Error when Bootstraping a Beta regression model in R with {betareg} - r

I need to bootstrap a beta regression model to check its robustness - because of a data point with a large cook's distance - with the boot package (other suggestions welcomed).
I have the following error:
Error in t.star[r, ] <- res[[r]] :
incorrect number of subscripts on matrix
Here's a reproductible example:
library(betareg)
library(boot)
fake_data <- data.frame(diet = as.factor(c(rep("A",10),rep("B",10))),
fat = c(runif(10,.1,.5),runif(10,.4,.9)) )
plot(fat~diet, data = fake_data)
my_beta_reg <- function(data,i){
data_i <- data[i,]
mod <- betareg(data_i[,"fat"] ~ data_i[,"diet"])
return(mod$coef)
}
b = boot(fake_data, statistic = my_beta_reg, R= 50)
Error in t.star[r, ] <- res[[r]] :
incorrect number of subscripts on matrix
What's the issue?
Thanks in advance.


The issue is that mod$coef is a list:
betareg(fat ~ diet, data = fake_data)$coef
#$mean
#(Intercept) dietB
# -1.275793 2.490126
#
#$precision
# (phi)
#20.59014
You need to unlist it or preferably use the function you are supposed to use for extraction of coefficients:
my_beta_reg <- function(data,i){
mod <- betareg(fat ~ diet, data = data[i,])
#unlist(mod$coef)
coef(mod)
}
b = boot(fake_data, statistic = my_beta_reg, R= 50)
print(b)
#ORDINARY NONPARAMETRIC BOOTSTRAP
#
#
#Call:
#boot(data = fake_data, statistic = my_beta_reg, R = 50)
#
#
#Bootstrap Statistics :
# original bias std. error
#t1* -1.275793 -0.019847377 0.2003523
#t2* 2.490126 0.009008892 0.2314521
#t3* 20.590142 8.265394485 17.2271497

Related

How to use the replicate function in R to repeat the function

I have a problem when using replicate to repeat the function.
I tried to use the bootstrap to fit
a quadratic model using concentration as the predictor and Total_lignin as the response and going to report an estimate of the maximum with a corresponding standard error.
My idea is to create a function called bootFun that essentially did everything within one iteration of a for loop. bootFun took in only the data set the predictor, and the response to use (both variable names in quotes).
However, the SD is 0, not correct. I do not know where is the wrong place. Could you please help me with it?
# Load the libraries
library(dplyr)
library(tidyverse)
# Read the .csv and only use M.giganteus and S.ravennae.
dat <- read_csv('concentration.csv') %>%
filter(variety == 'M.giganteus' | variety == 'S.ravennae') %>%
arrange(variety)
# Check the data
head(dat)
# sample size
n <- nrow(dat)
# A function to do one iteration
bootFun <- function(dat, pred, resp){
# Draw the sample size from the dataset
sample <- sample_n(dat, n, replace = TRUE)
# A quadratic model fit
formula <- paste0('resp', '~', 'pred', '+', 'I(pred^2)')
fit <- lm(formula, data = sample)
# Derive the max of the value of concentration
max <- -fit$coefficients[2]/(2*fit$coefficients[3])
return(max)
}
max <- bootFun(dat = dat, pred = 'concentration', resp = 'Total_lignin' )
# Iterated times
N <- 5000
# Use 'replicate' function to do a loop
maxs <- replicate(N, max)
# An estimate of the max of predictor and corresponding SE
mean(maxs)
sd(maxs)

