I'm using bayesglm for a logistic regression problem. It's a dataset of 150 rows and 2000 variables. I'm trying to do variable selection and usually look at glmnet in caret::rfe. However there isn't a method for bayesglm.
Is there anyway to manually define a method for rfe?
As for the the question I can only think of rewriting lmFuncs$fit function, for example:
lmFuncs$fit<-function (x, y, first, last, ...){
tmp <- as.data.frame(x)
tmp$y <- y
bayesglm (y ~ ., family = gaussian, data = tmp)
}
and then do your rfe.fit with rfeControl(functions = lmFuncs)
Related
I am fitting a model using a random site-level effect using a generalized additive model, implemented in the mgcv package for R. I had been doing this using the function gam() however, to speed things up I need to shift to the bam() framework, which is basically the same as gam(), but faster. I further sped up fitting by passing the options bam(nthreads = N, discrete=T), where nthreads is the number of cores on my machine. However, when I use the discretization option, and then try to make predictions with my model on new data, while ignoring the random effect, I consistent get an error.
Here is code to generate example data and reproduce the error.
library(mgcv)
#generate data.
N <- 10000
x <- runif(N,0,1)
y <- (0.5*x / (x + 0.2)) + rnorm(N)*0.1 #non-linear relationship between x and y.
#uninformative random effect.
random.x <- as.factor(do.call(paste0, replicate(2, sample(LETTERS, N, TRUE), FALSE)))
#fit models.
fit1 <- gam(y ~ s(x) + s(random.x, bs = 're')) #this one takes ~1 minute to fit, rest faster.
fit2 <- bam(y ~ s(x) + s(random.x, bs = 're'))
fit3 <- bam(y ~ s(x) + s(random.x, bs = 're'), discrete = T, nthreads = 2)
#make predictions on new data.
newdat <- data.frame(runif(200, 0, 1))
colnames(newdat) <- 'x'
test1 <- predict(fit1, newdata=newdat, exclude = c("s(random.x)"), newdata.guaranteed = T)
test2 <- predict(fit2, newdata=newdat, exclude = c("s(random.x)"), newdata.guaranteed = T)
test3 <- predict(fit3, newdata=newdat, exclude = c("s(random.x)"), newdata.guaranteed = T)
Making predictions with the third model which uses discretization throws this error (which the other two do not):
Error in model.frame.default(object$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 need to cross-validate several glmer models on the same data so I've made a function to do this (I'm not interested in preexisting functions for doing this). I want to pass an arbitrary glmer model to my function as the only argument. Sadly, I can't figure out how to do this, and the interwebz won't tell me.
Ideally, I would like to do something like:
model = glmer(y ~ x + (1|z), data = train_folds, family = "binomial"
model2 = glmer(y ~ x2 + (1|z), data = train_folds, family = "binomial"
And then call cross_validation_function(model) and cross_validation_function(model2). The training data within the function is called train_fold.
However, I suspect I need to pass the model formula in different way using reformulate.
Here is an example of my function. The project is about predicting autism(ASD) from behavioral features. The data variable is da.
library(pacman)
p_load(tidyverse, stringr, lmerTest, MuMIn, psych, corrgram, ModelMetrics,
caret, boot)
cross_validation_function <- function(model){
#creating folds
participants = unique(da$participant)
folds <- createFolds(participants, 10)
cross_val <- sapply(seq_along(folds), function(x) {
train_folds = filter(da, !(as.numeric(participant) %in% folds[[x]]))
predict_fold = filter(da, as.numeric(participant) %in% folds[[x]])
#model to be tested should be passed as an argument here
train_model <- model
predict_fold <- predict_fold %>%
mutate(predictions_perc = predict(train_model, predict_fold, allow.new.levels = T),
predictions_perc = inv.logit(predictions_perc),
predictions = ifelse(predictions_perc > 0.5, "ASD","control"))
conf_mat <- caret::confusionMatrix(data = predict_fold$predictions, reference = predict_fold$diagnosis, positive = "ASD")
accuracy <- conf_mat$overall[1]
sensitivity <- conf_mat$byClass[1]
specificity <- conf_mat$byClass[2]
fixed_ef <- fixef(train_model)
output <- c(accuracy, sensitivity, specificity, fixed_ef)
})
cross_df <- t(cross_val)
return(cross_df)
}
Solution developed from the comment: Using as.formula strings can be converted into a formula which can passed as arguments to my function in the following way:
cross_validation_function <- function(model_formula){
...
train_model <- glmer(model_formula, data = da, family = "binomial")
...}
formula <- as.formula( "y~ x + (1|z"))
cross_validation_function(formula)
If you aim is to extract the model formula from a fitted model, the you can use
attributes(model)$call[[2]]. Then you can use this formula when fitting model with the cv folds.
mod_formula <- attributes(model)$call[[2]]
train_model = glmer(mod_formula , data = train_data,
family = "binomial")
I wonder if I can use such as for loop or apply function to do the linear regression in R. I have a data frame containing variables such as crim, rm, ad, wd. I want to do simple linear regression of crim on each of other variable.
