I have a data set that looks like this:
I'm interested in the best possible multilinear regression, that's why I'm trying this LASSO method.
R, which stands for stock market returns, should be the dependent variable, whereas all the others (except D/Date and P/Price) are independent variables.
Here's what I've tried so far:
library(Matrix)
library(foreach)
library(glmnet)
trainX <- spxdata[c(4:11)]
trainY <- spxdata[c(3)]
CV = cv.glmnet(x = trainX, y = trainY, alpha = 1, nlambda = 100)
and this gives me the following error message:
Error in storage.mode(y) <- "double" : (list) object cannot be coerced to type 'double'
I'm not accustomed to R and only use it rarely, so I'm not sure how to go about this problem. I guess it has something to do with the format of my trainX and trainY subset, but what exactly have I done wrong here?
The predictor matrix should be a matrix, and not a data frame, which is what you have there. Similarly, the response should be a vector, and not a one-column data frame.
You can get these with
trainX <- as.matrix(spxdata[4:11])
trainY <- spxdata[[3]] # not [3]
But in general, you may want to avoid these and other issues by using my glmnetUtils package, which implements a formula interface to glmnet. This lets you use it the same way you'd use glm or rpart or other modelling functions.
Related
Trying to use a more obscure package in my field of study. I am attempting to add a matrix of predictor variables. Wording in the package is as follows:
"An optional matrix of predictor variables for the time-intensity
parameters, where the columns represent the predictor variables. Cat-
egorical predictor variables need to be dummy coded."
Thus I convert into factors/characters which gives the following error which I attempt to solve in the code chunk (Xit_form being a 18x2 matrix of categorical variables)
"Error in crossprod(x, y) :
requires numeric/complex matrix/vector arguments "
xit_form[,1] <- as.factor(xit_form[,1])
xit_form[,2] <- as.factor(xit_form[,2])
mode(xit_form) <- "numeric"
however when i run this, the package does not seem to be treating it as a factor. Any way I can solve this?
Maybe this is completely off, but try model.matrix.
xit_form[] <- apply(xit_form, 2, factor)
model.matrix(~ 0 + ., data = as.data.frame(xit_form))
Data
set.seed(2021)
xit_form <- matrix(sample(letters[1:4], 20, TRUE), ncol = 2)
We may use dummies from fastDummies
library(fastDummies)
library(dplyr)
xit_form %>%
dummy_cols()
I have a csv file (single column, numeric values) called "y" that consists of zeros and ones where the rows with the value 1 indicate the target variable for logistic regression, and another file called "x" with the same number of rows and with columns of numeric predictor values. How do I load these so that I can then use cv.glmnet, i.e.
x <- read.csv('x',header=FALSE,sep=",")
y <- read.csv('y',header=FALSE )
is throwing an error
Error in y %*% rep(1, nc) :
requires numeric/complex matrix/vector arguments
when I call
cvfit = cv.glmnet(x, y, family = "binomial")
I know that "y" should be loaded as a "factor," but how do I do this? My online searches have found all sorts of approaches that have just confused me. What is the simple one-liner to just load this data ready for glmnet?
The cv.glmnet requires data to be provided in vector or matrix format. You can use the following code
xmat = as.matrix(x)
yvec = as.vector(y)
Then use
cvfit = cv.glmnet(xmat, yvec, family = "binomial")
If you can provide your data in dput() format, I can give a try.
I am trying to solve the digit Recognizer competition in Kaggle and I run in to this error.
I loaded the training data and adjusted the values of it by dividing it with the maximum pixel value which is 255. After that, I am trying to build my model.
Here Goes my code,
Given_Training_data <- get(load("Given_Training_data.RData"))
Given_Testing_data <- get(load("Given_Testing_data.RData"))
Maximum_Pixel_value = max(Given_Training_data)
Tot_Col_Train_data = ncol(Given_Training_data)
training_data_adjusted <- Given_Training_data[, 2:ncol(Given_Training_data)]/Maximum_Pixel_value
testing_data_adjusted <- Given_Testing_data[, 2:ncol(Given_Testing_data)]/Maximum_Pixel_value
label_training_data <- Given_Training_data$label
final_training_data <- cbind(label_training_data, training_data_adjusted)
smp_size <- floor(0.75 * nrow(final_training_data))
set.seed(100)
training_ind <- sample(seq_len(nrow(final_training_data)), size = smp_size)
training_data1 <- final_training_data[training_ind, ]
train_no_label1 <- as.data.frame(training_data1[,-1])
train_label1 <-as.data.frame(training_data1[,1])
svm_model1 <- svm(train_label1,train_no_label1) #This line is throwing an error
Error : Error in predict.svm(ret, xhold, decision.values = TRUE) : Model is empty!
