CRAN Package Check Results for Package CFC

Last updated on 2024-04-25 07:59:48 CEST.

Flavor Version Tinstall Tcheck Ttotal Status Flags
r-devel-linux-x86_64-debian-clang 1.2.0 43.59 78.67 122.26 NOTE
r-devel-linux-x86_64-debian-gcc 1.2.0 35.11 58.91 94.02 NOTE
r-devel-linux-x86_64-fedora-clang 1.2.0 162.74 NOTE
r-devel-linux-x86_64-fedora-gcc 1.2.0 162.96 NOTE
r-devel-windows-x86_64 1.2.0 35.00 99.00 134.00 NOTE
r-patched-linux-x86_64 1.2.0 37.45 74.90 112.35 NOTE
r-release-linux-x86_64 1.2.0 44.98 78.88 123.86 OK
r-release-macos-arm64 1.2.0 42.00 NOTE
r-release-windows-x86_64 1.2.0 34.00 97.00 131.00 NOTE
r-oldrel-macos-arm64 1.2.0 46.00 OK
r-oldrel-macos-x86_64 1.2.0 84.00 OK
r-oldrel-windows-x86_64 1.2.0 44.00 115.00 159.00 OK

Check Details

Version: 1.2.0
Check: Rd files
Result: NOTE checkRd: (-1) cfc.tbasis.Rd:23: Lost braces 23 | Assuming one-dimensional \code{p1} and \code{p2} for clarity, the algorithm calculates cumulative incidence function for cuase 1 using a recursive formula: \code{ci1[n+1] = ci1[n] + dci1[n]}, where \code{dci1[n] = 0.5*(p2[n] + p2[n+1])*(p1[n] - p1[n+1])}. The increment in cumulative incidence function for cause 2 is similarly calculated, \code{dci2[n] = 0.5*(p1[n] + p1[n+1])*(p2[n] - p2[n+1])}. These equations guarantee that \code{dci1[n] + dci2[n] = p1[n]*p2[n] - p1[n+1]*p2[n+1]}. Event-free probability is simply calculated as code{efp[n] = p1[n]*p2[n]}. Taken together, this numerical integration ensures that \code{efp[n+1] - efp[n] + dci1[n] + dci2[n] = 0}. | ^ Flavors: r-devel-linux-x86_64-debian-clang, r-devel-linux-x86_64-debian-gcc, r-devel-linux-x86_64-fedora-clang, r-devel-linux-x86_64-fedora-gcc, r-devel-windows-x86_64, r-patched-linux-x86_64, r-release-macos-arm64, r-release-windows-x86_64