Re: [R] Rcmdr X11 protocol error message
Michael Bibo [EMAIL PROTECTED] writes: What's happening? Both Java and R are reading events from the X11 event loop ``at the same time''. As a result, R is reading events meant for Java and vice-versa and bad things happen! There are two solutions: * use the Java graphics device which uses the Java graphics facilities to do the rendering; * don't run the GUI and the X11 device at the same time. The second solution is clearly not ideal. What we can do is to compile the X11 device code to be thread-safe. However, the same effect will be seen if we load another X11 event handler (e.g. the Tcl/Tk library). I don't think that is the issue. The X11 device and Tk sit on different connections to the X device, so they shouldn't be seeing eachother's events (the event loops can starve eachother though since they are multiplexed). I'd rather suspect a race condition with the Tk code itself - sending events to a window that is in the process of being destroyed and things like that. Which Tk version are we talking about, by the way? -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [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] D'agostino test
Alexandre,
I found once a code to calculate Shapiro_Francia and D'Agostino tests of normality
written long time ago by Peter B. Mandeville. Below his code and references (It is in
spanish, I assume):
Gregor
#--
I coded the Shapiro-Francia test which handles from 5 to 5000 observations
from
Patrick Royston
1993 A Pocket-Calculator Algorithm for the Shapiro-Francia Test for
Non-Normality: An Application to Medicine. Statistics in Medicine vol
12, 181-184
1991 Estimating Departure from Normality. Statistics in Medicine.
vol 10, 1283-1293
1983 A Simple Method for Evaluating the Shapiro-Francia W' Test of
Non-Normality. The Statistician 32 (1983) 297-300
and the D'Agostino tests from
Ralph B. D'Agostino, Albert Belanger, and Ralph B. D'Agostino, Jr.
1990 A Suggestion for Using Powerful and Informative Tests of Normality.
The American Statistician, November 1990, Vol. 44, No. 4
for testing normality in R.
# Lee 1996:33-34
MOMENTS - function(data,r) sum((data-mean(data))^r)/length(data)
# Peter
SKEW - function(data) MOMENTS(data,3)/(MOMENTS(data,2)*sqrt(MOMENTS(data,2)))
# Peter
KURTOSIS - function(data) MOMENTS(data,4)/(MOMENTS(data,2)*MOMENTS(data,2))
# D'Agostino, Belanger and D'Agostino 1990:316-321
DAGOSTINO - function(data){
cat( PRUEBAS DE D'AGOSTINO\n)
n - length(data)
cat(SIMETRIA\n)
cat( COEFICIENTE DE SIMETRIA:,sqrtb1 - SKEW(data),\n)
if(n8){
y - sqrtb1*sqrt((n+1)*(n+3)/(6*(n-2)))
beta2 - 3*(n*n+27*n-70)*(n+1)*(n+3)/((n-2)*(n+5)*(n+7)*(n+9))
w - sqrt(-1+sqrt(2*(beta2-1)))
delta - 1/sqrt(log(w))
ALPHA - sqrt(2/(w*w-1))
cat( ZETA CALCULADA:,zb1 -
delta*log(y/ALPHA+sqrt((y/ALPHA)^2+1)),\n)
cat( PROBABILIDAD:,2*(1-pnorm(abs(zb1))),\n)
}else
cat(LA PRUEBA DE SESGO REQUIERE POR LO MENOS 9 REPETICIONES\n)
cat(KURTOSIS\n)
cat( COEFICIENTE DE KURTOSIS:,b2 - KURTOSIS(data),\n)
if(n19){
meanb2 - 3*(n-1)/(n+1)
varb2 - 24*n*(n-2)*(n-3)/((n+1)*(n+1)*(n+3)*(n+5))
x - (b2-meanb2)/sqrt(varb2)
moment -
6*(n*n-5*n+2)/((n+7)*(n+9))*sqrt(6*(n+3)*(n+5)/(n*(n-2)*(n-3)))
a - 5+8/moment*(2/moment+sqrt(1+4/(moment*moment)))
cat( ZETA CALCULADA:,zb2 -
(1-2/(9*a)-((1-2/a)/(1+x*sqrt(2/(a-4^(1/3))/sqrt(2/(9*a)),\n)
cat( PROBABILIDAD:,2*(1-pnorm(abs(zb2))),\n)
cat(OMNIBUS\n)
cat( JI-CUADRADA CALCULADA:,k2 - zb1*zb1+zb2*zb2,\n)
cat( GRADOS DE LIBERTAD: 2\n)
cat( PROBABILIDAD:,probji2 - 1-pchisq(k2,2),\n)
}else
cat(LAS PRUEBAS DE KURTOSIS Y OMNIBUS REQUIEREN POR LO MENOS 20
REPETICIONES\n)
}
# Royston 1993:183-184
SHAPIRO.FRANCIA - function(data){
cat( PRUEBA DE NORMALIDAD DE SHAPIRO-FRANCIA\n)
n - length(data)
if(n5 || n5000)
cat(REQUIERE ENTRE 5 Y 5000 REPETICIONES\n)
else{
xbar - mean(data)
sdata - sort(data)
resid - sdata-xbar
uniform - seq(1,n)
np - qnorm((uniform-0.375)/(n+0.25))
cat(W':,w -
(sum(np*sdata))^2/(sum(np*np)*sum(resid*resid)),\n)
u - log(n)
v - log(u)
muy - -1.2725+1.0521*(v-u)
sigmay - 1.0308-0.26758*(v+2/u)
cat(ZETA CALCULADA:,zeta - (log(1-w)-muy)/sigmay,\n)
cat(PROBABILIDAD:,probz - 1-pnorm(zeta),\n)
}
}
Peter B.
