Suppose, there is some data.frame foo_data_frame and one wants to find regression of the target column Y by some others columns. For that purpose usualy some formula and model are used. For example:
linear_model <- lm(Y ~ FACTOR_NAME_1 + FACTOR_NAME_2, foo_data_frame)
That does job well if the formula is coded statically. If it is desired to root over several models with the constant number of dependent variables (say, 2) it can be treated like that:
for (i in seq_len(factor_number)) {
for (j in seq(i + 1, factor_number)) {
linear_model <- lm(Y ~ F1 + F2, list(Y=foo_data_frame$Y,
F1=foo_data_frame[[i]],
F2=foo_data_frame[[j]]))
# linear_model further analyzing...
}
}
My question is how to do the same affect when the number of variables is changing dynamically during program running?
for (number_of_factors in seq_len(5)) {
# Then root over subsets with #number_of_factors cardinality.
for (factors_subset in all_subsets_with_fixed_cardinality) {
# Here I want to fit model with factors from factors_subset.
linear_model <- lm(Does R provide smth to write here?)
}
}
See ?as.formula, e.g.:
factors <- c("factor1", "factor2")
as.formula(paste("y~", paste(factors, collapse="+")))
# y ~ factor1 + factor2
where factors is a character vector containing the names of the factors you want to use in the model. This you can paste into an lm model, e.g.:
set.seed(0)
y <- rnorm(100)
factor1 <- rep(1:2, each=50)
factor2 <- rep(3:4, 50)
lm(as.formula(paste("y~", paste(factors, collapse="+"))))
# Call:
# lm(formula = as.formula(paste("y~", paste(factors, collapse = "+"))))
# Coefficients:
# (Intercept) factor1 factor2
# 0.542471 -0.002525 -0.147433
An oft forgotten function is reformulate. From ?reformulate:
reformulate creates a formula from a character vector.
A simple example:
listoffactors <- c("factor1","factor2")
reformulate(termlabels = listoffactors, response = 'y')
will yield this formula:
y ~ factor1 + factor2
Although not explicitly documented, you can also add interaction terms:
listofintfactors <- c("(factor3","factor4)^2")
reformulate(termlabels = c(listoffactors, listofintfactors),
response = 'y')
will yield:
y ~ factor1 + factor2 + (factor3 + factor4)^2
Another option could be to use a matrix in the formula:
Y = rnorm(10)
foo = matrix(rnorm(100),10,10)
factors=c(1,5,8)
lm(Y ~ foo[,factors])
You don't actually need a formula. This works:
lm(data_frame[c("Y", "factor1", "factor2")])
as does this:
v <- c("Y", "factor1", "factor2")
do.call("lm", list(bquote(data_frame[.(v)])))
I generally solve this by changing the name of my response column. It is easier to do dynamically, and possibly cleaner.
model_response <- "response_field_name"
setnames(model_data_train, c(model_response), "response") #if using data.table
model_gbm <- gbm(response ~ ., data=model_data_train, ...)
Related
Let's consider data following:
set.seed(42)
y <- runif(100)
df <- data.frame("Exp" = rexp(100), "Norm" = rnorm(100), "Wei" = rweibull(100, 1))
I want to perform linear regression but when formula is a string in format:
form <- "Exp + Norm + Wei"
I thought that I only have to use:
as.formula(lm(y~form, data = df))
However it's not working. The error is about variety in length of variables. (it seems like it still treats form as a string vector of length 1, but I have no idea why).
Do you know how I can do it ?
We can use paste to construct the formula, and use it directly on lm
lm(paste('y ~', form), data = df)
-output
#Call:
#lm(formula = paste("y ~", form), data = df)
#Coefficients:
#(Intercept) Exp Norm Wei
# 0.495861 0.026988 0.046689 0.003612
I have a linear model with lots of explaining variables (independent variables)
model <- lm(y ~ x1 + x2 + x3 + ... + x100)
some of which are linear depended on each other (multicollinearity).
I want the machine to search for the name of the explaining variable which has the highest VIF coefficient (x2 for example), delete it from the formula and then run the old lm function with the new formula
model <- lm(y ~ x1 + x3 + ... + x100)
I already learned how to retrieve the name of the explaining variable which has the highest VIF coefficient:
max_vif <- function(x) {
vifac <- data.frame(vif(x))
nameofmax <- rownames(which(vifac == max(vifac), arr.ind = TRUE))
return(nameofmax)
}
But I still don't understand how to search the needed explaining variable, delete it from the formula and run the function again.
We can use the update function and paste in the column that needs to be removed. We first can fit a model, and then use update to change that model's formula. The model formula can be expressed as a character string, which allows you to concatenate the general formula .~. and whatever variable(s) you'd like removed (using the minus sign -).
