How does the 'with' function work with glm model output?

Question

I'm trying to understand the with function in context of a glm function output from here

First, a simple data set

set.seed(101)
x.test <- runif(50,2,8)
y.test <- 0.5^(x.test)
df <- data.frame(x.test, y.test)

I want to fit the function of form y.test ~ lambda/(1+ aii*x.test) using the inverse link function in glm

g1 <- glm(y.test ~ x.test, family = gaussian(link = "inverse"))
coef(g1)

(Intercept) x.test
-10.884889 6.946893

s0 <- with(as.list(coef(g1)), 
      list(lambda = 1/`(Intercept)`, aii = x.test/`(Intercept)`))

This produces a single value for $aii: [1] -0.6382144

Why does the with function do that? Should aii not produce 50 values given that x.test is a vector containing 50 values?

@Roland Ohh gotcha! I thought that s0 was evaluating x.test from the dataset for some reason. Thank you. — Rspacer, Jan 10 '22 at 13:39

score 0 · Answer 1 · answered Jan 10 '22 at 15:45

This is a bit confusing because x.test can be found in three different places:

as a variable in the global workspace (a vector)
as an element of df (a vector)
as an element of as.list(coef(g1)) (a single value giving the coefficient of x.test in the regression equation)

When with(as.list(coef(g1)), ...) is used, the third location is the one searched first.

1 Answers1