Base package boot, function boot, can ease the job of calling the bootstrap function repeatedly. The first argument must be the data set, the second argument is an indices argument, that the user does not set and other arguments can also be passed toit. In this case those other arguments are the predictor and the response names.
library(boot)
bootFun <- function(dat, indices, pred, resp){
# Draw the sample size from the dataset
dat.sample <- dat[indices, ]
# A quadratic model fit
formula <- paste0(resp, '~', pred, '+', 'I(', pred, '^2)')
formula <- as.formula(formula)
fit <- lm(formula, data = dat.sample)
# Derive the max of the value of concentration
max <- -fit$coefficients[2]/(2*fit$coefficients[3])
return(max)
}
N <- 5000
set.seed(1234) # Make the bootstrap results reproducible
results <- boot(dat, bootFun, R = N, pred = 'concentration', resp = 'Total_lignin')
results
#
#ORDINARY NONPARAMETRIC BOOTSTRAP
#
#
#Call:
#boot(data = dat, statistic = bootFun, R = N, pred = "concentration",
# resp = "Total_lignin")
#
#
#Bootstrap Statistics :
# original bias std. error
#t1* -0.4629808 -0.0004433889 0.03014259
#
results$t0 # this is the statistic, not bootstrapped
#concentration
# -0.4629808
mean(results$t) # bootstrap value
#[1] -0.4633233
Note that to fit a polynomial, function poly is much simpler than to explicitly write down the polynomial terms one by one.
formula <- paste0(resp, '~ poly(', pred, ',2, raw = TRUE)')
Check the distribution of the bootstrapped statistic.
op <- par(mfrow = c(1, 2))
hist(results$t)
qqnorm(results$t)
qqline(results$t)
par(op)
Test data
set.seed(2020) # Make the results reproducible
x <- cumsum(rnorm(100))
y <- x + x^2 + rnorm(100)
dat <- data.frame(concentration = x, Total_lignin = y)

Confidence interval for quantile regression using bootstrap

I am trying to get the five types of bootstrap intervals for linear and quantile regression. I was able to bootstrap and find the 5 boostrap intervals (Quantile,Normal,Basic,Studentized and BCa) for the linear regression using Boot from car and boot.ci from boot. When i tried to do the same for quantile regression using rq from quantreg, it throws up an error. Here is the sample code
Creating the model
library(car)
library(quantreg)
library(boot)
newdata = Prestige[,c(1:4)]
education.c = scale(newdata$education, center=TRUE, scale=FALSE)
prestige.c = scale(newdata$prestige, center=TRUE, scale=FALSE)
women.c = scale(newdata$women, center=TRUE, scale=FALSE)
new.c.vars = cbind(education.c, prestige.c, women.c)
newdata = cbind(newdata, new.c.vars)
names(newdata)[5:7] = c("education.c", "prestige.c", "women.c" )
mod1 = lm(income ~ education.c + prestige.c + women.c, data=newdata)
mod2 = rq(income ~ education.c + prestige.c + women.c, data=newdata)
Booting linear and quantile regression
mod1.boot <- Boot(mod1, R=999)
boot.ci(mod1.boot, level = .95, type = "all")
dat2 <- newdata[5:7]
mod2.boot <- boot.rq(cbind(1,dat2),newdata$income,tau=0.5, R=10000)
boot.ci(mod2.boot, level = .95, type = "all")
Error in if (ncol(boot.out$t) < max(index)) { :
argument is of length zero
1) Why does boot.ci not work for quantile regression
2)Using this solution I got from stackexchange, I was able to find the quantile CI.
Solution for quantile(percentile CI) for rq
t(apply(mod2.boot$B, 2, quantile, c(0.025,0.975)))
how do i obtain other CI for bootstrap (normal, basic, studentized, BCa).
3) Also, my boot.ci command for linear regression produces this warning
Warning message:
In sqrt(tv[, 2L]) : NaNs produced
What does this signify?

Using summary.rq you can calculate boostrap standard errors of model coefficients.
Five boostrap methods (bsmethods) are available (see ?boot.rq).
summary(mod2, se = "boot", bsmethod= "xy")
# Call: rq(formula = income ~ education.c + prestige.c + women.c, data = newdata)
#
# tau: [1] 0.5
#
# Coefficients:
# Value Std. Error t value Pr(>|t|)
# (Intercept) 6542.83599 139.54002 46.88860 0.00000
# education.c 291.57468 117.03314 2.49139 0.01440
# prestige.c 89.68050 22.03406 4.07009 0.00010
# women.c -48.94856 5.79470 -8.44712 0.00000
To calculate bootstrap confidence intervals, you can use the following trick:
mod1.boot <- Boot(mod1, R=999)
set.seed(1234)
boot.ci(mod1.boot, level = .95, type = "all")
dat2 <- newdata[5:7]
set.seed(1234)
mod2.boot <- boot.rq(cbind(1,dat2),newdata$income,tau=0.5, R=10000)
# Create an object with the same structure of mod1.boot
# but with boostrap replicates given by boot.rq
mod3.boot <- mod1.boot
mod3.boot$R <- 10000
mod3.boot$t0 <- coef(mod2)
mod3.boot$t <- mod2.boot$B
boot.ci(mod3.boot, level = .95, type = "all")
# BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
# Based on 10000 bootstrap replicates
#
# CALL :
# boot.ci(boot.out = mod3.boot, type = "all", level = 0.95)
#
# Intervals :
# Level Normal Basic Studentized
# 95% (6293, 6838 ) (6313, 6827 ) (6289, 6941 )
#
# Level Percentile BCa
# 95% (6258, 6772 ) (6275, 6801 )

