I am trying to perform classification using Support Vector Machines in R using e1071 package. Using the following code, and specifying the cost and gamma parameters, I could train the models successfully.
svm_models <- lapply(training_data,
function(data)
{
svm(label~., data=data,
method="C-classification", kernel="radial",
cost=10, gamma=0.1)
})
But If I perform paramter tuning within the above function as the following code,
svmmodels <- lapply(trainingdata,
function(data)
{
params <- tune.svm(label~., data=data,
gamma=10^(-6:-2), cost=10^(1:2))
svm(label~., data=data,
method="C-classification", kernel="radial",
cost=params$best.parameter[[2]], gamma=params$best.parameter[[1]])
})
then I get the following error:
Error in predict.svm(ret, xhold, decision.values = TRUE) (from #4) :
Model is empty!
What could be the possible cause of this issue?
Thanks.
According to ?tune, it should be best.parameters, not best.parameter. Try adding the 's' at the end of both instances in your code, and see if it works.
Very difficult to say much definitive with no data for testing, (or even a description of the data). However, it is possible to say that your calling svm after tune.svm is not in keeping with the example in the e1071::tune help page. Furthermore the formal parameter that the "cost" and "price" parmeters should be given as list elements is "range". You should not need to run svm on the output.
Related
I am trying to use MLflow in R. According to https://www.mlflow.org/docs/latest/models.html#r-function-crate, the crate flavor needs to be used for the model. My model uses the Random Forest function implemented in the ranger package:
model <- ranger::ranger(formula = model_formula,
data = trainset,
importance = "impurity",
probability=T,
num.trees = 500,
mtry = 10)
The model itself works and I can do the prediction on a testset:
test_prediction <- predict(model, testset)
As a next step, I try to bring the model in the crate flavor. I follow here the approach shown in https://docs.databricks.com/_static/notebooks/mlflow/mlflow-quick-start-r.html.
predictor <- crate(function(x) predict(model,.x))
This results however in an error, when I apply the "predictor" on the testset
predictor(testset)
Error in predict(model, .x) : could not find function "predict"
Does anyone know how to solve this issue? To I have to transfer the prediction function differently in the crate function? Any help is highly appreciated ;-)
In my experience, that Databricks quickstart guide is wrong.
According to the Carrier documentation, you need to use explicit namespaces when calling non-base functions inside of crate. Since predict is actually part of the stats package, you'd need to specify stats::predict. Also, since your crate function depends on the global object named model, you'd need to pass that as an argument to the crate function as well.
Your code would end up looking something like this (I can't test it on your exact use case, since I don't have your data, but this works for me on MLflow in Databricks):
model <- ranger::ranger(formula = model_formula,
data = trainset,
importance = "impurity",
probability=T,
num.trees = 500,
mtry = 10)
predictor <- crate(function(x) {
stats::predict(model,x)
}, model = model)
predictor(testset)
I am very new to R, and this is my first time of encountering the eval() function. So I am trying to use the med and boot.med function from the following package: mma. I am using it to conduct mediation analysis. med and boot.med take in models such as linear models, and dataframes that specify mediators and predictors and then estimate the mediation effect of each mediator.
The author of the package gives the flexible option of specifying one's own custom.function. From the source code of med, it can be seen that the custom.function is passed to the eval(). So I tried insert the gbmt function as the custom function. However, R kept giving me error message: Error during wrapup: Number of trees to be used in prediction must be provided. I have been searching online for days and tried many ways of specifying the number of trees parameter n.trees, but nothing works (I believe others have raised similar issues: post 1, post 2).