Here is an example:
fit1 <- lm(wt ~ mpg + cyl + am, data = mtcars)
coef(fit1)
# (Intercept) mpg cyl am
# 4.83597190 -0.09470611 0.08015745 -0.52182463
rm_var <- "am"
fit2 <- update(fit1, paste0(".~. - ", rm_var))
coef(fit2)
# (Intercept) mpg cyl
# 5.07595833 -0.11908115 0.08625557
Using max_vif we can wrap this into a function:
rm_max_vif <- function(x){
# find variable(s) needing to be removed
rm_var <- max_vif(x)
# concatenate with "-" to remove variable(s) from formula
rm_var <- paste(paste0("-", rm_var), collapse = " ")
# update model
update(x, paste0(".~.", rm_var))
}
Problem solved!
I created a list containing all variables for lm model:
Price <- list(y,x1,...,x100)
Then I used different way for setting lm model:
model <- lm(y ~ ., data = Price)
So we can just delete variable with the highest VIF from Price list.
With the function i already came up the code will be:
Price <- list(y,x1,x2,...,x100)
model <- lm(y ~ ., data = Price)
max_vif <- function(x) { # Function for finding name of variable with the highest VIF
vifac <- data.frame(vif(x))
nameofmax <- rownames(which(vifac == max(vifac), arr.ind = TRUE))
return(nameofmax)
}
n <- max(data.frame(vif(model)))
while(n >= 5) { # Loop for deleting variable with the highest VIF from `Price` list one after another, untill there is no VIF equal or higher then 5
Price[[m]] <- NULL
model_auto <- lm(y ~ ., data = Price)
m <- max_vif(model)
n <- max(data.frame(vif(model)))
}
Suppose, there is some data.frame foo_data_frame and one wants to find regression of the target column Y by some others columns. For that purpose usualy some formula and model are used. For example:
linear_model <- lm(Y ~ FACTOR_NAME_1 + FACTOR_NAME_2, foo_data_frame)
That does job well if the formula is coded statically. If it is desired to root over several models with the constant number of dependent variables (say, 2) it can be treated like that:
for (i in seq_len(factor_number)) {
for (j in seq(i + 1, factor_number)) {
linear_model <- lm(Y ~ F1 + F2, list(Y=foo_data_frame$Y,
F1=foo_data_frame[[i]],
F2=foo_data_frame[[j]]))
# linear_model further analyzing...
}
}
My question is how to do the same affect when the number of variables is changing dynamically during program running?
for (number_of_factors in seq_len(5)) {
# Then root over subsets with #number_of_factors cardinality.
for (factors_subset in all_subsets_with_fixed_cardinality) {
# Here I want to fit model with factors from factors_subset.
linear_model <- lm(Does R provide smth to write here?)
}
}
See ?as.formula, e.g.:
factors <- c("factor1", "factor2")
as.formula(paste("y~", paste(factors, collapse="+")))
# y ~ factor1 + factor2
where factors is a character vector containing the names of the factors you want to use in the model. This you can paste into an lm model, e.g.:
set.seed(0)
y <- rnorm(100)
factor1 <- rep(1:2, each=50)
factor2 <- rep(3:4, 50)
lm(as.formula(paste("y~", paste(factors, collapse="+"))))
# Call:
# lm(formula = as.formula(paste("y~", paste(factors, collapse = "+"))))
# Coefficients:
# (Intercept) factor1 factor2
# 0.542471 -0.002525 -0.147433
An oft forgotten function is reformulate. From ?reformulate:
reformulate creates a formula from a character vector.
A simple example:
listoffactors <- c("factor1","factor2")
reformulate(termlabels = listoffactors, response = 'y')
will yield this formula:
y ~ factor1 + factor2
Although not explicitly documented, you can also add interaction terms:
listofintfactors <- c("(factor3","factor4)^2")
reformulate(termlabels = c(listoffactors, listofintfactors),
response = 'y')
will yield:
y ~ factor1 + factor2 + (factor3 + factor4)^2
Another option could be to use a matrix in the formula:
Y = rnorm(10)
foo = matrix(rnorm(100),10,10)
factors=c(1,5,8)
lm(Y ~ foo[,factors])
You don't actually need a formula. This works:
lm(data_frame[c("Y", "factor1", "factor2")])
as does this:
v <- c("Y", "factor1", "factor2")
do.call("lm", list(bquote(data_frame[.(v)])))
I generally solve this by changing the name of my response column. It is easier to do dynamically, and possibly cleaner.
model_response <- "response_field_name"
setnames(model_data_train, c(model_response), "response") #if using data.table
model_gbm <- gbm(response ~ ., data=model_data_train, ...)
Using different sources, I wrote a little function that creates a table with standard errors, t statistics and standard errors that are clustered according to a group variable "cluster" after a linear regression model. The code is as follows
cl1 <- function(modl,clust) {
# model is the regression model
# clust is the clustervariable
# id is a unique identifier in ids
library(plm)
library(lmtest)
# Get Formula
form <- formula(modl$call)
# Get Data frame
dat <- eval(modl$call$data)
dat$row <- rownames(dat)
dat$id <- ave(dat$row, dat[[deparse(substitute(clust))]], FUN =seq_along)
pdat <- pdata.frame(dat,
index=c("id", deparse(substitute(clust)))
, drop.index= F, row.names= T)
# # Regression
reg <- plm(form, data=pdat, model="pooling")
# # Adjustments
G <- length(unique(dat[, deparse(substitute(clust))]))
N <- length(dat[,deparse(substitute(clust))])
# # Resid degrees of freedom, adjusted
dfa <- (G/(G-1))*(N-1)/reg$df.residual
d.vcov <- dfa* vcovHC(reg, type="HC0", cluster="group", adjust=T)
table <- coeftest(reg, vcov=d.vcov)
# # Output: se, t-stat and p-val
cl1out <- data.frame(table[, 2:4])
names(cl1out) <- c("se", "tstat", "pval")
# # Cluster VCE
return(cl1out)
}
For a regression like reg1 <- lm (y ~ x1 + x2 , data= df), calling the function cl1(reg1, cluster) will work just fine.
