This is a chunk of the dataset,

structure(list(gender = structure(c(2L, 1L, 1L, 1L, 1L), .Label = c("1", 
"2"), class = "factor"), Work_less = c(0, 1, 0, 0, NA), happy = c(7, 
8, 7, 6, 7), lifestatisfied = c(8, 8, 9, 9, 7), country = c(8, 
8, 9, 9, 7)), row.names = c(NA, -5L), class = "data.frame")

whose I'm trying to fit different models as follows:

model <- list()
for (i in 2:ncol(dat)) {
  model[[i]] <- lm(lifestatisfied ~  dat[,i], dat)
}

In the output, you might see that the independent variable (dat[, i]) but not the correspondent name of the variable:

[[2]]

Call:
lm(formula = lifestatisfied ~ dat[, i], data = dat)

Coefficients:
(Intercept)     dat[, i]  
     8.6667      -0.6667  


[[3]]

Call:
lm(formula = lifestatisfied ~ dat[, i], data = dat)

Coefficients:
(Intercept)     dat[, i]  
       11.7         -0.5  


[[4]]

Call:
lm(formula = lifestatisfied ~ dat[, i], data = dat)

Coefficients:
(Intercept)     dat[, i]  
  1.589e-15    1.000e 00  


[[5]]

Call:
lm(formula = lifestatisfied ~ dat[, i], data = dat)

Coefficients:
(Intercept)     dat[, i]  
  1.589e-15    1.000e 00 

Could you please suggest a way to print into the output the name of independent to which the index of the model refers to?

CodePudding user response：

We can use:

xvar <- setdiff(names(dat), "lifestatisfied")
model <- vector("list", length(xvar))
for (i in 1:length(xvar)) {
  form <- reformulate(xvar[i], "lifestatisfied")
  model[[i]] <- do.call("lm", list(formula = form, data = quote(dat)))
}

And the output model is:

#[[1]]
#
#Call:
#lm(formula = lifestatisfied ~ gender, data = dat)
#
#Coefficients:
#(Intercept)      gender2  
#       8.25        -0.25  
#
#
#[[2]]
#
#Call:
#lm(formula = lifestatisfied ~ Work_less, data = dat)
#
#Coefficients:
#(Intercept)    Work_less  
#     8.6667      -0.6667  
#
#
#[[3]]
#
#Call:
#lm(formula = lifestatisfied ~ happy, data = dat)
#
#Coefficients:
#(Intercept)        happy  
#       11.7         -0.5  
#
#
#[[4]]
#
#Call:
#lm(formula = lifestatisfied ~ country, data = dat)
#
#Coefficients:
#(Intercept)      country  
#  1.589e-15    1.000e 00  

Remarks

1. reformulate is useful in a loop where we dynamically create model formulae.

2. The do.call part is also important. You can try

   model[[i]] <- lm(form, data = dat)
   

   instead to see the difference.

3. The other answer using broom::tidy only produces a summary of estimated coefficients. It does not return a list of fitted models which facilitate for example, prediction.

CodePudding user response：

A purrr option:

   library(tidyverse)
    df %>% 
      select(-lifestatisfied) %>% 
      imap(~lm(df$lifestatisfied ~ .x, data = df) %>% broom::tidy() %>% 
             mutate(term = ifelse(term == ".x", .y, term)))