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

Nicky Welton N|cky@We|ton @end|ng |rom br|@to|@@c@uk
Thu May 28 13:19:06 CEST 2020

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,


-----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 (%)")


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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