Thank you!
If you really want to do this, it's pretty trivial with lapply(), where we use it to "loop" over the other columns of df. A custom function takes each variable in turn as x and fits a model for that covariate.
df <- data.frame(crim = rnorm(20), rm = rnorm(20), ad = rnorm(20), wd = rnorm(20))
mods <- lapply(df[, -1], function(x, dat) lm(crim ~ x, data = dat))
mods is now a list of lm objects. The names of mods contains the names of the covariate used to fit the model. The main negative of this is that all the models are fitted using a variable x. More effort could probably solve this, but I doubt that effort is worth the time.
If you are just selecting models, which may be dubious, there are other ways to achieve this. For example via the leaps package and its regsubsets function:
library("leapls")
a <- regsubsets(crim ~ ., data = df, nvmax = 1, nbest = ncol(df) - 1)
summa <- summary(a)
Then plot(a) will show which of the models is "best", for example.
Original
If I understand what you want (crim is a covariate and the other variables are the responses you want to predict/model using crim), then you don't need a loop. You can do this using a matrix response in a standard lm().
Using some dummy data:
df <- data.frame(crim = rnorm(20), rm = rnorm(20), ad = rnorm(20), wd = rnorm(20))
we create a matrix or multivariate response via cbind(), passing it the three response variables we're interested in. The remaining parts of the call to lm are entirely the same as for a univariate response:
mods <- lm(cbind(rm, ad, wd) ~ crim, data = df)
mods
> mods
Call:
lm(formula = cbind(rm, ad, wd) ~ crim, data = df)
Coefficients:
rm ad wd
(Intercept) -0.12026 -0.47653 -0.26419
crim -0.26548 0.07145 0.68426
The summary() method produces a standard summary.lm output for each of the responses.
Suppose you want to have response variable fix as first column of your data frame and you want to run simple linear regression multiple times individually with other variable keeping first variable fix as response variable.
h=iris[,-5]
for (j in 2:ncol(h)){
assign(paste("a", j, sep = ""),lm(h[,1]~h[,j]))
}
Above is the code which will create multiple list of regression output and store it in a2,a3,....
I have an x-matrix of 8 columns. I want to run glmnet to do a lasso regression. I know I need to call:
glmnet(x, y, family = "binomial", ...).
However, how do I get x to consider all one way interactions as well? Do I have to manually remake the data frame: if so, is there an easier way? I suppose I was hoping to do something using an R formula.
Yes, there is a convenient way for that. Two steps in it are important.
library(glmnet)
# Sample data
data <- data.frame(matrix(rnorm(9 * 10), ncol = 9))
names(data) <- c(paste0("x", 1:8), "y")
# First step: using .*. for all interactions
f <- as.formula(y ~ .*.)
y <- data$y
# Second step: using model.matrix to take advantage of f
x <- model.matrix(f, data)[, -1]
glmnet(x, y)
f <- as.formula( ~ .^2) should also work for including main effects and all pairwise interactions
I would like to perform a likelihood ratio test to determine the power of a model term in a DOE. Till now I have been using the p-value from the glm fit to do this and things have been fine. As I started to use the anova function, I realized that there does not seem to be an anova function designed to accept the input from a glm.fit function, only a glm function. Here is an example of what I would like to do:
X # This is a model matrix from matrix.model
y # These are the y values for the fit
tfit = glm.fit(X, y, family = poisson())
anova(tfit, test = 'LRT')
Typically I would assume that the anova function call would just need to be altered to anova.glm, but that is not the case. How can I get the glm.fit function output to be compatible with an anova function input?
The problem is that glm.fit does not output of class glm, but a raw list with all kinds of data about the model. This cannot be fed to anova.glm since this function expects an object of class glm as produced by the glm function. If you have the raw data available (thus not turned in to a model matrix, you can apply the glm function to this to produce the desired outcome.
X <- matrix(c(runif(10), rnorm(10)), ncol = 2)
y <- round(runif(10, 1, 5))
X.mm <- model.matrix(y ~ X)
model.fit.1 <- glm.fit(X.mm, y, family = poisson())
class(model.fit.1)
model.fit.2 <- glm(y ~ X, family = "poisson")
class(model.fit.2)
anova(model.fit.2, test = "LRT")
If you can't use the glm function and must use the glm.fit then you can construct the LRT yourself from the glm.fit output. For a start take the following function
LRT.glm.fit <- function(glm.fit.mod){
df.null <- glm.fit.mod$df.null
df.mod <- glm.fit.mod$df.residual
dev.null <- glm.fit.mod$null.deviance
dev.mod <- glm.fit.mod$deviance
dev.diff <- dev.null - dev.mod
p.value <- 1 - pchisq(dev.null - dev.mod, df.null - df.mod)
output <- c(round(df.null), round(df.mod), dev.null, dev.mod, p.value)
names(output) <- c("df.null", "df.mod", "dev.null", "dev.mod", "p.value")
output
}