Peter B. Mandeville [EMAIL PROTECTED]
Jefe del Depto. de Informática y Bioestadística [EMAIL PROTECTED]
Facultad de MedicineTel: 48 26-23-45 ext. 232
Universidad Autónoma de San Luis Potosí Fax: 48 28-23-52
Av. V. Carranza 2405
Col. Los Filtros
Apartado Postal 145
San Luis Potosí, S.L.P.
78210 México
#--
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Alexandre Bournery
Sent: Montag, 30. August 2004 18:08
To: [EMAIL PROTECTED]
Subject: [R] D'agostino test
Hi, Does anyone know if the D'agostino test is available with R ? Alex
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[R] (no subject)
Hi this is a very simple question ! if a have a set of date like x=c(1,2,3,4,4,5,6,8,7,8,8) how can I find out the percentile of 7 ? thanks Paolo __ [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] (no subject)
sum(x=7)/length(x)*100 *** REPLY SEPARATOR *** On 8/31/2004 at 8:16 AM Paolo Tommasini wrote: Hi this is a very simple question ! if a have a set of date like x=c(1,2,3,4,4,5,6,8,7,8,8) how can I find out the percentile of 7 ? thanks Paolo __ [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 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] __ [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] Rcmdr X11 protocol error message
Dear Michael, A question: Do you observe these problems with tcltk in general or just with the Rcmdr package? Is it possible for you to test a Tcl/Tk program outside of R? Regards, John -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Peter Dalgaard Sent: Tuesday, August 31, 2004 2:32 AM To: Michael Bibo Cc: [EMAIL PROTECTED] Subject: Re: [R] Rcmdr X11 protocol error message Michael Bibo [EMAIL PROTECTED] writes: What's happening? Both Java and R are reading events from the X11 event loop ``at the same time''. As a result, R is reading events meant for Java and vice-versa and bad things happen! There are two solutions: * use the Java graphics device which uses the Java graphics facilities to do the rendering; * don't run the GUI and the X11 device at the same time. The second solution is clearly not ideal. What we can do is to compile the X11 device code to be thread-safe. However, the same effect will be seen if we load another X11 event handler (e.g. the Tcl/Tk library). I don't think that is the issue. The X11 device and Tk sit on different connections to the X device, so they shouldn't be seeing eachother's events (the event loops can starve eachother though since they are multiplexed). I'd rather suspect a race condition with the Tk code itself - sending events to a window that is in the process of being destroyed and things like that. Which Tk version are we talking about, by the way? -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [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 __ [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
[R] appending data to a dataframe
Dear R users,
I am sorry to ask you such a pathetic newbie question, but how does one
append data at the end of a data frame?
I am working with GRASS/R library, but the question is about R.
I have a data.frame containing the following variables
basinID, distoutlet, drainage_area, slope
These variables are stored for all pixels of Grass Raster objects. For each
drainage basin (basinID), I'd like to find the maximum of distoutlet. How
can I store the pair of info (bv$ID, max(bv$distoutlet)) at each step of the
for loop ?
I presume something like this could do
for (i 1:max(basinID)){
bid - i
length - max(distoutlet[which(basinID == I(i))])
}
But how do I handle the output and record bid and length in parallel in an
object???
Any hint is welcome
Thomas
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Re: [R] appending data to a dataframe
On Tue, 31 Aug 2004 14:05:59 +0200, Dewez Thomas [EMAIL PROTECTED]
wrote :
Dear R users,
I am sorry to ask you such a pathetic newbie question, but how does one
append data at the end of a data frame?
The rbind() function should work: it adds rows to matrices and
dataframes.
However, I'd do something different for the problem you give:
I am working with GRASS/R library, but the question is about R.
I have a data.frame containing the following variables
basinID, distoutlet, drainage_area, slope
These variables are stored for all pixels of Grass Raster objects. For each
drainage basin (basinID), I'd like to find the maximum of distoutlet. How
can I store the pair of info (bv$ID, max(bv$distoutlet)) at each step of the
for loop ?
I presume something like this could do
for (i 1:max(basinID)){
bid - i
length - max(distoutlet[which(basinID == I(i))])
}
But how do I handle the output and record bid and length in parallel in an
object???
Appending a row at a time is very slow. You'll find it much faster to
set up a matrix to hold the results in advance:
results - matrix(NA, max(basinID), 2)
for (i in 1:max(basinID)) {
bid - i
length - max(distoutlet[which(basinID == i)])
results[i,] - c(i, length)
}
Duncan Murdoch
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[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/ __ [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
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
[R] I've forgotten, why is box() the default?
I've searched on CRAN for axes, axis, and other terms
I've already forgotten, without (re)discovering the
reason for S using non-joining axes by default, instead
of box(l).
MASS points me towards Cleveland (1993) but I don't
have ready access to this any more. Could someone
give me a one-liner to justify this choice to a sceptic?
It's something to do with not mis-interpreting the axes
intersection as (0,0), isn't it?
TIA,
Simon
Simon Fear
Senior Statistician
Syne qua non Ltd
Tel: +44 (0) 1379 69
Fax: +44 (0) 1379 65
email: [EMAIL PROTECTED]
web: http://www.synequanon.com
Number of attachments included with this message: 0
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RE: [R] D'agostino test
-Original Message-
From: Alexandre Bournery [mailto:[EMAIL PROTECTED]
Sent: Monday, August 30, 2004 11:08 AM
To: [EMAIL PROTECTED]
Subject: [R] D'agostino test
Hi, Does anyone know if the D'agostino test is available with R ?
