I have a linear mixed effects model and I am trying to do variable selection. The model is testing the level of forest degradation in 1000 sampled points. Most points have no degradation, and so the dependent variable is highly skewed with many zeros. Therefore, I am using the Tweedie distribution to fit the model. My main question is: can the Tweedie distribution actually be used in the glmmLasso function? My second question is: do I even need to use this distribution in glmmLasso()? Any help is much appreciated!
When I run the function with family = tweedie(var.power=1.2,link.power=0) I get the following error:
Error in logLik.glmmLasso(y = y, yhelp = yhelp, mu = mu, family = family, :
object 'loglik' not found
If I change the link.power from 0 to 1 (which I think is not correct for my model, but just for the sake of figuring out the problem), I get a different error:
Error in grad.lasso[b.is.0] <- score.beta[b.is.0] - lambda.b * sign(score.beta[b.is.0]) :
NAs are not allowed in subscripted assignments
Here tweedie comes from the statmod package. A simple example:
library(tweedie)
library(tidyverse)
library(glmmLasso)
library(statmod)
power <- 2
mu <- 1
phi <- seq(2, 8, by=0.1)
set.seed(10000)
y <- rtweedie( 100, mu=mu, power=power, phi=3)
x <- rnorm(100)
z <- c(rep(1, 50), rep(2,50))
df = as.data.frame(cbind(y,x,z))
df$z = as.factor(df$z)
f = y ~ x
varSelect = glmmLasso(fix = f, rnd = list(z=~1), data = df,
lambda = 5, family = tweedie(var.power=1.2,link.power=0))
I created a hacked version of glmmLasso that incorporates the Tweedie distribution as an option and put it on Github. I had to change two aspects of the code:
add a clause to compute the log-likelihood if family$family == "Tweedie"
in a number of places where the code was essentially if (family$family in list_of_families) ..., add "Tweedie" as an option.
remotes::install_github("bbolker/glmmLasso-bmb")
packageVersion("glmmLasso")
## [1] ‘1.6.2.9000’
Your example runs for me now, but I haven't checked at all to see if the results are sensible.
Related
Consider this synthetic dataset for classification,
library(tidyverse)
library(iml)
library(randomForest)
# Generate data
set.seed(5)
x = matrix(rnorm(2000), nrow=500)
z = x %*% matrix(c(1,1,1,1), nrow=4)
y = round(1 / (1 + exp(-z)), 0) %>% as.integer()
x = cbind(x, rnorm(500))
y_factor = as.factor(y)
data = data.frame(x, y_factor)
# Train model
rf = randomForest(y_factor ~ X1+X2+X3+X4+X5, data=data, ntree = 50)
# Compute feature importance using iml package
x_df = data[,-6]
predictor_rf <- Predictor$new(rf, data=x_df, y=y_factor)
imp_rf <- FeatureImp$new(predictor_rf, loss = "ce")
plot(imp_rf)
Here, x is a matrix with 5 independent variables, 4 of them are related to the response, and the fith is just noise. Then I train a random forest algorithm and finally compute the variable importance using feature permutation from the iml package and obtain the output from the figure below. In the manual from the package says that:
The importance is measured as
the factor by which the model’s prediction error increases when the feature is shuffled.
So here, variable X4 obtained a feature importance value of 0.2, which means that the prediction error "increased" by a factor of 0.2. However, being 0.2 a factor smaller than 1 this means that the prediction error actually decreased when doing the permutation on X2, which makes no sense to me, because on one side, it would imply that just random shuffled numbers obtain better results than the actual variables, but on the other side, the current model with the original variable obtains an accuracy of 100%. Same interpretation could be seen in the rest of the variables, except for variable X5, which was noise and obtained an importance of 0.
So... what am I missing here? What is that 0.2 value?
