Re: [R] blockwise sums
Hi, all!!
From the help page for 'aggregate':
Splits the data into subsets, computes summary statistics for
each, and returns the result in a convenient form.
So here's the solution I found to this problem:
blocksums - function(x,n)
{
temp - 1:length(x)-1
temp - list((temp%/%n)+1)
aggregate(x,by=temp,sum)$x
}
For instance:
blocksums(1:10,3)
[1] 6 15 24 10
Hope this helps!!
Vicente.
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RE: [R] blockwise sums
If you insist, here's one way:
my.blockwisesum - function(x, n, ...) {
tapply(x, seq(1, length(x), by=n), sum, ...)
}
Andy
From: Lutz Prechelt
I am looking for a function like
my.blockwisesum(vector, n)
that computes sums of disjoint subsequences of length n from vector
and can work with vector lengths that are not a multiple of n.
It should give me for instance
my.blockwisesum(1:10, 3) == c(6, 15, 24, 10)
Is there a builtin function that can do this?
One could do it by coercing the vector into a matrix of width n,
and then use apply,
but that is cumbersome if the length is not divisible by n,
is it not?
Any other ideas?
Lutz
Prof. Dr. Lutz Prechelt; [EMAIL PROTECTED]
Institut fuer Informatik; Freie Universitaet Berlin
Takustr. 9; 14195 Berlin; Germany
+49 30 838 75115; http://www.inf.fu-berlin.de/inst/ag-se/
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Re: [R] blockwise sums
Liaw, Andy wrote:
If you insist, here's one way:
my.blockwisesum - function(x, n, ...) {
tapply(x, seq(1, length(x), by=n), sum, ...)
}
Did you test that? I get:
my.blockwisesum(1:10, 3)
Error in tapply(x, seq(1, length(x), by = n), sum, ...) :
arguments must have same length
Here's my solution with tapply and rep() to generate a vector like
c(1,1,1,2,2,2,3,3,3,4):
baz.blockwisesum=
function(v,n){tapply(v,rep(1:(1+length(v)/n),each=n)[1:length(v)],sum)}
baz.blockwisesum(1:10,3)
1 2 3 4
6 15 24 10
- just ignore the 1 to 4 names, they cant hurt you.
Baz
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Re: [R] blockwise sums
Hi!
ind-c(sort(rep(1:floor(length(x)/ 3 ) , 3 )) , floor(length(x)/ 3 )+1)
by(x,ind,sum)
my.blockwisesum-function(x,n,...)
{
ind-c(sort(rep(1:floor(length(x)/ n ) , n )) , floor(length(x)/ n )+1)
return(tapply(x,ind,sum))
}
/Eryk
*** REPLY SEPARATOR ***
On 8/31/2004 at 2:19 PM Lutz Prechelt wrote:
I am looking for a function like
my.blockwisesum(vector, n)
that computes sums of disjoint subsequences of length n from vector
and can work with vector lengths that are not a multiple of n.
It should give me for instance
my.blockwisesum(1:10, 3) == c(6, 15, 24, 10)
Is there a builtin function that can do this?
One could do it by coercing the vector into a matrix of width n,
and then use apply,
but that is cumbersome if the length is not divisible by n,
is it not?
Any other ideas?
Lutz
Prof. Dr. Lutz Prechelt; [EMAIL PROTECTED]
Institut fuer Informatik; Freie Universitaet Berlin
Takustr. 9; 14195 Berlin; Germany
+49 30 838 75115; http://www.inf.fu-berlin.de/inst/ag-se/
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Dipl. bio-chem. Witold Eryk Wolski @ MPI-Moleculare Genetic
Ihnestrasse 63-73 14195 Berlin'v'
tel: 0049-30-83875219/ \
mail: [EMAIL PROTECTED]---W-Whttp://www.molgen.mpg.de/~wolski
[EMAIL PROTECTED]
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RE: [R] blockwise sums
From: Barry Rowlingson
Liaw, Andy wrote:
If you insist, here's one way:
my.blockwisesum - function(x, n, ...) {
tapply(x, seq(1, length(x), by=n), sum, ...)
}
Did you test that? I get:
Of course not (slap on wrist)!! My apologies...