The following codes are part of the source code of the med function:
cf1 = gsub("responseY", "y[,j]", custom.function[j])
cf1 = gsub("dataset123", "x2", cf1)
cf1 = gsub("weights123", "w", cf1)
full.model[[j]] <- eval(parse(text = cf1))
One custom function example the author gives in the package documentation is as follows:
temp1<-med(data=data.bin,n=2,custom.function = 'glm(responseY~.,data=dataset123,family="quasibinomial",
weights=weights123)')
Here the glm is the custom function. This example code works and you can replicate it easily (if you have mma installed and loaded). However when I am trying to use the gbmt function on a survival object, I got errors and here is what my code looks like:
temp1 <- med(data = data.surv,n=2,type = "link",
custom.function = 'gbmt(responseY ~.,
data = dataset123,
distribution = dist,
train_params = start_stop,
cv_folds=10,
keep_gbm_data = TRUE,
)')
Anyone has any idea how the argument about number of trees n.trees can be added somewhere in the above code?
Many thanks in advance!
Update: in order to replicate the example code, please install mma and try the following:
library("mma")
data("weight_behavior") ##binary x #binary y
x=weight_behavior[,c(2,4:14)]
pred=weight_behavior[,3]
y=weight_behavior[,15]
data.bin<-data.org(x,y,pred=pred,contmed=c(7:9,11:12),binmed=c(6,10), binref=c(1,1),catmed=5,catref=1,predref="M",alpha=0.4,alpha2=0.4)
temp1<-med(data=data.bin,n=2) #or use self-defined final function
temp1<-med(data=data.bin,n=2, custom.function = 'glm(responseY~.,data=dataset123,family="quasibinomial",
weights=weights123)')
I changed the custom.function to gbmt and used a survival object as responseY and the error occurs. When I use the gbmt function on my data outside the med function, there is no error.
I have been trying to run an example code for supervised kohonen SOMs from https://clarkdatalabs.github.io/soms/SOM_NBA . When I tried to predict test set data I got the following error:
pos.prediction <- predict(NBA.SOM3, newdata = NBA.testing)
Error in FUN(X[[i]], ...) :
Data type not allowed: should be a matrix or a factor
I tried newdata = as.matrix(NBA.testing) but it did not help. Neither did as.factor().
Why does it happen? And how can I fix that?
You should put one more argument to the predict function, i.e. "whatmap", then set its value to 1.
The code would be like:
pos.prediction <- predict(NBA.SOM3, newdata = NBA.testing, whatmap = 1)
To verify the prediction result, you can check using:
table(NBA$Pos[-training_indices], pos.prediction$predictions[[2]], useNA = 'always')
The result may be different from that of the tutorial, since it did not declare the use of set.seed() function.
I suggest that the set.seed() with an arbitrary number in it was declared somewhere before the training phase.
For simplicity, put it once on the top most of your script, e.g.
set.seed(12345)
This will guarantee a reproducible result of your model next time you re-run your script.
Hope that will help.
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.
i trying do create a plot for my model create using SVM in e1071 package.
my code to build the model, predict and build confusion matrix is
ptm <- proc.time()
svm.classifier = svm(x = train.set.list[[0.999]][["0_0.1"]],
y = train.factor.list[[0.999]][["0_0.1"]],
kernel ="linear")
pred = predict(svm.classifier, test.set.list[[0.999]][["0_0.1"]], decision.values = TRUE)
time[["svm"]] = proc.time() - ptm
confmatrix = confusionMatrix(pred,test.factor.list[[0.999]][["0_0.1"]])
confmatrix
train.set.list and test.set.list contains the test and train set for several conditions. train and set factor has the true label for each set. Train.set and test.set are both documenttermmatrix.
Then i tried to see a plot of my data, i tried with
plot(svm.classifier, train.set.list[[0.999]][["0_0.1"]])
but i got the message:
"Error in plot.svm(svm.classifier, train.set.list[[0.999]][["0_0.1"]]) :
missing formula."
what i'm doing wrong? confusion matrix seems good to me even not using formula parameter in svm function
Without given code to run, it's hard to say exactly what the problem is. My guess, given
?plot.svm
which says
formula formula selecting the visualized two dimensions. Only needed if more than two input variables are used.
is that your data has more than two predictors. You should specify in your plot function:
plot(svm.classifier, train.set.list[[0.999]][["0_0.1"]], predictor1 ~ predictor2)