CRAN Package Check Results for Package CFC

Last updated on 2024-02-21 00:54:23 CET.

Flavor Version Tinstall Tcheck Ttotal Status Flags
r-devel-linux-x86_64-debian-clang 1.2.0 NOTE
r-devel-linux-x86_64-debian-gcc 1.2.0 30.68 60.86 91.54 NOTE
r-devel-linux-x86_64-fedora-clang 1.2.0 161.70 OK
r-devel-linux-x86_64-fedora-gcc 1.2.0 158.00 OK
r-devel-windows-x86_64 1.2.0 35.00 100.00 135.00 OK
r-patched-linux-x86_64 1.2.0 43.55 77.34 120.89 OK
r-release-linux-x86_64 1.2.0 44.28 78.69 122.97 OK
r-release-macos-arm64 1.2.0 46.00 OK
r-release-macos-x86_64 1.2.0 84.00 OK
r-release-windows-x86_64 1.2.0 45.00 115.00 160.00 OK
r-oldrel-macos-arm64 1.2.0 45.00 OK
r-oldrel-windows-x86_64 1.2.0 42.00 111.00 153.00 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