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

18 messages · Michael Dewey, Gerta Ruecker, ne gic +3 more

Original
ne gic
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Dear List,
instance from included studies - this I understand.

Does it make sense to perform a meta-analysis of the proportion (%) one
gets from overall time survival e.g. Overall 5y survival? imagine a
scenario where different studies are reporting different proportions of
patients surviving at this time point and I want to report a summary
proportion from all the studies at this time point.

If this is possible, does just collecting the proportion at that time point
e.g. 5 year suffice as the data to use for this calculation? Or what would
you suggest? Haven't seen a package that just takes a proportion.

Sincerely,
nelly
Gerta Ruecker
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Dear Nelly,

You could do this, at least in principle, if all proportions refer to 
the same timepoint, for example 5 years. The problem is that the data 
you obtain from studies with a time-to-event endpoint are different from 
those that directly provide a five-year survival proportion: The 
time-to-event analysis accounts for censoring, while the proportion of 
living after five years relatively to all patients at baseline usually 
does not account for censoring or missing data (and thus may 
underestimate the true proportion).

If I understand you correctly, you want to pool survival proportions 
(single-arm), not hazard ratios (comparing two arms).

The technical thing is that you have survival proportions with standard 
error from the time-to-event studies and single proportions (survived/n) 
from other studies. Survival proportions with standard errors can be 
pooled usingthe? generic inverse variance method. Proportions are best 
be pooled using generalized linear models. See, for example, the 
examples for function metaprop() in R package meta.

Best,

Gerta

Am 19.05.2020 um 14:15 schrieb ne gic:

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      ne gic
    

    

      
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Dear Gerta,

Thank you for your reply.

Correct, I want to pool survival proportions (single-arm), not hazard
ratios (comparing two arms).

But you raise some important points and I shall first have to re-evaluate
each study to see if they all adjusted for covariates, and which.

In case of further questions after re-checking them, I shall respond to
this email.

Thanks!

nelly

On Tue, May 19, 2020 at 2:32 PM Gerta Ruecker <ruecker at imbi.uni-freiburg.de>
wrote:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

  
  


  


      

    

    
4 days later

    

      
      

        


  

        

      ne gic
    

    

      
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Dear List and Gerta,

Once more am interested in overall survival and my aim is to analyse the
proportion(s) of patients left in the study at the 2 years time point as
reported by Kaplan-Meir (KM) curves. Of course there are those that are
censored and those that experience the event as time goes by as expected in
KM curves. I have now double checked all the studies to be included in my
meta-analysis dataset and I have selected all those that report the
proportion of patients left in the study at 2 years.
A number of those in the subset also included a risk table, thus I have
access to those at risk at this 2 year time point should I need them.

However, as I cannot directly infer the number of events(event) and total
at risk(n) from the curves at 2 years time point which would have been
convenient to plug into metaprop,
I thought that I could instead try Gerta's advice and see if I can use the
proportion (from each of the studies) and it's standard error (SE) -
manually calculated instead.

Questions:

   1. Is it correct to manually calculate the SE using the formula: SE =
   square root (p(1-p)/n). Where p = proportion, n = total at risk?
   2. Which R/Stata/SAS software function can then take in the proportion
   and SE and give me a pooled proportion with CI and forest plot?

I welcome any comments and hints. If this is not reasonable, anything else
I can do?

Sincerely,
nelly

On Tue, May 19, 2020 at 2:32 PM Gerta Ruecker <ruecker at imbi.uni-freiburg.de>
wrote:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Michael Dewey
    

    

      
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Dear Nelly

Comments in-line

        
On 23/05/2020 16:27, ne gic wrote:

            

      
      
      
    

        
But you said you do not have the n necessary to do this so it is not 
going to help I think.

            

      
      
      
    

        
The two most used R packages are meta and metafor either of which will 
do what you want.

            

      
      
      
    

        
I think you are going to have to restrict yourself to those studies 
which do give the number at risk at 2 years. I must say I would be 
rather nervous about doing this if the degree of and reasons for 
censoring were likely to be different between studies.

Michael

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

        

  
    
  


  


      

    

    

      
      

        


  

        

      Gerta Ruecker
    

    

      
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Dear Nelly, dear Michael,

Maybe I have misundersood something, but I do not understand why (as 
Michael said) the number at risk at two years should be relevant if you 
want to know the survival proportion at two years. The survival 
proportion, as I understand it, is the proportion who survived two 
years, relative to those who were there at baseline. By contrast, the 
number at risk at 2 years are those that are living just before the 2 
years date.

The problem is this:

 1. For studies that provide proportions (absolute numbers, or
    two-by-two tables) for the 2 years time point you know the number
    (per group) at baseline (n_0) and the number living after two years.
    However, the proportion calculated therefrom ignores that some
    individuals may have been censored during the two years and are
    perhaps still alive, but not known, and thus the survival proportion
    is underestimated.
 2. For studies providing a Kaplan-Meier estimate at 2 years (and also
    n_0 at baseline), you have an unbiased estimate of the survival
    proportion (because censoring is accounted for, provided the
    censoring assumptions are valid), and you can simply estimate the
    absolute number of surviving as n_t = n_0*S(t).

