How can I use the simulate function on a pooled GLM from MICE?
I have my MICE-imputed data in the object miceData. On this data i made a model as such:
form3 <- survived ~ sex + age * pclass - 1
glm3 <- glm.mids(data=miceData, form3, family = binomial(link=logit))
glm_pooled <- pool(glm3)
I now which to simulate data from this model in an equlivivant way of the following:
form3 <- survived ~ sex + age * pclass - 1
glm3_ref <- glm(form3, family = binomial(link=logit), data = titanic)
yNew <- simulate(glm3_ref)[,1]
How can i do this?
So you need is a glm-object that has the coefficients you obtained in glm_pool but otherwise looks like a usual (i.e. not imputation-based) glm output.
You can do that by first creating a glm-object from one complete dataset (one imputation) and then changing the coefficients:
glm_sim <- glm(form3, family = binomial(link=logit), data = mice::complete(miceData, action = 1) )
glm_sim$coefficients <- glm_pooled$pooled$estimate #please check if the order of estimates is correct!
Then use glm_sim as you intended. However, as others (who know a lot more than I do) have pointed out, be very cautious with this because anything derived from other components of glm_sim is still only based on a single imputation sample and thus not valid. For instance, model diagnostics will be useless.
Related
I am attempting to get clustered SEs (at the school level in my data) with data that is both imputed (MICE) and weighted (CBPS). I have tried a couple different approaches that have thrown different errors.
This is what I have to start, which works fine:
library(tidyverse)
library(mice)
library(MatchThem)
library(CBPS)
tempdata <- mice(d, m = 10, maxit = 50, meth = "pmm", seed = 99)
weighted_data <- weightthem(trtmnt ~ x1 + x2 + x3,
data = tempdata,
method = "cbps",
estimand = "ATT")
Using this (https://www.r-bloggers.com/2021/05/clustered-standard-errors-with-r/) as a guide, I attempted all 3, which all resulted in various types of error messages.
My data is in a restricted server so unfortunately I can't bring it into here to reproduce things exactly, although if it's useful I could attempt to recreate some sample data.
So attempting with estimatr first, I get this error:
m1 <- estimatr::lm_robust(outcome ~ trtmnt + x1 + x2 + x3,
clusters = schoolID,
data = weighted_data)
Error in eval_tidy(mfargs[[da]], data = data) :
object 'schoolID' not found
I have no clue where the schoolID variable would have dropped out/not be recognized. It isn't part of the weighting procedure but it should still be in the data frame...if I use it as a covariate in a standard model without clustering, it's there.
I also attempted with miceadds and got this error:
m2 <- miceadds::lm.cluster(outcome ~ trtmnt + x1 + x2 + x3,
cluster = "schoolID",
data = weighted_data)
Error in as.data.frame.default(data) :
cannot coerce class `"wimids"` to a data.frame
And finally, with sandwich and lmtest:
library(sandwich)
library(lmtest)
m3 <- weighted_models <- with(weighted_data,
exp=lm(outcome ~ trtmnt + x1 + x2 + x3))
msandwich <- coeftest(m3, vcov = vcovCL, cluster = ~schoolID)
Error in UseMethod("estfun") :
no applicable method for `estfun` applied to an object of class "c(`mimira`, `mira`)"
Any ideas on any of the above methods, or where to go next?
You were really close. You need to use with(weighted_data, .) to fit a model in your weighted datasets, and you need to use estimatr::lm_robust() to get the clustered standard errors. So try the following:
weighted_models <- with(weighted_data,
estimatr::lm_robust(outcome ~ trtmnt + x1 + x2 + x3,
cluster = schoolID))
Your first and second approaches were incorrect because you supplied weighted_data to a single model as if it were a data frame, but it's not; it's a complicated wimids object. You need to use the with() infrastructure to fit a model to the imputed weighted data.
Your third approach was close, but coeftest() needs to be used on a single model, not a mimira object, which contains all the models fit the imputed datasets. Although you can use coeftest() inside with() with mira objects, you cannot do so with mimira objects from MatchThem. This is where estimatr::lm_robust() comes in since it is able to apply the clustering within each imputed dataset.
