A question about the trimfill function in the "meta" package in R I was doing a meta-analysis of single proportion using the meta package in R.
I performed a double arcsine transformation to my data. I also wanted to do a trim and fill procedure. However, when backtransf= in the trimfill function is TRUE, the result came back as NA. I was not able to get a transformed back proportion. Interestingly, when I applied other transformations, including PRAW, PLOGIT, and PAS, everything worked fine. So, my question is how to get the original proportion after you perform a PFT in this situation?
Below is my code: 
proportion = read.table("D:\\...\\Example.csv", header=T, sep=",")
metaproportion = metaprop(cases, total, author, data=proportion, 
                          sm="PFT", method.tau="DL", method.ci="CP",
                          incr=0.5, allincr=FALSE, addincr=FALSE, title="")
taf = trimfill(metaproportion, backtransf=TRUE, comb.fixed=FALSE, 
               comb.random=TRUE, ma.fixed=FALSE, type="R")
taf

The results:
          proportion     95%-CI            z  p-value
Random effects model         NA        16.10 < 0.0001

I uploaded my data into my Google drive.
 A: The issue here is that it is difficult to work out the back transformation for the summary although relatively easy for the individual studies.
Sometimes also referred to as the double
arcsin(e) transformation described by Freeman and Tukey here 
$$
y_i = \sin^{-1} \sqrt{\frac{r_i}{n_i+1}} +
   \sin^{-1} \sqrt{\frac{r_i+1}{n_i+1}}
$$
Where the $i$th study has $r_i$ successes out of $n_i$ trials.
We can weight each $y_i$ by the 
inverse of its variance.
$$
V(y_i) = (n_i + 0.5)^{-1}
$$
The weights $w_i=(n_i+0.5)$ are then
applied in the usual way.
Having formed the weighted average $y$
it is usual to back--transform it onto the original
scale by noting that for large $r_i, n_i$
reduces to
$$
y = 2 \sin^{-1} \sqrt{\frac{r}{n}}
$$
and using
$$
p = \left(\sin \frac{y}{2}\right)^2
$$
Incidentally Miller gives a more accurate formula
for small $n$.
$$
p = 0.5\left(1-\mathrm{sign}(\cos y)
\sqrt{1 - \left(\sin y + \frac{\sin y - \frac{1}{\sin y}}{\tilde{n}}\right)^2}\right)
$$
where $\tilde{n}$ is the harmonic mean of the study sizes,
$$
\tilde{n} = \frac{k}{\sum{n_i^{-1}}}
$$
The problem for software is that this demands the step of forming the harmonic mean and the required information has not usually been supplied to the function. So the fairly simple calculations need to be done separately outside the function. There is an extensive example of how to do it using a different package (metafor) available here which you should be able to use to work out the code you need to do this using your preferred package, or indeed using any other software system..
Note that the Freeman--Tukey transformation although
often referred to as variance stabilising
only stabilises the variance for equal
$n_i$ hence the need for the weighting
step.
