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$type)
# define the training range
training.range = sample(500)[1:300]
# formula to use
tr_form = formula(type ~ x + y + z)
# Function that includes the randomForest function
train_rf = function(form, all_data, tr_subset) {
p = randomForest(
formula = form,
data = all_data,
subset = tr_subset,
na.action = na.omit
)
return(p)
}
# test the new defined function
test_tree = train_rf(form = tr_form, all_data = train.data, tr_subset = training.range)
Running this gives the error:
Error in eval(substitute(subset), data, env) :
object 'tr_subset' not found
If, however, subset = tr_subset is removed from the randomForest function, and tr_subset is removed from the train_rf function, this code runs fine, however the whole data set is used for training!
It should be noted that using the subset argument in randomForest when not defined in another function works completely fine, and is the intended method for the function, as described in the vignette linked above.
I know in the mean time I could just define another training set that has just the row numbers required, and train using all of that, but is there a reason why my original code doesn't work please?
Thanks.
EDIT: I conjecture that, as subset() is a base R function, R is getting confused and thinking you're wanting to use the base R function rather than defining an argument of the randomForest function. I'm not an expert, though, so I may be wrong.
Related
I am trying to extract the relative influence of each variable from a gbm.fit object but it is coming up with the error below:
> summary(boost_cox, plotit = FALSE)
Error in data.frame(var = object$var.names[i], rel.inf = rel.inf[i]) :
row names contain missing values
The boost_cox object itself is fitted as follows:
boost_cox = gbm.fit(x = x,
y = y,
distribution="coxph",
verbose = FALSE,
keep.data = TRUE)
I have to use the gbm.fit function rather than the standard gbm function due to the large number of predictors (26k+)
I have solve this issue now myself.
The relative.influence() function can be used and works for objects created using both gbm() and gbm.fit(). However, it does not provide the plots as in the summary() function.
I hope this helps anyone else looking in the future.
I am trying to de-clutter some scripts by creating functions to complete repetitive tasks in R. One task I complete repeatedly is fitting a linear model to a set of data and creating predictions from that linear model fit. The data I am working with is concentration and flow data from streams, and flow is always the explanatory variable but the response variable changes and therefore I would like to include it as a function input. However, I receive a "non-numeric argument to mathematical function" error when I run the function. I have tried both with and without quotes since the lm() call does not require quotes but that results in the classis "object 'myobject' not found". Here's a simple example.
Update
flows <- seq(0,7,0.01)
dat <- tibble(flow=sample(flows,30),
parameter1_conc=rnorm(30,15,4),
parameter2_conc=rnorm(30,50,8))
regr_func <- function(modeldata,parameter,pred_maxflow,pred_flowint) {
mod <- lm(as.formula(paste('log(', parameter, ') ~ log(', flow, ')')), data=modeldata)
newflow <- data.frame(flow = seq(0, pred_maxflow, pred_flowint))
preds <<- predict(mod, newdata = newflow,
interval = 'prediction')
}
regr_func(modeldata = dat,
parameter = 'parameter1_conc',
pred_maxflow = 20,
pred_flowint = 0.001)
Original Example Error
flows <- seq(0,7,0.01)
dat <- tibble(flow=sample(flows,30),
parameter1_conc=rnorm(30,15,4),
parameter1_conc=rnorm(30,50,8))
regr_func <- function(modeldata,parameter,pred_maxflow,pred_flowint) {
mod <- lm(log(parameter)~log(flow), data = modeldata)
newflow <- data.frame(flow = seq(0, maxflow, flowint))
preds <<- predict(mod, newdata = newflow,
interval = 'prediction')
}
regr_func(modeldata = dat,
parameter = 'parameter1_conc',
pred_maxflow = 20,
pred_flowint = 0.001)
There are 3 issues here. The main one is that log(parameter) in your lm formula does not get substituted for the variable passed in as parameter. That means lm is literally looking for a column called parameter in your data, which doesn't exist. You can fix this by creating a formula with the name substituted in. Although doing this with strings is the most commonly used method to do this, it is a bit more efficient and safer to use substitute. This also allows you to pass your column name without quotes.
The second issue is that the arguments maxflow and flowint should probably be pred_maxflow and pred_flowint to match your function parameters.
Thirdly, using the <<- operator to write to a variable in the calling frame is bad practice. R users expect functions not to have such side effects, and know to store the output of function calls to variables under their control. Only in very rare circumstances should this be done within the function.
Putting all this together, we have:
regr_func <- function(modeldata, parameter, pred_maxflow, pred_flowint) {
f <- `[[<-`(x ~ log(flow), 2, substitute(log(parameter)))
mod <- lm(f, data = modeldata)
newflow <- data.frame(flow = seq(0, pred_maxflow, pred_flowint))
predict(mod, newdata = newflow, interval = 'prediction')
}
And we would call the function like this:
preds <- regr_func(modeldata = dat,
parameter = parameter1_conc,
pred_maxflow = 20,
pred_flowint = 0.001)
resulting in:
head(preds)
#> fit lwr upr
#> 1 Inf NaN NaN
#> 2 3.365491 2.188942 4.542041
#> 3 3.312636 2.219223 4.406049
#> 4 3.281717 2.236294 4.327140
#> 5 3.259780 2.248073 4.271488
#> 6 3.242765 2.256998 4.228531
Created on 2022-06-03 by the reprex package (v2.0.1)
I am trying to input matrix data into the brm() function to run a signal regression. brm is from the brms package, which provides an interface to fit Bayesian models using Stan. Signal regression is when you model one covariate using another within the bigger model, and you use the by parameter like this: model <- brm(response ~ s(matrix1, by = matrix2) + ..., data = Data). The problem is, I cannot input my matrices using the 'data' parameter because it only allows one data.frame object to be inputted.