Thanks for everyone who helped. I was able to figure out the solution myself. I ran a loop calculating the coefficients of the quantile regression and then used boot and boot.ci respectively. Here is the code
Booting commands only, model creation from question
mod3 <- formula(income ~ education.c + prestige.c + women.c)
coefsf <- function(data,ind){
rq(mod3, data=newdata[ind,])$coef
}
boot.mod <- boot(newdata,coefsf,R=10000)
myboot.ci <- list()
for (i in 1:ncol(boot.mod$t)){
myboot.ci[[i]] <- boot.ci(boot.mod, level = .95, type =
c("norm","basic","perc", "bca"),index = i)
}
I did this as I wanted CI on all variables not just the intercept.

How to compute log loss in machine learning

The following code are used to produce the probability output of binary classification with Random Forest.
library(randomForest)
rf <- randomForest(train, train_label,importance=TRUE,proximity=TRUE)
prediction<-predict(rf, test, type="prob")
Then the result about prediction is as follows:
The true label about test data are known (named test_label). Now I want to compute logarithmic loss for probability output of binary classification. The function about LogLoss is as follows.
LogLoss=function(actual, predicted)
{
result=-1/length(actual)*(sum((actual*log(predicted)+(1-actual)*log(1-predicted))))
return(result)
}
How to compute logarithmic loss with probability output of binary classification. Thank you.

library(randomForest)
rf <- randomForest(Species~., data = iris, importance=TRUE, proximity=TRUE)
prediction <- predict(rf, iris, type="prob")
#bound the results, otherwise you might get infinity results
prediction <- apply(prediction, c(1,2), function(x) min(max(x, 1E-15), 1-1E-15))
#model.matrix generates a true probabilities matrix, where an element is either 1 or 0
#we subtract the prediction, and, if the result is bigger than 0 that's the correct class
logLoss = function(pred, actual){
-1*mean(log(pred[model.matrix(~ actual + 0) - pred > 0]))
}
logLoss(prediction, iris$Species)

I think the logLoss formula is a little bit wrong.
model <- glm(vs ~ mpg, data = mtcars, family = "binomial")
### OP's formula (Wrong)
logLoss1 <- function(pred, actual){
-1*mean(log(pred[model.matrix(~ actual + 0) - pred > 0]))
}
logLoss1(actual = model$y, pred = model$fitted.values)
# [1] 0.4466049
### Correct formula in native R
logLoss2 <- function(pred, actual){
-mean(actual * log(pred) + (1 - actual) * log(1 - pred))
}
logLoss2(actual = model$y, pred = model$fitted.values)
# [1] 0.3989584
## Results from various packages to verify the correct answer
### From ModelMetrics package
ModelMetrics::logLoss(actual = model$y, pred = model$fitted.values)
# [1] 0.3989584
### From MLmetrics package
MLmetrics::LogLoss(y_pred = model$fitted.values, y_true = model$y)
# [1] 0.3989584
### From reticulate package
sklearn.metrics <- import("sklearn.metrics")
sklearn.metrics$log_loss(y_true = model$y, y_pred = model$fitted.values)
# [1] 0.3989584
I used the R version 4.1.0 (2021-05-18).

predict() R function caret package errors: "newdata" rows different, "type" not accepted

