Change SMOTE parameters inside CARET k-fold cross-validation classification - r

I have a classification problem with a very skewed class to predict (e.g. 90% / 10% unbalanced binary variable to predict).
In order to deal with that issue, I want to use the SMOTE method to oversample this class variable. However, as I read here (http://www.marcoaltini.com/blog/dealing-with-imbalanced-data-undersampling-oversampling-and-proper-cross-validation) it is best practice to use SMOTE inside the k-fold loop to avoid overfitting.
As I'm using the caret package to perform my analysis, I'm referring to this link (http://topepo.github.io/caret/sampling.html). I undestand everything perfectly but the last part where it explains how to change the SMOTE parameters:
smotest <- list(name = "SMOTE with more neighbors!",
func = function (x, y) {
library(DMwR)
dat <- if (is.data.frame(x)) x else as.data.frame(x)
dat$.y <- y
dat <- SMOTE(.y ~ ., data = dat, k = 10)
list(x = dat[, !grepl(".y", colnames(dat), fixed = TRUE)],
y = dat$.y)
},
first = TRUE)
I simply don't understand this. Someone care to explain? Let's say I want to include the SMOTE parameters perc.over, k and perc.under, how would I do that?
Thank you very much.
EDIT:
Actually I realized I could probably just add these parameters inside the "SMOTE" expression in the above function, this would for instance give something like:
smotest <- list(name = "SMOTE with more neighbors!",
func = function (x, y) {
library(DMwR)
dat <- if (is.data.frame(x)) x else as.data.frame(x)
dat$.y <- y
dat <- SMOTE(.y ~ ., data = dat, k = 10, perc.over = 1200, perc.under = 100)
list(x = dat[, !grepl(".y", colnames(dat), fixed = TRUE)],
y = dat$.y)
},
first = TRUE)

I am not sure to have understood what you do not understand but here is an attempt to clarify what is done in this piece of code.
The smotest object is created as list because it is the way the argument sampling of trainControl function must be represented. The first element of this list is a name used only for display purposes. The second, func, is the actual sampling function. The third, first, is a logical value indicating whether samplin must be done before or after the pre-processing step.
The element func is here only a wrapper of SMOTE function. In this wrapper, line 3 is here because only a data.frame can be passed to SMOTE function. Line 4 is added because a formula combined to a data.frame is used in SMOTE rather than a couple x y. Line 6 is here to ensure that the appropriate format is returned to trainControl.
And, to answer you last question: yes, you can do what you have proposed to set additional parameters to SMOTE.

Related

Caret train function for muliple data frames as function

there has been a similar question to mine 6 years+ ago and it hasn't been solve (R -- Can I apply the train function in caret to a list of data frames?)
This is why I am bringing up this topic again.
I'm writing my own functions for my big R project at the moment and I'm wondering if there is an opportunity to sum up the model training function train() of the pakage caret for different dataframes with different predictors.
My function should look like this:
lda_ex <- function(data, predictor){
model <- train(predictor ~., data,
method = "lda",
trControl = trainControl(method = "none"),
preProc = c("center","scale"))
return(model)
}
Using it afterwards should work like this:
data_iris <- iris
predictor_iris <- "Species"
iris_res <- lda_ex(data = data_iris, predictor = predictor_iris)
Unfortunately the R formula is not able to deal with a variable as input as far as I tried.
Is there something I am missing?
Thank you in advance for helping me out!
Solving this would help me A LOT to keep my function sheet clean and safe work for sure.
By writing predictor_iris <- "Species", you are basically saving a string object in predictor_iris. Thus, when you run lda_ex, I guess you incur in some error concerning the formula object in train(), since you are trying to predict a string using vectors of covariates.
Indeed, I tried the following toy example:
X = rnorm(1000)
Y = runif(1000)
predictor = "Y"
lm(predictor ~ X)
which gives an error about differences in the lengths of variables.
Let me modify your function:
lda_ex <- function(data, formula){
model <- train(formula, data,
method = "lda",
trControl = trainControl(method = "none"),
preProc = c("center","scale"))
return(model)
}
The key difference is that now we must pass in the whole formula, instead of the predictor only. In that way, we avoid the string-related problem.
library(caret) # Recall to specify the packages needed to reproduce your examples!
data_iris <- iris
formula_iris = Species ~ . # Key difference!
iris_res <- lda_ex(data = data_iris, formula = formula_iris)

