Re: [R] Artifacts in pdf() of image() (w/o comments)
It may be worth outputting postscript and converting to PDF from
there. Although Preview can do this, it may be worth looking at
Ghostscript which *may* not have simila problems.
I have also had PDFs which have displayed well in Preview and open
source tools and have been garbled in Adobe Acrobat, so the problems
aren't limited to Preview.
Best wishes,
Mark
On 13/08/07, Duncan Murdoch [EMAIL PROTECTED] wrote:
Michael Kubovy wrote:
On Aug 12, 2007, at 6:24 AM, Duncan Murdoch wrote:
Michael Kubovy wrote:
Dear r-helpers,
In my previous message there were comments in the code that may
have made cutting and pasting awkward. Here it is w/o them.
I have two questions:
(1) The following produces a pdf with artifacts. How do I prevent
them?
What artifacts do you see? It looks like a smoothly varying field
when produced by R 2.5.1 and viewed in Acrobat Reader 6.0 on Windows.
Duncan Murdoch
require(grDevices)
imSize - 200
lambda - 10
theta - 15
sigma - 40
x - 1:imSize
x0 - x / imSize -.5
freq = imSize/lambda
xf = x0 * freq * 2 * pi
f - function(x, y){r - -((x^2 + y^2)/(sigma ^2)); exp(r)}
z - outer(xf, xf, f)
f1 - function(x, y){cos(.1 * x)}
z1 - outer(xf, xf, f1)
pdf('gabor.pdf')
image(xf, xf, z * z1, col = gray(250:1000/1000),
xlab = '', ylab = '', bty = 'n', axes = FALSE, asp = 1)
dev.off()
I'm working on a Mac. You're right, Acrobat 6.05 renders the figure
nicely, but when it's included in a LaTeX-produced pdf or viewed with
the Mac Preview program, a grid of fine white lines is superimposed
on the figure. So I believe that it's a matter of aliasing, which I
might be able to prevent by adjusting the parameters of the figures.
I just don't know enough to figure this out, and would appreciate
guidance.
I see the artifacts in Preview on a Mac too. So it looks to me like a
Mac bug.
Preview is actually pretty poor at graphics display; see
http://www.geuz.org/pipermail/gl2ps/2007/000223.html.
My only suggestion is not to use Preview.
Duncan Murdoch
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[R] p value statistic for lm object
Hi,
I conduct a univariate regression with lm function. I would like to get
the p value for the regression. Is there a method that would enable me to
extract the p value into a variable.
Thanks.
Arjun Bhandari
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Re: [R] A clean way to initialize class slot of type numeric vector
Well, c() is NULL, so R did as you asked it to. See ?integer: an integer vector of length 0 can be gotten by integer(0) (and other ways). If you want integers, why have a slot which is numeric? setClass(foo, representation(members=integer)) [1] foo new(foo) An object of class foo Slot members: integer(0) is the natural and simpler way to do this. On Mon, 13 Aug 2007, [EMAIL PROTECTED] wrote: Hi, I have a class definition like this: setClass(foo, representation(members=numeric), prototype(members=c())) I intend my class to have members, a slot whose value should be a vector of integer. When I initialize this class, I don't have any member yet. So my member is blank. But if I run the above definition into R, it will complain that my slot members is assigned to NULL which does not extend class numeric. So how can I fix this? Is there any clean way to do this? This is quite a common situation but I can't seem to find a way out. Any help would be really appreciated. Thank you. - adschai __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- 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 __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] creating unique and sorted values by combining columns
Hi List,
I am combining column values of a dataframe to create a new variable
that is the sorted and unqiue combination of the columns (and excluding
0). The following works, but surely there is a more elegant way of doing
this?
t1-NULL
for(i in 1:nrow(tmp)) {
if(i == 1){
sort(c(tmp[i,1], tmp[i,2],tmp[i,3],tmp[i,4],tmp[i,5]),
decreasing=TRUE)-t1
}
else {
rbind(t1,sort(c(tmp[i,1],
tmp[i,2],tmp[i,3],tmp[i,4],tmp[i,5]),decreasing=TRUE))-t1
}
}
t2-NULL
for(i in 1:nrow(t1)){
if(i == 1) paste(unique(t1[i,t1[i,]0]),collapse=_)-t2
else cbind(t2,paste(unique(t1[i,t1[i,]0]),collapse=_))-t2
}
tmp
t1
t2
Any hints appreciated.
