Extract Formula From lm with Coefficients (R)

Question

I have an lm object and want to get the formula extracted with coefficients. I know how to extract the formula without coefficients, and how to get the coefficients without the formula, but not how to get eg. y ~ 10 + 1.25b as opposed to y~b or a table of what intercept, b etc. equal

This is the code I'm working with currently:

a = c(1, 2, 5)
b = c(12, 15, 20)

model = lm(a~b)
summary(model)
formula = formula(model)
formula
coefficients(model)

What I'd like to get from the above is y ~ -5.326 + .51b

Thanks

Edit: In my actual code I'm working with over 63 predictors and 18 different models, so I'd like something that can scale up without too much work.

Copt

If you add a second predictor, the code will add it into the formula automatically. Try it out. – Overissue

Oh, I thought I'd tried this out when I first saw it, but I forgot to edit the actual model. Cheers :-) – Copt

Answer 1

as.formula(
  paste0("y ~ ", round(coefficients(model)[1],2), " + ", 
    paste(sprintf("%.2f * %s", 
                  coefficients(model)[-1],  
                  names(coefficients(model)[-1])), 
          collapse=" + ")
  )
)
# y ~ -5.33 + 0.51 * b

Overissue

Hi, this works for the code I supplied so thanks. I'm wondering if there's a way that will scale up easier though? Like if I add a second predictor, c, I assume I'd need to add another line or two in telling it to grab the relevant bits for that too, which could get quite tricky with upwards of 30 predictors for most models (and I'm not entirely sure where I'd put it in, still new to r so this code's a little opaque to me). – Copt

@Copt I made a new answer about your scalability question. See below. – Rocray

Answer 2

Might I suggest an edit to lukeA's excellent answer:

as.formula(
  paste0("y ~ ", round(coefficients(model)[1],2), "", 
    paste(sprintf(" %+.2f*%s ", 
                  coefficients(model)[-1],  
                  names(coefficients(model)[-1])), 
          collapse="")
  )
)

This will make sure that negative coefficients are printed correctly

Suppose you land up with a negative coefficient for b, then the output will be

# y ~ -5.33 + -0.51 * b

instead of

# y ~ -5.33 - 0.51 * b

Derangement

Answer 3

I figured out a versatile way of creating the model formula with coefficients using substitution. It's a lot more versatile than manually building a string with paste0.

e.g.

I have a model that already has the optimized coefficients:

> model
Nonlinear regression model
  model: players ~ pop * (decay^days_from_start) + ycept
   data: data
      pop     decay     ycept 
6.896e+06 2.633e-01 4.300e+05 
 residual sum-of-squares: 1.64e+08

Number of iterations to convergence: 12 
Achieved convergence tolerance: 1.49e-08

These were the coefficients:

> coef(model)
         pop        decay        ycept 
6.896421e+06 2.632545e-01 4.300453e+05 

Putting it all together:

> newFormula = as.formula(substituteDirect(formula(model), as.list(coef(model))))
> newFormula
players ~ 6896421.4399627 * (0.263254460933212^days_from_start) + 430045.26142703

Bambibambie

Answer 4

An alternative with mixing broom and dplyr libraries:

    get_formula <- function(model) {
  
  broom::tidy(model)[, 1:2] %>%
    mutate(sign = ifelse(sign(estimate) == 1, ' + ', ' - ')) %>% #coeff signs
    mutate_if(is.numeric, ~ abs(round(., 2))) %>% #for improving formatting
    mutate(a = ifelse(term == '(Intercept)', paste0('y ~ ', estimate), paste0(sign, estimate, ' * ', term))) %>%
    summarise(formula = paste(a, collapse = '')) %>%
    as.character
  
}

It works for both simple and multiple linear regressions:

model1 <- lm(hp ~ disp, data = mtcars)
model2 <- lm(hp ~ mpg, data = mtcars)
model3 <- lm(hp ~ disp + mpg, data = mtcars)

# > get_formula(model1)
# [1] "y ~ 45.73 + 0.44 * disp"
# > get_formula(model2)
# [1] "y ~ 324.08 - 8.83 * mpg"
# > get_formula(model3)
# [1] "y ~ 172.22 + 0.26 * disp - 4.27 * mpg"

Burgenland