Andy
my.blockwisesum(1:10, 3)
Error in tapply(x, seq(1, length(x), by = n), sum, ...) :
arguments must have same length
Here's my solution with tapply and rep() to generate a vector like
c(1,1,1,2,2,2,3,3,3,4):
baz.blockwisesum=
function(v,n){tapply(v,rep(1:(1+length(v)/n),each=n)[1:length(
v)],sum)}
baz.blockwisesum(1:10,3)
1 2 3 4
6 15 24 10
- just ignore the 1 to 4 names, they cant hurt you.
Baz
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Re: [R] blockwise sums
Hi Lutz,
you could try the following:
blockwisesum - function(x, n){
nx - length(x)
if(nx%%n) x. - c(x, rep(0., n*ceiling(nx/n)-nx)) else x. - x
x. - matrix(x., ncol=n, byrow=TRUE)
rowSums(x.)
}
blockwisesum(1:10, 3)
I hope this helps.
Best,
Dimitris
Dimitris Rizopoulos
Doctoral Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/16/396887
Fax: +32/16/337015
Web: http://www.med.kuleuven.ac.be/biostat/
http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm
- Original Message -
From: Lutz Prechelt [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, August 31, 2004 2:19 PM
Subject: [R] blockwise sums
I am looking for a function like
my.blockwisesum(vector, n)
that computes sums of disjoint subsequences of length n from vector
and can work with vector lengths that are not a multiple of n.
It should give me for instance
my.blockwisesum(1:10, 3) == c(6, 15, 24, 10)
Is there a builtin function that can do this?
One could do it by coercing the vector into a matrix of width n,
and then use apply,
but that is cumbersome if the length is not divisible by n,
is it not?
Any other ideas?
Lutz
Prof. Dr. Lutz Prechelt; [EMAIL PROTECTED]
Institut fuer Informatik; Freie Universitaet Berlin
Takustr. 9; 14195 Berlin; Germany
+49 30 838 75115; http://www.inf.fu-berlin.de/inst/ag-se/
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RE: [R] blockwise sums
Liaw, Andy wrote:
If you insist, here's one way:
my.blockwisesum - function(x, n, ...) {
tapply(x, seq(1, length(x), by=n), sum, ...)
}
Did you test that? I get:
my.blockwisesum(1:10, 3)
Error in tapply(x, seq(1, length(x), by = n), sum, ...) :
arguments must have same length
Here's my solution with tapply and rep() to generate a vector like
c(1,1,1,2,2,2,3,3,3,4):
baz.blockwisesum=
function(v,n){tapply(v,rep(1:(1+length(v)/n),each=n)[1:length(
v)],sum)}
baz.blockwisesum(1:10,3)
1 2 3 4
6 15 24 10
- just ignore the 1 to 4 names, they cant hurt you.
Baz
To complete the picture: here is another one:
my.blockwisesum - function(vec, n){
vec - as.vector(vec)
n - as.integer(n)
total - length(vec)
if(total = n){
stop(\nn should be smaller than length of vector.\n)
}
start - seq(1, total, n)
end - start + n - 1
end[end total] - max(start)
index - 1 : length(start)
return(sapply(index, function(x)sum(test[start[x]:end[x]])))
}
test - 1:150
ptn - proc.time()
baz.blockwisesum(test,3)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
20
6 15 24 33 42 51 60 69 78 87 96 105 114 123 132 141 150 159 168
177
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39
40
186 195 204 213 222 231 240 249 258 267 276 285 294 303 312 321 330 339 348
357
41 42 43 44 45 46 47 48 49 50
366 375 384 393 402 411 420 429 438 447
proc.time()-ptn
[1] 0.00 0.00 0.22 NA NA
ptn - proc.time()
my.blockwisesum(test,3)
[1] 6 15 24 33 42 51 60 69 78 87 96 105 114 123 132 141 150 159
168
[20] 177 186 195 204 213 222 231 240 249 258 267 276 285 294 303 312 321 330
339
[39] 348 357 366 375 384 393 402 411 420 429 438 447
proc.time()-ptn
[1] 0.00 0.00 0.19 NA NA
HTH,
Bernhard
The information contained herein is confidential and is inte...{{dropped}}
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Re: [R] blockwise sums
Barry Rowlingson [EMAIL PROTECTED] writes:
Liaw, Andy wrote:
If you insist, here's one way:
my.blockwisesum - function(x, n, ...) {
tapply(x, seq(1, length(x), by=n), sum, ...)