Thanks
Herry
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Re: [R] binomial simulation
Hi,
The probability of false detection is: P(T+ | D-)=1-P(T+ | D+)=0.05.
and I want to find the joint probability P(T+,D+)=P(T+|D+)*P(D+)
Thank you for your reply,
Sigalit.
On 8/13/07, Moshe Olshansky [EMAIL PROTECTED] wrote:
Hi Sigalit,
Do you want to find the probability P(T+ = t AND D+ =
d) for all the combinations of t and d (for ICU and
Reg.)?
Is the probability of false detection (when there is
no disease) always 0?
Regards,
Moshe.
--- sigalit mangut-leiba [EMAIL PROTECTED] wrote:
hello,
I asked about this simulation a few days ago, but
still i can't get what i
need.
I have 2 units: icu and regular. from icu I want to
take 200 observations
from binomial distribution, when probability for
disease is: p=0.6.
from regular I want to take 300 observation with the
same probability: p=0.6
.
the distribution to detect disease when disease
occurred- *for someone from
icu* - is: p(T+ | D+)=0.95.
the distribution to detect disease when disease
occurred- *for someone from
reg.unit* - is: p(T+ | D+)=0.8.
I want to compute the joint distribution for each
unit: p(T+,D+) for icu,
and the same for reg.
I tried:
pdeti - 0
pdetr - 0
picu - pdeti*.6
preg - pdetr*.6
dept - c(icu,reg)
icu - rbinom(200, 1, .6)
reg - rbinom(300, 1, .6)
for(i in 1:300) {
if(dept==icu) pdeti==0.95
if (dept==reg) pdetr==0.80
}
print(picu)
print(preg)
and got 50 warnings:
the condition has length 1 and only the first
element will be used in: if
(dept == icu) pdeti == 0.95
the condition has length 1 and only the first
element will be used in: if
(dept == reg) pdetr == 0.8
I would appreciate any suggestions,
thank you,
Sigalit.
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Re: [R] need help with pdf-plot
Thank you both! The hint with the pty=s was very useful. I did not know it before. Ciao, Antje Ivar Herfindal schrieb: Dear Antje I cannot see that you have got any replies yet, so I will make and attempt. However, I am sure other have more formally correct solutions. When you call the pdf(), you can set paper=a4 (or a4r for landscape). However, the width and the height of your plot should then not exceed the size of the paper (which is approximately 8.27*11.69 inches for a4). Try (I have only tested on windows XP, R 2.5.0): pdf(test1.pdf, width=10, heigh=5, paper=a4r) par(mfrow=c(1,3), pty=s) #pty=s gives square plotting regions plot(rnorm(100)) plot(rnorm(100)) plot(rnorm(100)) dev.off() Hope this helps Ivar Antje skrev: I still have this problem. Does anybody know any solution? Antje Antje schrieb: Hello, I'm trying to plot a set of barplots like a matrix (2 rows, 10 columns fromreduced_mat) to a pdf. It works with the following parameters: pdf(test.pdf,width=ncol(reduced_mat)*2, height=nrow(reduced_mat)*2, pointsize = 12) par(mfcol = c(nrow(reduced_mat),ncol(reduced_mat)), oma = c(0,0,0,0), lwd=48/96, cex.axis = 0.5, las = 2, cex.main = 1.0) The I get a long narrow page format with the quadratic barplots. But I would like to have a A4 format in the end and the plots not filling the whole page (they should stay somehow quadratic and not be stretched...). What shall I look for to achieve this? Antje __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] need help to manipulate function and time interval
Hi,
data are extracted from MS Access with the format: (%d/%m%Y %H%M%S); ex:
16/09/2006 03:38:37
I still have error messneger
Thank you for your help.
Lassana KOITA
Chargé d'Etudes de Sécurité Aéroportuaire et d'Analyse Statistique /
Project Engineer Airport Safety Studies Statistical analysis
Service Technique de l'Aviation Civile (STAC) / Civil Aviation Technical
Department
Direction Générale de l'Aviation Civile (DGAC) / French Civil Aviation
Headquarters
Tel: 01 49 56 80 60
Fax: 01 49 56 82 14
E-mail: [EMAIL PROTECTED]
http://www.stac.aviation-civile.gouv.fr/
Uwe Ligges [EMAIL PROTECTED]
12/08/2007 15:24
A
Matthew Walker [EMAIL PROTECTED]
cc
Henrique Dallazuanna [EMAIL PROTECTED], r-help@stat.math.ethz.ch, KOITA
Lassana - STAC/ACE [EMAIL PROTECTED]
Objet
Re: [R] need help to manipulate function and time interval
Matthew Walker wrote:
Uwe Ligges wrote:
Henrique Dallazuanna wrote:
Hi,
Try whit:
if(time[j] = 18:00:00 23:59:59)
This code is obviously wrong and does not help for the next few lines
in the questioner's message, please do not post unsensible stuff.