Alex
Hi,
See below, taken pretty much verbatim from Zar (1999)
Douglas G. Scofield, PhDDepartment of Biology
[EMAIL PROTECTED]Indiana University
off: (812) 856-0115 1001 E. 3rd St.
fax: (812) 855-6705 Bloomington, IN 47405-3700
cell: (786) 514-9141
---
dagostino.pearson.test - function(x) {
# from Zar (1999), implemented by Doug Scofield,
[EMAIL PROTECTED]
DNAME - deparse(substitute(x))
n - length(x)
x2 - x * x
x3 - x * x2
x4 - x * x3
# compute Z_g1
k3 - ((n*sum(x3)) - (3*sum(x)*sum(x2)) + (2*(sum(x)^3)/n)) /
((n-1)*(n-2))
g1 - k3 / sqrt(var(x)^3)
sqrtb1 - ((n - 2)*g1) / sqrt(n*(n - 1))
A - sqrtb1 * sqrt(((n + 1)*(n + 3)) / (6*(n - 2)))
B - (3*(n*n + 27*n - 70)*(n+1)*(n+3)) / ((n-2)*(n+5)*(n+7)*(n+9))
C - sqrt(2*(B - 1)) - 1
D - sqrt(C)
E - 1 / sqrt(log(D))
F - A / sqrt(2/(C - 1))
Zg1 - E * log(F + sqrt(F*F + 1))
# compute Z_g2
G - (24*n*(n-2)*(n-3)) / (((n+1)^2)*(n+3)*(n+5))
k4 - (((n*n*n + n*n)*sum(x4)) - (4*(n*n + n)*sum(x3)*sum(x)) -
(3*(n*n - n)*sum(x2)^2) + (12*n*sum(x2)*sum(x)^2) - (6*sum(x)^4)) /
(n*(n-1)*(n-2)*(n-3))
g2 - k4 / var(x)^2
H - ((n-2)*(n-3)*abs(g2)) / ((n+1)*(n-1)*sqrt(G))
J - ((6*(n*n - 5*n + 2)) / ((n+7)*(n+9))) * sqrt((6*(n+3)*(n+5)) /
(n*(n-2)*(n-3)))
K - 6 + (8/J)*(2/J + sqrt(1 + 4/(J*J)))
L - (1 - 2/K) / (1 + H*sqrt(2/(K-4)))
Zg2 - (1 - 2/(9*K) - (L^(1/3))) / (sqrt(2/(9*K)))
K2 - Zg1*Zg1 + Zg2*Zg2
pk2 - pchisq(K2, 2, lower.tail=FALSE)
RVAL - list(statistic = c(K2 = K2), p.value = pk2, method =
D'Agostino-Pearson normality test\n\nK2 is distributed as Chi-squared
with df=2, alternative = distribution is not normal, data.name =
DNAME)
class(RVAL) - htest
return(RVAL)
}
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RE: [R] (no subject)
On Mon, 30 Aug 2004, Austin, Matt wrote: A correction. You either need to open the plotting device prior to the simulation that includes your plotting commands and close it after the simulation or have the name change dynamically in your simulation so that the runs go in separate files. An example of the first method would be postscript(file=/where/to/put/file/filename.ps) It's also worth pointing out that many people will find pdf() more convenient than postscript() -thomas ## your simulation commands dev.off() for the second method more info would be needed to know how you are running your simulation to help with dynamic naming. --Matt -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Austin, Matt Sent: Monday, August 30, 2004 18:56 PM To: 'Loke Chok Kang'; [EMAIL PROTECTED] Subject: RE: [R] (no subject) You need to use a device to print to such as postscript(file=/where/to/put/file/filename.ps) ##your plotting code here dev.off() Writing a plot at each iteration of your simulation can impact the runtime greatly. Please read the posting guide at the bottom of the e-mail, it can help you get more helpful . . . help --Matt -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Loke Chok Kang Sent: Monday, August 30, 2004 18:42 PM To: [EMAIL PROTECTED] Subject: [R] (no subject) Hi, Sorry, I have a problem that require some help. As I am doing a project with R and this project requires me to do a lot of plotting as I run my simulation, I need R to help me store my plots automatically as the simulation is run. Could anyone advise me on how this can be done? If possible I need all the plots to be written to a file. PS:In all I have over 500 over plots, thus it will not be possible to manually plot all the plots out. From yours faithfully, Chok Kang [EMAIL PROTECTED] __ [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 __ [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 __ [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 Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle __ [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] I've forgotten, why is box() the default?
On Tue, 31 Aug 2004 14:13:49 +0100, Simon Fear [EMAIL PROTECTED] wrote : I've searched on CRAN for axes, axis, and other terms I've already forgotten, without (re)discovering the reason for S using non-joining axes by default, instead of box(l). MASS points me towards Cleveland (1993) but I don't have ready access to this any more. Could someone give me a one-liner to justify this choice to a sceptic? It's something to do with not mis-interpreting the axes intersection as (0,0), isn't it? I'm not sure I understand your question: the default in S is a full box unless you ask for no box (bty=n), isn't it? The justification for this is The four scale lines also provide a clearly defined region where our eyes can search for data. With just two, data can be camouflaged by virtue of where they lie (p. 35). There's also advice to keep the data away from the axes and to put the ticks outside the box to avoid hiding extreme points. I can't spot a discussion in Cleveland of the reason the axes don't join when bty=n is specified. Duncan Murdoch __ [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] I've forgotten, why is box() the default?
Not helped by my typo and being in a rush!
It's true that plot.default and surely many others do a box, since
par(bty)
[1] o
But this isn't what happens if you do things via axis() ...
(see first example in help(axis)).
Anyway, point taken, I couldn't find it because it is NOT what
is recommended!
-Original Message-
From: Duncan Murdoch [mailto:[EMAIL PROTECTED]
Sent: 31 August 2004 14:45
To: Simon Fear
Cc: R-help
Subject: Re: [R] I've forgotten, why is box() the default?
