Hi All,
I am counting the number of occurrences of the terms listed in one
vector in another vector.
My code runs:
for( i in 1:length(vector3)){
vector3[i] = sum(1*is.element(vector2, vector1[i]))
}
where
vector1 = vector containing the terms whose occurrences I want to count
vector2 = made up of a number of repetitions of all the elements of
vector1
vector3 = a vector of NAs that is meant to get the result of the
counting
My problem is that vector1 is about 60000 terms, and vector2 is
620000... can anyone suggest a faster code than the one I wrote?
Cheers,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St. Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 75941602 Fax +44 (0)20 75943193
f.calboli [.a.t] imperial.ac.uk
f.calboli [.a.t] gmail.com
On 20 Jun 2005, at 21:24, Erin Hodgess wrote:> Hello, Federico! > > I'm a bit confused about your question, please: > > What sorts of things are in Vector1, please?numbers (as in "numeric") that code individuals> > Why are you counting NAs in Vector3, please?I am counting how many times the code for each individual (as listed in vector1) is present in vector2, and save the number of counts in vector3 Cheers, Federico -- Federico C. F. Calboli Department of Epidemiology and Public Health Imperial College, St. Mary's Campus Norfolk Place, London W2 1PG Tel +44 (0)20 75941602 Fax +44 (0)20 75943193 f.calboli [.a.t] imperial.ac.uk f.calboli [.a.t] gmail.com
v3 <- numeric()
v3[v1] <- table(v2)[v1]
Jim
__________________________________________________________
James Holtman "What is the problem you are trying to solve?"
Executive Technical Consultant -- Convergys Labs
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Federico Calboli
<f.calboli at imperial.a To: r-help
<r-help at stat.math.ethz.ch>
c.uk> cc:
Sent by: Subject: [R] vectorisation
suggestion
r-help-bounces at stat.m
ath.ethz.ch
06/20/2005 16:15
Hi All,
I am counting the number of occurrences of the terms listed in one
vector in another vector.
My code runs:
for( i in 1:length(vector3)){
vector3[i] = sum(1*is.element(vector2, vector1[i]))
}
where
vector1 = vector containing the terms whose occurrences I want to count
vector2 = made up of a number of repetitions of all the elements of
vector1
vector3 = a vector of NAs that is meant to get the result of the
counting
My problem is that vector1 is about 60000 terms, and vector2 is
620000... can anyone suggest a faster code than the one I wrote?
Cheers,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St. Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 75941602 Fax +44 (0)20 75943193
f.calboli [.a.t] imperial.ac.uk
f.calboli [.a.t] gmail.com
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Federico - "match" will give you the (first) index of each element of its first argument in its second argument. So match(vector.1, vector.2) tells you where each element of vector.1 appears in vector.2. So if you use "table" on that vector, you'll see how many times each element of vector.2 appears. Something like: first.occur <- match(vector.1, vector.2) table(factor(vector.2[first.occur], levels = sort(unique(vector.2))) (changing it into a factor means you won't lose values of vector.2 that never appear in vector.1.) An example:> vec1 <- sample(1:10, 500, replace = TRUE) > table(vec1)vec1 1 2 3 4 5 6 7 8 9 10 61 37 43 47 49 59 51 48 53 52> vec2 <- 0:11 > vec3 <- match(vec1, vec2) > table(factor(vec2[vec3], levels = sort(unique(vec2))))0 1 2 3 4 5 6 7 8 9 10 11 0 61 37 43 47 49 59 51 48 53 52 0>This also works if one of the members of vec1 is not in vec2 -- that member simply gets ignored. (As you can see if add, say, a "20" at the end of vec1.) Hope this helps, Matt Wiener -----Original Message----- From: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Federico Calboli Sent: Monday, June 20, 2005 4:16 PM To: r-help Subject: [R] vectorisation suggestion Hi All, I am counting the number of occurrences of the terms listed in one vector in another vector. My code runs: for( i in 1:length(vector3)){ vector3[i] = sum(1*is.element(vector2, vector1[i])) } where vector1 = vector containing the terms whose occurrences I want to count vector2 = made up of a number of repetitions of all the elements of vector1 vector3 = a vector of NAs that is meant to get the result of the counting My problem is that vector1 is about 60000 terms, and vector2 is 620000... can anyone suggest a faster code than the one I wrote? Cheers, Federico Calboli -- Federico C. F. Calboli Department of Epidemiology and Public Health Imperial College, St. Mary's Campus Norfolk Place, London W2 1PG Tel +44 (0)20 75941602 Fax +44 (0)20 75943193 f.calboli [.a.t] imperial.ac.uk f.calboli [.a.t] gmail.com ______________________________________________ R-help at stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html