Uwe Ligges
Actually, I would have said that was quite an interesting solution. If
the time is guaranteed to be in this 8 character format, then that
idea's quite easily implemented:
# Returns true if time is between 18:00:00 and 23:59:59
check_time - function(time_string) {
if (nchar(time_string)!=8) stop (Incorrect format)
time_string = 18:00:00 time_string = 23:59:59
}
check_time(19:59:00)
[1] TRUE
check_time(16:59:00)
[1] FALSE
check_time(18:00:00)
[1] TRUE
check_time(23:59:59)
[1] TRUE
check_time(24:00:00)
[1] FALSE
check_time(18:05)
Error in check_time(18:05) : Incorrect format
Perhaps there is an issue if the locale does not sort character-based
numbers in the same way as ASCII? But Otherwise, I can't see why this
solution wouldn't do the job.
A more robust solution solution would parse the strings (?strptime) and
then check their days/hours/mins/seconds (?DateTimeClasses). But
perhaps the above is sufficient?
Well, the interesting part of the original question that everynody seems
to omit now was the second part:
if (time[j] is between 22:00:00 and 05:59:59)
hence the answer is still not sufficient (even if the syntax error has
been corrected) - and hence I asked for the format the time is
originally in.
Uwe Ligges
Cheers,
Matthew
[[alternative HTML version deleted]]
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Re: [R] creating unique and sorted values by combining columns
apply(tmp, 1, function(x) paste(sort(unique(x[x!=0])),collapse=_))
Jacques VESLOT
INRA - Biostatistique Processus Spatiaux
Site Agroparc 84914 Avignon Cedex 9, France
Tel: +33 (0) 4 32 72 21 58
Fax: +33 (0) 4 32 72 21 84
[EMAIL PROTECTED] a écrit :
Hi List,
I am combining column values of a dataframe to create a new variable
that is the sorted and unqiue combination of the columns (and excluding
0). The following works, but surely there is a more elegant way of doing
this?
t1-NULL
for(i in 1:nrow(tmp)) {
if(i == 1){
sort(c(tmp[i,1], tmp[i,2],tmp[i,3],tmp[i,4],tmp[i,5]),
decreasing=TRUE)-t1
}
else {
rbind(t1,sort(c(tmp[i,1],
tmp[i,2],tmp[i,3],tmp[i,4],tmp[i,5]),decreasing=TRUE))-t1
}
}
t2-NULL
for(i in 1:nrow(t1)){
if(i == 1) paste(unique(t1[i,t1[i,]0]),collapse=_)-t2
else cbind(t2,paste(unique(t1[i,t1[i,]0]),collapse=_))-t2
}
tmp
t1
t2
Any hints appreciated.
Thanks
Herry
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Re: [R] p value statistic for lm object
Arjun Bhandari arb.em at adia.ae writes: I conduct a univariate regression with lm function. I would like to get the p value for the regression. Is there a method that would enable me to extract the p value into a variable. # From lm docs (Sorry, Annette, for misusing your data) ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) lm.D90 - lm(ctl ~ trt) # let's look if it is in lm.. str(lm.D90) # mmm.. looks like there are no p-test results # try summary sum.lm = summary(lm.D90) sum.lm # .. coming closer. # Colum Pr(|t|), row trt is what we are looking for # Let's look into the structure str(sum.lm) # $coeffients has the wanted sum.lm$coefficients # looks good. We could use this, but better use the accessor function # Check the documentation, but coef is always a good try cf = coef(sum.lm) #now pick the right column pslope = cf[trt,Pr(|t|)] pslope ## # Could also use index, but less safe in general, even if the # Pr(.. is definitively ugly cf[2,4] # In general, the $ approach is not recommened, if there is an # accessor funtion, use it. The closest (with no p-values) is confint(lm.D90) # Which probably tells more than the beloved p-values. # The better the journal's referees, the more they prefer confints over p __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] RServe
yoo wrote: Yea, I found the shutdown function in the java interface as well.. but is there a way I can send a shutdown command through linux shell? (something that I can cron?) Write a minimal java program that sends the shutdown command, then run that from your shell... /obvious Barry __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] need help to manipulate function and time interval
KOITA Lassana - STAC/ACE wrote:
Hi,
data are extracted from MS Access with the format: (%d/%m%Y %H%M%S); ex:
16/09/2006 03:38:37
1. the example you gave in your first message is still not reproducible,
you gave the following:
myfunc - function(mytab, Time, Level)
{
vect - rep(0, length(mytab))
for(i in 1:length(vect))
{
for(j in 1:length(Time))
if(time[j] is between 18:00:00 and 23:59:59)
L[i] - L[j]+5
vect[i] - 10^((L[i])/10
if (time[j] is between 22:00:00 and 05:59:59)
L[i] - L[j]+10
vect[i] - 10^((L[i])/10
else
L[i] = L[j]
vect[i] - 10^((L[i])/10
}
}
a) Please check the parentheses (some closing ) are missing, you also
probably want to group some code {} that is executed in some of the
if conditions rather than executing the (syntactically erroneous) line
vect[i] - 10^((L[i])/10
three times for each inner loop. Particularly, the else is
syntactically incorrect with not preceding if.