Security Warning:
If you are not sure an attachment is safe to open contact
Andy on x234.
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On Tue, 31 Aug 2004 14:13:49 +0100, Simon Fear
[EMAIL PROTECTED] wrote :
I've searched on CRAN for axes, axis, and other terms
I've already forgotten, without (re)discovering the
reason for S using non-joining axes by default, instead
of box(l).
MASS points me towards Cleveland (1993) but I don't
have ready access to this any more. Could someone
give me a one-liner to justify this choice to a sceptic?
It's something to do with not mis-interpreting the axes
intersection as (0,0), isn't it?
I'm not sure I understand your question: the default in S is a full
box unless you ask for no box (bty=n), isn't it? The justification
for this is The four scale lines also provide a clearly defined
region where our eyes can search for data. With just two, data can be
camouflaged by virtue of where they lie (p. 35). There's also advice
to keep the data away from the axes and to put the ticks outside the
box to avoid hiding extreme points.
I can't spot a discussion in Cleveland of the reason the axes don't
join when bty=n is specified.
Duncan Murdoch
Simon Fear
Senior Statistician
Syne qua non Ltd
Tel: +44 (0) 1379 69
Fax: +44 (0) 1379 65
email: [EMAIL PROTECTED]
web: http://www.synequanon.com
Number of attachments included with this message: 0
This message (and any associated files) is confidential and\...{{dropped}}
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[R] Restore function in R
R-help, Any function in R that restore an object to its original state so that if not satisfied with the modifications made to it there is no need to redefine again ( 'undo' action) I believe there is something like that in S-plus. Thank you __ [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
[R] N-dimensional delaunay tesselation voronoi diagrams
Hi, I've been looking for functions that can do delaunay tesselation and generate voronoi cells. I came across deldir and tripack but both seem to be restricted to 2D points. Are there any packages that can do a tesselation in N dimensions? I know that Matlab and Mathematica use the qhull package to provide functions for this. Does anybody know of any R packages that do this (maybe by calling on qhull)? Thanks, --- Rajarshi Guha [EMAIL PROTECTED] http://jijo.cjb.net GPG Fingerprint: 0CCA 8EE2 2EEB 25E2 AB04 06F7 1BB9 E634 9B87 56EE --- Q: What do you get when you cross a mosquito with a mountain climber? A: Nothing. You can't cross a vector with a scaler. __ [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
[R] More efficient matrix computation
I have a 20x3 matrix as follows: m - replicate(3, matrix(rnorm(20),20,1)) I need to compute, say, 95th and 99th percentiles of each column such that the resulting matrix becomes 2x3 with each row representing the respective percentile. My best effort is to compute one column at a time as follows: quantile(m[,1], c(0.95, 0.99)) To do the same for columns 2 and 3, I would simply change the column number accordingly. Clearly, this is not very efficient as I may have a large matrix (e.g., 100,000x500) to work with. Any help with the code is appreciated. Jack Wang __ [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] More efficient matrix computation
JTW wrote: I have a 20x3 matrix as follows: m - replicate(3, matrix(rnorm(20),20,1)) Why not m - matrix(rnorm(60), 20, 3) I need to compute, say, 95th and 99th percentiles of each column such that the resulting matrix becomes 2x3 with each row representing the respective percentile. My best effort is to compute one column at a time as follows: quantile(m[,1], c(0.95, 0.99)) That's what apply() is made for: apply(m, 2, quantile, c(0.95, 0.99)) Uwe Ligges To do the same for columns 2 and 3, I would simply change the column number accordingly. Clearly, this is not very efficient as I may have a large matrix (e.g., 100,000x500) to work with. Any help with the code is appreciated. Jack Wang __ [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 __ [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] More efficient matrix computation
Hi! apply(m,2,quantile,c(0.95,0.99)) /Eryk *** REPLY SEPARATOR *** On 8/31/2004 at 9:16 AM JTW wrote: I have a 20x3 matrix as follows: m - replicate(3, matrix(rnorm(20),20,1)) I need to compute, say, 95th and 99th percentiles of each column such that the resulting matrix becomes 2x3 with each row representing the respective percentile. My best effort is to compute one column at a time as follows: quantile(m[,1], c(0.95, 0.99)) To do the same for columns 2 and 3, I would simply change the column number accordingly. Clearly, this is not very efficient as I may have a large matrix (e.g., 100,000x500) to work with. Any help with the code is appreciated. Jack Wang __ [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 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] __ [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
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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Re: [R] More efficient matrix computation
A million thanks to all for the prompt reply: apply() works! Jack Wang __ [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] Rcmdr X11 protocol error message
