On Wed, 25 Jun 2008, Marc Schwartz wrote:
on 06/25/2008 11:19 AM Daren Tan wrote:
unique(c(1:10,1)) gives 1:10 (i.e. unique values), is there any
method to get only 2:10 (i.e. values that are unique) ?
The easiest might be:
Vec
[1] 1 2 3 4 5 6 7 8 9 10 1
Vec[table(Vec) == 1]
[1] 2 3 4 5 6 7 8 9 10
I don't think that is right: you are relying on recycling indices. Try
Vec <- c(1,1:10)
Vec[table(Vec) == 1]
which should be the same.
I was about to write
tab <- table(Vec)
names(tab)[tab==1]
but that gives a character vector. Here's a different way:
Vec[rowSums(outer(Vec, Vec, "=="))==1]
--
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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