the data set is rather simple,
I was trying to see what the prediction of the result would be if i consider result1 as a numeric variable.
so I subsetted the dataset into two part, one for training and the other one for testing randomly, run the svm code, and predict the testing set
wh.train = wh[1:1106,]
wh.test = wh[1107:1120,]
model1 = svm(result1 ~., wh.train)
pred1 = predict(model1, wh.test)
then result come out to me is really weird,
named numeric(0)
Related
I am trying to create a for loop to index thorugh each individual response variable I have and train a model using the train() funciton within the Caret Package. I have about 30 response variable and 43 predictor variables. I can train each model individually but I would like to automate the process and have a for loop run through a model (I would like to eventually upscale to multiple models if possible, i.e. lm, rf, cubist, etc.). I then want to save each model to a dataframe along with R-squared values and RMSE values. The individual models that I currenlty have that will run for me goes as follows, with column 11 being the response variable and column 35-68 being predictor variables.
data_Mg <- subset(data_3, !is.na(Mg))
mg.lm <- train(Mg~., data=data_Mg[,c(11,35:68)], method="lm", trControl=control)
mg.cubist <- train(Mg~., data=data_Mg[,c(11,35:68)], method="cubist", trControl=control)
mg.rf <- train(Mg~., data=data_Mg[,c(11,35:68)], method="rf", trControl=control, na.action = na.roughfix)
max(mg.lm$results$Rsquared)
min(mg.lm$results$RMSE)
max(mg.cubist$results$Rsquared)
min(mg.cubist$results$RMSE)
max(mg.rf$results$Rsquared) #Highest R squared
min(mg.rf$results$RMSE)
This gives me 3 models with everything the relevant information that I need. Now for the for loop. I've only tried the lm model so far for this.
bucket <- list()
for(i in 1:ncol(data_4)) { # for-loop response variables, need to end it at response variables, rn will run through all variables
data_y<-subset(data_4, !is.na(i))#get rid of NA's in the "i" column
predictors_i <- colnames(data_4)[i] # Create vector of predictor names
predictors_1.1 <- noquote(predictors_i)
i.lm <- train(predictors_1.1~., data=data_4[,c(i,35:68)], method="lm", trControl=control)
bucket <- i.lm
#mod_summaries[[i - 1]] <- summary(lm(y ~ ., data_y[ , c("i.lm", predictors_i)]))
#data_y <- data_4
}
Below is the error code that I am getting, with Bulk_Densi being the first variable in predictors_1.1. The error code is that variable lengths differ so I originally thought that my issue was that quotes were being added around "Bulk_Densi" but after trying the NoQuote() function I have not gotten anywehre so I am unsure of where I am going wrong.
Error code that I am getting
Please let me know if I can provide any extra info and thanks in advance for the help! I've already tried the info in How to train several models within a loop for and was struggling with that as well.
I am new to R and trying to save my svm model in R and have read the documentation but still do not understand what is wrong.
I am getting the error "object is not a matrix" which would seem to mean that my data is not a matrix, but it is... so something is missing.
My data is defined as:
data = read.table("data.csv")
trainSet = as.data.frame(data[,1:(ncol(data)-1)])
Where the last line is my label
I am trying to define my model as:
svm.model <- svm(type ~ ., data=trainSet, type='C-classification', kernel='polynomial',scale=FALSE)
This seems like it should be correct but I am having trouble finding other examples.
Here is my code so far:
# load libraries
require(e1071)
require(pracma)
require(kernlab)
options(warn=-1)
# load dataset
SVMtimes = 1
KERNEL="polynomial"
DEGREE = 2
data = read.table("head.csv")
results10foldAll=c()
# Cross Fold for training and validation datasets
for(timesRun in 1:SVMtimes) {
cat("Running SVM = ",timesRun," result = ")
trainSet = as.data.frame(data[,1:(ncol(data)-1)])
trainClasses = as.factor(data[,ncol(data)])
model = svm(trainSet, trainClasses, type="C-classification",
kernel = KERNEL, degree = DEGREE, coef0=1, cost=1,
cachesize = 10000, cross = 10)
accAll = model$accuracies
cat(mean(accAll), "/", sd(accAll),"\n")
results10foldAll = rbind(results10foldAll, c(mean(accAll),sd(accAll)))
}
# create model
svm.model <- svm(type ~ ., data = trainSet, type='C-classification', kernel='polynomial',scale=FALSE)
An example of one of my samples would be:
10.135338 7.214543 5.758917 6.361316 0.000000 18.455875 14.082668 31
Here, trainSet is a data frame but in the svm.model function it expects data to be a matrix(where you are assigning trainSet to data). Hence, set data = as.matrix(trainSet). This should work fine.
Indeed as pointed out by #user5196900 you need a matrix to run the svm(). However beware that matrix object means all columns have same datatypes, all numeric or all categorical/factors. If this is true for your data as.matrix() may be fine.
In practice more than often people want to model.matrix() or sparse.model.matrix() (from package Matrix) which gives dummy columns for categorical variables, while having single column for numerical variables. But a matrix indeed.