John Fox jfox at mcmaster.ca writes: Dear Michael, A question: Do you observe these problems with tcltk in general or just with the Rcmdr package? Is it possible for you to test a Tcl/Tk program outside of R? And Peter asked which version of Tcl/Tk. Apparently my system has version 8.4 installed, specifically: Tcl-8.4.5-3-mdk (tclsh8.4) and Tk-8.4.5-3-mdk (libtk8.4.so). As I understand it, these are installed from RPMs on the installation DVD, and I note that they are mandrake specific. John - I wasn't sure if I had any other Tcl/Tk applications installed (it's not always obvious when installing from RPM's). I have certainly not encountered these error messages with any other application. I quickly downloaded WISH Supernotepad 1.2.3. This application requires Tcl/Tk 8.4 or greater. There are no graphics window in this application, but plenty of dialogue boxes. It gave no errors. Is this an appropriate test? If this is a mandrake-configuration-specific problem, it may not be worth spending too much time investigating, as R Commandr still works. I can always try re-installing Tcl/Tk from source when/if I have time. Regards, Michael __ [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] after lm-fit: equality of two regression coefficients
Dear Christoph, You can use the package gregmisc require(gregmisc) C-c(0,1,-1) glh.test(my.lm,C) I hope that helps Alexandre Galvão Hi Let's assume, we have a multiple linear regression, such as the one using the Scottish hills data (MASS, data(hills)): one dependent variable: time two independent var (metric): dist, climb if I am interested, after (!) fitting a lm: my. lm - lm(time ~ dist + climb, data = hills) in the equivalence (or non-equivalence) of the two predictors dist and climb: H0: dist = climb I think you intend to ask if the *coefficients* in the fit should be equal, which is nonsense in this example of course. Is there any function in R, which lets me calculate this, in just giving the lm-object my.lm and e.g. a vector such as c(1, -1), operationalizing the hypothesis H0: t(c(1, -1)) %*% c(dist, climb) = 0 ? library(car) ?linear.hypothesis Christoph, If you are interested in testing for b1=b2 in a regression model, say y=b0+b1*x1+b2*x2+e you can compare the two models o1 and o2 o1-lm(y~x1+x2) o2-lm(y~I(x1+x2)) best, vito __ [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
[R] enter browser on error
Is there a way I can get R to automatically enter the browser inside a user-defined
function on the generation of an error? Specifically, I'm trying to debug this:
Error in as.double.default(sapply(lis, FUN)) :
(list) object cannot be coerced to double
In addition: There were 38 warnings (use warnings() to see them)
traceback()
8: as.double.default(sapply(lis, FUN))
7: as.numeric(sapply(lis, FUN))
6: numeric.sapply(function(x) {
[EMAIL PROTECTED]
})
On detection of the error, I would like browser() to be called at the level of
numeric.sapply(), so that I can examine x. I'm wondering if this can be done by
modifying the default error handling. Using try() with browser() didn't work.
Thanks,
David
_
David Bickel http://davidbickel.com
Research Scientist
Pioneer Hi-Bred International
Bioinformatics Exploratory Research
7250 NW 62nd Ave., PO Box 552
Johnston, Iowa 50131-0552
515-334-4739 Tel
515-334-6634 Fax
[EMAIL PROTECTED], [EMAIL PROTECTED]
This communication is for use by the intended recipient and ...{{dropped}}
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Re: [R] enter browser on error
Have you tried options(error=recover)?
On Tue, 31 Aug 2004, Bickel, David wrote:
Is there a way I can get R to automatically enter the browser inside a user-defined
function on the generation of an error? Specifically, I'm trying to debug this:
Error in as.double.default(sapply(lis, FUN)) :
(list) object cannot be coerced to double
In addition: There were 38 warnings (use warnings() to see them)
traceback()
8: as.double.default(sapply(lis, FUN))
7: as.numeric(sapply(lis, FUN))
6: numeric.sapply(function(x) {
[EMAIL PROTECTED]
})
On detection of the error, I would like browser() to be called at the
level of numeric.sapply(), so that I can examine x. I'm wondering if
this can be done by modifying the default error handling. Using try()
with browser() didn't work.
--
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, UKFax: +44 1865 272595
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Re: [R] enter browser on error
use options(error=recover), e.g.:
remove(x)
NULL
Warning message:
remove: variable x was not found
(function() {x})()
Error in (function() { : Object x not found
options(error=recover)
(function(y=1) {x})(2)
Error in (function(y = 1) { : Object x not found
Enter a frame number, or 0 to exit
1:(function(y = 1) {
Selection: 1
Called from: eval(expr, envir, enclos)
Browse[1] y
[1] 2
Browse[1]
Enter a frame number, or 0 to exit
1:(function(y = 1) {
Selection: 0
At Tuesday 11:26 AM 8/31/2004, Bickel, David wrote:
Is there a way I can get R to automatically enter the browser inside a
user-defined function on the generation of an error? Specifically, I'm
trying to debug this:
Error in as.double.default(sapply(lis, FUN)) :
(list) object cannot be coerced to double
In addition: There were 38 warnings (use warnings() to see them)
traceback()
8: as.double.default(sapply(lis, FUN))
7: as.numeric(sapply(lis, FUN))
6: numeric.sapply(function(x) {
[EMAIL PROTECTED]
})
On detection of the error, I would like browser() to be called at the
level of numeric.sapply(), so that I can examine x. I'm wondering if this
can be done by modifying the default error handling. Using try() with
browser() didn't work.
