better than sapply
I don't know if its faster but you could try timing this to find out: r$seid <- merge(h, r, by = "cid")[,2]
On 8/27/05, Omar Lakkis <uofiowa at gmail.com> wrote:
I have the following two mapping data frames (r) and (h). I want to fill teh value of r$seid with the value of r$seid where r$cid==h$cid. I can do it with sapply as such:
r$seid = sapply(r$cid, function(cid) h[h$cid==cid,]$seid)
Is ther a better (faster) way to do this?
r <- data.frame(seid=NA, cid= c(2181,2221,2222)) r
seid cid 1 NA 2181 2 NA 2221 3 NA 2222
h <- data.frame(seid= c(5598,5609,4931,5611,8123,8122), cid= c(2219,2222,2181,2190,2817,2221)) h
cid seid 1 5598 2219 2 5609 2222 3 4931 2181 4 5611 2190 5 8123 2817 6 8122 2221 to get the desired result of:
r
seid cid 1 4931 2181 2 8122 2221 3 5609 2222
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