I also recommend you take a look at this blog post on estimating treatment effects after weighting with multiply imputed data. The only difference in your case to the code presented in the post is that you would change vcov = "HC3" to vcov = ~schoolID in whichever function you use.
I am trying to create a function to run a k-fold cross validation on a glmer object.
This is just data I got online (my dataset is quite large) so the model isn't the best but if I can get this to work using this data I should be able to switch it to my dataset quite easily.
I want to do a LOOCV(Leave One Out Cross-Validation)
"LOOCV(Leave One Out Cross-Validation) is a type of cross-validation approach in which each observation is considered as the validation set and the rest (N-1) observations are considered as the training set."
The outline I got was from Caroline's answer on this researchgate thread.
https://www.researchgate.net/post/Does_R_code_for_k-fold_cross_validation_of_a_nested_glmer_model_exist
#load libraries
library(tidyverse)
library(optimx)
library(lme4)
#add example data
Data <- read.csv("https://stats.idre.ucla.edu/stat/data/hdp.csv")
Data <- select(Data, remission, IL6, CRP, DID)
Data
Data$remission<- as.factor(Data$remission)
Data$DID<- as.factor(Data$DID)
#add ROW column
Data <- Data %>% mutate(ROW = row_number())
head(Data)
PTOT=NULL
for (i in 1:8825) { # i in total number of observations in dataset
##Data that will be predicted
DataC1=Data[unique(Data$ROW)==i,]
###To train the model
DataCV=Data[unique(DataC1$ROW)!=i,]
M1 <- glmer(remission ~ 1 + IL6 + CRP + ( 1 | DID ), data = DataCV, family = binomial, control = glmerControl(optimizer ='optimx', optCtrl=list(method='L-BFGS-B')))
P1=predict(M1, DataC1)
names(P1)=NULL
P1
PTOT= c(PTOT, P1)
}
R2cv=1-(sum((remission-PTOT)^2)/(length(PTOT))/(var(remission)))
This is the error I get
"Error: Invalid grouping factor specification, DID"
DataCV is empty.
For example:
i <- 1 ## first time through the loop
DataCV=Data[unique(DataC1$ROW)!=i,]
I think that should have been DataC$ROW), not DataC1$ROW.
A few other comments: a more compact version of your code would look something like this:
## fit the full model
M1 <- glmer(remission ~ 1 + IL6 + CRP + ( 1 | DID ), data = DataC,
family = binomial, control = glmerControl(optimizer ='optimx', optCtrl=list(method='L-BFGS-B')))
res <- numeric(nrow(DataCV))
for (i in 1:nrow(DataCV)) {
new_fit <- update(M1, data = dataC[-i,]
res[i] <- (predict(new_fit, newdata=dataC[i,]) - remission[i])^2
}
For a well-specified model LOOCV is asymptotically equivalent to AIC, so you might be doing a lot of work to get something that's not very different from the AIC (which you can get directly from a single model fit) ...
EDITED:
I'm trying to assess the effect of variables (e.g. presence of severe trauma) on a continous variable (here energy expenditure (=REE) in calories) over time (Day). The dataframe is called my_data. Amongst the variables
Following I would like to display the results using the mixed linear model for each assessed variable in one large file.
General concept:
REE ~ Time*predictor + (1 + Time | Case identifier)
(1) Starting creating the lmer model:
library(tidyverse)
library(ggpmisc)
library(sjPlot)
library(lme4)
mixed.modelloop <- function(x) {
lmer(REE ~ Day*(x) + (1 + Day | Studynumber),
data=my_data,
REML=FALSE,
na.action=na.omit,
control = lmerControl(check.nobs.vs.nRE = "ignore"))
}
(2) Then creating the predictors (x)
cols <- c(colnames(my_data))
(3) And then generating the overall purrr function:
output <- purrr::map(cols, ~ mixed.modelloop(.x) %>% tab_model)
(4) generating the file which should include all separate univariate mixed model analyses:
pdf(file="mixed linear models.pdf" )
output
dev.off()
Unfortunately currently after step (3) I'm getting the following error message:
Error in model.frame.default(data = my_data, na.action = na.omit, drop.unused.levels = TRUE, :
variable lengths differ (found for 'x')
Any idea on how to adapt the function to resolve this issue?