I am trying to run a glmm with a beta distribution using the glmmTMB function (package glmmTMB). My response variable has a lot of 0 observations so I get this error when running the model
Error in eval(family$initialize) : y values must be 0 < y < 1
I have attached what my response variable looks like regular and also normalized (see image)
Zero values cannot occur in data that are truly Beta-distributed (the probability density of y==0 is either zero or infinite unless the first shape parameter is exactly 1.0). You can fit a zero-inflated Beta response by specifying ziformula. For example:
simulate data
set.seed(101)
y <- rbeta(1000, shape1 = 1, shape2 = 5)
y[sample(1000, replace= FALSE, size = 100)] <- 0
dd <- data.frame(y)
fit
library(glmmTMB)
glmmTMB(y ~ 1, ziformula = ~1, data = dd, family = beta_family)
This example doesn't have a random-effects component, but that doesn't change anything important.
I am fitting a model using a random site-level effect using a generalized additive model, implemented in the mgcv package for R. I had been doing this using the function gam() however, to speed things up I need to shift to the bam() framework, which is basically the same as gam(), but faster. I further sped up fitting by passing the options bam(nthreads = N, discrete=T), where nthreads is the number of cores on my machine. However, when I use the discretization option, and then try to make predictions with my model on new data, while ignoring the random effect, I consistent get an error.
Here is code to generate example data and reproduce the error.
library(mgcv)
#generate data.
N <- 10000
x <- runif(N,0,1)
y <- (0.5*x / (x + 0.2)) + rnorm(N)*0.1 #non-linear relationship between x and y.
#uninformative random effect.
random.x <- as.factor(do.call(paste0, replicate(2, sample(LETTERS, N, TRUE), FALSE)))
#fit models.
fit1 <- gam(y ~ s(x) + s(random.x, bs = 're')) #this one takes ~1 minute to fit, rest faster.
fit2 <- bam(y ~ s(x) + s(random.x, bs = 're'))
fit3 <- bam(y ~ s(x) + s(random.x, bs = 're'), discrete = T, nthreads = 2)
#make predictions on new data.
newdat <- data.frame(runif(200, 0, 1))
colnames(newdat) <- 'x'
test1 <- predict(fit1, newdata=newdat, exclude = c("s(random.x)"), newdata.guaranteed = T)
test2 <- predict(fit2, newdata=newdat, exclude = c("s(random.x)"), newdata.guaranteed = T)
test3 <- predict(fit3, newdata=newdat, exclude = c("s(random.x)"), newdata.guaranteed = T)
Making predictions with the third model which uses discretization throws this error (which the other two do not):
Error in model.frame.default(object$dinfo$gp$fake.formula[-2], newdata) :
variable lengths differ (found for 'random.x')
In addition: Warning message:
'newdata' had 200 rows but variables found have 10000 rows
How can I go about making predictions for a new dataset using the model fit with discretization?
newdata.gauranteed doesn't seem to be working for bam() models with discrete = TRUE. You could email the author and maintainer of mgcv and send him the reproducible example so he can take a look. See ?bug.reports.mgcv.
You probably want
names(newdat) <- "x"
as data frames have names.
But the workaround is just to pass in something for random.x
newdat <- data.frame(x = runif(200, 0, 1), random.x = random.x[[1]])
and then do your call to generate test3 and it will work.
The warning message and error are the result of you not specifying random.x in the newdata and then mgcv looking for random.x and finding it in the global environment. You should really gather that variables into a data frame and use the data argument when you are fitting your models, and try not to leave similarly named objects lying around in your global environment.
I'd like to use the CausalImpact package in R to estimate the impact of an intervention on infectious disease case counts. We typically characterize the distributions of case counts as either Poisson or negative binomial. The bsts() function allows us to specify the Poisson family. However this encountered an error in CausalImpact()
set.seed(1)
x1 <- 100 + arima.sim(model = list(ar = 0.999), n = 100)
y <- rpois(100, 1.2 * x1)
y[71:100] <- y[71:100] + 10
data <- cbind(y, x1)
pre.period <- c(1, 70)
post.period <- c(71, 100)
post.period.response <- y[post.period[1] : post.period[2]]
y[post.period[1] : post.period[2]] <- NA
ss <- AddLocalLevel(list(), y)
bsts.model <- bsts(y ~ x1, ss, family="poisson", niter = 1000)
impact <- CausalImpact(bsts.model = bsts.model,
post.period.response = post.period.response)
Error in rnorm(prod(dim(state.samples)), 0, sigma.obs) : invalid arguments
This is due to the fact that bsts.model has no sigma.obs slot when generated using family="poisson".
