Question

This is a saga which began with the problem of how to do survey weighting. Now that I appear to be doing that correctly, I have hit a bit of a wall (see previous post for details on the import process and where the strata variable came from):

> require(foreign)
> ipums <- read.dta('/path/to/data.dta')
> require(survey)
> ipums.design <- svydesign(id=~serial, strata=~strata, data=ipums, weights=perwt)
Error in if (nbins > .Machine$integer.max) stop("attempt to make a table with >= 2^31 elements") : 
  missing value where TRUE/FALSE needed
In addition: Warning messages:
1: In pd * (as.integer(cat) - 1L) : NAs produced by integer overflow
2: In pd * nl : NAs produced by integer overflow
> traceback()
9: tabulate(bin, pd)
8: as.vector(data)
7: array(tabulate(bin, pd), dims, dimnames = dn)
6: table(ids[, 1], strata[, 1])
5: inherits(x, "data.frame")
4: is.data.frame(x)
3: rowSums(table(ids[, 1], strata[, 1]) > 0)
2: svydesign.default(id = ~serial, weights = ~perwt, strata = ~strata, 
       data = ipums)
1: svydesign(id = ~serial, weights = ~perwt, strata = ~strata, data = ipums)

This error seems to come from the tabulate function, which I hoped would be straightforward enough to circumvent, first by changing .Machine$integer.max

> .Machine$integer.max <- 2^40

and when that didn’t work the whole source code of tabulate:

> tabulate <- function(bin, nbins = max(1L, bin, na.rm=TRUE))
{
    if(!is.numeric(bin) && !is.factor(bin))
    stop("'bin' must be numeric or a factor")
    #if (nbins > .Machine$integer.max)
    if (nbins > 2^40) #replacement line
        stop("attempt to make a table with >= 2^31 elements")
    .C("R_tabulate",
       as.integer(bin),
       as.integer(length(bin)),
       as.integer(nbins),
       ans = integer(nbins),
       NAOK = TRUE,
       PACKAGE="base")$ans
}

Neither circumvented the problem. Apparently this is one reason why the ff package was created, but what worries me is the extent to which this is a problem I cannot avoid in R. This post seems to indicate that even if I were to use a package that would avoid this problem, I would only be able to access 2^31 elements at a time. My hope was to use sql (either sqlite or postgresql) to get around the memory problems, but I’m afraid I’ll spend a while getting that to work, only to run into the same fundamental limit.

Attempting to switch back to Stata doesn’t solve the problem either. Again see the previous post for how I use svyset, but the calculation I would like to run causes Stata to hang:

svy: mean age, over(strata)

Whether throwing more memory at it will solve the problem I don’t know. I run R on my desktop which has 16 gigs, and I use Stata through a Windows server, currently setting memory allocation to 2000MB, but I could theoretically experiment with increasing that.

So in sum:

1. Is this a hard limit in R?
2. Would sql solve my R problems?
3. If I split it up into many separate files would that fix it (a lot of work…)?
4. Would throwing a lot of memory at Stata do it?
5. Am I seriously barking up the wrong tree somehow?

Answer 1

Both @Gavin and @Martin deserve credit for this answer, or at least leading me in the right direction. I’m mostly answering it separately to make it easier to read.

In the order I asked:

1. Yes 2^31 is a hard limit in R, though it seems to matter what type it is (which is a bit strange given it is the length of the vector, rather than the amount of memory (which I have plenty of) which is the stated problem. Do not convert strata or id variables to factors, that will just fix their length and nullify the effects of subsetting (which is the way to get around this problem).

2. sql could probably help, provided I learn how to use it correctly. I did the following test:

   library(multicore) # make svy fast!
   ri.ny <- subset(ipums, statefips_num %in% c(36, 44))
   ri.ny.design <- svydesign(id=~serial, weights=~perwt, strata=~strata, data=ri.ny)
   svyby(~incwage, ~strata, ri.ny.design, svymean, data=ri.ny, na.rm=TRUE, multicore=TRUE)
   
   ri <- subset(ri.ny, statefips_num==44)
   ri.design <- svydesign(id=~serial, weights=~perwt, strata=~strata, data=ri)
   ri.mean <- svymean(~incwage, ri.design, data=ri, na.rm=TRUE)
   
   ny <- subset(ri.ny, statefips_num==36)
   ny.design <- svydesign(id=~serial, weights=~perwt, strata=~strata, data=ny)
   ny.mean <- svymean(~incwage, ny.design, data=ny, na.rm=TRUE, multicore=TRUE)
   

   And found the means to be the same, which seems like a reasonable test.

   So: in theory, provided I can split up the calculation by either using plyr or sql, the results should still be fine.

3. See 2.

4. Throwing a lot of memory at Stata definitely helps, but now I’m running into annoying formatting issues. I seem to be able to perform most of the calculation I want (much quicker and with more stability as well) but I can’t figure out how to get it into the form I want. Will probably ask a separate question on this. I think the short version here is that for big survey data, Stata is much better out of the box.

5. In many ways yes. Trying to do analysis with data this big is not something I should have taken on lightly, and I’m far from figuring it out even now. I was using the svydesign function correctly, but I didn’t really know what’s going on. I have a (very slightly) better grasp now, and it’s heartening to know I was generally correct about how to solve the problem. @Gavin’s general suggestion of trying out small data with external results to compare to is invaluable, something I should have started ages ago. Many thanks to both @Gavin and @Martin.