Thanks,
David
_
David Bickel http://davidbickel.com
Research Scientist
Pioneer Hi-Bred International
Bioinformatics Exploratory Research
7250 NW 62nd Ave., PO Box 552
Johnston, Iowa 50131-0552
515-334-4739 Tel
515-334-6634 Fax
[EMAIL PROTECTED], [EMAIL PROTECTED]
This communication is for use by the intended recipient and ...{{dropped}}
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[R] sample size for t-tests
Dear all, Could any one please tell me the exact formula R uses to calculate the sample size for one-sample and two-sample t-tests? Thanks, Caimiao [[alternative HTML version deleted]] __ [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] More efficient matrix computation
JTW wrote: I have a 20x3 matrix as follows: m - replicate(3, matrix(rnorm(20),20,1)) I need to compute, say, 95th and 99th percentiles of each column such that the resulting matrix becomes 2x3 with each row representing the respective percentile. My best effort is to compute one column at a time as follows: quantile(m[,1], c(0.95, 0.99)) To do the same for columns 2 and 3, I would simply change the column number accordingly. Clearly, this is not very efficient as I may have a large matrix (e.g., 100,000x500) to work with. Any help with the code is appreciated. Jack Wang How about: apply(m, 2, quantile, c(0.95, 0.99)) --sundar __ [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] sample size for t-tests
On Tue, 31 Aug 2004, Caimiao Wei wrote: Dear all, Could any one please tell me the exact formula R uses to calculate the sample size for one-sample and two-sample t-tests? Thanks, There isn't a formula in closed form. The exact procedure is in the code. In words: the critical value for the t-test comes from a t-distribution. The distribution of the t-statistic under the alternative is a non-central t-distribution, and the sample size is adjusted to get the requested power. -thomas __ [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
[R] Obtaining Least Square Means with lme (mixed models)
Hello, I was wondering how one could obtain LSM for fixed factors in a mixed-effect model using lme. In other words, if I have model such as: model-lme(yield~period+treatment+period*treatment, data=data,random=~1|cow) how can I get LSM for period, treatment and their interaction? Thanks a lot! Alex __ [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
[R] spatial autcorrelation in glmmPQL
I am unable to specify error spatial autocorrelation structure in glmmPQL: DATA x and y coordinates for sample points at which presence/absence of seedlings and canopy openness were recorded in different forest stands. QUESTION Does seedling density increase with canopy openness? ANALYSIS Initial analysis using glmmPQL with binomial errors and stand as random effect result - glmmPQL(seedlings ~ canopy, random = ~1|stand, family=binomial, data=regen) Semivariogram of residuals from this fit using geoR showed spatial autocorrelation with range 34.9 m and relative nugget of 75%. Therefore I tried to create a corStruct object: corel - corSpher(value = c(34.9,0.75), form = ~x+y, nugget=TRUE) corel - Initialize(corel, data = regen) And specify the structure in a new glmmPQL result2 - glmmPQL(seedlings ~ canopy, random = ~|stand, family = binomial, correlation = corel, data=regen) PROBLEM I get the error message: iteration 1 Error in eval(expr, envir, enclos) : Object x not found (Running R 1.9.1 on Windows ME) I appear to be implementing corSpher and Initialize incorrectly- any help greatly appreciated. Dr. Daniel P. Bebber Department of Plant Sciences University of Oxford South Parks Road Oxford OX1 3RB Tel. 01865 275060 --- __ [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
[R] add single contour line to levelplot
Hello,
I want to add a single contour line to a levelplot but can't figure out
how to do it 'on-the-fly'. When I include the last line in the code below,
I get the following error:
Error in NextMethod([) : Argument subscripts is missing, with no default
Any tips on how to fix this are greatly appreciated!
Ian Jonsen
levelplot(dens~nu*sigma,data=tsurflikB,at=c(-20,-10,-5,-4,-3,-2,-1),
col.regions=gray(seq(0.5,0.9,length=14)),region=T,colorkey=F,aspect=1,
panel=function(x,y,z,...){
panel.levelplot(x=x,y=y,z=z,...)
lpoints(x[z==0],y[z==0],pch=16,cex=0.5)
panel.levelplot(z~x*y,at=-3,contour=T,aspect=1)
}
)
___
Ian Jonsen, Postdoctoral Fellow
Dept Biology, Dalhousie University, Halifax, NS, CAN
Phone: 902 494 3910; Fax: 902 494 3736
[EMAIL PROTECTED]
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Re: [R] Rcmdr X11 protocol error message
Michael Bibo [EMAIL PROTECTED] writes: John Fox jfox at mcmaster.ca writes: Dear Michael, A question: Do you observe these problems with tcltk in general or just with the Rcmdr package? Is it possible for you to test a Tcl/Tk program outside of R? And Peter asked which version of Tcl/Tk. Apparently my system has version 8.4 installed, specifically: Tcl-8.4.5-3-mdk (tclsh8.4) and Tk-8.4.5-3-mdk (libtk8.4.so). As I understand it, these are installed from RPMs on the installation DVD, and I note that they are mandrake specific. John - I wasn't sure if I had any other Tcl/Tk applications installed (it's not always obvious when installing from RPM's). I have certainly not encountered these error messages with any other application. I quickly downloaded WISH Supernotepad 1.2.3. This application requires Tcl/Tk 8.4 or greater. There are no graphics window in this application, but plenty of dialogue boxes. It gave no errors. Is this an appropriate test? If this is a mandrake-configuration-specific problem, it may not be worth spending too much time investigating, as R Commandr still works. I can always try re-installing Tcl/Tk from source when/if I have time. I don't think we have evidence that it's the Tcl installation, although it could be (Google suggests that there have been problems with at least some versions, although most references seem rather old). I can't seem to reproduce the effect with SUSE's tk-8.4.6-28 either. If it is a bug in Rcmdr, then we'd want to find it and you have the only handle on it BTW, sometimes Tk errors allow you to see a trace of the execution. Would this happen to be one of those situations? -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [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] I've forgotten, why is box() the default?
On Tue, 31 Aug 2004 15:02:13 +0100, Simon Fear [EMAIL PROTECTED] wrote : Not helped by my typo and being in a rush! It's true that plot.default and surely many others do a box, since par(bty) [1] o But this isn't what happens if you do things via axis() ... (see first example in help(axis)). I suspect it's just done that way for programming convenience. Since it's hard or impossible to delete a line once drawn, axis draws the minimal sensible one; if you want more, you use box() to draw it. Duncan Murdoch __ [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
[R] Problem (bug?) with vector indices
Hi all,
I found a problem on R that looks like a bug to me. I am working with R
for windows verson 1.9.0, but the problem also happens on the linux
version 1.9.0. I hope that the problem is on my reasoning, and I would be
grateful if someone could explain me why this happens.
