I have a matrix (x) containing 100 samples (rows) and 10000 independent features (columns). The observations are binary, either the sample is good or bad {0,1} (stored in vector y). I want to perform leave one out cross-validation and determine the Area Under Curve (AUC) for each feature separately (something like colAUC from CAtools package). I tried to use glmnet, but it didn't work. As it is said in manual, I tried to set the nfold parameter to be equal to the number of observations (100).
>result=cv.glmnet(x,y,nfolds=100,type.measure="auc",family="binomial")
And I'm getting these warnings:
>"Warning messages:
1: Too few (< 10) observations per fold for type.measure='auc' in
cv.lognet; changed to type.measure='deviance'. Alternatively, use smaller
value for nfolds
2: Option grouped=FALSE enforced in cv.glmnet, since < 3 observations per
fold"
Any ideas what I'm doing wrong? And is there any other way or R package to obtain LOO-balanced AUC values for each of the features?
I'll really appreciate any help. Thank you!
When you do a LOO-CV, you have a test set with only 1 sample in it, and you can of course not build an AUC with that. However, you can loop and store the predictions at each step:
k <- dim(x)[1]
predictions <- c()
for (i in 1:k) {
model <- glmnet(x[-i,], y[-i], family="binomial")
predictions <- c(predictions, predict(model, newx=x[i,]))
}
So that in the end you can make a ROC curve, for example:
library(pROC)
roc(y, predictions)
Related
I'm using the caret package with the leaps package to get the number of variables to use in a linear regression. How do I extract the model with the lowest RMSE that uses mdl$bestTune number of variables? If this can't be done are there functions in other packages you would recommend that allow for loocv of a stepwise linear regression and actually allow me to find the final model?
Below is reproducible code. From it, I can tell from mdl$bestTune that the number of variables should be 4 (even though I would have hoped for 3). It seems like I should be able to extract the variables from the third row of summary(mdl$finalModel) but I'm not sure how I would do this in a general case and not just this example.
library(caret)
set.seed(101)
x <- matrix(rnorm(36*5), nrow=36)
colnames(x) <- paste0("V", 1:5)
y <- 0.2*x[,1] + 0.3*x[,3] + 0.5*x[,4] + rnorm(36) * .0001
train.control <- trainControl(method="LOOCV")
mdl <- train(x=x, y=y, method="leapSeq", trControl = train.control, trace=FALSE)
coef(mdl$finalModel, as.double(mdl$bestTune))
mdl$bestTune
summary(mdl$finalModel)
mdl$results
Here's the context behind my question in case it's of interest. I have historical monthly returns hundreds of mutual fund. Each fund's returns will be a dependent variable that I'd like to regress against a set of returns on a handful (e.g. 5) factors. For each fund I want to run a stepwise regression. I expect only 1 to 3 of the five factors to be significant for any fund.
you can use:
coef(mdl$finalModel,unlist(mdl$bestTune))
I have the following data:
# actual value:
a <- c(26.77814,29.34224,10.39203,29.66659,20.79306,20.73860,22.71488,29.93678,10.14384,32.63233,24.82544,38.14778,25.12343,23.07767,14.60789)
# predicted value
p <- c(27.238142,27.492240,13.542026,32.266587,20.473063,20.508603,21.414882,28.536775,18.313844,32.082333,24.545438,30.877776,25.703430,22.397666,15.627892)
I already calculated MSE and RMSE for these two, but they're asking for AUC and ROC curve. How can I calculate it from this data using R? I thought AUC is for classification problems, was I mistaken? Can we still calculate AUC for numeric values like above?
Question:
I thought AUC is for classification problems, was I mistaken?
You are not mistaken. The area under the receiver operating characteristic curve can't be computed for two numeric vectors like in your example. It's used to determine how well your binary classifier stands up to a gold standard binary classifier. You need a vector of cases vs. controls, or levels for the a vector that put each value in one of two categories.
Here's an example of how you'd do this with the pROC package:
library(pROC)
# actual value
a <- c(26.77814,29.34224,10.39203,29.66659,20.79306,20.73860,22.71488,29.93678,10.14384,32.63233,24.82544,38.14778,25.12343,23.07767,14.60789)
# predicted value
p <- c(27.238142,27.492240,13.542026,32.266587,20.473063,20.508603,21.414882,28.536775,18.313844,32.082333,24.545438,30.877776,25.703430,22.397666,15.627892)
df <- data.frame(a = a, p = p)
# order the data frame according to the actual values
odf <- df[order(df$a),]
# convert the actual values to an ordered binary classification
odf$a <- odf$a > 12 # arbitrarily decided to use 12 as the threshold
# construct the roc object
roc_obj <- roc(odf$a, odf$p)
auc(roc_obj)
# Area under the curve: 0.9615
Here, we have arbitrarily decided that threshold for the gold standard (a) is 12. If that's the case, than observations that have a lower value than 12 are controls. The prediction (p) classifies very well, with an AUC of 0.9615. We don't have to decide on the threshold for our prediction classifier in order to determine the AUC, because it's independent of the threshold decision. We can slide up and down depending on whether it's more important to find cases or to not misclassify a control.