Here are my code and the errors I obtained from trying to get around that constraint...
First off, my reproducible code leading up to the model-building:
library(brms)
#100 rows, 4 columns. Each cell contains a number between 1 and 10
Data <- data.frame(runif(100,1,10),runif(100,1,10),runif(100,1,10),runif(100,1,10))
#Assign names to the columns
names(Data) <- c("d0_10","d0_100","d0_1000","d0_10000")
Data$Density <- as.matrix(Data)%*%c(-1,10,5,1)
#the coefficients we are modelling
d <- c(-1,10,5,1)
#Made a matrix with 4 columns with values 10, 100, 1000, 10000 which are evaluation points. Rows are repeats of the same column numbers
Bins <- 10^matrix(rep(1:4,times = dim(Data)[1]),ncol = 4,byrow =T)
Bins
As mentioned above, since 'data' only allows one data.frame object to be inputted, I've tried other ways of inputting my matrix data. These methods include:
1) making the matrix within the brm() function using as.matrix()
signalregression.brms <- brm(Density ~ s(Bins,by=as.matrix(Data[,c(c("d0_10","d0_100","d0_1000","d0_10000"))])),data = Data)
#Error in is(sexpr, "try-error") :
argument "sexpr" is missing, with no default
2) making the matrix outside the formula, storing it in a variable, then calling that variable inside the brm() function
Donuts <- as.matrix(Data[,c(c("d0_10","d0_100","d0_1000","d0_10000"))])
signalregression.brms <- brm(Density ~ s(Bins,by=Donuts),data = Data)
#Error: The following variables can neither be found in 'data' nor in 'data2':
'Bins', 'Donuts'
3) inputting a list containing the matrix using the 'data2' parameter
signalregression.brms <- brm(Density ~ s(Bins,by=donuts),data = Data,data2=list(Bins = 10^matrix(rep(1:4,times = dim(Data)[1]),ncol = 4,byrow =T),donuts=as.matrix(Data[,c(c("d0_10","d0_100","d0_1000","d0_10000"))])))
#Error in names(dat) <- object$term :
'names' attribute [1] must be the same length as the vector [0]
None of the above worked; each had their own errors and it was difficult troubleshooting them because I couldn't find answers or examples online that were of a similar nature in the context of brms.
I was able to use the above techniques just fine for gam(), in the mgcv package - you don't have to define a data.frame using 'data', you can call on variables defined outside of the gam() formula, and you can make matrices inside the gam() function itself. See below:
library(mgcv)
signalregression2 <- gam(Data$Density ~ s(Bins,by = as.matrix(Data[,c("d0_10","d0_100","d0_1000","d0_10000")]),k=3))
#Works!
It seems like brms is less flexible... :(
My question: does anyone have any suggestions on how to make my brm() function run?
Thank you very much!
My understanding of signal regression is limited enough that I'm not convinced this is correct, but I think it's at least a step in the right direction. The problem seems to be that brm() expects everything in its formula to be a column in data. So we can get the model to compile by ensuring all the things we want are present in data:
library(tidyverse)
signalregression.brms = brm(Density ~
s(cbind(d0_10_bin, d0_100_bin, d0_1000_bin, d0_10000_bin),
by = cbind(d0_10, d0_100, d0_1000, d0_10000),
k = 3),
data = Data %>%
mutate(d0_10_bin = 10,
d0_100_bin = 100,
d0_1000_bin = 1000,
d0_10000_bin = 10000))
Writing out each column by hand is a little annoying; I'm sure there are more general solutions.
For reference, here are my installed package versions:
map_chr(unname(unlist(pacman::p_depends(brms)[c("Depends", "Imports")])), ~ paste(., ": ", pacman::p_version(.), sep = ""))
[1] "Rcpp: 1.0.6" "methods: 4.0.3" "rstan: 2.21.2" "ggplot2: 3.3.3"
[5] "loo: 2.4.1" "Matrix: 1.2.18" "mgcv: 1.8.33" "rstantools: 2.1.1"
[9] "bayesplot: 1.8.0" "shinystan: 2.5.0" "projpred: 2.0.2" "bridgesampling: 1.1.2"
[13] "glue: 1.4.2" "future: 1.21.0" "matrixStats: 0.58.0" "nleqslv: 3.3.2"
[17] "nlme: 3.1.149" "coda: 0.19.4" "abind: 1.4.5" "stats: 4.0.3"
[21] "utils: 4.0.3" "parallel: 4.0.3" "grDevices: 4.0.3" "backports: 1.2.1"
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.
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.