So my guess is you want something like
myfunc - function(mytab, Time, Level)
{
vect - rep(0, length(mytab))
for(i in 1:length(vect))
{
for(j in 1:length(Time))
{
if(time[j] is between 18:00:00 and 23:59:59){
L[i] - L[j]+5
vect[i] - 10^(L[i]/10)
}
if (time[j] is between 22:00:00 and 05:59:59){
L[i] - L[j]+10
vect[i] - 10^(L[i]/10)
} else {
L[i] - L[j]
vect[i] - 10^(L[i]/10)
}
}
}
But I am really not sure about that and why you subsitute L[i] so many
times...!
b) What are mytab, Time and Level?
2. Which is the format do you have?
(%d/%m%Y %H%M%S)
*or*
16/09/2006 03:38:37
If the latter, you can use
dat - strptime(16/09/2006 03:38:37, %d/%m/%Y %H:%M:%S)
in order to get some POSIXlt object for later calculations.
3. What do you mean with
time[j] is between 22:00:00 and 05:59:59
Is the day of the date important or not for this comparison?
Uwe Ligges
I still have error messneger
Thank you for your help.
Lassana KOITA
Chargé d'Etudes de Sécurité Aéroportuaire et d'Analyse Statistique /
Project Engineer Airport Safety Studies Statistical analysis
Service Technique de l'Aviation Civile (STAC) / Civil Aviation Technical
Department
Direction Générale de l'Aviation Civile (DGAC) / French Civil Aviation
Headquarters
Tel: 01 49 56 80 60
Fax: 01 49 56 82 14
E-mail: [EMAIL PROTECTED]
http://www.stac.aviation-civile.gouv.fr/
Uwe Ligges [EMAIL PROTECTED]
12/08/2007 15:24
A
Matthew Walker [EMAIL PROTECTED]
cc
Henrique Dallazuanna [EMAIL PROTECTED], r-help@stat.math.ethz.ch, KOITA
Lassana - STAC/ACE [EMAIL PROTECTED]
Objet
Re: [R] need help to manipulate function and time interval
Matthew Walker wrote:
Uwe Ligges wrote:
Henrique Dallazuanna wrote:
Hi,
Try whit:
if(time[j] = 18:00:00 23:59:59)
This code is obviously wrong and does not help for the next few lines
in the questioner's message, please do not post unsensible stuff.
Uwe Ligges
Actually, I would have said that was quite an interesting solution. If
the time is guaranteed to be in this 8 character format, then that
idea's quite easily implemented:
# Returns true if time is between 18:00:00 and 23:59:59
check_time - function(time_string) {
if (nchar(time_string)!=8) stop (Incorrect format)
time_string = 18:00:00 time_string = 23:59:59
}
check_time(19:59:00)
[1] TRUE
check_time(16:59:00)
[1] FALSE
check_time(18:00:00)
[1] TRUE
check_time(23:59:59)
[1] TRUE
check_time(24:00:00)
[1] FALSE
check_time(18:05)
Error in check_time(18:05) : Incorrect format
Perhaps there is an issue if the locale does not sort character-based
numbers in the same way as ASCII? But Otherwise, I can't see why this
solution wouldn't do the job.
A more robust solution solution would parse the strings (?strptime) and
then check their days/hours/mins/seconds (?DateTimeClasses). But
perhaps the above is sufficient?