There goes the example:
x-seq(-2,2,0.001)
y-rep(0,length(x))
for (i in x){y[1000*i+2001]-i^2}
plot(x,y)# Here I found something wrong
which(y==0)
[1] 244 353 357 361 365 369 482 486 490 494 579 583 587 591
595
[16] 599 603 607 611 615 619 708 712 716 720 724 728 732 736
740
[31] 744 837 841 845 849 853 857 861 865 869 962 966 970 974
978
[46] 982 986 990 994 998 2001
x[244]
[1] -1.757
i-x[244]
1000*i+2001
[1] 244
y[1000*i+2001]
[1] 3.087049
y[244]
[1] 0
y[243]
[1] 3.087049
What means that the expression 1000*i+2001, which is equal to 244, inside
the brackets is evaluated as 243! The same occurs with every value of x
that results in a zero value of y, the right value is stored in y one
position before it should be, overwriting the previous value at this
position.
The bug is on R or on my mind?
Thanks a lot.
Rodrigo Drummond
Rodrigo D. Drummond
Laboratorio Genoma Funcional
Centro de Biologia Molecular e Eng. Genetica
Universidade Estadual de Campinas
Caixa Postal 6010
13083-875 - Campinas - SP - Brasil
Tel: xx-19-3788-1119 Fax: xx-19-3788-1089
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Re: [R] Problem (bug?) with vector indices
Rodrigo Drummond [EMAIL PROTECTED] writes:
Hi all,
I found a problem on R that looks like a bug to me. I am working with R
for windows verson 1.9.0, but the problem also happens on the linux
version 1.9.0. I hope that the problem is on my reasoning, and I would be
grateful if someone could explain me why this happens.
There goes the example:
x-seq(-2,2,0.001)
y-rep(0,length(x))
for (i in x){y[1000*i+2001]-i^2}
...
x[244]
[1] -1.757
i-x[244]
1000*i+2001
[1] 244
y[1000*i+2001]
[1] 3.087049
y[244]
[1] 0
y[243]
[1] 3.087049
What means that the expression 1000*i+2001, which is equal to 244, inside
the brackets is evaluated as 243! The same occurs with every value of x
that results in a zero value of y, the right value is stored in y one
position before it should be, overwriting the previous value at this
position.
The bug is on R or on my mind?
The latter:
Have a look at (1000*x[244]+2001)-244
Beware floating point arithmetic!
--
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] add single contour line to levelplot
Ian Jonsen wrote:
Hello,
I want to add a single contour line to a levelplot but can't figure out
how to do it 'on-the-fly'. When I include the last line in the code below,
I get the following error:
Error in NextMethod([) : Argument subscripts is missing, with no default
Any tips on how to fix this are greatly appreciated!
Ian Jonsen
levelplot(dens~nu*sigma,data=tsurflikB,at=c(-20,-10,-5,-4,-3,-2,-1),
col.regions=gray(seq(0.5,0.9,length=14)),region=T,colorkey=F,aspect=1,
panel=function(x,y,z,...){
panel.levelplot(x=x,y=y,z=z,...)
lpoints(x[z==0],y[z==0],pch=16,cex=0.5)
panel.levelplot(z~x*y,at=-3,contour=T,aspect=1)
}
)
You have misued panel.levelplot and also forgot (neglected?) to include
... in the second call. I think you want something like:
# from ?levelplot
library(lattice)
x - seq(pi/4, 5 * pi, length = 100)
y - seq(pi/4, 5 * pi, length = 100)
r - as.vector(sqrt(outer(x^2, y^2, +)))
grid - expand.grid(x = x, y = y)
grid$z - cos(r^2) * exp(-r/(pi^3))
levelplot(z ~ x * y, grid, cuts = 50, scales=list(log=e),
xlab = , ylab = , main = Weird Function,
sub = with log scales,
panel = function(x, y, z, at, contour, ...) {
panel.levelplot(x, y, z, at = 0, contour = TRUE, ...)
lpoints(x[z = -0.5],y[z = -0.5], pch = 16, cex = 0.5)
},
colorkey = FALSE, region = TRUE)
Note the usage of I also think you do not need two calls to
panel.levelplot, though I may be misunderstanding your intention. One
will suffice, as it does here.
--sundar
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[R] Sparse Matrices in R
I have data in i,j,r format, where r is the value in location A[i,j] for some imaginary matrix A. I need to build this matrix A, but given the sizes of i and j, I believe that using a sparse format would be most adequate. Hopefully this will allow me to perform some basic matrix manipulation such as multiplication, addition, rowsums, transpositions, subsetting etc etc. Is there any way to achieve this goal in R? Thanks, Danny [[alternative HTML version deleted]] __ [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] Sparse Matrices in R
Take a look at the package SparseM url:www.econ.uiuc.edu/~rogerRoger Koenker email [EMAIL PROTECTED] Department of Economics vox:217-333-4558University of Illinois fax:217-244-6678Champaign, IL 61820 On Aug 31, 2004, at 4:52 PM, Danny Heuman wrote: I have data in i,j,r format, where r is the value in location A[i,j] for some imaginary matrix A. I need to build this matrix A, but given the sizes of i and j, I believe that using a sparse format would be most adequate. Hopefully this will allow me to perform some basic matrix manipulation such as multiplication, addition, rowsums, transpositions, subsetting etc etc. Is there any way to achieve this goal in R? Thanks, Danny [[alternative HTML version deleted]] __ [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 __ [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] Sparse Matrices in R
