On 13/11/2018 12:58 PM, William Dunlap wrote:
You also asked about doing this for the rows of a matrix. unique() give
the unique rows but match operates on a per element, not per row,
basis. You can use split, which operates on rows of a matrix, to help.
> m <- cbind( A=c(i=5,ii=5,iii=5,iv=4,v=4,vi=4), B=c(2,3,2,2,2,2) )
> unique(m)
A B
i 5 2
ii 5 3
iv 4 2
> match(m, unique(m)) # bad
[1] 1 1 1 3 3 3 4 5 4 4 4 4
> asRows <- function(x) split(x, seq_len(NROW(x))) # convert to
list of rows
> match(asRows(m), unique(asRows(m)))
[1] 1 2 1 3 3 3
For data.frames unique works on rows but match works on columns, and
converting
to a list of rows does not quite work, because unique looks at the row
names. A
modification of asRoiws works around that:
> d <- data.frame(m)
> unique(d)
A B
i 5 2
ii 5 3
iv 4 2
> match(d, unique(d))
[1] NA NA
> asRows <- function(x) lapply(split(x, seq_len(NROW(x))), as.list)
> match(asRows(d), unique(asRows(d)))
[1] 1 2 1 3 3 3
Thanks! That's very nice.
Is this the sort of issue that Hadley's vectors package is addressing?
I don't know; hopefully someone else will respond...
Duncan Murdoch
Bill Dunlap
TIBCO Software
wdunlap tibco.com <http://tibco.com>
On Tue, Nov 13, 2018 at 2:15 AM, Duncan Murdoch
<murdoch.dun...@gmail.com <mailto:murdoch.dun...@gmail.com>> wrote:
On 13/11/2018 12:35 AM, Pages, Herve wrote:
Hi,
On 11/12/18 17:08, Duncan Murdoch wrote:
The duplicated() function gives TRUE if an item in a vector
(or row in
a matrix, etc.) is a duplicate of an earlier item. But what
I would
like to know is which item does it duplicate?
For example,
v <- c("a", "b", "b", "a")
duplicated(v)
returns
[1] FALSE FALSE TRUE TRUE
What I want is a fast way to calculate
[1] NA NA 2 1
or (equally useful to me)
[1] 1 2 2 1
The result should have the property that if result[i] == j,
then v[i]
== v[j], at least for i != j.
Does this already exist somewhere, or is it easy to write?
I generally use match() for that:
> v <- c("a", "b", "b", "a")
> match(v, v)
[1] 1 2 2 1
Yes, this is perfect. Thanks to you (and the private answer I
received that suggested the same).
Duncan Murdoch
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