2 messages · Omar Lakkis, Brian Ripley
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$f),]
s
date f 1 1999-01-01 1 4 1999-01-04 2
Rather than using a loop, how can I remove all consequentially repeated values as in this example? I am guessing using diff would help but not quite sure how.
get s
date f 1 1999-01-01 1 2 1999-01-02 1 3 1999-01-03 1 4 1999-01-04 2 5 1999-01-05 2
v <- s[1,'f']; for (i in 2:nrow(s)) { if (s[i,'f'] == v) s[i,'f'] <- NA else v <- s[i,'f'] }
s <- s[!is.na(s$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, ripley at stats.ox.ac.uk 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