp=rep(pred_temp0,n),lag=rep(pred_lag,each=m1),rain=rep(pred_rain,m2),humidity=rep(pred_humidity,m2),time=rep(pred_time,m2))

# make predictions
predictions <- predict.gam(mod, newdata=pd, type = "terms")
predictions0 <- predict.gam(mod, newdata=pd0, type = "terms")

# calculate RR (relative to reference temperature set above)
diff <- predictions[,2] - predictions0[,2]
rr <- as.vector(exp(diff))

# convert rr to matrix required for z argument for persp() plot
rr_mat <- matrix(rr, m1, m2)
persp(x=pred_temp,y=pred_lag,z=rr_mat,theta=30,phi=30,ticktype =
"detailed", col="blue",zlab="RR")

# create plots of overall RR (y-axis, summed over lags 0-6) vs
temperature (x-axis)
# convert predictions to matrix to allow summing over lags
pmat <- matrix(predictions[,2],m1,m2)
pmat0 <- matrix(predictions0[,2],m1,m2)

# sum predictions over lags 0-6 for each temperature
psummed <- rowSums(pmat)
psummed0 <-rowSums(pmat0)

# subtract the predictions for the reference temperature from the
predictions for each temperature
pref <- psummed-psummed0

# exponentiating a difference on log-scale gives RR
rr_overall <- exp(pref)
plot(pred_temp, rr_overall, type = "l")

# now create CIs for overall RR plot
Xp <- predict(a1a_high_heat,newdata=pd,type="lpmatrix")

# and how to procede now?????? So close to having results, and yet....



On Sat, 23 Jul 2022 at 21:26, Simon Wood <simon.wood using bath.edu> wrote:
>
> I doubt you want a 7 by 1000 grid for your persp plot. Here's an example
> of producing a persp plot using predict.gam and a custom grid. Since
> only the effects of x1 and x2 are being plotted it doesn't matter what
> x0 and x3 are set to (the model is additive after all). In your case
> only one smooth term is involved of course.
>
> library(mgcv)
> n <- 200
> sig <- 2
> dat <- gamSim(1,n=n,scale=sig)
>
> b <- gam(y~s(x0)+s(I(x1^2))+s(x2)+offset(x3),data=dat)
>
> m1 <- 20;m2 <- 30; n <- m1*m2
> x1 <- seq(.2,.8,length=m1);x2 <- seq(.2,.8,length=m2) ## marginal values
> for evaluation grid
> df <- data.frame(x0=rep(.5,n),x1=rep(x1,m2),x2=rep(x2,each=m1),x3=rep(0,n))
> pf <- predict(b,newdata=df,type="terms")
>
> persp(x1,x2,matrix(pf[,2]+pf[,3],m1,m2),theta=-130,col="blue",zlab="")
>
> On 23/07/2022 14:54, jade.shodan using googlemail.com wrote:
> > Hi Simon and all,
> >
> > I've corrected some mistakes in setting up the prediction data frame
> > (sorry, very stressed and sleep deprived due to closing in deadlines),
> > but am having problems getting the perspective plot using persp().
> >
> > I've calculated relative risk (RR) but am having trouble getting the
> > perspective (or contour) plot. persp() takes x and y vectors in
> > ascending order, and z as a matrix.  I've seen the outer() function
> > being used to facilitate perspective plots, but every example I've
> > seen treats z as a function, to be evaluated at combinations for x and
> > y. In my case, I already have observations for z. Not sure if I need
> > to use outer(), and if so, how?
> >
> > My code for making predictions, getting RR and trying to get the plot
> > is as follows (full reproducible example including model at the
> > bottom). Can someone tell me what I'm doing wrong?
> >
> > ########### CODE ########################
> >
> > # 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)  # prediction data for temperature
> > pred_lag <- rep(c(0, 1,2,3,4,5,6),each=N)
> > pred_rain <- seq(min(dat$rain, na.rm = T), max(dat$rain, na.rm = T), length = N)
> > pred_humidity <- seq(min(dat$humidity, na.rm = T), max(dat$humidity,
> > na.rm = T), length = N)
> > pred_time <- seq(min(dat$time, na.rm = T), max(dat$time, na.rm = T), length = N)
> >
> > pd <- data.frame(temp = pred_temp, lag = pred_lag, humidity =
> > pred_humidity, rain = pred_rain, time = pred_time)
> >
> > # create prediction data with reference temperature set to median temperature
> > # identical to pd but now with all temperatures set to median value of
> > temperature
> > pred_temp0 <- rep(median(dat$temp, na.rm=T), N)
> > pd0 <- data.frame(temp = pred_temp0, lag = pred_lag, humidity =
> > pred_humidity, rain = pred_rain, time = pred_time)
> >
> > # make predictions
> > predictions <- predict.gam(mod, pd, type = "terms")
> > predictions0 <- predict.gam(mod, pd0, type = "terms")
> >
> > # calculate RR