Thanks!
Formulas have special rules, you can't insert a string into them and expect them to work.
This should work, although you haven't given a reproducible example to test with ...
mixed.modelloop <- function(x) {
form <- reformulate(c(sprintf("Day*%s", x), "(1 + Day | Studynumber)"),
response = "REE")
lmer(form,
data=my_data,
REML=FALSE,
na.action=na.omit,
control = lmerControl(check.nobs.vs.nRE = "ignore"))
}
I have data from an Experience Sampling Study, which consists of 8140 observations nested in 106 participants. I want to test if there is a mediation, in which I also want to compare the predictors (X1= socialInteraction_tech, X2= socialInteraction_ftf, M = MPEE_int, Y= wellbeing). X1, X2, and M are person-mean centred in order to obtain the within-person effects. To account for the autocorrelation I have fit a model with an ARMA(2,1) structure. We control for time with the variable "obs".
This is the final model including all variables of interest:
fit_mainH1xmy <- lme(fixed = wellbeing ~ 1 + obs # Controls
+ MPEE_int_centred + socialInteraction_tech_centred + socialInteraction_ftf_centred,
random = ~ 1 + obs | ID, correlation = corARMA(form = ~ obs | ID, p = 2, q = 1),
data = file, method = "ML", na.action=na.exclude)
summary(fit_mainH1xmy)
The mediation is partial, as my predictor X still significantly predicts Y after adding M.
However, I can't find a way to calculate c'(cprime), the indirect effect.
I have found the mlma package, but it looks weird and requires me to do transformations to my data.
I have tried melting the data in a long format and using lmer() to fit the model (following https://quantdev.ssri.psu.edu/sites/qdev/files/ILD_Ch07_2017_Within-PersonMedationWithMLM.html), but lmer() does not let me take into account the moving average (MA-part of the ARMA(2,1) structure).
Does anyone know how I could now obtain the indirect effect?
http://www.statsci.org/data/oz/snails.txt
You can get data from here.
My data is 4*3*3*2 completely randomized design experiment data. I want to model the probability of survival in terms of the stimulus variables.
I tried ANOVA, but I'm not sure whether it's right or not.
Because I want to model the "probability", should I use logistic model??
(I also tried logistic model. But the data shows the sum of 0(Survived) and 1(Deaths). Even though it is not 0 and 1, can I use logistic??)
I want to put "probability" as Y variable.
So I used logit but it's not working.
The program says that y is Inf.
How can I use logit as Y variable in aov?
glm_a <- glm(Deaths ~ Exposure + Rel.Hum + Temp + Species, data = data,
family = binomial)
prob <- Deaths / 20
logitt <- log(prob / (1 - prob))
logmodel <- lm(logitt ~ data$Species + data$Exposure + data$Rel.Hum + data$Temp)
summary(logmodel)
A <- factor(data$Species, levels = c("A", "B"), labels = c(-1, 1))
glm_a <- glm(Y ~ data$Species * data$Exposure * data$Rel.Hum * data$Temp,
data=data, family = binomial)
summary(glm_a)
help("glm") should direct you to help("family"), which reveals the following
For the binomial and quasibinomial families the response can be specified in one of three ways:
As a factor: ‘success’ is interpreted as the factor not having the first level (and hence usually of having the second level).
As a numerical vector with values between 0 and 1, interpreted as the proportion of successful cases (with the total number of cases given by the weights).
As a two-column integer matrix: the first column gives the number of successes and the second the number of failures.
So for the question "How can I make logistic model with this data?", we can go with route #3 quite easily:
data <- read.table("http://www.statsci.org/data/oz/snails.txt", header = TRUE)
glm_a <- glm(cbind(Deaths, N - Deaths) ~ Species * Exposure * Rel.Hum * Temp,
data = data, family = binomial)
summary(glm_a)
# [output omitted]
As for the question "I tried ANOVA, but I'm not sure whether it's right or not. Because I want to model the "probability", should I use logistic model?", it's better to ask on Cross Validated