[R-meta] metafor::vcalc() for multiple sources of dependence
Thanks, Wolfgang. If I only intend to assume that the r_i for the same repeatedly measured group on the same outcome, as well as on multiple outcomes is constant (r_i = .6) and at the same time factor in the shared control group, which arguments should be simultaneously used for the Brown study below? I did see the help page but wasn't able to find an example that was similar to my study below. Thank you, Simon m=" study yi vi group time outcome Brown -0.10 0.05 1 0 1 Brown 0.24 0.05 1 0 2 Brown -0.11 0.05 1 0 3 Brown -0.01 0.05 1 0 4 Brown -0.12 0.05 2 0 1 Brown 0.38 0.05 2 0 2 Brown -0.19 0.05 2 0 3 Brown 0.30 0.05 2 0 4 Brown 0.45 0.05 1 1 1 Brown 0.66 0.05 1 1 2 Brown 0.27 0.05 1 1 3 Brown 0.13 0.05 1 1 4 Brown 0.08 0.05 1 2 1 Brown 0.25 0.05 1 2 2 Brown -0.34 0.05 1 2 3 Brown -0.06 0.05 1 2 4 Brown 0.48 0.05 2 1 1 Brown 0.28 0.05 2 1 2 Brown 0.10 0.05 2 1 3 Brown 0.25 0.05 2 1 4 Brown 0.11 0.05 2 2 1 Brown 0.65 0.05 2 2 2 Brown 0.39 0.05 2 2 3 Brown 0.13 0.05 2 2 4" data <- read.table(text=m,h=T) On Fri, Mar 4, 2022 at 4:45 AM Viechtbauer, Wolfgang (SP)
<wolfgang.viechtbauer at maastrichtuniversity.nl> wrote:
Hi Simon, Yes, that's what vcalc() is for. Best, Wolfgang
-----Original Message----- From: Simon Harmel [mailto:sim.harmel at gmail.com] Sent: Thursday, 03 March, 2022 17:35 To: Viechtbauer, Wolfgang (SP); R meta Subject: metafor::vcalc() for multiple sources of dependence Hello Wolfgang, Thank you for adding such an important function! In my studies, I have quite a few studies that are structured as shown below. Can I use vcalc() to construct a V matrix for the three sources of dependence (i.e., multiple groups, times, outcomes) in say the study below? Thank you for your guidance, Simon m=" study yi vi group time outcome Brown -0.10 0.05 1 0 1 Brown 0.24 0.05 1 0 2 Brown -0.11 0.05 1 0 3 Brown -0.01 0.05 1 0 4 Brown -0.12 0.05 2 0 1 Brown 0.38 0.05 2 0 2 Brown -0.19 0.05 2 0 3 Brown 0.30 0.05 2 0 4 Brown 0.45 0.05 1 1 1 Brown 0.66 0.05 1 1 2 Brown 0.27 0.05 1 1 3 Brown 0.13 0.05 1 1 4 Brown 0.08 0.05 1 2 1 Brown 0.25 0.05 1 2 2 Brown -0.34 0.05 1 2 3 Brown -0.06 0.05 1 2 4 Brown 0.48 0.05 2 1 1 Brown 0.28 0.05 2 1 2 Brown 0.10 0.05 2 1 3 Brown 0.25 0.05 2 1 4 Brown 0.11 0.05 2 2 1 Brown 0.65 0.05 2 2 2 Brown 0.39 0.05 2 2 3 Brown 0.13 0.05 2 2 4" read.table(text=m,h=T)