[R-meta] Meta-Analysis: Proportion in overall survival rate

[ne gic]

Thank you Ken and Nicky for your inputs!

Sincerely,
nelly



On Thu, May 28, 2020 at 1:19 PM Nicky Welton <Nicky.Welton using bristol.ac.uk>
wrote:

> Dear Nelly,
>
> If you are obtaining survival proportions from a Kaplan-Meier curve, then
> this has adjusted for censoring (which is a good thing). It's best to
> obtain the standard error for the survival proportion from the Kaplan-Meier
> curve which accounts for censoring. Of course papers don't usually report
> this, but you can reconstruct the survival data by scanning in the curves
> and following the algorithm proposed by Guyot et al (
> https://bmcmedresmethodol.biomedcentral.com/articles/10.1186/1471-2288-12-9)
> to obtain an approximation to the data used to construct the Kaplan-meier
> curves. This reconstructed data can then be re-analysed to obtain survival
> proportions and their standard errors at any time point. There is R code to
> accompany the Guyot algorithm (I can send you this).There is a Stata
> version available now too (
> https://journals.sagepub.com/doi/abs/10.1177/1536867X1801700402).
>
> Best wishes,
>
> Nicky
>
>
> -----Original Message-----
> From: R-sig-meta-analysis <r-sig-meta-analysis-bounces using r-project.org> On
> Behalf Of ne gic
> Sent: 28 May 2020 12:02
> To: Viechtbauer, Wolfgang (SP) <
> wolfgang.viechtbauer using maastrichtuniversity.nl>
> Cc: r-sig-meta-analysis using r-project.org
> Subject: Re: [R-meta] Meta-Analysis: Proportion in overall survival rate
>
> Dear Wolfgang,
>
> A quick follow up. After meta-analyzing the proportions using a logit
> transformation using qlogis(p), how I can back transform the proportion to
> fit the normal range as I get some values below 0 on the forest plot when I
> directly use the rma object.
>
> forest(pes.da_30plus, xlab = "2-year survival (%)")
>
> Sincerely,
> nelly
>
> On Wed, May 27, 2020 at 8:17 PM Viechtbauer, Wolfgang (SP) <
> wolfgang.viechtbauer using maastrichtuniversity.nl> wrote:
>
> > Dear Nelly,
> >
> > Your equation for the SE assumes that the p behaves like a 'regular'
> > proportion computed from a binomial distribution. I am not sure if
> > this is correct when using the Kaplan-Meier estimator to derive such a
> proportion.
> >
> > As far as your input to rma() is concerned - that is correct. However,
> > I would consider not meta-analyzing the proportions directly, but
> > doing a logit transformation on p, so using qlogis(p) for yi and
> > sqrt(1/(p*n) +
> > 1/((1-p)*n)) for the SE.
> >
> > Best,
> > Wolfgang
> >
> > >-----Original Message-----
> > >From: R-sig-meta-analysis [mailto:
> > r-sig-meta-analysis-bounces using r-project.org]
> > >On Behalf Of ne gic
> > >Sent: Wednesday, 27 May, 2020 20:02
> > >To: Dr. Gerta Rücker
> > >Cc: r-sig-meta-analysis using r-project.org
> > >Subject: Re: [R-meta] Meta-Analysis: Proportion in overall survival
> > >rate
> > >
> > >Dear Michael, Gerta and List,
> > >
> > >I would like to cross-check with you what I have done.
> > >
> > >I have restricted myself to Kaplan-Meier studies which gave the
> > >number at risk at 2 years, and also n_0 at baseline.
> > >
> > >I then estimated the absolute number of those surviving as *n_t *=
> > n_0*S(t)
> > >following Gerta's idea. I took the reported proportions at 2 years to
> > >represent the S(t).
> > >
> > >I calculated the standard error (SE) using the formula: *se *= square
> > root (
> > >*p*(1-*p*)/n). Where *p* = proportion at 2 years i.e. S(t) , n =
> > >*n_t*, the estimated number of of those surviving.
> > >
> > >I then used the random effects model in metafor as follows:
> > >rma(yi = *p*, sei = *se*, data=mydata, method="REML")
> > >
> > >The resulting estimate seems reasonable to me. But I want to confirm
> > >with you if this is the way one would input SE and the proportion to
> > >the function.
> > >
> > >Welcome any comments.
> > >
> > >Sincerely,
> > >nelly
> >
>
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