Well, the interesting part of the original question that everynody seems
to omit now was the second part:
if (time[j] is between 22:00:00 and 05:59:59)
hence the answer is still not sufficient (even if the syntax error has
been corrected) - and hence I asked for the format the time is
originally in.
Uwe Ligges
Cheers,
Matthew
[[alternative HTML version deleted]]
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Re: [R] need help to manipulate function and time interval
a) I agreed with your suggestion about the condition else. L = Level of
noise fluctuates during the various moments of the day (i.e L= L+5 between
18:00:00 and 21:59:59; L=L+10 between 22:00:00 and 05:59:59 and L = L
else).
b)
mytab is a data table including Time column, Level column, ,
Time - Table[,time] and Level - Table[, Levele]
c) Time format is 16/09/2006 03:38:37
d)
the time interval [ 18:00:00 ; 21:59:59 ]and [ 22:00:00 ; 05:59:59] is
very important for me.
Best regards
Lassana KOITA
Chargé d'Etudes de Sécurité Aéroportuaire et d'Analyse Statistique /
Project Engineer Airport Safety Studies Statistical analysis
Service Technique de l'Aviation Civile (STAC) / Civil Aviation Technical
Department
Direction Générale de l'Aviation Civile (DGAC) / French Civil Aviation
Headquarters
Tel: 01 49 56 80 60
Fax: 01 49 56 82 14
E-mail: [EMAIL PROTECTED]
http://www.stac.aviation-civile.gouv.fr/
Uwe Ligges [EMAIL PROTECTED]
Envoyé par : [EMAIL PROTECTED]
13/08/2007 10:39
A
KOITA Lassana - STAC/ACE [EMAIL PROTECTED]
cc
r-help@stat.math.ethz.ch
Objet
Re: [R] need help to manipulate function and time interval
KOITA Lassana - STAC/ACE wrote:
Hi,
data are extracted from MS Access with the format: (%d/%m%Y %H%M%S);
ex:
16/09/2006 03:38:37
1. the example you gave in your first message is still not reproducible,
you gave the following:
myfunc - function(mytab, Time, Level)
{
vect - rep(0, length(mytab))
for(i in 1:length(vect))
{
for(j in 1:length(Time))
if(time[j] is between 18:00:00 and 23:59:59)
L[i] - L[j]+5
vect[i] - 10^((L[i])/10
if (time[j] is between 22:00:00 and 05:59:59)
L[i] - L[j]+10
vect[i] - 10^((L[i])/10
else
L[i] = L[j]
vect[i] - 10^((L[i])/10
}
}
a) Please check the parentheses (some closing ) are missing, you also
probably want to group some code {} that is executed in some of the
if conditions rather than executing the (syntactically erroneous) line
vect[i] - 10^((L[i])/10
three times for each inner loop. Particularly, the else is
syntactically incorrect with not preceding if.
So my guess is you want something like
myfunc - function(mytab, Time, Level)
{
vect - rep(0, length(mytab))
for(i in 1:length(vect))
{
for(j in 1:length(Time))
{
if(time[j] is between 18:00:00 and 23:59:59){
L[i] - L[j]+5
vect[i] - 10^(L[i]/10)
}
if (time[j] is between 22:00:00 and 05:59:59){
L[i] - L[j]+10
vect[i] - 10^(L[i]/10)
} else {
L[i] - L[j]
vect[i] - 10^(L[i]/10)
}
}
}
But I am really not sure about that and why you subsitute L[i] so many
times...!
b) What are mytab, Time and Level?
2. Which is the format do you have?
(%d/%m%Y %H%M%S)
*or*
16/09/2006 03:38:37
If the latter, you can use
dat - strptime(16/09/2006 03:38:37, %d/%m/%Y %H:%M:%S)
in order to get some POSIXlt object for later calculations.
3. What do you mean with
time[j] is between 22:00:00 and 05:59:59
Is the day of the date important or not for this comparison?
Uwe Ligges
I still have error messneger
Thank you for your help.
