Do a partition over using dplyr in R [duplicate] - r

Question

I have a really simple problem, but I'm probably not thinking vector-y enough to solve it efficiently. I tried two different approaches and they've been looping on two different computers for a long time now. I wish I could say the competition made it more exciting, but ... bleh.
rank observations in group
I have long data (many rows per person, one row per person-observation) and I basically want a variable, that tells me how often the person has been observed already.
I have the first two columns and want the third one:
person wave obs
pers1 1999 1
pers1 2000 2
pers1 2003 3
pers2 1998 1
pers2 2001 2
Now I'm using two loop-approaches. Both are excruciatingly slow (150k rows). I'm sure I'm missing something, but my search queries didn't really help me yet (hard to phrase the problem).
Thanks for any pointers!
# ordered dataset by persnr and year of observation
person.obs <- person.obs[order(person.obs$PERSNR,person.obs$wave) , ]
person.obs$n.obs = 0
# first approach: loop through people and assign range
unp = unique(person.obs$PERSNR)
unplength = length(unp)
for(i in 1:unplength) {
print(unp[i])
person.obs[which(person.obs$PERSNR==unp[i]),]$n.obs =
1:length(person.obs[which(person.obs$PERSNR==unp[i]),]$n.obs)
i=i+1
gc()
}
# second approach: loop through rows and reset counter at new person
pnr = 0
for(i in 1:length(person.obs[,2])) {
if(pnr!=person.obs[i,]$PERSNR) { pnr = person.obs[i,]$PERSNR
e = 0
}
e=e+1
person.obs[i,]$n.obs = e
i=i+1
gc()
}

Answer 1

The answer from Marek in this question has proven very useful in the past. I wrote it down and use it almost daily since it was fast and efficient. We'll use ave() and seq_along().
foo <-data.frame(person=c(rep("pers1",3),rep("pers2",2)),year=c(1999,2000,2003,1998,2011))
foo <- transform(foo, obs = ave(rep(NA, nrow(foo)), person, FUN = seq_along))
foo
person year obs
1 pers1 1999 1
2 pers1 2000 2
3 pers1 2003 3
4 pers2 1998 1
5 pers2 2011 2
Another option using plyr
library(plyr)
ddply(foo, "person", transform, obs2 = seq_along(person))
person year obs obs2
1 pers1 1999 1 1
2 pers1 2000 2 2
3 pers1 2003 3 3
4 pers2 1998 1 1
5 pers2 2011 2 2

Answer 2

A few alternatives with the data.table and dplyr packages.
data.table:
library(data.table)
# setDT(foo) is needed to convert to a data.table
# option 1:
setDT(foo)[, rn := rowid(person)]
# option 2:
setDT(foo)[, rn := 1:.N, by = person]
both give:
> foo
person year rn
1: pers1 1999 1
2: pers1 2000 2
3: pers1 2003 3
4: pers2 1998 1
5: pers2 2011 2
If you want a true rank, you should use the frank function:
setDT(foo)[, rn := frank(year, ties.method = 'dense'), by = person]
dplyr:
library(dplyr)
# method 1
foo <- foo %>% group_by(person) %>% mutate(rn = row_number())
# method 2
foo <- foo %>% group_by(person) %>% mutate(rn = 1:n())
both giving a similar result:
> foo
Source: local data frame [5 x 3]
Groups: person [2]
person year rn
(fctr) (dbl) (int)
1 pers1 1999 1
2 pers1 2000 2
3 pers1 2003 3
4 pers2 1998 1
5 pers2 2011 2

Answer 3

Would by do the trick?
> foo <-data.frame(person=c(rep("pers1",3),rep("pers2",2)),year=c(1999,2000,2003,1998,2011),obs=c(1,2,3,1,2))
> foo
person year obs
1 pers1 1999 1
2 pers1 2000 2
3 pers1 2003 3
4 pers2 1998 1
5 pers2 2011 2
> by(foo, foo$person, nrow)
foo$person: pers1
[1] 3
------------------------------------------------------------
foo$person: pers2
[1] 2

Answer 4

Another option using aggregate and rank in base R:
foo$obs <- unlist(aggregate(.~person, foo, rank)[,2])
# person year obs
# 1 pers1 1999 1
# 2 pers1 2000 2
# 3 pers1 2003 3
# 4 pers2 1998 1
# 5 pers2 2011 2