Error with svyglm function in survey package in R: "all variables must be in design=argument"

New to stackoverflow. I'm working on a project with NHIS data, but I cannot get the svyglm function to work even for a simple, unadjusted logistic regression with a binary predictor and binary outcome variable (ultimately I'd like to use multiple categorical predictors, but one step at a time).
El_under_glm<-svyglm(ElUnder~SO2, design=SAMPdesign, subset=NULL, family=binomial(link="logit"), rescale=FALSE, correlation=TRUE)
Error in eval(extras, data, env) :
object '.survey.prob.weights' not found
I changed the variables to 0 and 1 instead:
Under_narm$SO2REG<-ifelse(Under_narm$SO2=="Heterosexual", 0, 1)
Under_narm$ElUnderREG<-ifelse(Under_narm$ElUnder=="No", 0, 1)
But then get a different issue:
El_under_glm<-svyglm(ElUnderREG~SO2REG, design=SAMPdesign, subset=NULL, family=binomial(link="logit"), rescale=FALSE, correlation=TRUE)
Error in svyglm.survey.design(ElUnderREG ~ SO2REG, design = SAMPdesign, :
all variables must be in design= argument
This is the design I'm using to account for the weights -- I'm pretty sure it's correct:
SAMPdesign=svydesign(data=Under_narm, id= ~NHISPID, weight= ~SAMPWEIGHT)
Any and all assistance appreciated! I've got a good grasp of stats but am a slow coder. Let me know if I can provide any other information.
Using some make-believe sample data I was able to get your model to run by setting rescale = TRUE. The documentation states
Rescaling of weights, to improve numerical stability. The default
rescales weights to sum to the sample size. Use FALSE to not rescale
weights.
So, one solution maybe is just to set rescale = TRUE.
library(survey)
# sample data
Under_narm <- data.frame(SO2 = factor(rep(1:2, 1000)),
ElUnder = sample(0:1, 1000, replace = TRUE),
NHISPID = paste0("id", 1:1000),
SAMPWEIGHT = sample(c(0.5, 2), 1000, replace = TRUE))
# with 'rescale' = TRUE
SAMPdesign=svydesign(ids = ~NHISPID,
data=Under_narm,
weights = ~SAMPWEIGHT)
El_under_glm<-svyglm(formula = ElUnder~SO2,
design=SAMPdesign,
family=quasibinomial(), # this family avoids warnings
rescale=TRUE) # Weights rescaled to the sum of the sample size.
summary(El_under_glm, correlation = TRUE) # use correlation with summary()
Otherwise, looking code for this function's method with 'survey:::svyglm.survey.design', it seems like there may be a bug. I could be wrong, but by my read when 'rescale' is FALSE, .survey.prob.weights does not appear to get assigned a value.
if (is.null(g$weights))
g$weights <- quote(.survey.prob.weights)
else g$weights <- bquote(.survey.prob.weights * .(g$weights)) # bug?
g$data <- quote(data)
g[[1]] <- quote(glm)
if (rescale)
data$.survey.prob.weights <- (1/design$prob)/mean(1/design$prob)
There may be a work around if you assign a vector of numeric values to .survey.prob.weights in the global environment. No idea what these values should be, but your error goes away if you do something like the following. (.survey.prob.weights needs to be double the length of the data.)
SAMPdesign=svydesign(ids = ~NHISPID,
data=Under_narm,
weights = ~SAMPWEIGHT)
.survey.prob.weights <- rep(1, 2000)
El_under_glm<-svyglm(formula = ElUnder~SO2,
design=SAMPdesign,
family=quasibinomial(),
rescale=FALSE)
summary(El_under_glm, correlation = TRUE)

