predict

问题 If I use undersampling in case of an unbalanced binary target variable to train a model, the prediction method calculates probabilities under the assumption of a balanced data set. How can I convert these probabilities to actual probabilities for the unbalanced data? Is the a conversion argument/function implemented in the mlr package or another package? For example: a <- data.frame(y=factor(sample(0:1, prob = c(0.1,0.9), replace=T, size=100))) a$x <- as.numeric(a$y)+rnorm(n=100, sd=1) task <

问题 I am getting the following error: $ operator is invalid for atomic vectors . I am getting the error when trying to calculate the prediction error for a logistic regression model. Here is the code and data I am using: install.packages("ElemStatLearn") library(ElemStatLearn) # training data train = vowel.train # only looking at the first two classes train.new = train[1:3] # test data test = vowel.test test.new = test[1:3] # performing the logistic regression train.new$y <- as.factor(train.new$y

问题 I am doing a sentiment analysis on twitter comments (in Kazakh language) using below R script. 3000 (1500sad, 1500happy) comments for the training set and 1000 (happy sad mixed) comments for the test set. Everything works great but at the end, the predicted values are showing all happy, which is not right. I have checked every function and all are working up until the naiveBayes function. I checked classifier values and they are correct. I think either naiveBayes or predict is messing things

问题 This question already has answers here : Predict() - Maybe I'm not understanding it (4 answers) Closed 2 years ago . I loaded the inbuilt R data 'women' which has a tabular data of average American women's height and corresponding weight. this table has 15 rows. Using this data I am trying to predict the weight for specific values of height. I made a linear model first and gave new values to predict. But R still comes up with the 15 figures from the original data. I am a beginner in

问题 My question is quite the same with this thread, however, since that seems not to have an satisfying answer yet, I think it is still appropriate to ask again along with reproducible codes. training <- read.csv("https://d396qusza40orc.cloudfront.net/predmachlearn/pml-training.csv")[,-1] testing <- read.csv("https://d396qusza40orc.cloudfront.net/predmachlearn/pml-testing.csv")[,-1] # Importing data library(e1071) # Load the required package for SVM svm_model <- svm(classe ~ pitch_arm + pitch

问题 Is there any difference between the predict() and forecast() functions in R? If yes, in which specific cases should they be used? 回答1: Intro predict -- for many kinds of R objects (models). Part of the base library. forecast -- for time series. Part of the forecast package. (See example). Example #load training data trnData = read.csv("http://www.bodowinter.com/tutorial/politeness_data.csv") model <- lm(frequency ~ attitude + scenario, trnData) #create test data tstData <- t(cbind(c("H1", "H"

问题 I want to predict the future values for my simple moving average model. I used the following procedure: x <- c(14,10,11,7,10,9,11,19,7,10,21,9,8,16,21,14,6,7) df <- data.frame(x) dftimeseries <- ts(df) library(TTR) smadf <- SMA(dftimeseries, 4) # lag is 4 library(forecast) forecasteddf <- forecast(smadf, 4) # future 4 values When run the above code, my forecast values are the same for all the next 4 days. Am I coding it correctly? Or, am I conceptually wrong? The same is the case with

问题 I am suppose to use the predict function to predict when fjbjor is 5.5 and I always get this warning message and I have tried many ways but it always comes so is there anyone who can see what I am doing wrong here This is my code fit.lm <- lm(fjbjor~amagn, data=bjor) summary(fit.lm) new.bjor<- data.frame(fjbjor=5.5) predict(fit.lm,new.bjor) and this comes out 1 2 3 4 5 6 7 8 9 10 11 5.981287 2.864521 9.988559 5.758661 4.645530 2.419269 4.645530 5.313409 6.871792 3.309773 4.200278 12 13 14 15

问题 > reg.len <- lm(chao1.ave ~ lg.std.len, b.div) # b.div is my data frame imported from a CSV file > reg.len Call: lm(formula = chao1.ave ~ lg.std.len, data = b.div) Coefficients: (Intercept) lg.std.len 282.4 -115.7 > newx <- seq(0.6, 1.4, 0.01) > prd.len <- predict(reg.len, newdata=data.frame(x=newx), interval="confidence", level=0.90, type="response") Error in eval(expr, envir, enclos) : object 'lg.std.len' not found I've tried doing the lm like this: lm(b.div$chao1.ave ~ b.div$lg.std.len) ,

问题 Shall I convert the output of ar.ols to be some type that predict can accept? y=rnorm(100, 0,1) z=rnorm(100, 0,1) yz=cbind(y,z) > output = ar.ols(yz, aic = F, order.max = 2, demean = F, intercept = T) > predict(output, n.ahead = 2, se.fit = F) x=as.data.frame(yz) # x is a data frame, and otherwise same as yz. > output = ar.ols(x, aic = F, order.max = 2, demean = F, intercept = T) > predict(output, n.ahead = 2, se.fit = F) Error in array(STATS, dims[perm]) : 'dims' cannot be of length 0 Thanks