I am running a logistic regression analysis using the caret package.
Data is input as a 18x6 matrix
everything is fine so far except the predict() function.
R is telling me the type parameter is supposed to be raw or prob but raw just spits out an exact copy of the last column (the values of the binomial variable). prob gives me the following error:
"Error in dimnames(out)[[2]] <- modelFit$obsLevels :
length of 'dimnames' [2] not equal to array extent
In addition: Warning message:
'newdata' had 7 rows but variables found have 18 rows"
install.packages("pbkrtest")
install.packages("caret")
install.packages('e1071', dependencies=TRUE)
#install.packages('caret', dependencies = TRUE)
require(caret)
library(caret)
A=matrix(
c(
64830,18213,4677,24761,9845,17504,22137,12531,5842,28827,51840,4079,1000,2069,969,9173,11646,946,66161,18852,5581,27219,10159,17527,23402,11409,8115,31425,55993,0,0,1890,1430,7873,12779,627,68426,18274,5513,25687,10971,14104,19604,13438,6011,30055,57242,0,0,2190,1509,8434,10492,755,69716,18366,5735,26556,11733,16605,20644,15516,5750,31116,64330,0,0,1850,1679,9233,12000,500,73128,18906,5759,28555,11951,19810,22086,17425,6152,28469,72020,0,0,1400,1750,8599,12000,500,1,1,1,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1
),
nrow = 18,
ncol = 6,
byrow = FALSE) #"bycol" does NOT exist
################### data set as vectors
a<-c(64830,18213,4677,24761,9845,17504,22137,12531,5842,28827,51840,4079,1000,2069,969,9173,11646,946)
b<-c(66161,18852,5581,27219,10159,17527,23402,11409,8115,31425,55993,0,0,1890,1430,7873,12779,627)
c<-c(68426,18274,5513,25687,10971,14104,19604,13438,6011,30055,57242,0,0,2190,1509,8434,10492,755)
d<-c(69716,18366,5735,26556,11733,16605,20644,15516,5750,31116,64330,0,0,1850,1679,9233,12000,500)
e<-c(73128,18906,5759,28555,11951,19810,22086,17425,6152,28469,72020,0,0,1400,1750,8599,12000,500)
f<-c(1,1,1,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1)
######################
n<-nrow(A);
K<-ncol(A)-1;
Train <- createDataPartition(f, p=0.6, list=FALSE) #60% of data set is used as training.
training <- A[ Train, ]
testing <- A[ -Train, ]
nrow(training)
#this is the logistic formula:
#estimates from logistic regression characterize the relationship between the predictor and response variable on a log-odds scale
mod_fit <- train(f ~ a + b + c + d +e, data=training, method="glm", family="binomial")
mod_fit
#this isthe exponential function to calculate the odds ratios for each preditor:
exp(coef(mod_fit$finalModel))
predict(mod_fit, newdata=training)
predict(mod_fit, newdata=testing, type="prob")

I'm not very sure to understand, but A is a matrix of (a,b,c,d,e,f). So you don't need to create two objects.
install.packages("pbkrtest")
install.packages("caret")
install.packages('e1071', dependencies=TRUE)
#install.packages('caret', dependencies = TRUE)
require(caret)
library(caret)
A=matrix(
c(
64830,18213,4677,24761,9845,17504,22137,12531,5842,28827,51840,4079,1000,2069,969,9173,11646,946,66161,18852,5581,27219,10159,17527,23402,11409,8115,31425,55993,0,0,1890,1430,7873,12779,627,68426,18274,5513,25687,10971,14104,19604,13438,6011,30055,57242,0,0,2190,1509,8434,10492,755,69716,18366,5735,26556,11733,16605,20644,15516,5750,31116,64330,0,0,1850,1679,9233,12000,500,73128,18906,5759,28555,11951,19810,22086,17425,6152,28469,72020,0,0,1400,1750,8599,12000,500,1,1,1,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1
),
nrow = 18,
ncol = 6,
byrow = FALSE) #"bycol" does NOT exist
A <- data.frame(A)
colnames(A) <- c('a','b','c','d','e','f')
A$f <- as.factor(A$f)
Train <- createDataPartition(A$f, p=0.6, list=FALSE) #60% of data set is used as training.
training <- A[ Train, ]
testing <- A[ -Train, ]
nrow(training)
And to predict a variable you must enter the explanatory variables and not the variable to predict
mod_fit <- train(f ~ a + b + c + d +e, data=training, method="glm", family="binomial")
mod_fit
#this isthe exponential function to calculate the odds ratios for each preditor:
exp(coef(mod_fit$finalModel))
predict(mod_fit, newdata=training[,-which(colnames(training)=="f")])
predict(mod_fit, newdata=testing[,-which(colnames(testing)=="f")])

