This should be fairly straightforward (although it would me more straightforward with a reproducible example ...) If you have a fitted model land1, then ## I'm picking arbitrary values here since I don't ## know what's sensible for your system pframe <- data.frame(area_forage_uncult=200:210) predict(land1,newdata=pframe,re.form=~0) The argument re.form=~0 tells the predict()...

Hard to tell for sure given your code is not exactly minimally reproducible, but almost certainly the problem is from: sample(DF, dim(DF)[1], rep=T) The problem is you are sampling the columns of the data frame, not the rows. Consider: DF <- data.frame(a=1:4, b=5:8) sample(DF, dim(DF)[1], rep=T) Produces: b b.1 b.2...

found it by testing all its methods.. ibk.buildClassifier(dataSet); rez2 = ibk.distributionForInstance(i2); //distrib int result = (int)rez3[0]; //it goes tha same with Kstar Came to realize that classifiers in weka normaly run with discrete data (equal steps from min to max). And my data is not all discrete. Ibk and Kstar...

r,hidden-markov-models,predict

I'm not a user of this package and this is not really an answer, but a comment would obscure some of the structures. It appears that the "proportion" value of your model is missing (so the structures are different. The "mean" value looks like this: $ mean :List of 5...

There's no need to create a matrix. Stata has commands that facilitate the task. Try estimates store and estimates restore. An example: clear set more off sysuse auto // initial regression/predictions regress price weight estimates store myest predict double resid, residuals // second regression/prediction regress price mpg predict double residdiff,...

Try this. In this way, you can still keep everything dynamic. variable.list<-names(dat) lin <- lm(as.formula(paste(variable.list[1],variable.list[2], sep="~") ), data=dat) Let me know if it works...

If your purposes are related to just one prediction you can just grab your coefficient with coef(mod) Or you can just build a simple equation like this. coef(mod)[1] + "Your_Value"*coef(mod)[2] ...

You can simplify with: library(ISLR) Hitters <- na.omit(Hitters) # remove NA set.seed(1) train <- sample(1:nrow(Hitters), nrow(Hitters)/2) # random sampling test <- (1:nrow(Hitters))[-train] # your definition of test was incorrect lm.fit <- lm(Salary ~ ., data = Hitters, subset = train) lm.pred <- predict(lm.fit, newdata = Hitters[test,]) dim(Hitters[test,]) # output 132*20...

r,statistics,prediction,lm,predict

There are ways to transform your response variable, G in this occasion but there needs to be a good reason to do this. For example, if you want the output to be probabilities between 0 and 1 and your response variable is binary (0,1) then you need a logistic regression....

You haven't provided a reproducible example (i.e., data and code that allows others to reproduce your error), but I don't have a problem when I try something similar with a built-in data frame: m1 = lm(mpg ~ wt + carb + qsec*hp, data=mtcars) pred.dat=data.frame(carb=2, hp=120, qsec=10, wt=2.5) predict(m1, newdata=pred.dat) 1...

scikit-learn,cluster-analysis,data-mining,predict,dbscan

Clustering is not classification. Clustering is unlabeled. If you want to squeeze it into a prediction mindset (which is not the best idea), then it essentially predicts without learning. Because there is no labeled training data available for clustering. It has to make up new labels for the data, based...

There is a way to do it in weka. You should look into clustering: https://www.youtube.com/watch?v=zjYUYJ2b4r8 I would also suggest trying to get more features (more columns) for better results....

First, here's some sample data set.seed(15) train <- data.frame(x1=sample(0:1, 100, replace=T), x2=rpois(100,10), y=sample(0:1, 100, replace=T)) test <- data.frame(x1=sample(0:1, 10, replace=T), x2=rpois(10,10)) Now we can fit the models. Here I place them in a list to make it easier to keep them together, and I also remove x1 from the model...

The squiggly mess happens because line(...) draws lines between successive points in the data's original order. Try this at the end. p <- data.frame(x=F_Div$Obs_Richness,y=predict(poly.mod)) p <- p[order(p$x),] lines(p) ...

r,ggplot2,predict,confidence-interval

There's a difference between a prediction interval and a confidence interval. Observe predict(LinearModel.2,newdata50,interval="predict") # fit lwr upr # 1 82.24791 72.58054 91.91528 predict(LinearModel.2,newdata50,interval="confidence") # fit lwr upr # 1 82.24791 80.30089 84.19494 ggplot draws the confidence interval, not the prediction interval....

r,machine-learning,classification,missing-data,predict

There are a number of ways to go about this but here is one. I also tried using it on your dataset but it's either too small, has too many linear combinations or something else because it's not converging. Amelia - http://fastml.com/impute-missing-values-with-amelia/ data(mtcars) mtcars1<-mtcars[rep(row.names(mtcars),10),] #increasing dataset #inserting NAs into dataset...

Always pass a data.frame to lm if you want to predict: a <- mtcars$mpg x <- data.matrix(cbind(mtcars$wt, mtcars$hp)) DF <- data.frame(a, x) xTest <- x[2,] # We will use this for prediction later fitCar <-lm(a ~ ., data = DF) yPred <- predict(fitCar, newdata = data.frame(X1 = xTest[1], X2 =...

r,formula,random-forest,caret,predict

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...

Upon further thinking (and reading an old article by Nick Cox), it occurred to me that statsby can be used to avoid the loop and speed up the program. Here's a comparison of their speed. Let's first prepare example data. set more off timer clear webuse nlswork,clear keep idcode ln_wage...

If you read the documentation for predict.lm, you will see the following. So, use the newdata argument to pass the newmodel data you imported to get predictions. predict(object, newdata, se.fit = FALSE, scale = NULL, df = Inf, interval = c("none", "confidence", "prediction"), level = 0.95, type = c("response", "terms"),...

This is due to the fact 2011^3 is a very big number (greater tha and this is causing the coeffiicent to be returned as NA. If you had inspected the models, you would have noticed this. coef(lm(attend ~ year + I(year^2) + I(year^3),ds)) # (Intercept) year I(year^2) I(year^3) # -7.025524e+04...

r,statistics,svm,predict,kernlab

From the documentation: argument scaled: A logical vector indicating the variables to be scaled. If scaled is of length 1, the value is recycled as many times as needed and all non-binary variables are scaled. Per default, data are scaled internally (both x and y variables) to zero mean and...

r,dynamic,linear-regression,predict

Unfortunately, the dynlm package does not provide a predict() method. At the moment the package completely separates the data pre-processing (which knows about functions like d(), L(), trend(), season() etc.) and the model fitting (which itself is not aware of the functions). A predict() method has been on my wishlist...

You can predict into a new data set of whatever length you want, you just need to make sure you assign the results to an existing vector of appropriate size. This line causes a problem because stackloss$predict1[-1] <- predict(stackloss.lm, newdata) because you can't assign and subset a non-existing vector at...