When using predict() on an object returned by glmmPQL (MASS package in R), I appear not to be able to return the standard errors. Here's a representative example of my workflow with some dummy data:
#### simulate some representative data
set.seed(1986)
dep <- rbinom(200, 1, .5) # dependent binomial variable
set.seed(1987)
ind <- rnorm(200) # Gaussian independent variable
set.seed(1988)
ran <- rep(1:5, 40) # random factor
##### use PQL to run binomial GLMM
anTest <- glmmPQL(dep~ind, random=~1|ran, family=binomial)
##### specify values of *ind* at which to predict. expand.grid() is overkill here...
newData <- expand.grid(ind=seq(min(ind), max(ind), length.out=100))
#### and generate predictions
pred <- predict(anTest, newdata=newData, type="response", level=0, se.fit=TRUE)
(newData <- data.frame(newData, fit=pred))
However, as you can see, even though se.fit is set to TRUE, only predictions are generated by the function. What's going on here? I've tried doing this with simulated Poisson and Gaussian data and I still get no standard errors. Help please!
I'm running R Studio v 0.98.490 on Apple OSX v. 10.9.1
Related
I am trying to understand the way the e1071 package obtains its SVM predictions in a two-class classification framework. Consider the following toy example.
library(mvtnorm)
library(e1071)
n <- 50
### Gaussians
eps <- 0.05
data1 <- as.data.frame(rmvnorm(n, mean = c(0,0), sigma=diag(rep(eps,2))))
data2 <- as.data.frame(rmvnorm(n, mean = c(1,1), sigma=diag(rep(eps,2))))
### Train Model
data_df <- as.data.frame(rbind(data1, data2))
data <- as.matrix(data_df)
data_df$y <- as.factor(c(rep(-1,n), rep(1,n)))
svm <- svm(y ~ ., data = data_df, kernel = "radial", gamma=1, type = "C-classification", scale = FALSE)
Having trained the SVM, I would like to write a function that uses the coefficients and the intercept to predict on a new data point.
Recall that the kernel trick guarantees that we can write the prediction on a new point as the weighted sum of the kernel evaluated at the support vectors and the new point itself (plus some intercept).
In other words: How to combine the following three terms
supportv <- svm$SV
coefs <- svm$coefs
intercept <- svm$rho
to get the prediction associated with the corresponding SVM?
If this is not possible, or too complicated, I would also switch to a different package.
I am working with the wine quality database.
I am studying regression trees depending on different variables as:
library(rpart)
library(rpart.plot)
library(rattle)
library(naniar)
library(dplyr)
library(ggplot2)
vinos <- read.csv(file = 'Wine.csv', header = T)
arbol0<-rpart(formula=quality~chlorides, data=vinos, method="anova")
fancyRpartPlot(arbol0)
arbol1<-rpart(formula=quality~chlorides+density, data=vinos, method="anova")
fancyRpartPlot(arbol1)
I want to calculate the mean square error to see if arbol1 is better than arbol0. I will use my own dataset since no more data is available. I have tried to do it as
aaa<-predict(object=arbol0, newdata=data.frame(chlorides=vinos$chlorides), type="anova")
bbb<-predict(object=arbol1, newdata=data.frame(chlorides=vinos$chlorides, density=vinos$density), type="anova")
and then substract manually the last column of the dataframe from aaa and bbb. However, I am getting an error. Can someone please help me?
This website could be useful for you. It's very important to split your dataset into train and test subsets before training your models. In the following code, I've done it with base functions, but there's another function called sample.split from the caTools package that does the same procedure. I attach you this website where you can see all the ways to split data in R.
Remember that the function of the Mean Squared Error (MSE) is the following one:
So, it's very simple to apply it with R. You just have to compute the mean of the squared difference between the observed (i.e, the response variable from your test subset) and predicted values (i.e, the values you have predicted from the model with the predict function).
A solution for your wine dataset could be this one, based on the previous website.
library(rpart)
library(dplyr)
library(data.table)
vinos <- fread(file = 'Winequality-red.csv', header = TRUE)
# Split data into train and test subsets
sample_index <- sample(nrow(vinos), size = nrow(vinos)*0.75)
train <- vinos[sample_index, ]
test <- vinos[-sample_index, ]
# Train regression trees models
arbol0 <- rpart(formula = quality ~ chlorides, data = train, method = "anova")
arbol1 <- rpart(formula = quality ~ chlorides + density, data = train, method = "anova")
# Make predictions for each model
pred0 <- predict(arbol0, newdata = test)
pred1 <- predict(arbol1, newdata = test)
# Calculate MSE for each model
mean((pred0 - test$quality)^2)
mean((pred1 - test$quality)^2)
i would like to develop a cox proportional hazard model with r, use it to predict input and evaluate the accuracy of the model. For the evaluation I would like to use the Brior score.
