rsq <- function(formula, Data1, indices) {
d <- Data1[indices,] # allows boot to select sample
fit <- lm(formula, Data1=d)
return(summary(fit)$r.square)
}
results = boot(data = Data1, statistic = rsq, R = 500)
When I execute the code, I get the following error:
Error in Data1[indices,] : incorrect number of dimensions
Background info: I am creating a predictive model using Linear Regressions. I would like to test my Predictive Model and through some research, I decided to use the Bootstrapping Method.
Credit goes to #Rui Barradas, check comments for original post.
If you read the help page for function boot::boot you will see that the function it calls has first argument data, then indices, then others. So change the order of your function definition to rsq <- function(Data1, indices, formula)
Another problem that I had was that I didn't define the Function.
Related
I would like to do something with MLFlow but I do not find any solution on Internet. I am working with MLFlow and R, and I want to save a regression model. The thing is that by the time I want to predict the testing data, I want to do some transformation of that data. Then I have:
data <- #some data with numeric regressors and dependent variable called 'y'
# Divide into train and test
ind <- sample(nrow(data), 0.8*nrow(data), replace = FALSE)
dataTrain <- data[ind,]
dataTest <- data[-ind,]
# Run model in the mlflow framework
with(mlflow_start_run(), {
model <- lm(y ~ ., data = dataTrain)
predict_fun <- function(model, data_to_predict){
data_to_predict[,3] <- data_to_predict[,3]/2
data_to_predict[,4] <- data_to_predict[,4] + 1
return(predict(model, data_to_predict))
}
predictor <- crate(~predict_fun(model,dataTest),model)
### Some code to use the predictor to get the predictions and measure the accuracy as a log_metric
##################
##################
##################
mlflow_log_model(predictor,'model')
}
As you can notice, my prediction function not only consists in predict the new data you are evaluating, but it also makes some transformations in the third and fourth columns. All examples I saw on the web use the function predict in the crate as the default function of R.
Once I save this model, when I run it in another notebook with some Test data, I get the error: "predict_fun" doesn't exist. That is because my algorithm has not saved this specific function. Do you know what can I do to save and specific prediction function that I have created instead of the default functions that are in R?
This is not the real example I am working with, but it is an approximation of it. The fact is that I want to save extra functions apart from the model itself.
Thank you very much!
The question is more or less as the title indicates. I would like to use the caret::train function with beta-binomial models made with glmmTMB package (although I am not opposed to other functions capable of fitting beta-binomial models) to calculate median absolute error (MdAE) estimates through jack-knife (leave-one-out) cross-validation. The glmmTMBControl function is already capable of estimating the optimal dispersion parameter but I was hoping to retain this information somehow as well... or having caret do the calculation possibly?
The dataset I am working with looks like this:
df <- data.frame(Effect = rep(seq(from = 0.05, to = 1, by = 0.05), each = 5), Time = rep(seq(1:20), each = 5))
Ideally I would be able to pass the glmmTMB function to trainControl like so:
BB.glmm1 <- train(Time ~ Effect,
data = df, method = "glmmTMB",
method = "", metric = "MAD")
The output would be as per the examples contained in train, although possibly with estimates for the dispersion parameter.
Although I am in no way opposed to work arounds - Thank you in advance!
I am unsure how to perform the required operation with caret without creating a custom method but I trust it is fairly easy to implement it with a for (lapply) loop.
In the example I will use the sleepstudy data set since your example data throws a bunch of warnings.
library(glmmTMB)
to perform LOOCV - for every row, create a model without that row and predict on that row:
data(sleepstudy,package="lme4")
LOOCV <- lapply(1:nrow(sleepstudy), function(x){
m1 <- glmmTMB(Reaction ~ Days + (Days|Subject),
data = sleepstudy[-x,])
return(predict(m1, sleepstudy[x,], type = "response"))
})
get the median of the residuals (I think this is MdAE? if not post a comment on how its calculated):
median(abs(unlist(LOOCV) - sleepstudy$Reaction))
AIM: The aim here was to find a suitable fit, using step functions, which uses age to describe wage, in the Wage dataset in the library ISLR.
PLAN:
To find a suitable fit, I'll try multiple fits, which will have different cut points. I'll use the glm() function (of the boot library) for the fitting purpose. In order to check which fit is the best, I'll use the cv.glm() function to perform cross-validation over the fitted model.
PROBLEM:
In order to do so, I did the following:
all.cvs = rep(NA, 10)
for (i in 2:10) {
lm.fit = glm(wage~cut(Wage$age,i), data=Wage)
all.cvs[i] = cv.glm(Wage, lm.fit, K=10)$delta[2]
}
But this gives an error:
Error in model.frame.default(formula = wage ~ cut(Wage$age, i), data =
list( : variable lengths differ (found for 'cut(Wage$age, i)')
Whereas, when I run the code given below, it runs.(It can be found here)
all.cvs = rep(NA, 10)
for (i in 2:10) {
Wage$age.cut = cut(Wage$age, i)
lm.fit = glm(wage~age.cut, data=Wage)
all.cvs[i] = cv.glm(Wage, lm.fit, K=10)$delta[2]
}
Hypotheses and Results:
Well, it might be possible that cut() and glm() might not work together. But this works:
glm(wage~cut(age,4),data=Wage)
Question:
So, basically we're using the cut() function, saving it's results in a variable, then using that variable in the glm() function. But we can't put the cut function inside the glm() function. And that too, only if the code is in a loop.
So, why is the first version of the code not working?
This is confusing. Any help appreciated.
I am trying to find the optimal "lambda" parameter for the Box-Cox transformation.
I am using the implementation from the MASS package, so I only need to create the model and extract the lambda.
Here is the code for the function:
library(MASS)
find_lambda <- function(x) {
# Function to find the best lambda for the Box-Cox transform
my_tmp <- data.frame(x = x) # Create a temporary data frame, to use it with the lm
str(my_tmp) # Gives the expected output
the_lm <- lm(x ~ 1, data = my_tmp) # Creates the linear model, no error here
print(summary(the_lm)) # Prints the summary, as expected
out <- boxcox(the_lm, plotit=FALSE) # Gives the error
best_lambda <- out$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))
I'm fitting GARCH model to the residuals of and ARIMA, and trying to apply ARCH(p) for p from 1 to 10 to compare the fitness. Here is my code. Errors are returned in the for loop part but I cannot figure out the reason why. Could anyone give some tips?
So for the single value p=1 the codes are as below and it's no problem.
fitone<- garchFit(~garch(1,0),data=logprice)
coef(fitone)
summary(fitone)
And for the for loop my codes go like
for (n in 1:10) {
fit [[n]]<- garchFit(~garch(n,0),data=logprice)
coef(fit[[n]])
summary(fit[[n]])
}
Error in .garchArgsParser(formula = formula, data = data, trace = FALSE) :
Formula and data units do not match.
I never wrote a loop code before. Can someone help me with the codes?
The problem is that generally one tries to evaluate all the variables in a formula in the context of the data= parameter, but your n variable isn't coming from logprice, it's coming from the global environment. You will need to dynamically create the formula. Here's one way to run all the models with lapply rather than a for look would be
library(fGarch)
#sample data
x.vec = as.vector(garchSim(garchSpec(rseed = 1985), n = 200)[,1])
fits <- lapply(1:10, function(n) {
garchFit(bquote(~garch(.(n),0)), data = x.vec, trace = FALSE)
})
and then we can get the coefs with
lapply(fits, coef)