Please Kindly share your thoughts. I am not looking for an answer but rather some idea that guides me in the right direction as I am in a learning phase.
Thanks.
Update to the question :
trainlabel1 <- train_label1[sapply(train_label1, function(x) !is.factor(x) | length(unique(x))>1 )]
trainnolabel1 <- train_no_label1[sapply(train_no_label1, function(x) !is.factor(x) | length(unique(x))>1 )]
svm_model2 <- svm(trainlabel1,trainnolabel1,scale = F)
It didn't help either.
Read the manual (https://cran.r-project.org/web/packages/e1071/e1071.pdf):
svm(x, y = NULL, scale = TRUE, type = NULL, ...)
...
Arguments:
...
x a data matrix, a vector, or a sparse matrix (object of class
Matrix provided by the Matrix package, or of class matrix.csr
provided by the SparseM package,
or of class simple_triplet_matrix provided by the slam package).
y a response vector with one label for each row/component of x.
Can be either a factor (for classification tasks) or a numeric vector
(for regression).
Therefore, the mains problems are that your call to svm is switching the data matrix and the response vector, and that you are passing the response vector as integer, resulting in a regression model. Furthermore, you are also passing the response vector as a single-column data-frame, which is not exactly how you are supposed to do it. Hence, if you change the call to:
svm_model1 <- svm(train_no_label1, as.factor(train_label1[, 1]))
it will work as expected. Note that training will take some minutes to run.
You may also want to remove features that are constant (where the values in the respective column of the training data matrix are all identical) in the training data, since these will not influence the classification.
I don't think you need to scale it manually since svm itself will do it unlike most neural network package.
You can also use the formula version of svm instead of the matrix and vectors which is
svm(result~.,data = your_training_set)
in your case, I guess you want to make sure the result to be used as factor，because you want a label like 1,2,3 not 1.5467 which is a regression
I can debug it if you can share the data:Given_Training_data.RData
Using R 3.2.0 with caret 6.0-41 and randomForest 4.6-10 on a 64-bit Linux machine.
When trying to use the predict() method on a randomForest object trained with the train() function from the caret package using a formula, the function returns an error.
When training via randomForest() and/or using x= and y= rather than a formula, it all runs smoothly.
Here is a working example:
library(randomForest)
library(caret)
data(imports85)
imp85 <- imports85[, c("stroke", "price", "fuelType", "numOfDoors")]
imp85 <- imp85[complete.cases(imp85), ]
imp85[] <- lapply(imp85, function(x) if (is.factor(x)) x[,drop=TRUE] else x) ## Drop empty levels for factors.
modRf1 <- randomForest(numOfDoors~., data=imp85)
caretRf <- train( numOfDoors~., data=imp85, method = "rf" )
modRf2 <- caretRf$finalModel
modRf3 <- randomForest(x=imp85[,c("stroke", "price", "fuelType")], y=imp85[, "numOfDoors"])
caretRf <- train(x=imp85[,c("stroke", "price", "fuelType")], y=imp85[, "numOfDoors"], method = "rf")
modRf4 <- caretRf$finalModel
p1 <- predict(modRf1, newdata=imp85)
p2 <- predict(modRf2, newdata=imp85)
p3 <- predict(modRf3, newdata=imp85)
p4 <- predict(modRf4, newdata=imp85)
Among the last 4 lines, only the second one p2 <- predict(modRf2, newdata=imp85) returns the following error:
Error in predict.randomForest(modRf2, newdata = imp85) :
variables in the training data missing in newdata
It seems that the reason for this error is that the predict.randomForest method uses rownames(object$importance) to determine the name of the variables used to train the random forest object. And when looking at
rownames(modRf1$importance)
rownames(modRf2$importance)
rownames(modRf3$importance)
rownames(modRf4$importance)
We see:
[1] "stroke" "price" "fuelType"
[1] "stroke" "price" "fuelTypegas"
[1] "stroke" "price" "fuelType"
[1] "stroke" "price" "fuelType"
So somehow, when using the caret train() function with a formula changes the name of the (factor) variables in the importance field of the randomForest object.
