Log response ratios with 'Metafor' with zeros - r

Question

I'm using the 'metafor' package in R to perform log response ratios. Some of my means are zero, which seems to be the cause of a warning after my escalc command (since log(0) is -inf). The metafor package provides a method of adding a small value to zero to avoid this. The documentation states:
"Cell entries with a zero can be problematic especially for the relative risk and the odds ratio. Adding a small constant to the cells of the 2 × 2 tables is a common solution to this problem [...] When to = "only0", the value of add is added to each cell of the 2 × 2 tables only in those tables with at least one cell equal to 0."
For some reason this is not resolving my error, perhaps because my data is not a 2x2 table? (It is output from summarise with ddply from the ply package, similar to the formatting in this example). Must I replace the zero values with a small number manually or is there a more elegant way? (Note that in this example the rows with zero also have sample size of 1 and thus no variance and will be dropped from the analysis anyway. I just want to know how this works for the future).
Reproducible example:
dat<-dput(Bin_Y_count_summary_wide)
structure(list(Species.ID = c("CAFERANA", "TR11", "TR118", "TR500",
"TR504", "TR9", "TR9_US1"), Y_num_mean.early = c(2, 147.375,
4.5, 0.5, 12.5, 93.4523809523809, 5), N.early = c(1L, 4L, 2L,
4L, 4L, 7L, 2L), sd.early = c(NA, 174.699444284558, 6.36396103067893,
1, 22.4127939653523, 137.506118190001, 7.07106781186548), se.early = c(NA,
87.3497221422789, 4.5, 0.5, 11.2063969826762, 51.9724274972283,
5), Y_num_mean.late = c(0, 3.625, 2.98482142857143, 0.8, 3, 47.2,
0), N.late = c(1L, 4L, 7L, 10L, 10L, 8L, 1L), sd.late = c(NA,
7.25, 5.10407804830748, 1.75119007154183, 8.03118920210451, 40.7351024477486,
NA), se.late = c(NA, 3.625, 1.9291601697265, 0.553774924194538,
2.53968501984006, 14.4020335865659, NA), Y_num_mean.wet = c(NA,
71.5, 0, 12, 27, 0, NA), N.wet = c(NA, 2L, 1L, 2L, 2L, 2L, NA
), sd.wet = c(NA, 17.6776695296637, NA, 9.89949493661167, 38.1837661840736,
0, NA), se.wet = c(NA, 12.5, NA, 7, 27, 0, NA)), row.names = c(NA,
7L), .Names = c("Species.ID", "Y_num_mean.early", "N.early",
"sd.early", "se.early", "Y_num_mean.late", "N.late", "sd.late",
"se.late", "Y_num_mean.wet", "N.wet", "sd.wet", "se.wet"), class = "data.frame", reshapeWide = structure(list(
v.names = c("Y_num_mean", "N", "sd", "se"), timevar = "early_or_late",
idvar = "Species.ID", times = c("early", "late", "wet"),
varying = structure(c("Y_num_mean.early", "N.early", "sd.early",
"se.early", "Y_num_mean.late", "N.late", "sd.late", "se.late",
"Y_num_mean.wet", "N.wet", "sd.wet", "se.wet"), .Dim = c(4L,
3L))), .Names = c("v.names", "timevar", "idvar", "times",
"varying")))
# Warning produced from this command
test <- escalc(measure="ROM", m1i=Y_num_mean.early, sd1i=sd.early, n1i=N.early, m2i=Y_num_mean.late, sd2i=sd.late, n2i=N.late, data=dat, add=1/2, to="only0")

Answer 1

The paragraph you are quoting applies to measures that one can calculate based on 2x2 tables (i.e., RR, OR, RD, AS, and PETO). The add and to arguments do not have any effect for measures such as SMD and ROM.
The only way you can get a mean of 0 for a ratio scale variable (which is what use of response ratios assumes) is if every value is equal to 0. Therefore, by definition, the variance must also be 0. This applies whether the sample size is 1 (in which case the variance is of course also 0) or whether you have a larger sample size.
In general, whenever at least one of the two means is 0, one cannot calculate the log response ratio. Of course, one could start adding some kind of constant to the means manually (and the same for the SDs), but this seems rather arbitrary. The adjustments we can do to counts in 2x2 tables are motivated based on statistical theory (those adjustments are actually bias reductions, which also happen to make the calculation of certain measures possible when there is a 0 count).