However, if I use a model like reg2 <- lm(y ~ . , data=df), I will get the error message:
Error in terms.formula(object) : '.' in formula and no 'data' argument
After some tests, I am guessing that I can't use "." to signal "use all variables in the data frame" for {plm}. Is there a way I can do this with {plm}? Otherwise, any ideas on how I could improve my function in a way that does not use {plm} and that accepts all possible specifications of a linear model?
Indeed you can't use . notation for formula within plm pacakge.
data("Produc", package = "plm")
plm(gsp ~ .,data=Produc)
Error in terms.formula(object) : '.' in formula and no 'data' argument
One idea is to expand the formula when you have a .. Here is a custom function that does the job (surely is done within other packages):
expand_formula <-
function(form="A ~.",varNames=c("A","B","C")){
has_dot <- any(grepl('.',form,fixed=TRUE))
if(has_dot){
ii <- intersect(as.character(as.formula(form)),
varNames)
varNames <- varNames[!grepl(paste0(ii,collapse='|'),varNames)]
exp <- paste0(varNames,collapse='+')
as.formula(gsub('.',exp,form,fixed=TRUE))
}
else as.formula(form)
}
Now test it :
(eform = expand_formula("gsp ~ .",names(Produc)))
# gsp ~ state + year + pcap + hwy + water + util + pc + emp + unemp
plm(eform,data=Produc)
# Model Formula: gsp ~ state + year + pcap + hwy + water + util + pc + emp + unemp
# <environment: 0x0000000014c3f3c0>
Suppose, there is some data.frame foo_data_frame and one wants to find regression of the target column Y by some others columns. For that purpose usualy some formula and model are used. For example:
linear_model <- lm(Y ~ FACTOR_NAME_1 + FACTOR_NAME_2, foo_data_frame)
That does job well if the formula is coded statically. If it is desired to root over several models with the constant number of dependent variables (say, 2) it can be treated like that:
for (i in seq_len(factor_number)) {
for (j in seq(i + 1, factor_number)) {
linear_model <- lm(Y ~ F1 + F2, list(Y=foo_data_frame$Y,
F1=foo_data_frame[[i]],
F2=foo_data_frame[[j]]))
# linear_model further analyzing...
}
}
My question is how to do the same affect when the number of variables is changing dynamically during program running?
for (number_of_factors in seq_len(5)) {
# Then root over subsets with #number_of_factors cardinality.
for (factors_subset in all_subsets_with_fixed_cardinality) {
# Here I want to fit model with factors from factors_subset.
linear_model <- lm(Does R provide smth to write here?)
}
}
See ?as.formula, e.g.:
factors <- c("factor1", "factor2")
as.formula(paste("y~", paste(factors, collapse="+")))
# y ~ factor1 + factor2
where factors is a character vector containing the names of the factors you want to use in the model. This you can paste into an lm model, e.g.:
set.seed(0)
y <- rnorm(100)
factor1 <- rep(1:2, each=50)
factor2 <- rep(3:4, 50)
lm(as.formula(paste("y~", paste(factors, collapse="+"))))
# Call:
# lm(formula = as.formula(paste("y~", paste(factors, collapse = "+"))))
# Coefficients:
# (Intercept) factor1 factor2
# 0.542471 -0.002525 -0.147433
An oft forgotten function is reformulate. From ?reformulate:
reformulate creates a formula from a character vector.
A simple example:
listoffactors <- c("factor1","factor2")
reformulate(termlabels = listoffactors, response = 'y')
will yield this formula:
y ~ factor1 + factor2
Although not explicitly documented, you can also add interaction terms:
listofintfactors <- c("(factor3","factor4)^2")
reformulate(termlabels = c(listoffactors, listofintfactors),
response = 'y')
will yield:
y ~ factor1 + factor2 + (factor3 + factor4)^2
Another option could be to use a matrix in the formula:
Y = rnorm(10)
foo = matrix(rnorm(100),10,10)
factors=c(1,5,8)
lm(Y ~ foo[,factors])
You don't actually need a formula. This works:
lm(data_frame[c("Y", "factor1", "factor2")])
as does this:
v <- c("Y", "factor1", "factor2")
do.call("lm", list(bquote(data_frame[.(v)])))
I generally solve this by changing the name of my response column. It is easier to do dynamically, and possibly cleaner.
model_response <- "response_field_name"
setnames(model_data_train, c(model_response), "response") #if using data.table
model_gbm <- gbm(response ~ ., data=model_data_train, ...)