mgcv: relative risk from GAM with distributed lag

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 at googlemail.com
<jade.shodan at 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 at googlemail.com
<jade.shodan at 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 at 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 at googlemail.com
<jade.shodan at 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

--
Simon Wood, School of Mathematics, University of Edinburgh,
https://www.maths.ed.ac.uk/~swood34/