CRAN Package Check Results for Package CFC

Last updated on 2024-05-02 01:56:54 CEST.

Flavor	Version	Tinstall	Tcheck	Ttotal	Status	Flags
r-devel-linux-x86_64-debian-clang	1.2.0	45.91	80.77	126.68	NOTE
r-devel-linux-x86_64-debian-gcc	1.2.0	33.74	59.21	92.95	NOTE
r-devel-linux-x86_64-fedora-clang	1.2.0			165.55	NOTE
r-devel-linux-x86_64-fedora-gcc	1.2.0			162.96	NOTE
r-devel-windows-x86_64	1.2.0	36.00	101.00	137.00	NOTE
r-patched-linux-x86_64	1.2.0	40.62	75.46	116.08	NOTE
r-release-linux-x86_64	1.2.0	44.72	75.47	120.19	NOTE
r-release-macos-arm64	1.2.0			42.00	NOTE
r-release-windows-x86_64	1.2.0	36.00	98.00	134.00	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	46.00	118.00	164.00	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-linux-x86_64, r-release-macos-arm64, r-release-windows-x86_64