Important Note
I completely made up the threshold for the gold standard classifier. If you choose a different threshold (for the gold standard), you'll get a different AUC. For example, if we chose 28, the AUC would be 1. The AUC is independent of the threshold for the predictor, but absolutely depends on the threshold for the gold standard.
EDIT
To clarify the above note, which was apparently misunderstood, you were not mistaken. This kind of analysis is for classification problems. You cannot use it here without more information. In order to do it, you need a threshold for your a vector, which you don't have. You CAN'T make one up and expect to get a non made up result for the AUC. Because the AUC depends on the threshold for the gold standard classifier, if you just make up the threshold, as we did in the exercise above, you are also just making up the AUC.
I'm using the package glmnet, I need to run several LASSO analysis for the calibration of a large number of variables (%reflectance for each wavelength throughout the spectrum) against one dependent variable. I have a couple of doubts on the procedure and on the results I wish to solve. I show my provisional code below:
First I split my data in training (70% of n) and testing sets.
smp_size <- floor(0.70 * nrow(mydata))
set.seed(123)
train_ind <- sample(seq_len(nrow(mydata)), size = smp_size)
train <- mydata[train_ind, ]
test <- mydata[-train_ind, ]
Then I separate the target trait (y) and the independent variables (x) for each set as follows:
vars.train <- train[3:2153]
vars.test <- test[3:2153]
x.train <- data.matrix(vars.train)
x.test <- data.matrix(vars.test)
y.train <- train$X1
y.test <- test$X1
Afterwords, I run a cross-validated LASSO model for the training set and extract and writte the non-zero coefficients for lambdamin. This is because one of my concerns here is to note which variables (wavebands of the reflectance spectrum) are selected by the model.
install.packages("glmnet")
library(glmnet)
cv.lasso.1 <- cv.glmnet(y=y.train, x= x.train, family="gaussian", nfolds =
5, standardize=TRUE, alpha=1)
coef(cv.lasso.1,s=cv.lasso.1$lambda.min) # Using lambda min.
(cv.lasso.1)
install.packages("broom")
library(broom)
c <- tidy(coef(cv.lasso.1, s="lambda.min"))
write.csv(c, file = "results")
Finally, I use the function “predict” and apply the object “cv.lasso1” (the model obtained previously) to the variables of the testing set (x.2) in order to get the prediction of the variable and I run the correlation between the predicted and the actual values of Y for the testing set.
predict.1.2 <- predict(cv.lasso.1, newx=x.2, type = "response", s =
"lambda.min")
cor.test(x=c(predict.1.2), y=c(y.2))
This is a simplified code and had no problem so far, the point is that I would like to make a loop (of one hundred repetitions) of the whole code and get the non-zero coefficients of the cross-validated model as well as the correlation coefficient of the predicted vs actual values (for the testing set) for each repetition. I've tried but couldn't get any clear results. Can someone give me some hint?
thanks!
In general, running repeated analyses of the same type over and over on the same data can be tricky. And in your case, may not be necessary the way in which you have outlined it.
If you are trying to find the variables most predictive, you can use PCA, Principal Component Analysis to select variables with the most variation within the a variable AND between variables, but it does not consider your outcome at all, so if you have poor model design it will pick the least correlated data in your repository but it may not be predictive. So you should be very aware of all variables in the set. This would be a way of reducing the dimensionality in your data for a linear or logistic regression of some sort.
You can read about it here
yourPCA <- prcomp(yourData,
center = TRUE,
scale. = TRUE)
Scaling and centering are essential to making these models work right, by removing the distance between your various variables setting means to 0 and standard deviations to 1. Unless you know what you are doing, I would leave those as they are. And if you have skewed or kurtotic data, you might need to address this prior to PCA. Run this ONLY on your predictors...keep your target/outcome variable out of the data set.
If you have a classification problem you are looking to resolve with much data, try an LDA, Linear Discriminant Analysis which looks to reduce variables by optimizing the variance of each predictor with respect to the OUTCOME variable...it specifically considers your outcome.
require(MASS)
yourLDA =r <- lda(formula = outcome ~ .,
data = yourdata)
You can also set the prior probabilities in LDA if you know what a global probability for each class is, or you can leave it out, and R/ lda will assign the probabilities of the actual classes from a training set. You can read about that here:
LDA from MASS package
So this gets you headed in the right direction for reducing the complexity of data via feature selection in a computationally solid method. In looking to build the most robust model via repeated model building, this is known as crossvalidation. There is a cv.glm method in boot package which can help you get this taken care of in a safe way.
