Running Multiple Linear Regression Models in for-Loop - r

Question

The logic is similar to the content-based recommender,
content
undesirable
desirable
user_1
...
user_10
1
3.00
2.77
0.11
NA
...
5000
2.50
2.11
NA
0.12
I need to run the model for undesirable and desirable as independent values and each user as the dependent value, thus I need run 10 times to fit the model and predict each user's NA value.
This is the code that I hard coding, but I wonder how to use for loop, I just searched for several methods but they do not work for me...
the data as 'test'
hard code
#fit model
fit_1 = lm(user_1 ~ undesirable + desirable, data = test)
...
fit_10 = lm(user_10 ~ undesirable + desirable, data = test)
#prediction
u_1_na = test[is.na(test$user_1), c('user_1', 'undesirable', 'desirable')]
result1 = predict(fit_1, newdata = u_1_na)
which(result1 == max(result1))
max(result1)
...
u_10_na = test[is.na(test$user_10), c('user_10', 'undesirable', 'desirable')]
result10 = predict(fit_10, newdata = u_10_na)
which(result10 == max(result10))
max(result10)
#make to csv file
apply each max predict value to csv.
this is what I try for now(for loop)
mod_summaries <- list()
for(i in 1:10) {
predictors_i <- colnames(data)[1:10]
mod_summaries[[i - 1]] <- summary(
lm(predictors_i ~ ., test[ , c("undesirable", 'desirable')]))
}

Answer 1

An apply method:
mod_summaries_lapply <-
lapply(
colnames(mtcars),
FUN = function(x)
summary(lm(reformulate(".", response = x), data = mtcars))
)
A for loop method to make linear models for each column. The key is the reformulate() function, which creates the formula from strings. In the question, the function is made of a string and results in error invalid term in model formula. The string needs to be evaluated with eval() . This example uses the mtcars dataset.
mod_summaries <- list()
for(i in 1:11) {
predictors_i <- colnames(mtcars)[i]
mod_summaries[[i]] <- summary(lm(reformulate(".", response = predictors_i), data=mtcars))
#summary(lm(reformulate(". -1", response = predictors_i), data=mtcars)) # -1 to exclude intercept
#summary(lm(as.formula(paste(predictors_i, "~ .")), data=mtcars)) # a "paste as formula" method
}

Answer 2

You could use the function as.formula together with the paste function to create your formula. Following is an example
formula_lm <- as.formula(
paste(response_var,
paste(expl_var, collapse = " + "),
sep = " ~ "))
This implies that you have more than one explanatory variable (separated in the paste with +). If you only have one, omit the second paste.
With the created formula, you can use the lm funciton like this:
lm(formula_lm, data)
Edit: the vector expl_var would in your case include the undesirable and desirable variable.

Answer 3

Avoid the loop. Make your data tidy. Something like:
library(tidyverse)
test %>%
select(-content) %>%
pivot_longer(
starts_with("user"),
names_to="user",
values_to="value"
) %>%
group_by(user) %>%
group_map(
function(.x, .y) {
summary(lm(user ~ ., data=.x))
}
)
Untested code since your example is not reproducible.