In the past few days I have developed multiple PLS models in R for spectral data (wavebands as explanatory variables) and various vegetation parameters (as individual response variables). In total, the dataset comprises of 56. The first 28 (training set) have been used for model calibration, now all I want to do is to predict the response values for the remaining 28 observations in the tesset. For some reason, however, R keeps on the returning the fitted values of the calibration set for a given number of components rather than predictions for the independent test set. Here is what the model looks like in short.
# first simulate some data
set.seed(123)
bands=101
data <- data.frame(matrix(runif(56*bands),ncol=bands))
colnames(data) <- paste0(1:bands)
data$height <- rpois(56,10)
data$fbm <- rpois(56,10)
data$nitrogen <- rpois(56,10)
data$carbon <- rpois(56,10)
data$chl <- rpois(56,10)
data$ID <- 1:56
data <- as.data.frame(data)
caldata <- data[1:28,] # define model training set
valdata <- data[29:56,] # define model testing set
# define explanatory variables (x)
spectra <- caldata[,1:101]
# build PLS model using training data only
library(pls)
refl.pls <- plsr(height ~ spectra, data = caldata, ncomp = 10, validation =
"LOO", jackknife = TRUE)
It was then identified that a model comprising of 3 components yielded the best performance without over-fitting. Hence, the following command was used to predict the values of the 28 observations in the testing set using the above calibrated PLS model with 3 components:
predict(refl.pls, ncomp = 3, newdata = valdata)
Sensible as the output may seem, I soon discovered that all this piece of code generates are the fitted values of the PLS model for the calibration/training data, rather than predictions. I discovered this because the below code, in which newdata = is omitted, yields identical results.
predict(refl.pls, ncomp = 3)
Surely something must be going wrong, although I cannot seem to find out what specifically is. Is there someone out there who can, and is willing to help me move in the right direction?
I think the problem is with the nature of the input data. Looking at ?plsr and str(yarn) that goes with the example, plsr requires a very specific data frame that I find tricky to work with. The input data frame should have a matrix as one of its elements (in your case, the spectral data). I think the following works correctly (note I changed the size of the training set so that it wasn't half the original data, for troubleshooting):
library("pls")
set.seed(123)
bands=101
spectra = matrix(runif(56*bands),ncol=bands)
DF <- data.frame(spectra = I(spectra),
height = rpois(56,10),
fbm = rpois(56,10),
nitrogen = rpois(56,10),
carbon = rpois(56,10),
chl = rpois(56,10),
ID = 1:56)
class(DF$spectra) <- "matrix" # just to be certain, it was "AsIs"
str(DF)
DF$train <- rep(FALSE, 56)
DF$train[1:20] <- TRUE
refl.pls <- plsr(height ~ spectra, data = DF, ncomp = 10, validation =
"LOO", jackknife = TRUE, subset = train)
res <- predict(refl.pls, ncomp = 3, newdata = DF[!DF$train,])
Note that I got the spectral data into the data frame as a matrix by protecting it with I which equates to AsIs. There might be a more standard way to do this, but it works. As I said, to me a matrix inside of a data frame is not completely intuitive or easy to grok.
As to why your version didn't work quite right, I think the best explanation is that everything needs to be in the one data frame you pass to plsr for the data sources to be completely unambiguous.
I am currently taking the "Practical Machine Learning" course from Coursera right now and am having run across some strange behavior with the predict function. The question that has been asked was to train a tree and then make some predictions. So that I am not posting the answer here I have changed the dataset used for the problem. The code is as follows:
rm(list = ls())
library(rattle)
data(mtcars)
mtcars$vs = as.factor(mtcars$vs)
set.seed(125)
model = train(am ~ ., method = 'rpart', data = mtcars)
print(model)
fancyRpartPlot(model$finalModel)
sampleData = mtcars[1,]
sampleData[1,names(sampleData)] = rep(NA, length(names(sampleData)))
sampleData[1, c('wt')] = c(4)
predict(model, sampleData[1,], verbose = TRUE)
In the above code, there are two primary sections. The first builds the tree and the second (where sampleData starts) creates a small sample set of data to apply the model to. To make sure that I have the exact same structure as the original data I simply copy the first row of the training dataset and then set all the columns to NA. I then put data in only the columns that the decision tree needs (in this case the wt variable).
When I execute the above code, I get the following result:
Number of training samples: 32
Number of test samples: 0
rpart : 0 unknown predictions were added
numeric(0)
For reference, the following is the structure of the tree:
fancyRpartPlot(model$finalModel)
Can somebody help me to understand why the predict function is not returning a predicted value for the sampleData that I provided?
Unfortunately, even though rpart only used the wt variable in splits, prediction still requires the others to be present. Use a data set with the sample columns:
> predict(model, mtcars[1,])
[1] 0.8571429
Max
I am doing just a regular logistic regression using the caret package in R. I have a binomial response variable coded 1 or 0 that is called a SALES_FLAG and 140 numeric response variables that I used dummyVars function in R to transform to dummy variables.
data <- dummyVars(~., data = data_2, fullRank=TRUE,sep="_",levelsOnly = FALSE )
dummies<-(predict(data, data_2))
model_data<- as.data.frame(dummies)
This gives me a data frame to work with. All of the variables are numeric. Next I split into training and testing:
trainIndex <- createDataPartition(model_data$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