Short answer, you should not include the explained variable, which is f in your predict equation. So you should do:
predict(mod_fit, newdata=training[, -ncol(training])
predict(mod_fit, newdata=testing[, -ncol(testing])
The issue with the warning message 'newdata' had 11 rows but variables found have 18 rows is because you run the regression using the whole data set (18 observations), but predict using just part of it (either 11 or 7).
EDIT: To simplify the data creation and glm processes we can do:
library(caret)
A <- data.frame(a = c(64830,18213,4677,24761,9845,17504,22137,12531,5842,28827,51840,4079,1000,2069,969,9173,11646,946),
b = c(66161,18852,5581,27219,10159,17527,23402,11409,8115,31425,55993,0,0,1890,1430,7873,12779,627),
c = c(68426,18274,5513,25687,10971,14104,19604,13438,6011,30055,57242,0,0,2190,1509,8434,10492,755),
d = c(69716,18366,5735,26556,11733,16605,20644,15516,5750,31116,64330,0,0,1850,1679,9233,12000,500),
e = c(73128,18906,5759,28555,11951,19810,22086,17425,6152,28469,72020,0,0,1400,1750,8599,12000,500),
f = c(1,1,1,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1))
Train <- createDataPartition(f, p=0.6, list=FALSE) #60% of data set is used as training.
training <- A[ Train, ]
testing <- A[ -Train, ]
mod_fit <- train(f ~ a + b + c + d + e, data=training, method="glm", family="binomial")

I try to run logistic regression model. I wrote this code:
install.packages('caret')
library(caret)
setwd('C:\\Users\\BAHOZ\\Documents\\')
D<-read.csv(file = "D.csv",header = T)
D<-read.csv(file = 'DataSet.csv',header=T)
names(D)
set.seed(111134)
Train<-createDataPartition(D$X, p=0.7,list = FALSE)
training<-D[Train,]
length(training$age)
testing<-D[-Train,]
length(testing$age)
mod_fit<-train(X~age + gender + total.Bilirubin + direct.Bilirubin + total.proteins + albumin + A.G.ratio+SGPT + SGOT + Alkphos,data=training,method="glm", family="binomial")
summary(mod_fit)
exp(coef(mod_fit$finalModel))
And I recived this message for last command:
(Intercept) age gender total.Bilirubin direct.Bilirubin total.proteins albumin A.G.ratio
0.01475027 1.01596886 1.03857883 1.00022899 1.78188072 1.00065332 1.01380334 1.00115742
SGPT SGOT Alkphos
3.93498241 0.05616662 38.29760014
By running this command I could predict my data,
predict(mod_fit , newdata=testing)
But if I set type="prob" or type="raw"
predict(mod_fit , newdata=testing, type = "prob")
it falls in error:
Error in dimnames(out) <- *vtmp* :
length of 'dimnames' [2] not equal to array extent

formula error inside function

I want use survfit() and basehaz() inside a function, but they do not work. Could you take a look at this problem. Thanks for your help. The following code leads to the error:
library(survival)
n <- 50 # total sample size
nclust <- 5 # number of clusters
clusters <- rep(1:nclust,each=n/nclust)
beta0 <- c(1,2)
set.seed(13)
#generate phmm data set
Z <- cbind(Z1=sample(0:1,n,replace=TRUE),
Z2=sample(0:1,n,replace=TRUE),
Z3=sample(0:1,n,replace=TRUE))
b <- cbind(rep(rnorm(nclust),each=n/nclust),rep(rnorm(nclust),each=n/nclust))
Wb <- matrix(0,n,2)
for( j in 1:2) Wb[,j] <- Z[,j]*b[,j]
Wb <- apply(Wb,1,sum)
T <- -log(runif(n,0,1))*exp(-Z[,c('Z1','Z2')]%*%beta0-Wb)
C <- runif(n,0,1)
time <- ifelse(T<C,T,C)
event <- ifelse(T<=C,1,0)
mean(event)
phmmd <- data.frame(Z)
phmmd$cluster <- clusters
phmmd$time <- time
phmmd$event <- event
fmla <- as.formula("Surv(time, event) ~ Z1 + Z2")
BaseFun <- function(x){
start.coxph <- coxph(x, phmmd)
print(start.coxph)
betahat <- start.coxph$coefficient
print(betahat)
print(333)
print(survfit(start.coxph))
m <- basehaz(start.coxph)
print(m)
}
BaseFun(fmla)
Error in formula.default(object, env = baseenv()) : invalid formula
But the following function works:
fit <- coxph(fmla, phmmd)
basehaz(fit)

