I have the following pipeline:
model = make_pipeline(
OneHotEncoder(use_cat_names=True),
LinearRegression()
)
model.fit(X_train, y_train)
When I return X_train or y_train, the dataset is still the original dataset. May i know how can i access the columns that have been transformed?
Related
I have been stumped on this problem for a very long time and cannot figure it out. I believe the issue stems from subsets of data.frame objects retaining information of the parent but I also feel it's causing issues when training h2o.deeplearning models on what I think is just my training set (though this may not be true). See below for sample code. I included comments to clarify what I'm doing but it's fairly short code:
dataset = read.csv("dataset.csv")[,-1] # Read dataset in but omit the first column (it's just an index from the original data)
y = dataset[,1] # Create response
X = dataset[,-1] # Create regressors
X = model.matrix(y~.,data=dataset) # Automatically create dummy variables
y=as.factor(y) # Ensure y has factor data type
dataset = data.frame(y,X) # Create final data.frame dataset
train = sample(length(y),length(y)/1.66) # Create training indices -- A boolean
test = (-train) # Create testing indices
h2o.init(nthreads=2) # Initiate h2o
# BELOW: Create h2o.deeplearning model with subset of dataset.
mlModel = h2o.deeplearning(y='y',training_frame=as.h2o(dataset[train,,drop=TRUE]),activation="Rectifier",
hidden=c(6,6),epochs=10,train_samples_per_iteration = -2)
predictions = h2o.predict(mlModel,newdata=as.h2o(dataset[test,-1])) # Predict using mlModel
predictions = as.data.frame(predictions) # Convert predictions to dataframe object. as.vector() caused issues for me
predictions = predictions[,1] # Extract predictions
mean(predictions!=y[test])
The problem is that if I evaluate this against my test subset I get almost 0% error:
[1] 0.0007531255
Has anyone encountered this issue? Have an idea of how to alleviate this problem?
It will be more efficient to use the H2O functions to load the data and split it.
data = h2o.importFile("dataset.csv")
y = 2 #Response is 2nd column, first is an index
x = 3:(ncol(data)) #Learn from all the other columns
data[,y] = as.factor(data[,y])
parts = h2o.splitFrame(data, 0.8) #Split 80/20
train = parts[[1]]
test = parts[[2]]
# BELOW: Create h2o.deeplearning model with subset of dataset.
mlModel = h2o.deeplearning(x=x, y=y, training_frame=train,activation="Rectifier",
hidden=c(6,6),epochs=10,train_samples_per_iteration = -2)
h2o.performance(mlModel, test)
It is hard to say what the problem with your original code is, without seeing the contents of dataset.csv and being able to try it. My guess is that train and test are not being split, and it is actually being trained on the test data.
I have R code for importing text data into R, remove stop words, stem words and then create a matrix. Below is the code for
Using the create container function to split the matrix into training and test data sets.
Use train_models function to create a model based on SVM.
Execute the model on test.
Then I save the model.
library("RTextTools")
container = create_container(matrix, as.numeric(as.factor(data[, 2])),
trainSize = 1:2800,testSize = 2801:3162, virgin = FALSE)
models = train_models(container,"SVM", kernel = "linear",cost =1)
results = classify_models(container, models)
save(models, file = "my_model1.rda")
I am not able to use the saved model for prediction on new data(matrix_new) using predict function.
p <- predict(models,matrix_new)
#Error in predict.svm(X[[1L]], ...) : test data does not match model !
My question is: Is it feasible to use saved models on new data to predict sentiment ? From the error it looks like there is mismatch between the words that were used while creating the model and the new data. Please clarify if my understanding is correct.
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)
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 caret::train there are many pre-processing options that can be passed via the 'preProcessing' argument. This makes life super-simple because the test data is then auto-magically pre-processed in the same manner as the training data when calling 'predict.train'. Is it possible to do the same with 'findCorrelation' and 'nearZeroVar' in some manner?
I clearly understand from the documentation why the following code does not work, but I am hoping this clarifies my question. Ideally, I could do the following.
library("caret")
set.seed (1234)
data (iris)
# split test vs training
train.index <- createDataPartition (y = iris[,5], p = 0.80, list = F)
train <- iris [ train.index, ]
test <- iris [-train.index, ]
# train the model after imputing the missing data
fit <- train (Species ~ .,
train,
preProcess = c("findCorrelation", "nearZeroVar"),
method = "rpart" )
predict (fit, test)
Right now, you are tied to whatever preProcess will do.
However, the next version (around the start of the year, I hope) will allow you to more easily write custom models and pre-processing. For example, you might want to down-sample the data etc.
Let me know if you would like to test that version when we have a beta availible.
Max