-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf
Of William Dunlap
Sent: Wednesday, October 17, 2012 2:11 PM
To: sds at gnu.org; r-help at r-project.org
Subject: Re: [R] uniq -c
Note that the relative speeds of these, which all use basically the same run-length-
encoding
algorithm, depend on the nature of the dataset. I made a million row data.frame with
10,000
unique users, 26 unique countries, and 6 unique languages with c. 3/4 million unique
rows. Then the times for methods 1, 2, and 3 were 0.7, 6.2, and 10.5 seconds,
respectively. With a million row data.frame with 100, 10, and 4 unique users, countries,
and languages, with 4000 unique rows, the times were 0.3, 1.4, and 0.7.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf
Of Sam Steingold
Sent: Wednesday, October 17, 2012 12:58 PM
To: r-help at r-project.org
Subject: Re: [R] uniq -c
* Sam Steingold <fqf at tah.bet> [2012-10-16 11:03:27 -0400]:
I need an analogue of "uniq -c" for a data frame.
Summary of options:
1. William:
isFirstInRun <- function(x) UseMethod("isFirstInRun")
isFirstInRun.default <- function(x) c(TRUE, x[-1] != x[-length(x)])
isFirstInRun.data.frame <- function(x) {
stopifnot(ncol(x)>0)
retval <- isFirstInRun(x[[1]])
for(column in x) {
retval <- retval | isFirstInRun(column)
}
retval
}
row.count.1 <- function (x) {
i <- which(isFirstInRun(x))
data.frame(x[i,], count=diff(c(i, 1L+nrow(x))))
}
147 seconds
2. http://orgmode.org/worg/org-contrib/babel/examples/Rpackage.html#sec-6-1
row.count.2 <- function (x) {
equal.to.previous <- rowSums( x[2:nrow(x),] != x[1:(nrow(x)-1),] )==0
tf.runs <- rle(equal.to.previous)
counts <- c(1, unlist(mapply(function(x,y) if (y) x+1 else (rep(1,x)),
tf.runs$length, tf.runs$value)))
counts <- counts[ c( diff( counts ) <= 0, TRUE ) ]
unique.rows <- which( c(TRUE, !equal.to.previous ) )
cbind(x[ unique.rows, ,drop=FALSE ], counts)
}
136 seconds
3. Micael: paste/strsplit
row.count.3 <- function (x) {
pa <- do.call(paste,x)
rl <- rle(p)
sp <- strsplit(as.character(rl$values)," ")
data.frame(user = sapply(sp,"[",1),
country = sapply(sp,"[",2),
language = sapply(sp,"[",3),
count = rl$length)
}
here I know the columns and rely on absense of spaces in values.
27 seconds.
Thanks to all who answered.
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
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