Is it really an inconsistency between the formula and and non-formula version of the caret train() function? Or am I missing something?
First, almost never use the $finalModel object for prediction. Use predict.train. This is one good example of why.
There is some inconsistency between how some functions (including randomForest and train) handle dummy variables. Most functions in R that use the formula method will convert factor predictors to dummy variables because their models require numerical representations of the data. The exceptions to this are tree- and rule-based models (that can split on categorical predictors), naive Bayes, and a few others.
So randomForest will not create dummy variables when you use randomForest(y ~ ., data = dat) but train (and most others) will using a call like train(y ~ ., data = dat).
The error occurs because fuelType is a factor. The dummy variables created by train don't have the same names so predict.randomForest can't find them.
Using the non-formula method with train will pass the factor predictors to randomForest and everything will work.
TL;DR
Use the non-formula method with train if you want the same levels or use predict.train
There can be two reasons why you get this error.
1. The categories of the categorical variables in the train and test sets don't match. To check that, you can run something like the following.
Well, first of all, it is good practice to keep the independent variables/features in a list. Say that list is "vars". And say, you separated "Data" into "Train" and "Test". Let's go:
for (v in vars){
if (class(Data[,v]) == 'factor'){
print(v)
# print(levels(Train[,v]))
# print(levels(Test[,v]))
print(all.equal(levels(Train[,v]) , levels(Test[,v])))
}
}
Once you find the non-matching categorical variables, you can go back, and impose the categories of Test data onto Train data, and then re-build your model. In a loop similar to above, for each nonMatchingVar, you can do
levels(Test$nonMatchingVar) <- levels(Train$nonMatchingVar)
2. A silly one. If you accidentally leave the dependent variable in the set of independent variables, you may run into this error message. I have done that mistake. Solution: Just be more careful.
Another way is to explicitly code the testing data using model.matrix, e.g.
p2 <- predict(modRf2, newdata=model.matrix(~., imp85))
I am doing just a regular logistic regression using the caret package in R. I have a binomial response variable coded 1 or 0 that is called a SALES_FLAG and 140 numeric response variables that I used dummyVars function in R to transform to dummy variables.
data <- dummyVars(~., data = data_2, fullRank=TRUE,sep="_",levelsOnly = FALSE )
dummies<-(predict(data, data_2))
model_data<- as.data.frame(dummies)
This gives me a data frame to work with. All of the variables are numeric. Next I split into training and testing:
trainIndex <- createDataPartition(model_data$SALE_FLAG, p = .80,list = FALSE)
train <- model_data[ trainIndex,]
test <- model_data[-trainIndex,]
Time to train my model using the train function:
model <- train(SALE_FLAG~. data=train,method = "glm")
Everything runs nice and I get a model. But when I run the predict function it does not give me what I need:
predict(model, newdata =test,type="prob")
and I get an ERROR:
Error in dimnames(out)[[2]] <- modelFit$obsLevels :
length of 'dimnames' [2] not equal to array extent
On the other hand when I replace "prob" with "raw" for type inside of the predict function I get prediction but I need probabilities so I can code them into binary variable given my threshold.
Not sure why this happens. I did the same thing without using the caret package and it worked how it should:
model2 <- glm(SALE_FLAG ~ ., family = binomial(logit), data = train)
predict(model2, newdata =test, type="response")
I spend some time looking at this but not sure what is going on and it seems very weird to me. I have tried many variations of the train function meaning I didn't use the formula and used X and Y. I used method = 'bayesglm' as well to check and id gave me the same error. I hope someone can help me out. I don't need to use it since the train function to get what I need but caret package is a good package with lots of tools and I would like to be able to figure this out.
Show us str(train) and str(test). I suspect the outcome variable is numeric, which makes train think that you are doing regression. That should also be apparent from printing model. Make it a factor if you want to do classification.
Max