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[R-meta] metafor::vcalc() for multiple sources of dependence

4 messages · Simon Harmel, Wolfgang Viechtbauer

Simon Harmel
# 
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)
Wolfgang Viechtbauer
# 
Hi Simon,

Yes, that's what vcalc() is for.

Best,
Wolfgang
Simon Harmel
# 
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:
Wolfgang Viechtbauer
# 
Dear Simon,

Of course the help pages won't have examples for every possible use case that is out there. But the documentation is very thorough in explaining the purpose of the various arguments. Please give this a proper try and do not use the mailing list as a way to get other people to do the work for you.

Best,
Wolfgang