Lassana KOITA
Chargé d'Etudes de Sécurité Aéroportuaire et d'Analyse Statistique /
Project Engineer Airport Safety Studies Statistical analysis
Service Technique de l'Aviation Civile (STAC) / Civil Aviation Technical
Department
Direction Générale de l'Aviation Civile (DGAC) / French Civil Aviation
Headquarters
Tel: 01 49 56 80 60
Fax: 01 49 56 82 14
E-mail: [EMAIL PROTECTED]
http://www.stac.aviation-civile.gouv.fr/
Uwe Ligges [EMAIL PROTECTED]
12/08/2007 15:24
A
Matthew Walker [EMAIL PROTECTED]
cc
Henrique Dallazuanna [EMAIL PROTECTED], r-help@stat.math.ethz.ch, KOITA
Lassana - STAC/ACE [EMAIL PROTECTED]
Objet
Re: [R] need help to manipulate function and time interval
Matthew Walker wrote:
Uwe Ligges wrote:
Henrique Dallazuanna wrote:
Hi,
Try whit:
if(time[j] = 18:00:00 23:59:59)
This code is obviously wrong and does not help for the next few lines
in the questioner's message, please do not post unsensible stuff.
Uwe Ligges
Actually, I would have said that was quite an interesting solution. If
the time is guaranteed to be in this 8 character format, then that
idea's quite easily implemented:
# Returns true if time is between 18:00:00 and 23:59:59
check_time - function(time_string) {
if (nchar(time_string)!=8) stop (Incorrect format)
time_string = 18:00:00 time_string = 23:59:59
}
check_time(19:59:00)
[1] TRUE
check_time(16:59:00)
[1] FALSE
check_time(18:00:00)
[1] TRUE
check_time(23:59:59)
[1] TRUE
check_time(24:00:00)
[1] FALSE
check_time(18:05)
Error in check_time(18:05) : Incorrect format
Perhaps there is an issue if the locale does not sort character-based
numbers in the same
Re: [R] How to store the parameter estimated by nls( ) to a variable?
On Sun, Aug 12, 2007 at 10:50:59AM -0700, Yuchen Luo wrote: Dear Professor Murdoch. Thank you so much Best Wishes Yuchen Luo On 8/12/07, Duncan Murdoch [EMAIL PROTECTED] wrote: Yuchen Luo wrote: Dear Colleagues. I believe this should be a problem encountered by many: nls( ) is a very useful and efficient function to use if we are just to display the estimated value on screen. What if we need R to store the estimated parameter in a variable? For example: x=rnorm(10, mean=1000, sd=10) y=x^2+100+rnorm(10) a=nls(y~(x^2+para),control=list(maxiter = 1000, minFactor=0.5 ^1024),start=list(para=0.0)) How to store the estimated value of para in this case, in a variable, say, b? It is easy to display a and find all the information. How ever, I need to fit a different set of x and y in every loop of my code and I need to store the estimated values for further use. I have checked both the online manual and several S-plus books but no example as such showed up. Your help will be highly appreciated! coef(a) will get what you want. coef() works for most modelling functions where it makes sense. Duncan Murdoch further information might be extracted from the `summary' return value. e.g., if you need not only the parameters but also their error estimates, you could get them via summary(a)$parameters[, Std. Error] (question: is this the canonical way to do it or is their a better one?) __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to write to a table column by column?
Assuming that the daily.incomes are the same lengths, then your loop could be:
Lst - list()
for (i in 1:count) Lst[[i]] - list(..)
Lst.col - do.call('cbind', Lst)
On 8/12/07, Yuchen Luo [EMAIL PROTECTED] wrote:
Dear friends.
Every loop of my program will result in a list that is very long, with a
structure similar to the one below:
Lst - list(name=Fred, wife=Mary, daily.incomes=c(1:850))
Please notice the large size of daily.incomes.
I need to store all such lists in a csv file so that I can easily view them
in Excel. Excel cannot display a row of more than 300 elements, therefore, I
have to store the lists as columns. It is not hard to store one list as a
column in the csv file. The problem is how to store the second list as a
second column, so that the two columns will lie side by side to each other
and I can easily compare their elements. ( If I use 'appened=TRUE', the
second time series will be stored in the same column. )
Thank you for your tine and your help will be highly appreciated!!
