R Output of fGarch - r

I am modelling a time series as a GARCH(1,1)-process:
And the z_t are t-distributed.
In R, I do this in the fGarch-package via
model <- garchFit(formula = ~garch(1,1), cond.dist = "std", data=r)
Is this correct?
Now, I would like to understand the output of this to check my formula.
Obviously, model#fit$coefs gives me the coefficients and model#fitted gives me the fitted r_t.
But how do I get the fitted sigma_t and z_t?


I believe that the best way is to define extractor functions when generics are not available and methods when generics already exist.
The first two functions extract the values of interest from the fitted objects.
get_sigma_t <- function(x, ...){
x#sigma.t
}
get_z_t <- function(x, ...){
x#fit$series$z
}
Here a logLik method for objects of class "fGARCH" is defined.
logLik.fGARCH <- function(x, ...){
x#fit$value
}
Now use the functions, including the method. The data comes from the first example in help("garchFit").
N <- 200
r <- as.vector(garchSim(garchSpec(rseed = 1985), n = N)[,1])
model <- garchFit(~ garch(1, 1), data = r, trace = FALSE)
get_sigma_t(model) # output not shown
get_z_t(model) # output not shown
logLik(model)
#LogLikelihood
# -861.9494
Note also that methods coef and fitted exist, there is no need for model#fitted or model#fit$coefs, like is written in the question.
fitted(model) # much simpler
coef(model)
# mu omega alpha1 beta1
#3.541769e-05 1.081941e-06 8.885493e-02 8.120038e-01


It is a list structure. Can find the structure with
str(model)
From the structure, it is easier to extract with $ or #
model#fit$series$z
model#sigma.t

Related

Looping over objects in R

I am trying to loop over objects in R.
myfunc.linear.pred <- function(x){
linear.pred <- predict(object = x)
w <- exp(linear.pred)/(1+exp(linear.pred))
as.vector(w)
}
The function here works perfectly as it should. It returns a vector of 48 rows and it comes from the object x. Now 'x' is nothing but the full regression model from a GLM function (think: mod.fit <- glm (dep~indep, data = data)). The problem is that I have 20 different such ('mod.fit') objects and need to find predictions for each of these. I could literally repeat the code, but I was looking to find a neater solution. So what I want is a matrix with 48 rows and 20 columns for the above function. This is probably basic for an advanced user, but I have only ever used "apply" and "for" loops for numbers and never objects. I looked into lapply but couldn't figure it out.
I tried: (and this is probably dumb)
allmodels <- c(mod.fit, mod.fit2, mod.fit3)
lpred.matrix <- matrix(data=NA, nrow=48, ncol=20)
for(i in allmodels){
lpred.matrix[i,] <- myfunc.linear.pred(i)
}
which obviously won't work because allmodels has a class of "list" and it contains all the stuff from the GLM function. Hope someone can help. Thanks!

In order to use lapply, you must have a list object not a vector object. Something like this should work:
## Load data
data("mtcars")
# fit models
mod.fit1 <- glm (mpg~disp, data = mtcars)
mod.fit2 <- glm (mpg~drat, data = mtcars)
mod.fit3 <- glm (mpg~wt, data = mtcars)
# build function
myfunc.linear.pred <- function(x){
linear.pred <- predict(object = x)
w <- exp(linear.pred)/(1+exp(linear.pred))
as.vector(w)
}
# put models in a list
allmodels <- list("mod1" = mod.fit1, "mod2" = mod.fit2, "mod2" =
mod.fit3)
# use lapply and do.call to generate matrix of prediction results
df <- do.call('cbind', lapply(allmodels, function(x){
a <- myfunc.linear.pred(x)
}))
Hope this helps

Pasting object names inside functions

This is a follow-up question to this (see data and previous commands).
Starting with a list of models in mods, i am now able to find the model with the least AIC (corresponds to the best model):
mods <- lapply(methods, function(m)
update(amod.null, correlation = getFunction(m)(1, form = ~ x + y), method="ML"))
names(mods) <- methods
list.AIC <- lapply(mods, function(x) AIC(x))
best.mod <- names(which.min(list.AIC))
Now, i need to do some testing on the model, e.g. Tukey between dates. The syntax is very simple, e.g. for amod.null
library(multcomp)
res <- glht(amod.null, mcp(Date = "Tukey"))
The tricky part is, how can i tell glht to use the model which was put into best.mod (note: this is all happening within a loop). I tried
res <- glht(paste("mods$", as.factor(best.mod),sep = "") , mcp(Date = "Tukey"))
but to no avail, as glht needs to find a model-object in the first argument.
/edit:
Possibly useful:
names(mods)
[1] "corExp" "corGaus" "corLin" "corRatio" "corSpher"

Since the models are stored in the list mods, you can access the "best model" by using the index of which.min(list.AIC):
list.AIC <- sapply(mods, AIC)
best.mod <- mods[which.min(list.AIC)]
best.mod[[1]]

How to extract p-values from lmekin objects in coxme package

I want to be able to view the p-values for the lmekin object produced by the coxme package.
eg.
model= lmekin(formula = height ~ score + sex + age + (1 | IID), data = phenotype_df, varlist = kinship_matrix)
I tried:
summary(model)
coef(summary(model))
summary(model$coefficient$fixed)
fixef(model)/ sqrt(diag(vcov(model)) #(Calculates Z-scores but not p-values)
But these did not work. So how do I view the p-values for this linear mixed model?

