Version 3.2-0

pbreheny · Sep 27, 2018 · 768056f · 768056f
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diff --git a/DESCRIPTION b/DESCRIPTION
@@ -1,7 +1,7 @@
 Package: grpreg
 Title: Regularization Paths for Regression Models with Grouped Covariates
-Version: 3.2-000
-Date: 2018-06-15
+Version: 3.2-0
+Date: 2018-09-27
 Authors@R: c(
   person("Patrick", "Breheny", role=c("aut","cre"), email="[email protected]", comment=c(ORCID="000-0002-0650-1119")),
   person("Yaohui", "Zeng", role="ctb"))

diff --git a/man/Birthwt.Rd b/man/Birthwt.Rd
@@ -43,8 +43,9 @@ data(Birthwt)
       category.}
   }
 }
-\source{\code{MASS}.  \code{R} package.  Available at
-  \code{http://cran.r-project.org}.}
+\source{\code{MASS}.  \code{R} package.
+  \url{https://cran.r-project.org/package=MASS}
+}
 \references{
   \itemize{
     \item Venables, W. N. and Ripley, B. D. (2002) \emph{Modern Applied

diff --git a/man/predict.Rd b/man/predict.Rd
@@ -61,17 +61,17 @@ fit <- grpreg(X, y, group, penalty="grLasso", family="binomial")
 
 # Coef and predict methods
 coef(fit, lambda=.001)
-predict(fit, X, type="link", lambda=.001)
-predict(fit, X, type="response", lambda=.001)
-predict(fit, X, type="class", lambda=.001)
+predict(fit, X, type="link", lambda=.07)[1:10]
+predict(fit, X, type="response", lambda=.07)[1:10]
+predict(fit, X, type="class", lambda=.01)[1:15]
 predict(fit, type="vars", lambda=.07)
 predict(fit, type="groups", lambda=.07)
 predict(fit, type="norm", lambda=.07)
 
 # Coef and predict methods for cross-validation
 cvfit <- cv.grpreg(X, y, group, family="binomial", penalty="grMCP")
 coef(cvfit)
-predict(cvfit, X)
-predict(cvfit, X, type="response")
+predict(cvfit, X)[1:10]
+predict(cvfit, X, type="response")[1:10]
 predict(cvfit, type="groups")
 }
diff --git a/man/select-grpreg.Rd b/man/select-grpreg.Rd
@@ -27,7 +27,7 @@ select(obj,\dots)
   freedom, and \eqn{n}{n} is the sample size:
 
   \deqn{AIC = L + 2\nu}{AIC = L + 2*df}
-  \deqn{BIC = L + log(n)\nu}{BIC = L + log(n)*df}
+  \deqn{BIC = L + \log(n)\nu}{BIC = L + log(n)*df}
   \deqn{GCV = \frac{L}{(1-\nu/n)^2}}{GCV= L/((1-df/n)^2)}
   \deqn{AICc = AIC + 2\frac{\nu(\nu+1)}{n-\nu-1}}{AICc = AIC + 2*df*(df+1)/(n-df-1)}
   \deqn{EBIC = BIC + 2 \log{p \choose \nu}}{EBIC = BIC + 2*log(p choose df)}