Can't give a subset when using randomForest inside a function

I'm wanting to create a function that uses within it the randomForest function from the randomForest package. This takes the "subset" argument, which is a vector of row numbers of the data frame to use for training. However, if I use this argument when calling the randomForest function in another defined function, I get the error:
Error in eval(substitute(subset), data, env) :
object 'tr_subset' not found
Here is a reproducible example, where we attempt to train a random forest to classify a response "type" either "A" or "B", based on three numerical predictors:
library(randomForest)
# define a random data frame to train with
test.data = data.frame(
type = rep(NA, times = 500),
x = runif(500),
y = runif(500),
z = runif(500)
)
train.data$type[runif(500) >= 0.5] = "A"
train.data$type[is.na(test.data$type)] = "B"
train.data$type = as.factor(test.data$type)
# define the training range
training.range = sample(500)[1:300]
# formula to use
tr_form = formula(type ~ x + y + z)
# Function that includes the randomForest function
train_rf = function(form, all_data, tr_subset) {
p = randomForest(
formula = form,
data = all_data,
subset = tr_subset,
na.action = na.omit
)
return(p)
}
# test the new defined function
test_tree = train_rf(form = tr_form, all_data = train.data, tr_subset = training.range)
Running this gives the error:
Error in eval(substitute(subset), data, env) :
object 'tr_subset' not found
If, however, subset = tr_subset is removed from the randomForest function, and tr_subset is removed from the train_rf function, this code runs fine, however the whole data set is used for training!
It should be noted that using the subset argument in randomForest when not defined in another function works completely fine, and is the intended method for the function, as described in the vignette linked above.
I know in the mean time I could just define another training set that has just the row numbers required, and train using all of that, but is there a reason why my original code doesn't work please?
Thanks.
EDIT: I conjecture that, as subset() is a base R function, R is getting confused and thinking you're wanting to use the base R function rather than defining an argument of the randomForest function. I'm not an expert, though, so I may be wrong.

"nrow(x) == n is not TRUE" when using train in Caret; already set as factors

I'm using the dataset found here: http://archive.ics.uci.edu/ml/datasets/Qualitative_Bankruptcy
When running code:
library(caret)
bank <- read.csv("Qualitative_Bankruptcy.data.txt", header=FALSE, na.strings = "?",
strip.white = TRUE)
x=bank[1:6]
y=bank[7]
bank.knn <- train(x, y, method= "knn", trControl = trainControl(method = "cv"))
I get the following error:
Error: nrow(x) == n is not TRUE
The only example I've found is Error: nrow(x) == n is not TRUE when using Train in Caret ; my Y is already a factor vector with two classes, all the X features are factors as well. I've tried using as.matrix and as.data.frame on both the X and Y without success.
nrow(x) is equal to 250, but I'm not sure what the n is referring to in the package.
y is not actually a vector, but a data.frame with one column because bank[7] does not convert the 7th column into a vector, so length(y) is 1. Use bank[, 7] instead. It does not make a difference for x but it could as well be generated by bank[, 1:6].
Additionally to make KNN work you probably have to convert the x data.frame that consists of factor variables to numeric dummy variables.
x=model.matrix(~. - 1, bank[, 1:6])
y=bank[, 7]
bank.knn <- train(x, y, method= "knn",
trControl = trainControl(method = "cv"))
I'm not a caret user but I think you have two problems. The extraction method you used did not deliver an atomic vector but rahter a list that contained a vector. If you asked for length(y) you get 1 rather than 250. The first error is easily solved by changing to this definition of y:
y <- bank[[7]] # extract a vector rather than a sublist
Then things get messy. The KNN method expects continuous data (and the error messages you get indicate the caret's author considers it a "regression method" and you are passing factor data, so you therefore need to choose a classification method instead.

