CRAN Package Check Results for Package CFC

Last updated on 2025-02-23 19:51:01 CET.

Flavor	Version	Tinstall	Tcheck	Ttotal	Status	Flags
r-devel-linux-x86_64-debian-clang	1.2.0	29.98	62.20	92.18	NOTE
r-devel-linux-x86_64-debian-gcc	1.2.0	20.82	46.28	67.10	NOTE
r-devel-linux-x86_64-fedora-clang	1.2.0			165.79	NOTE
r-devel-linux-x86_64-fedora-gcc	1.2.0			147.03	NOTE
r-devel-macos-arm64	1.2.0			41.00	NOTE
r-devel-macos-x86_64	1.2.0			87.00	NOTE
r-devel-windows-x86_64	1.2.0	31.00	95.00	126.00	NOTE
r-patched-linux-x86_64	1.2.0	29.11	58.68	87.79	NOTE
r-release-linux-x86_64	1.2.0	26.69	58.58	85.27	NOTE
r-release-macos-arm64	1.2.0			42.00	NOTE
r-release-macos-x86_64	1.2.0			62.00	NOTE
r-release-windows-x86_64	1.2.0	31.00	95.00	126.00	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	38.00	116.00	154.00	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-macos-arm64, r-devel-macos-x86_64, r-devel-windows-x86_64, r-patched-linux-x86_64, r-release-linux-x86_64, r-release-macos-arm64, r-release-macos-x86_64, r-release-windows-x86_64