haziqj/iprior

Does the function call show the right information when using `iprior.ipriorMod()`?

haziqj opened this issue · 1 comments

haziqj commented

Does the function call show the right information when using iprior.ipriorMod()?

haziqj commented

Apparently not:

> mod <- kernL(stack.loss ~ ., stackloss)
> mod.fit <- iprior(mod)
> mod.fit$call
## iprior(y = object, method = method, control = control)