It took me ages of searching to figure this out, but I noticed a lot of other similar questions without proper answers, so I wanted to answer this.
You use:
library(coxme)
print(model)
Note it is important to load the coxme package beforehand or it will not work.
I've also noticed a lot of posts about how to extract the p-value from lmekin objects, or how to extract the p-value from coxme objects in general. I wrote this function, which is based on the coxme:::print.coxme function code (to view code type coxme:::print.coxme directly into R). print calculates p-values on the fly - this function allows the extraction of p-values and saves them to an object.
Note that mod is your model, eg. mod <- lmekin(y~x+a+b)
Use print(mod) to double check that the tables match.
extract_coxme_table <- function (mod){
beta <- mod$coefficients$fixed
nvar <- length(beta)
nfrail <- nrow(mod$var) - nvar
se <- sqrt(diag(mod$var)[nfrail + 1:nvar])
z<- round(beta/se, 2)
p<- signif(1 - pchisq((beta/se)^2, 1), 2)
table=data.frame(cbind(beta,se,z,p))
return(table)
}

I arrived at this topic because I was looking for the same thing for just the coxme object. The function of IcedCoffee works with a micro adjustment:
extract_coxme_table <- function (mod){
beta <- mod$coefficients #$fixed is not needed
nvar <- length(beta)
nfrail <- nrow(mod$var) - nvar
se <- sqrt(diag(mod$var)[nfrail + 1:nvar])
z<- round(beta/se, 2)
p<- signif(1 - pchisq((beta/se)^2, 1), 2)
table=data.frame(cbind(beta,se,z,p))
return(table)
}

can mice() handle crr()? Fine-Gray model

My doubt is if it is possible to pool multiple imputation data set, from "mice()", on a fit model of Fine-Gray from "crr()", and if it is statistically correct...
example
library(survival)
library(mice)
library(cmprsk)
test1 <- as.data.frame(list(time=c(4,3,1,1,2,2,3,5,2,4,5,1, 4,3,1,1,2,2,3,5,2,4,5,1),
status=c(1,1,1,0,2,2,0,0,1,1,2,0, 1,1,1,0,2,2,0,0,1,1,2,0),
x=c(0,2,1,1,NA,NA,0,1,1,2,0,1, 0,2,1,1,NA,NA,0,1,1,2,0,1),
sex=c(0,0,0,NA,1,1,1,1,NA,1,0,0, 0,0,0,NA,1,1,1,1,NA,1,0,0)))
dat <- mice(test1,m=10, seed=1982)
#Cox regression: cause 1
models.cox1 <- with(dat,coxph(Surv(time, status==1) ~ x +sex ))
summary(pool(models.cox1))
#Cox regression: cause 1 or 2
models.cox <- with(dat,coxph(Surv(time, status==1 | status==2) ~ x +sex ))
models.cox
summary(pool(models.cox))
#### crr()
#Fine-Gray model
models.FG<- with(dat,crr(ftime=time, fstatus=status, cov1=test1[,c( "x","sex")], failcode=1, cencode=0, variance=TRUE))
summary(pool(models.FG))
#Error in pool(models.FG) : Object has no vcov() method.
models.FG