Hi! help.search(sparse matrix) graph2SparseM(graph)Coercion methods between graphs and sparse matrices tripletMatrix-class(Matrix) Class tripletMatrix sparse matrices in triplet form SparseM.hb(SparseM) Harwell-Boeing Format Sparse Matrices image,matrix.csr-method(SparseM) Image Plot for Sparse Matrices etc . /E *** REPLY SEPARATOR *** On 8/31/2004 at 5:52 PM Danny Heuman wrote: I have data in i,j,r format, where r is the value in location A[i,j] for some imaginary matrix A. I need to build this matrix A, but given the sizes of i and j, I believe that using a sparse format would be most adequate. Hopefully this will allow me to perform some basic matrix manipulation such as multiplication, addition, rowsums, transpositions, subsetting etc etc. Is there any way to achieve this goal in R? Thanks, Danny [[alternative HTML version deleted]] __ [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 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] __ [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
[R] Problem with seq.dates in chron
I get faulty output from seq.dates() if I specify a length that is too long. For example, I ask for 129 months in the following call to the function, but it returns 131: R.version.string [1] R version 1.9.1, 2004-06-21 startdatetime - chron(dates=01/01/1995, times=00:00:00) beg.month.datetimes - seq.dates(from=startdatetime, by=months, length=129) beg.month.datetimes [1] 01/01/95 02/01/95 03/01/95 04/01/95 05/01/95 06/01/95 07/01/95 08/01/95 [9] 09/01/95 10/01/95 11/01/95 12/01/95 01/01/96 02/01/96 03/01/96 04/01/96 [17] 05/01/96 06/01/96 07/01/96 08/01/96 09/01/96 10/01/96 11/01/96 12/01/96 [25] 01/01/97 02/01/97 03/01/97 04/01/97 05/01/97 06/01/97 07/01/97 08/01/97 [33] 09/01/97 10/01/97 11/01/97 12/01/97 01/01/98 02/01/98 03/01/98 04/01/98 [41] 05/01/98 06/01/98 07/01/98 08/01/98 09/01/98 10/01/98 11/01/98 12/01/98 [49] 01/01/99 02/01/99 03/01/99 04/01/99 05/01/99 06/01/99 07/01/99 08/01/99 [57] 09/01/99 10/01/99 11/01/99 12/01/99 01/01/00 02/01/00 03/01/00 04/01/00 [65] 05/01/00 06/01/00 07/01/00 08/01/00 09/01/00 10/01/00 11/01/00 12/01/00 [73] 01/01/01 02/01/01 03/01/01 04/01/01 05/01/01 06/01/01 07/01/01 08/01/01 [81] 09/01/01 10/01/01 11/01/01 12/01/01 01/01/02 02/01/02 03/01/02 04/01/02 [89] 05/01/02 06/01/02 07/01/02 08/01/02 09/01/02 10/01/02 11/01/02 12/01/02 [97] 01/01/03 02/01/03 03/01/03 04/01/03 05/01/03 06/01/03 07/01/03 08/01/03 [105] 09/01/03 10/01/03 11/01/03 12/01/03 01/01/04 02/01/04 03/01/04 04/01/04 [113] 05/01/04 06/01/04 07/01/04 08/01/04 09/01/04 10/01/04 11/01/04 12/01/04 [121] 01/01/05 02/01/05 03/01/05 04/01/05 05/01/05 06/01/05 07/01/05 08/01/05 [129] 09/01/05 10/01/05 11/01/05 Is this a bug, and can it be fixed? Scott Waichler Pacific Northwest National Laboratory scott.waichleratpnl.gov __ [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
[R] subselect install problem
Trying to install subselect v0.8 on Redhat 7.3 and R 1.8.1 fails (below). Any help is greatly appreciated. Xiao-Jun * Installing *source* package 'subselect' ... ** libs f2c anneal.f anneal.c anneal: Error on line 263: Declaration error for fica: adjustable dimension on non-argument Error on line 263: Declaration error for valp: adjustable dimension on non-argument Error on line 263: Declaration error for auxw: adjustable dimension on non-argument Error on line 263: wr_ardecls: nonconstant array size Error on line 263: wr_ardecls: nonconstant array size Error on line 263: wr_ardecls: nonconstant array size make: *** [anneal.o] Error 1 ERROR: compilation failed for package 'subselect' __ [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] Problem with seq.dates in chron
Waichler, Scott R Scott.Waichler at pnl.gov writes: I get faulty output from seq.dates() if I specify a length that is too long. For example, I ask for 129 months in the following call to the function, but it returns 131: R.version.string [1] R version 1.9.1, 2004-06-21 startdatetime - chron(dates=01/01/1995, times=00:00:00) beg.month.datetimes - seq.dates(from=startdatetime, by=months, length=129) If you look at the seq.dates R source, it sets `to' like this: to - from + (length. - 1) * c(1, 7, 31, 366)[i] where in your case where i is 3 for months. It then picks out those days between from and to having the same day of the month as the from date (or uses a more complex algorithm if the day of the month exceeds 28). The multiplication by 31 gives an overestimate which is OK for a small lengths but overflows to give too many months if length is too large. Since we know the error will always be on the high side, until it is fixed as a workaround you could just subset it: seq.dates(from=startdatetime, by=months, length=129)[1:129] (Perhaps you could forward this to the maintainer of chron.) __ [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
[R] simpleboot, pairs.boot
Hi. I am a new R user trying to obtain bootstrap confidence intervals around vector correlations of Principal components. My data is a data frame of eigenvector loadings from PC1 in the columns the rows are different species. I want to calculated the vector correlation and bootstrap confidence intervals for all species pairs. I have tried using simpleboot as pairs.boot seems to be an easy way of sampling from two vectors of data. The function I wrote veccor below works outside of the pairs.boot function. However in this function it seems to calculate all 1000 vector correlations as -1. veccor function(x,y) ((sum((x)*(y)))/((sqrt(sum(y^2)))*(sqrt(sum(x^2) x-as.vector(faVC[1,]) y-as.vector(faVC[2,]) boot-pairs.boot(x,y,veccor,1000) # I have also tried using the boot function but only end up sampling from on of my vectors of data. Any suggestions would be greatly appriciated! Thanks! Best, Becca __ [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