Best
Yuchen Luo
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] question regarding is.factor()
Dear all, please help with what must be a straightforward question which I can't answer. I add a column of my dataframe as factor of an existing column e.g. df[,5] - factor(df[,2]) and can test that it is by is.factor() but if I did not know in advance what types the columns were, is there a function to tell me what they are. i.e. instead of is.factor(), is.matrix(), is.list(), a function more like what.is() - [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] question regarding is.factor()
typeof() for 'types'. However, factor is not a type but a class, so class() is probably what you want. On Mon, 13 Aug 2007, Jabez Wilson wrote: Dear all, please help with what must be a straightforward question which I can't answer. But 'An Introduction to R' could. I add a column of my dataframe as factor of an existing column e.g. df[,5] - factor(df[,2]) and can test that it is by is.factor() but if I did not know in advance what types the columns were, is there a function to tell me what they are. i.e. instead of is.factor(), is.matrix(), is.list(), a function more like what.is() - [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- 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 __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] question regarding is.factor()
See ?class -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O On 13/08/07, Jabez Wilson [EMAIL PROTECTED] wrote: Dear all, please help with what must be a straightforward question which I can't answer. I add a column of my dataframe as factor of an existing column e.g. df[,5] - factor(df[,2]) and can test that it is by is.factor() but if I did not know in advance what types the columns were, is there a function to tell me what they are. i.e. instead of is.factor(), is.matrix(), is.list(), a function more like what.is() - [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Values in horizontal versus vertical position on 'y' axe
hi, When I do a graph. the values on y axe are vertical position. How can I put the values in horizontal position? thanks [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Values in horizontal versus vertical position on 'y' axe
? par# then scroll down to look at 'las' You probably want par(las=1) HTH, Mike. akki wrote: hi, When I do a graph. the values on y axe are vertical position. How can I put the values in horizontal position? thanks [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Values-in-horizontal-versus-vertical-position-on-%27y%27-axe-tf4260726.html#a12124985 Sent from the R help mailing list archive at Nabble.com. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Very new - beginners questions
Dear all, I have 4 sites and want to determine how different they are from each other. For this I have decided to use R though it seems a bit daunting to learn. I have read data in from a CSV the structure is : Species1 Species2 Species3 Site1 4 4 7 Site2 3 1 0 Site3 0 99 6 Site4 75 3 33 There are many more species than shown above this is just an example. Here are the questions. How do I read one row of data so as to load site2 into a variable called site2? Once I plot a graph using ordiplot how do I extract it from R so that I can put it into a Word for Windows document? Once I have the data in varables I hope to use designdist and Sørensen to discover diversity indices. I had a crack at this once but because I had sites as the columns it didn't work. Now that I think I have the data correct I can proceed. x Input data (this will be the whole data set that I read into my variable 'allSites' from a CSV.)? The variables for Sørensen will contain terms J for shared quantity, A and B for totals, N for the number of rows (sites) and P for the number of columns (species) and 'Binary' as the term. How do I get the shared number of species for each row? Probably very beginner type questions but I want to get on an haven't yet found the answers in my trawl throgh the help. Is there a book that I can buy to learn R? All the best, Richard Price MSc student University Birmingham. [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Values in horizontal versus vertical position on 'y' axe
On Mon, 2007-08-13 at 13:37 +0200, akki wrote: hi, When I do a graph. the values on y axe are vertical position. How can I put the values in horizontal position? thanks If you mean How do I rotate the tick labels? then look at ?par and parameter 'las'. E.g. this shows the options available: opar - par(mfrow = c(2,2)) plot(1:10, main = expression(las == 0)) ## las = 0, default plot(1:10, las = 1, main = expression(las == 1)) plot(1:10, las = 2, main = expression(las == 2)) plot(1:10, las = 3, main = expression(las == 3)) par(opar) If not, try rephrasing your question, and provide an example of what you can plot at the moment and what is wrong with it that you'd like to change. HTH G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Robust estimators for ordered logistic regression
Hello, I am trying to estimate an ordered logistic regression in R, and I wish to use robust estimators. Can this be (simply) done? More specifically, can this be done using polr (in MASS) together with sandwich. I am using R 2.5.1 on Mac OS 10.4.10. If not, is there any other way? I can fit a model using polr, and then try: coeftest(fit8.polr, df=NULL, vcovHC(fit8.polr, type=HC3)) When I do this R returns the error: Error in estfun(x) : no applicable error for estfun This is the first time I have tried to use sandwich, so I wouldn't be surprised if I was doing something dumb--but I can't see what. I have been working through Achim Zeileis's Econometric Computing with HC and HAC Covariance Matrix Estimators. Any help greatly appreciated. Lindsay Stirton __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Problem with axes plotting
