On Feb 26, 2010, at 8:17 AM, Henrique Dallazuanna wrote:
Try this:
unique(c(X))
I did but it returned NA as did: unique(as.vector(X)).
To get rid of the NA's I needed to do:
X[!duplicated(as.vector(X)) & !is.na(X)]
(Logical indexing and does need as.vector() , or c() , to "straighten
out" the index expression.)
Not sure why applying the straightening to the second logical term is
not equivalent:
> X[!duplicated(X) & !is.na(as.vector(X))]
[1] 1 2 4 3 1 7 2
This also:
> unique(c(X))[!is.na(unique(c(X)))]
[1] 1 2 4 3 7
On Fri, Feb 26, 2010 at 10:06 AM, Todd DeWees
<t.dew...@cpse.dundee.ac.uk> wrote:
I have a 280,000 x 11 matrix with various values and many NA
values. What I would like to do is get a vector of every unique
value in the matrix.
For example:
X = [ 1 2 NA
4 3 1
7 NA 2 ]
Returns:
Unique_X = [ 1, 2, 3, 4, 7]
Thanks,
Todd
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O
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David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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