Am I doing this correctly or is there another way to use CausalImpact with Poisson data? (I'd also love to be able to use negative binomial data, but I won't get too greedy).
Last, is this the best place for coding issues for CausalImpact? I didn't see an Issues tab on the GitHub page.
The working data looks like:
set.seed(1234)
df <- data.frame(y = rnorm(1:30),
fac1 = as.factor(sample(c("A","B","C","D","E"),30, replace = T)),
fac2 = as.factor(sample(c("NY","NC","CA"),30,replace = T)),
x = rnorm(1:30))
The lme model is fitted as:
library(lme4)
mixed <- lmer(y ~ x + (1|fac1) + (1|fac2), data = df)
I used bootMer to run the parametric bootstrapping and I can successfully obtain the coefficients (intercept) and SEs for fixed&random effects:
mixed_boot_sum <- function(data){s <- sigma(data)
c(beta = getME(data, "fixef"), theta = getME(data, "theta"), sigma = s)}
mixed_boot <- bootMer(mixed, FUN = mixed_boot_sum, nsim = 100, type = "parametric", use.u = FALSE)
My first question is how to obtain the coefficients(slope) of each individual levels of the two random effects from the bootstrapping results mixed_boot ?
I have no problem extracting the coefficients(slope) from mixed model by using augment function from broom package, see below:
library(broom)
mixed.coef <- augment(mixed, df)
However, it seems like broom can't deal with boot class object. I can't use above functions directly on mixed_boot.
I also tried to modify the mixed_boot_sum by adding mmList( I thought this would be what I am looking for), but R complains as：
Error in bootMer(mixed, FUN = mixed_boot_sum, nsim = 100, type = "parametric", :
bootMer currently only handles functions that return numeric vectors
Furthermore, is it possible to obtain CI of both fixed&random effects by specifying FUN as well?
Now, I am very confused about the correct specifications for the FUN in order to achieve my needs. Any help regarding to my question would be greatly appreciated!
My first question is how to obtain the coefficients(slope) of each individual levels of the two random effects from the bootstrapping results mixed_boot ?
I'm not sure what you mean by "coefficients(slope) of each individual level". broom::augment(mixed, df) gives the predictions (residuals, etc.) for every observation. If you want the predicted coefficients at each level I would try
mixed_boot_coefs <- function(fit){
unlist(coef(fit))
}
which for the original model gives
mixed_boot_coefs(mixed)
## fac1.(Intercept)1 fac1.(Intercept)2 fac1.(Intercept)3 fac1.(Intercept)4
## -0.4973925 -0.1210432 -0.3260958 0.2645979
## fac1.(Intercept)5 fac1.x1 fac1.x2 fac1.x3
## -0.6288728 0.2187408 0.2187408 0.2187408
## fac1.x4 fac1.x5 fac2.(Intercept)1 fac2.(Intercept)2
## 0.2187408 0.2187408 -0.2617613 -0.2617613
## ...
If you want the resulting object to be more clearly named you can use:
flatten <- function(cc) setNames(unlist(cc),
outer(rownames(cc),colnames(cc),
function(x,y) paste0(y,x)))
mixed_boot_coefs <- function(fit){
unlist(lapply(coef(fit),flatten))
}
When run through bootMer/confint/boot::boot.ci these functions will give confidence intervals for each of these values (note that all of the slopes facW.xZ are identical across groups because the model assumes random variation in the intercept only). In other words, whatever information you know how to extract from a fitted model (conditional modes/BLUPs [ranef], predicted intercepts and slopes for each level of the grouping variable [coef], parameter estimates [fixef, getME], random-effects variances [VarCorr], predictions under specific conditions [predict] ...) can be used in bootMer's FUN argument, as long as you can flatten its structure into a simple numeric vector.