You can use the following as a rough guide:
require(boot)
yourCVGLM<- cv.glmnet(y = outcomeVariable, x = allPredictorVariables, family="gaussian", K=100) .
Here K=100 specifies that you are creating 100 randomly sampled models from your current data OBSERVATIONS not variables.
So the process is two fold, reduce variables using one of the two methods above, then use cross validation to build a single model from repeated trials without cumbersome loops!
Read about cv.glm here
Try starting on page 41, but look over the whole thing. The repeated sampling you are after is called booting and it is powerful and available in many different model types.
Not as much code and you might hope for, but pointing you in a decent direction.
I'm trying to boost a classification tree using the gbm package in R and I'm a little bit confused about the kind of predictions I obtain from the predict function.
Here is my code:
#Load packages, set random seed
library(gbm)
set.seed(1)
#Generate random data
N<-1000
x<-rnorm(N)
y<-0.6^2*x+sqrt(1-0.6^2)*rnorm(N)
z<-rep(0,N)
for(i in 1:N){
if(x[i]-y[i]+0.2*rnorm(1)>1.0){
z[i]=1
}
}
#Create data frame
myData<-data.frame(x,y,z)
#Split data set into train and test
train<-sample(N,800,replace=FALSE)
test<-(-train)
#Boosting
boost.myData<-gbm(z~.,data=myData[train,],distribution="bernoulli",n.trees=5000,interaction.depth=4)
pred.boost<-predict(boost.myData,newdata=myData[test,],n.trees=5000,type="response")
pred.boost
pred.boost is a vector with elements from the interval (0,1).
I would have expected the predicted values to be either 0 or 1, as my response variable z also consists of dichotomous values - either 0 or 1 - and I'm using distribution="bernoulli".
How should I proceed with my prediction to obtain a real classification of my test data set? Should I simply round the pred.boost values or is there anything I'm doing wrong with the predict function?
Your observed behavior is correct. From documentation:
If type="response" then gbm converts back to the same scale as the
outcome. Currently the only effect this will have is returning
probabilities for bernoulli.
So you should be getting probabilities when using type="response" which is correct. Plus distribution="bernoulli" merely tells that labels follows bernoulli (0/1) pattern. You can omit that and still model will run fine.
To proceed do predict_class <- pred.boost > 0.5 (cutoff = 0.5) or else plot ROC curve to decide on cutoff yourself.
Try using adabag. Class, probabilities, votes and error are inbuilt in adabag which makes it easy to interpret, and of course less lines of codes.
I am trying to create a ROC curve off the below. I get an error that states Error in prediction(bc_rf_predict_prob, bc_test$Class) :
Number of cross-validation runs must be equal for predictions and labels.
library(mlbench) #has the Breast Cancer dataset in it
library(caret)
data(BreastCancer) #two class model
bc_changed<-BreastCancer[2:11] #removes variables not to be used
#Create train and test/holdout samples (works fine)
set.seed(59)
bc_rand <- bc_changed[order(runif(699)), ] #699 observations
bc_rand <- sample(1:699, 499)
bc_train <- bc_changed[ bc_rand,]
bc_test <- bc_changed[-bc_rand,]
#random forest decision tree (works fine)
library(caret)
library(randomForest)
set.seed(59)
bc_rf <- randomForest(Class ~.,data=bc_train, ntree=500,na.action = na.omit, importance=TRUE)
#ROC
library(ROCR)
actual <- bc_test$Class
bc_rf_predict_prob<-predict(bc_rf, type="prob", bc_test)
bc.pred = prediction(bc_rf_predict_prob,bc_test$Class) #not work- error
Error-Error in prediction(bc_rf_predict_prob, bc_test$Class) :
Number of cross-validation runs must be equal for predictions and labels.
I think it is coming from the fact when I do the:
bc_rf_predict_prob<-predict(bc_rf, type="prob", bc_test)
I get a matrix as the result with two columns Benign and a list of its probabilities and a second column of Malignant and its list of probabilities. My logic tells me I should only have a vector of probabilities.
According to page 9 of the ROCR Library documentation, the prediction function has two required inputs, predictions and labels, which must have the same dimensions.
In the case of a matrix or data frame, all cross-validation runs must have the same length.
Since str(bc_rf_predict_prob) > [1] matrix [1:200, 1:2], this means str(bc_test$Class) should have a matching dimension.
It sounds like you only want the first column vector of bc_rf_predict_prob, but I can't be certain without looking at the data.