In other words, the problem is not calculation, but the difference in 
interpretation of both kinds of numbers: The studies of type 1 do not 
account for censoring, while those of type 2 do.

Best,

Gerta

Am 24.05.2020 um 12:25 schrieb Michael Dewey:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      ne gic
    

    

      
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Dear Gerta and Michael,

I thank both of you very much for your insights.

Sincerely,
nelly

On Sun, May 24, 2020 at 12:47 PM Dr. Gerta R?cker <

        
ruecker at imbi.uni-freiburg.de> wrote:

        

            

      
      
      
    

        

      
      
    

        

  
  


  


      

    

    
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      ne gic
    

    

      
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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

        
On Mon, May 25, 2020 at 9:34 AM ne gic <negic4 at gmail.com> wrote:

        

            

      
      
      
    

        

      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Wolfgang Viechtbauer
    

    

      
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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

            

      
      
      
    

  
  


  


      

    

    

      
      

        


  

        

      ne gic
    

    

      
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Many thanks for the insights, WoIfgang!

I concur, the proportion is likely not from a binomial distribution, so I
take your advice.

Sincerely,
nelly


On Wed, May 27, 2020 at 8:17 PM Viechtbauer, Wolfgang (SP) <

        
wolfgang.viechtbauer at maastrichtuniversity.nl> wrote:

        

            

      
      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Gerta Ruecker
    

    

      
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Dear Nelly, dear all,

I have another problem with this approach, and this is the denominator, 
"n". You insert n_t for n, but n_t = n_0*S(t) estimates the number of 
patients surviving at time t which is the numerator of the ratio. The 
denominator is n_0, not n_t. And n_0 should be used as n.

(Suppose a study without any censoring. Then the Kaplan-Meier estimator 
in t gives exactly p = n_t/n_0 which is the same as if the study reports 
simply the 2-year survival proportion relative to the sample size, n_0.)

Thus the n in your calculation should be n_0, irrespectively of the 
transformation method (I agree with Wolfgang that the logit 
transformation is preferable).

I don't see a principle problem with this approach, because n_t is an 
unbiased estimate of the numerator and sample size n_0 is - for sure - 
the denominator.

Best,

Gerta

Am 27.05.2020 um 21:07 schrieb ne gic:

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Gerta Ruecker
    

    

      
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Dear Nelly,

I forgot to add that my suggestion was thought as an approach when 
mixing directly reported proportions and proportions estimated from 
Kaplan-Meier curves (with the caveat I mentioned before in my first post).

However, if you really restrict your analysis to those studies that 
offer a Kaplan-Meier estimator, you may not need this proportion 
approach at all, if standard errors/confidence intervals for the 
Kaplan-Meier curves at time point 2 years are available. Then you could 
take these standard errors as se in the rma call.

Best,

Gerta

Am 27.05.2020 um 22:05 schrieb Dr. Gerta R?cker:

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      ne gic
    

    

      
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Dear Gerta,

Thank you, I take note. For simplicity, I restricted the meta-analysis to
proportions estimated from Kaplan-Meier curves, which was about 96% of the
studies.

I greatly appreciate the clarification.

Sincerely,
nelly

On Wed, May 27, 2020 at 10:16 PM Dr. Gerta R?cker <

        
ruecker at imbi.uni-freiburg.de> wrote:

        

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      ne gic
    

    

      
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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 at maastrichtuniversity.nl> wrote:

        

            

      
      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Wolfgang Viechtbauer
    

    

      
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Use atransf=plogis or transf=plogis. help(forest.rma) to read about the purpose of these arguments. But this won't give you percent, but proportions. If you really want percent, then use atransf=function(x) plogis(x)*100 or the same for transf.

Best,
Wolfgang

            

      
      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Ken Beath
    

    

      
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It won?t be correct. To calculate CI for Kaplan-Meier what is usually done is to use log(-log(S)) and there is then a formula for the SE of this which means that 95% CI can be calculated. If you have 95% CI calculated this way then it should be possible to work back to log(-log(S) and SE which can then be used for a meta-analysis. What also is sometimes used is Greenwoods estimate of the SE, which has the problem that it doesn?t guarantee that confidence intervals are constrained between 0 and 1. Most modern programs don?t use it except to give a SE on the survival but use log(-log(S)) for confidence intervals.

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Nicky Welton
    

    

      
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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 at r-project.org> On Behalf Of ne gic
Sent: 28 May 2020 12:02
To: Viechtbauer, Wolfgang (SP) <wolfgang.viechtbauer at maastrichtuniversity.nl>
Cc: r-sig-meta-analysis at 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 at maastrichtuniversity.nl> wrote:

        

            

      
      
      
    

        
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      ne gic
    

    

      
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Thank you Ken and Nicky for your inputs!

Sincerely,
nelly



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