On Wed, 18 May 2005, Omar Lakkis wrote:
Rather than using a loop, how can I remove all consequentially
repeated values as in this example?
I am guessing using diff would help but not quite sure how.
get
s
date f
1 1999-01-01 1
2 1999-01-02 1
3 1999-01-03 1
4 1999-01-04 2
5 1999-01-05 2
v <- s[1,'f']; for (i in 2:nrow(s)) { if (s[i,'f'] == v) s[i,'f'] <- NA else v
<- s[i,'f'] }
s <- s[!is.na(s$f),]
s
date f
1 1999-01-01 1
4 1999-01-04 2
I am not sure I see the pattern, but it might be one of
s[!duplicated(s$f), ]
s[diff(c(0, s$f)) != 0, ]
(which differ).
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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