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haziqj / iprior

An R package for I-prior regression
GNU General Public License v3.0
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Does the function call show the right information when using `iprior.ipriorMod()`? #42

Closed haziqj closed 6 years ago

haziqj commented 6 years ago

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

haziqj commented 6 years ago

Apparently not:

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