Object not found error when passing model formula to another function

I have a weird problem with R that I can't seem to work out.
I've tried to write a function that performs K-fold cross validation for a model chosen by the stepwise procedure in R. (I'm aware of the issues with stepwise procedures, it's purely for comparison purposes) :)
Now the issue is, that if I define the function parameters (linmod,k,direction) and run the contents of the function, it works flawlessly. BUT, if I run it as a function, I get an error saying the datas.train object can't be found.
I've tried stepping through the function with debug() and the object clearly exists, but R says it doesn't when I actually run the function. If I just fit a model using lm() it works fine, so I believe it's a problem with the step function in the loop, while inside a function. (try commenting out the step command, and set the predictions to those from the ordinary linear model.)
#CREATE A LINEAR MODEL TO TEST FUNCTION
lm.cars <- lm(mpg~.,data=mtcars,x=TRUE,y=TRUE)
#THE FUNCTION
cv.step <- function(linmod,k=10,direction="both"){
response <- linmod$y
dmatrix <- linmod$x
n <- length(response)
datas <- linmod$model
form <- formula(linmod$call)
# generate indices for cross validation
rar <- n/k
xval.idx <- list()
s <- sample(1:n, n) # permutation of 1:n
for (i in 1:k) {
xval.idx[[i]] <- s[(ceiling(rar*(i-1))+1):(ceiling(rar*i))]
}
#error calculation
errors <- R2 <- 0
for (j in 1:k){
datas.test <- datas[xval.idx[[j]],]
datas.train <- datas[-xval.idx[[j]],]
test.idx <- xval.idx[[j]]
#THE MODELS+
lm.1 <- lm(form,data= datas.train)
lm.step <- step(lm.1,direction=direction,trace=0)
step.pred <- predict(lm.step,newdata= datas.test)
step.error <- sum((step.pred-response[test.idx])^2)
errors[j] <- step.error/length(response[test.idx])
SS.tot <- sum((response[test.idx] - mean(response[test.idx]))^2)
R2[j] <- 1 - step.error/SS.tot
}
CVerror <- sum(errors)/k
CV.R2 <- sum(R2)/k
res <- list()
res$CV.error <- CVerror
res$CV.R2 <- CV.R2
return(res)
}
#TESTING OUT THE FUNCTION
cv.step(lm.cars)
Any thoughts?
When you created your formula, lm.cars, in was assigned its own environment. This environment stays with the formula unless you explicitly change it. So when you extract the formula with the formula function, the original environment of the model is included.
I don't know if I'm using the correct terminology here, but I think you need to explicitly change the environment for the formula inside your function:
cv.step <- function(linmod,k=10,direction="both"){
response <- linmod$y
dmatrix <- linmod$x
n <- length(response)
datas <- linmod$model
.env <- environment() ## identify the environment of cv.step
## extract the formula in the environment of cv.step
form <- as.formula(linmod$call, env = .env)
## The rest of your function follows
Another problem that can cause this is if one passes a character (string vector) to lm instead of a formula. vectors have no environment, and so when lm converts the character to a formula, it apparently also has no environment instead of being automatically assigned the local environment. If one then uses an object as weights that is not in the data argument data.frame, but is in the local function argument, one gets a not found error. This behavior is not very easy to understand. It is probably a bug.
Here's a minimal reproducible example. This function takes a data.frame, two variable names and a vector of weights to use.
residualizer = function(data, x, y, wtds) {
#the formula to use
f = "x ~ y"
#residualize
resid(lm(formula = f, data = data, weights = wtds))
}
residualizer2 = function(data, x, y, wtds) {
#the formula to use
f = as.formula("x ~ y")
#residualize
resid(lm(formula = f, data = data, weights = wtds))
}
d_example = data.frame(x = rnorm(10), y = rnorm(10))
weightsvar = runif(10)
And test:
> residualizer(data = d_example, x = "x", y = "y", wtds = weightsvar)
Error in eval(expr, envir, enclos) : object 'wtds' not found
> residualizer2(data = d_example, x = "x", y = "y", wtds = weightsvar)
1 2 3 4 5 6 7 8 9 10
0.8986584 -1.1218003 0.6215950 -0.1106144 0.1042559 0.9997725 -1.1634717 0.4540855 -0.4207622 -0.8774290
It is a very subtle bug. If one goes into the function environment with browser, one can see the weights vector just fine, but it somehow is not found in the lm call!
The bug becomes even harder to debug if one used the name weights for the weights variable. In this case, since lm can't find the weights object, it defaults to the function weights() from base thus throwing an even stranger error:
Error in model.frame.default(formula = f, data = data, weights = weights, :
invalid type (closure) for variable '(weights)'
Don't ask me how many hours it took me to figure this out.

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