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dataframe manipulation

6 messages · Gang Chen, arun, Sarah Goslee +1 more

#
Hi,
Try:
?d[match(unique(d$fac),d$fac),]
A.K.
On Friday, December 13, 2013 4:17 PM, Gang Chen <gangchen6 at gmail.com> wrote:
Suppose I have a dataframe defined as

? ?  L3 <- LETTERS[1:3]
? ?  (d <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace
= TRUE)))

?  x? y fac
1? 1? 1?  C
2? 1? 2?  A
3? 1? 3?  B
4? 1? 4?  C
5? 1? 5?  B
6? 1? 6?  B
7? 1? 7?  A
8? 1? 8?  A
9? 1? 9?  B
10 1 10?  A

I want to extract those rows that are the first occurrences for each level
of factor 'fac', which are basically the first three rows above. How can I
achieve that? The real dataframe is more complicated than the example
above, and I can't simply list all the levels of factor 'fac' by
exhaustibly listing all the levels like the following

d[d$fac=='A' | d$fac=='B' | d$fac=='C', ]

Thanks,
Gang

??? [[alternative HTML version deleted]]

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#
What about:

lapply(levels(d$fac), function(x)head(d[d$fac == x,], 1))


Thanks for the reproducible example. If you put set.seed(123) before
the call to sample, then everyone who tries it will get the same data
frame d.

Sarah
On Fri, Dec 13, 2013 at 4:15 PM, Gang Chen <gangchen6 at gmail.com> wrote:

  
    
#
The following does the same thing a little more directly (and quickly)
   d[ !duplicated(d$fac), ]

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com