There are a couple of things that need to be done to get this to work.
Your initial data and imputation.
library(survival)
library(mice)
library(cmprsk)
test1 <- as.data.frame(list(time=c(4,3,1,1,2,2,3,5,2,4,5,1, 4,3,1,1,2,2,3,5,2,4,5,1),
status=c(1,1,1,0,2,2,0,0,1,1,2,0, 1,1,1,0,2,2,0,0,1,1,2,0),
x=c(0,2,1,1,NA,NA,0,1,1,2,0,1, 0,2,1,1,NA,NA,0,1,1,2,0,1),
sex=c(0,0,0,NA,1,1,1,1,NA,1,0,0, 0,0,0,NA,1,1,1,1,NA,1,0,0)))
dat <- mice(test1,m=10, print=FALSE)
There is no vcov method for crr models which mice requires, however,
we can access the covariance matrix using the model$var returned value.
So write own vcov method to extract, and also need a coef method.
vcov.crr <- function(object, ...) object$var # or getS3method('vcov','coxph')
coef.crr <- function(object, ...) object$coef
There is also an error in how the model is passed to with.mids: your code has cov1=test1[,c( "x","sex")], but really you want cov1 to use the imputed data. I am not sure how to correctly write this as an expression due to the cov1 requiring a matrix with relevant variables, but you can easily hard code a function.
# This function comes from mice:::with.mids
Andreus_with <-
function (data, ...) {
call <- match.call()
if (!is.mids(data))
stop("The data must have class mids")
analyses <- as.list(1:data$m)
for (i in 1:data$m) {
data.i <- complete(data, i)
analyses[[i]] <- crr(ftime=data.i[,'time'], fstatus=data.i[,'status'],
cov1=data.i[,c( "x","sex")],
failcode=1, cencode=0, variance=TRUE)
}
object <- list(call = call, call1 = data$call, nmis = data$nmis,
analyses = analyses)
oldClass(object) <- c("mira", "matrix")
return(object)
}
EDIT:
The mice internals have changed since this answer; it now uses the broom package to extract elements from the fitted crr model. So tidy and glance methods for crr models are required:
tidy.crr <- function(x, ...) {
co = coef(x)
data.frame(term = names(co),
estimate = unname(co),
std.error=sqrt(diag(x$var)),
stringsAsFactors = FALSE)
}
glance.crr <- function(x, ...){ }
The above code then allows the data to be pooled.
models.FG <- Andreus_with(dat)
summary(pool(models.FG))
Note that this gives warnings over df.residual not being defined, and so large samples are assumed. I'm not familiar with crr so a more sensible value can perhaps be extracted -- this would then be added to the tidy method. (mice version ‘3.6.0’)

R Passing linear model to another function inside a function

I am trying to find the optimal "lambda" parameter for the Box-Cox transformation.
I am using the implementation from the MASS package, so I only need to create the model and extract the lambda.
Here is the code for the function:
library(MASS)
find_lambda <- function(x) {
# Function to find the best lambda for the Box-Cox transform
my_tmp <- data.frame(x = x) # Create a temporary data frame, to use it with the lm
str(my_tmp) # Gives the expected output
the_lm <- lm(x ~ 1, data = my_tmp) # Creates the linear model, no error here
print(summary(the_lm)) # Prints the summary, as expected
out <- boxcox(the_lm, plotit=FALSE) # Gives the error
best_lambda <- out$x[which.max(out$y)] # Extracting the best fitting lambda
return(best_lambda)
}
find_lambda(runif(100))
It gives the following error:
Error in is.data.frame(data) : object 'my_tmp' not found
The interesting thing is that the very same code is working outside the function. In other words, for some reason, the boxcox function from the MASS package is looking for the variable in the global environment.
I don't really understand, what exactly is going on... Do you have any ideas?
P.S. I do not provide a software/hardware specification, since this error was sucessfully replicated on a number of my friends' laptops.
P.P.S. I have found the way to solve the initial problem in the forecast package, but I still would like to know, why this code is not working.

Sometimes user contributed packages don't always do a great job tracking the environments where calls were executed when manipulating functions calls. The quickest fix for you would be to change the line from
the_lm <- lm(x ~ 1, data = my_tmp)
to
the_lm <- lm(x ~ 1, data = my_tmp, y=True, qr=True)
Because if the y and qr are not requested from the lm call, the boxcox function tries to re-run lm with those parameters via an update call and things get mucked up inside a function scope.

Why don't let box-cox do the fitting?
find_lambda <- function(x) {
# Function to find the best lambda for the Box-Cox transform
my_tmp <- data.frame(x = x) # Create a temporary data frame, to use it with the lm
out <- boxcox(x ~ 1, data = my_tmp, plotit=FALSE) # Gives the error
best_lambda <- out$x[which.max(out$y)] # Extracting the best fitting lambda
return(best_lambda)
}
I think your scoping issue is with update.default which calls eval(call, parent.frame()) and my_tmp doesn't exist in the boxcox environment. Please correct me if I'm wrong on this.

boxcox cannot find your data. This maybe because of some scoping issue.
You can feed data in to boxcox function.
find_lambda <- function(x) {
# Function to find the best lambda for the Box-Cox transform
my_tmp <- data.frame(x = x) # Create a temporary data frame, to use it with the lm
str(my_tmp) # Gives the expected output
the_lm <- lm(x ~ 1, data = my_tmp) # Creates the linear model, no error here
print(summary(the_lm)) # Prints the summary, as expected
out <- boxcox(the_lm, plotit=FALSE, data = my_tmp) # feed data in here
best_lambda <- out$x[which.max(out$y)] # Extracting the best fitting lambda
return(best_lambda)
}
find_lambda(runif(100))

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