> > diff <- predictions[,2] - predictions0[,2]
> > rr <- as.vector(exp(diff))
> >
> > # convert rr to matrix required for z argument for persp()
> > rr_mat <- matrix(rr, nrow = N)
> > persp(pred_temp, pred_lag, rr_mat)
> >
> > ######### ERROR MESSAGE ##################################
> >
> > The persp call results in the follow error:
> >
> > Error in persp.default(pred_temp, pred_lag, rr_mat) :
> >    increasing 'x' and 'y' values expected
> >
> > I don't understand this because pred_temp and pred_lag ARE in ascending order.
> >
> > ############################################################
> > FULL REPRODUCIBLE EXAMPLE
> > ############################################################
> >
> > library(mgcv)
> > set.seed(3) # make reproducible example
> > simdat <- gamSim(1,400)
> > g <- exp(simdat$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)
> >
> > summary(mod)
> > plot(mod, scheme = 1)  # perspective
> > plot(mod, scheme = 2)  # contour
> >
> > # 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_rain <- seq(min(dat$rain, na.rm = T), max(dat$rain, na.rm = T), length = N)
> > pred_humidity <- seq(min(dat$humidity, na.rm = T), max(dat$humidity,
> > na.rm = T), length = N)
> > pred_time <- seq(min(dat$time, na.rm = T), max(dat$time, na.rm = T), length = N)
> >
> > pd <- data.frame(temp = pred_temp, lag = pred_lag, humidity =
> > pred_humidity, rain = pred_rain, time = pred_time)
> >
> > # create prediction data with reference temperature set to median temperature
> > # identical to pd but now with all temperatures set to median value of
> > temperature
> > pred_temp0 <- rep(median(dat$temp, na.rm=T), N)
> > pd0 <- data.frame(temp = pred_temp0, lag = pred_lag, humidity =
> > pred_humidity, rain = pred_rain, time = pred_time)
> >
> > # make predictions
> > predictions <- predict.gam(mod, pd, type = "terms")
> > predictions0 <- predict.gam(mod, pd0, type = "terms")
> >
> > # calculate RR
> > diff <- predictions[,2] - predictions0[,2]
> > rr <- as.vector(exp(diff))
> >
> > # convert rr to matrix required for z argument for persp()
> > rr_mat <- matrix(rr, nrow = N)
> > persp(pred_temp, pred_lag, rr_mat)
> >
> >
> >
> >
> > On Fri, 22 Jul 2022 at 13:29, jade.shodan using googlemail.com
> > <jade.shodan using googlemail.com> wrote:
> >> I made a small error in the code below (not checking for NAs which are
> >> introduced by the lag function). However, this doesn't solve the issue
> >> I raised).
> >>
> >> So here's the problem (with corrected code) again:
> >>
> >> I'm not sure what to do with the 'time' variable. (I don't want to
> >> make predictions for specific points in time). I coded as follows
> >> (full reproducible example at bottom of email), but get a warning and
> >> error:
> >>
> >>
> >> N <- 1000     # number of points for smooth to be predicted
> >> # new temperatures and lags for prediction
> >> pred_temp <- seq(min(dat$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, na.rm = T), N)
> >> pred_rain <- rep(median(dat$rain, na.rm = T), N)
> >> pd <- data.frame(temp = pred_temp, lag = pred_lag, humidity =
> >> pred_humidity, rain = pred_rain)
> >>
> >> predictions <- predict(mod, pd, type = "terms")
> >>
> >>
> >> The predict line creates the following warning and error:
> >>
> >> Warning in predict.gam(mod, pd, type = "terms") :
> >>    not all required variables have been supplied in  newdata!
> >>
> >> Error in model.frame.default(ff, data = newdata, na.action = na.act) :
> >>    object is not a matrix
> >>
> >>
> >> For ease of reference, I've (re)included the full reproducible example:
> >>
> >> library(mgcv)
> >> set.seed(3) # make reproducible example
> >> simdat <- gamSim(1,400)
> >> g <- exp(simdat$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(simda
> >> t),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 12:47, jade.shodan using googlemail.com
> >> <jade.shodan using googlemail.com> wrote:
> >>> 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.wood using 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.shodan using googlemail.com
> >>>>> <jade.shodan using 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 using 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/
> >>>>
> --
> Simon Wood, School of Mathematics, University of Edinburgh,
> https://www.maths.ed.ac.uk/~swood34/
>