On Tue, 23 Nov 2010, seeliger.c...@epamail.epa.gov wrote:
Is there any similar function in R to the first. in SAS?
?duplicated
a$d <- ifelse( duplicated( a$a ), 0 , 1 )
a$d.2 <- as.numeric( !duplicated( a$a ) )
Actually, duplicated does not duplicate SAS' first. operator, though it
may suffice for the OP's needs.
To illustrate, let's start with a dataframe of 3 key columns and some data
in x:
tt <- data.frame(k1 = rep(1:3, each=10), k2 = rep(1:5, each=2, times=3),
k3=rep(1:2, times=15), x = 1:30)
# Try to mimic what the following SAS datastep would do,
# assuming 'tt' is already sorted:
# data foo;
# set tt;
# by k1, k2;
# put first.k1=, first.k2=;
# run;
# SAS' first. operations would result in these values:
tt$sas.first.k1 <- rep(c(1, rep(0,9)), 3)
tt$sas.first.k2 <- rep(1:0, 15)
# R duplicated() returns these values. You can see they
# are the same for k1, but dissimilar after row 10 for k2.
tt$duplicated.k1 <- 0+!duplicated(tt$k1)
tt$duplicated.k2 <- 0+!duplicated(tt$k2)
It depends on how you use duplicated()
all.equal( tt$sas.first.k2, 0+!duplicated( tt[, c("k1","k2") ] ) )
[1] TRUE
Chuck
# I've found I need to lag a column to mimic SAS' first.
# operator, thusly, though perhaps someone else knows
# differently. Note this does not work on unordered
# dataframes!
lag.k1 <- c(NA, tt$k1[1:(nrow(tt) - 1)])
tt$r.first.k1 <- ifelse(is.na(lag.k1), 1, tt$k1 != lag.k1)
lag.k2 <- c(NA, tt$k2[1:(nrow(tt) - 1)])
tt$r.first.k2 <- ifelse(is.na(lag.k2), 1, tt$k2 != lag.k2)
Mimicking SAS' last. operation can be done in a similar manner, by
anti-laging the column of interest and changing the comparisons somewhat.
Enjoy the days,
cur
--
Curt Seeliger, Data Ranger
Raytheon Information Services - Contractor to ORD
seeliger.c...@epa.gov
541/754-4638
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Charles C. Berry Dept of Family/Preventive Medicine
cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
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