}
Did you test that? I get:
my.blockwisesum(1:10, 3)
Error in tapply(x, seq(1, length(x), by = n), sum, ...) :
arguments must have same length
Here's my solution with tapply and rep() to generate a vector like
c(1,1,1,2,2,2,3,3,3,4):
baz.blockwisesum=
function(v,n){tapply(v,rep(1:(1+length(v)/n),each=n)[1:length(v)],sum)}
baz.blockwisesum(1:10,3)
1 2 3 4
6 15 24 10
Slight variant
pd.blockwisesum -
function(v,n){N - length(v); tapply(x,gl(ceiling(N/n), n, N), sum)}
pd.blockwisesum(1:10,3)
1 2 3 4
6 15 24 10
--
O__ Peter Dalgaard Blegdamsvej 3
c/ /'_ --- Dept. of Biostatistics 2200 Cph. N
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [R] blockwise sums
Lutz Prechelt prechelt at pcpool.mi.fu-berlin.de writes: : : I am looking for a function like : my.blockwisesum(vector, n) : that computes sums of disjoint subsequences of length n from vector : and can work with vector lengths that are not a multiple of n. : : It should give me for instance : my.blockwisesum(1:10, 3) == c(6, 15, 24, 10) tapply(v, (seq(v)-1)%/%n, sum) __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] blockwise sums
On Tue, 31 Aug 2004, Liaw, Andy wrote:
If you insist, here's one way:
my.blockwisesum - function(x, n, ...) {
tapply(x, seq(1, length(x), by=n), sum, ...)
}
If x is large, rowsum() should be faster than tapply().
-thomas
Andy
From: Lutz Prechelt
I am looking for a function like
my.blockwisesum(vector, n)
that computes sums of disjoint subsequences of length n from vector
and can work with vector lengths that are not a multiple of n.
It should give me for instance
my.blockwisesum(1:10, 3) == c(6, 15, 24, 10)
Is there a builtin function that can do this?
One could do it by coercing the vector into a matrix of width n,
and then use apply,
but that is cumbersome if the length is not divisible by n,
is it not?
Any other ideas?
Lutz
Prof. Dr. Lutz Prechelt; [EMAIL PROTECTED]
Institut fuer Informatik; Freie Universitaet Berlin
Takustr. 9; 14195 Berlin; Germany
+49 30 838 75115; http://www.inf.fu-berlin.de/inst/ag-se/
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Thomas Lumley Assoc. Professor, Biostatistics
[EMAIL PROTECTED] University of Washington, Seattle
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RE: [R] blockwise sums
This is a cute problem, so how about a cute algorithm? Since all that is
wanted is the sum, matrix multiplication can be used. If n=vector length and
b = blocksize, then the issue comes down to constructing a block diagonal
matrix with nblk=n%/%b columns with b 1's in each block to multiply the
first b*nblk elements of the vector by (the sum of any remaining elements
can be appended to the result). This muliplier matrix is easily constructed
using diag():
matrix(rep(diag(nblk),each=b),ncol=nblk)
There may even be slicker ways to do it. Anyway, as no implicit looping
(tapply) is used, I suspect this would run considerably faster,too, although
this may be unimportant, and I haven't tested it.
Of course, this approach doesn't generalize. But it is a nice example of
some of R's little tricks.
Cheers,
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
The business of the statistician is to catalyze the scientific learning
process. - George E. P. Box
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Liaw, Andy
Sent: Tuesday, August 31, 2004 5:42 AM
To: 'Barry Rowlingson'
Cc: [EMAIL PROTECTED]
Subject: RE: [R] blockwise sums
From: Barry Rowlingson
Liaw, Andy wrote:
If you insist, here's one way:
my.blockwisesum - function(x, n, ...) {
tapply(x, seq(1, length(x), by=n), sum, ...)
}
Did you test that? I get:
Of course not (slap on wrist)!! My apologies...
Andy
my.blockwisesum(1:10, 3)
Error in tapply(x, seq(1, length(x), by = n), sum, ...) :
arguments must have same length
Here's my solution with tapply and rep() to generate a
vector like
c(1,1,1,2,2,2,3,3,3,4):
baz.blockwisesum=
function(v,n){tapply(v,rep(1:(1+length(v)/n),each=n)[1:length(
v)],sum)}
baz.blockwisesum(1:10,3)
1 2 3 4
6 15 24 10
- just ignore the 1 to 4 names, they cant hurt you.
Baz
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