# import various packages, needed at some point of the script
library("survival")
library("survminer")
library("prodlim")
library("randomForestSRC")
library("pec")
library("rpart")
library("mlr")
library("Hmisc")
library("ipred")
# load lung cancer data
data("lung")
head(lung)
# recode status variable
lung$status <- lung$status-1
# Delete rows with missing values
lung <- na.omit(lung)
# split data into training and testing
## 80% of the sample size
smp_size <- floor(0.8 * nrow(lung))
## set the seed to make your partition reproducible
set.seed(123)
train_ind <- sample(seq_len(nrow(lung)), size = smp_size)
# training and testing data
train.lung <- lung[train_ind, ]
test.lung <- lung[-train_ind, ]
# time and failure event
s <- Surv(train.lung$time, train.lung$status)
# create model
cox.ph2 <- coxph(s~age+meal.cal+wt.loss, data=train.lung)
# predict
pred <- predict(cox.ph2, newdata = train.lung)
# evaluate
sbrier(s, pred)
as an outcome of the prediction I would expect the time (as in "when does this individuum experience failure). Instead I get values like this
[1] 0.017576359 -0.135928959 -0.347553969 0.112509137 -0.229301199 -0.131861582 0.044589175 0.002634008
[9] 0.345966978 0.209488560 0.002418358
What does that mean?
Furthermore sbrier does not work. Apparently it can not work with the prediction pred (no surprise there)
How do I solve this? How do I make a prediction with cox.ph2? How can I evaluate the model afterwards?
The predict() function won't return a time value, you have to specify the argument type = c("lp", "risk","expected","terms","survival") in the predict() function.
If you want to get the hazard ratios :
predict(cox.ph2, newdata = test.lung, type = "risk")
Note that you want to predict the values on the test set not the training set.
I have read that you can use AFT models in your case :
https://stats.stackexchange.com/questions/79362/how-to-get-predictions-in-terms-of-survival-time-from-a-cox-ph-model
You also can read this post :
Calculate the Survival prediction using Cox Proportional Hazard model in R
Hope it will help
I have been struggling with the following problem for some time and would be very grateful for any help. I am running a logit model in R using the mlogit function and am able to generate the predicted probability of choosing each alternative for a given value of the predictors as follows:
library(mlogit)
data("Fishing", package = "mlogit")
Fish <- mlogit.data(Fishing, varying = c(2:9), shape = "wide", choice = "mode")
Fish_fit<-Fish[-(1:4),]
Fish_test<-Fish[1:4,]
m <- mlogit(mode ~price+ catch | income, data = Fish_fit)
predict(m,newdata=Fish_test,)
I cannot, however, work out how to add confidence intervals to the predicted probability estimates. I have already tried adding arguments to the predict function, but none seem to generate them. Any ideas on how it can be achieved would be much appreciated.
One approach here is Monte Carlo simulation. You'd simulate repeated draws from a multivariate-normal sampling distribution whose parameters are given by your model results.
For each simulation, estimate your predicted probabilities, and use their empirical distribution over simulations to get your confidence intervals.
library(MASS)
est_betas <- m$coefficients
est_preds <- predict(m, newdata = Fish_test)
sim_betas <- mvrnorm(1000, m$coefficients, vcov(m))
sim_preds <- apply(sim_betas, 1, function(x) {
m$coefficients <- x
predict(m, newdata = Fish_test)
})
sim_ci <- apply(sim_preds, 1, quantile, c(.025, .975))
cbind(prob = est_preds, t(sim_ci))
# prob 2.5% 97.5%
# beach 0.1414336 0.10403634 0.1920795
# boat 0.3869535 0.33521346 0.4406527
# charter 0.3363766 0.28751240 0.3894717
# pier 0.1352363 0.09858375 0.1823240
I am trying to predict fitted values over data containing NAs, and based on a model generated by plm. Here's some sample code:
require(plm)
test.data <- data.frame(id=c(1,1,2,2,3), time=c(1,2,1,2,1),
y=c(1,3,5,10,8), x=c(1, NA, 3,4,5))
model <- plm(y ~ x, data=test.data, index=c("id", "time"),
model="pooling", na.action=na.exclude)
yhat <- predict(model, test.data, na.action=na.pass)
test.data$yhat <- yhat
When I run the last line I get an error stating that the replacement has 4 rows while data has 5 rows.
I have no idea how to get predict return a vector of length 5...
If instead of running a plm I run an lm (as in the line below) I get the expected result.
model <- lm(y ~ x, data=test.data, na.action=na.exclude)
As of version 2.6.2 of plm (2022-08-16), this should work out of the box: Predict out of sample on fixed effects model (from the NEWS file:
prediction implemented for fixed effects models incl. support for argument newdata and out-of-sample prediction. Help page (?predict.plm) added to specifically explain the prediction for fixed effects models and the out-of-sample case.
I think this is something that predict.plm ought to handle for you -- seems like an oversight on the package authors' part -- but you can use ?napredict to implement it for yourself:
pp <- predict(model, test.data)
na.stuff <- attr(model$model,"na.action")
(yhat <- napredict(na.stuff,pp))
## [1] 1.371429 NA 5.485714 7.542857 9.600000