It is a problem of scoping.
Notice that the environment of basehaz is:
environment(basehaz)
<environment: namespace:survival>
meanwhile:
environment(BaseFun)
<environment: R_GlobalEnv>
Therefore that is why the function basehaz cannot find the local variable inside the function.
A possible solution is to send x to the top using assign:
BaseFun <- function(x){
assign('x',x,pos=.GlobalEnv)
start.coxph <- coxph(x, phmmd)
print(start.coxph)
betahat <- start.coxph$coefficient
print(betahat)
print(333)
print(survfit(start.coxph))
m <- basehaz(start.coxph)
print(m)
rm(x)
}
BaseFun(fmla)
Other solutions may involved dealing with the environments more directly.

I'm following up on #moli's comment to #aatrujillob's answer. They were helpful so I thought I would explain how it solved things for me and a similar problem with the rpart and partykit packages.
Some toy data:
N <- 200
data <- data.frame(X = rnorm(N),W = rbinom(N,1,0.5))
data <- within( data, expr = {
trtprob <- 0.4 + 0.08*X + 0.2*W -0.05*X*W
Trt <- rbinom(N, 1, trtprob)
outprob <- 0.55 + 0.03*X -0.1*W - 0.3*Trt
Outcome <- rbinom(N,1,outprob)
rm(outprob, trtprob)
})
I want to split the data to training (train_data) and testing sets, and train the classification tree on train_data.
Here's the formula I want to use, and the issue with the following example. When I define this formula, the train_data object does not yet exist.
my_formula <- Trt~W+X
exists("train_data")
# [1] FALSE
exists("train_data", envir = environment(my_formula))
# [1] FALSE
Here's my function, which is similar to the original function. Again,
badFunc <- function(data, my_formula){
train_data <- data[1:100,]
ct_train <- rpart::rpart(
data= train_data,
formula = my_formula,
method = "class")
ct_party <- partykit::as.party(ct_train)
}
Trying to run this function throws an error similar to OP's.
library(rpart)
library(partykit)
bad_out <- badFunc(data=data, my_formula = my_formula)
# Error in is.data.frame(data) : object 'train_data' not found
# 10. is.data.frame(data)
# 9. model.frame.default(formula = Trt ~ W + X, data = train_data,
# na.action = function (x) {Terms <- attr(x, "terms") ...
# 8. stats::model.frame(formula = Trt ~ W + X, data = train_data,
# na.action = function (x) {Terms <- attr(x, "terms") ...
# 7. eval(expr, envir, enclos)
# 6. eval(mf, env)
# 5. model.frame.rpart(obj)
# 4. model.frame(obj)
# 3. as.party.rpart(ct_train)
# 2. partykit::as.party(ct_train)
# 1. badFunc(data = data, my_formula = my_formula)
print(bad_out)
# Error in print(bad_out) : object 'bad_out' not found
Luckily, rpart() is like coxph() in that you can specify the argument model=TRUE to solve these issues. Here it is again, with that extra argument.
goodFunc <- function(data, my_formula){
train_data <- data[1:100,]
ct_train <- rpart::rpart(
data= train_data,
## This solved it for me
model=TRUE,
##
formula = my_formula,
method = "class")
ct_party <- partykit::as.party(ct_train)
}
good_out <- goodFunc(data=data, my_formula = my_formula)
print(good_out)
# Model formula:
# Trt ~ W + X
#
# Fitted party:
# [1] root
# | [2] X >= 1.59791: 0.143 (n = 7, err = 0.9)
##### etc
documentation for model argument in rpart():
model:
if logical: keep a copy of the model frame in the result? If
the input value for model is a model frame (likely from an earlier
call to the rpart function), then this frame is used rather than
constructing new data.
Formulas can be tricky as they use lexical scoping and environments in a way that is not always natural (to me). Thank goodness Terry Therneau has made our lives easier with model=TRUE in these two packages!

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