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Does the function call show the right information when using iprior.ipriorMod()? #42

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haziqj opened this issue Dec 5, 2017 · 1 comment
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Does the function call show the right information when using iprior.ipriorMod()? #42

haziqj opened this issue Dec 5, 2017 · 1 comment

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@haziqj
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haziqj commented Dec 5, 2017

Does the function call show the right information when using iprior.ipriorMod()?
@haziqj
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haziqj commented Dec 6, 2017

Apparently not:

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

@haziqj haziqj closed this as completed in 32bc7d7 Dec 6, 2017
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