Hi, I've 2 little questions: Is possible change the size of the values of x and y axes, when I do plot?? How can I put a range of 0.0001 to 5.5000, although my max value is 4 on 'y' axe log scale (without pattern 1e-0'X')? Thanks. [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] statdataml question
Hi, I was wondering if Statdataml is currently the preferred way to represent statistical data in XML in R. And also if the Statdataml api provides ways to load the XML as a HTTP GET? If so can you give me an easy example of this. Thanks. Cheers, Bryan Rasmussen __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Very new - beginners questions
Hi. You should study An Introduction to R manual. It is installed with R in PDF format and is accessible from the menu (Help - Manuals (in PDF) - ) There are several links in the R web site. Go to http://www.r-project.org/ and see links under the word Documentation in the left frame. There are some contributed documents: http://cran.r-project.org/other-docs.html Array manipulations are described in the chapter 5 of the Introduction. The plot window has its own menu allowing using of the Windows clipboard (you didn't describe your system, but Word for Windows suggests you're on Windows). Richard Price-4 wrote: Dear all, I have 4 sites and want to determine how different they are from each other. For this I have decided to use R though it seems a bit daunting to learn. I have read data in from a CSV the structure is : Species1 Species2 Species3 Site1 4 4 7 Site2 3 1 0 Site3 0 99 6 Site4 75 3 33 There are many more species than shown above this is just an example. Here are the questions. How do I read one row of data so as to load site2 into a variable called site2? Once I plot a graph using ordiplot how do I extract it from R so that I can put it into a Word for Windows document? Once I have the data in varables I hope to use designdist and Sørensen to discover diversity indices. I had a crack at this once but because I had sites as the columns it didn't work. Now that I think I have the data correct I can proceed. x Input data (this will be the whole data set that I read into my variable 'allSites' from a CSV.)? The variables for Sørensen will contain terms J for shared quantity, A and B for totals, N for the number of rows (sites) and P for the number of columns (species) and 'Binary' as the term. How do I get the shared number of species for each row? Probably very beginner type questions but I want to get on an haven't yet found the answers in my trawl throgh the help. Is there a book that I can buy to learn R? All the best, Richard Price MSc student University Birmingham. -- View this message in context: http://www.nabble.com/Very-new---beginners-questions-tf4260866.html#a12125845 Sent from the R help mailing list archive at Nabble.com. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Very new - beginners questions
Hi [EMAIL PROTECTED] napsal dne 13.08.2007 14:04:50: Dear all, I have 4 sites and want to determine how different they are from each other. For this I have decided to use R though it seems a bit daunting to learn. I have read data in from a CSV the structure is : Species1 Species2 Species3 Site1 4 4 7 Site2 3 1 0 Site3 0 99 6 Site4 75 3 33 There are many more species than shown above this is just an example. Here are the questions. How do I read one row of data so as to load site2 into a variable called site2? ?read.table and other read. commands. Do not work with variables but with data frames. Once I plot a graph using ordiplot how do I extract it from R so that I can put it into a Word for Windows document? 1. save a graph from menu 2. use ?png, ?jpeg and others to see how to output graph to files (devices) Once I have the data in varables I hope to use designdist and Sřrensen to discover diversity indices. I had a crack at this once but because I had sites as the columns it didn't work. Now that I think I have the data correct I can proceed. x Input data (this will be the whole data set that I read into my variable 'allSites' from a CSV.)? The variables for Sřrensen will contain terms J for shared quantity, A and B for totals, N for the number of rows (sites) and P for the number of columns (species) and 'Binary' as the term. How do I get the shared number of species for each row? Probably very beginner type questions but I want to get on an haven't yet found the answers in my trawl throgh the help. Is there a book that I can buy to learn R? look at CRAN. You can find various books from beginners to quite advanced. e.g. P.Dalgaard Introductory statistics with R isd quite good for first steps. Or you can consult Rtips from Paul Johnson. (Just type statsrus into Google). Good idea is also to look into posting guide. Regards Petr All the best, Richard Price MSc student University Birmingham. [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Very new - beginners questions
Hi Richard, No specific answers for your data questions but for learning R there are any number of resources. There are links to references for books and a large number of on-line materials on the R website. Have a look at the list on the bottom left of the screen for books and other which leads to contributed documentation. Books by Dalgaard, Verzani and Crawley all get good reviews. In the contributed documents area Using R for Data Analysis and Graphics - Introduction, Examples and Commentary by John Maindonald and Kickstarting R by Jim Lemon are both very useful. I have also found that while An Introduction to R is invaluable it is best to download it and keep it as a pdf or a hard copy. Also it is better as a 'second intro' after you have looked at some of the books or other supporting documentation. Don't forget to browse the FAQs. They answer some of the more vexing questions about R very well. (Oh if you want to move a graph from R to Word in Windows just right click on the graph and copy it.) --- Richard Price [EMAIL PROTECTED] wrote: Dear all, I have 4 sites and want to determine how different they are from each other. For this I have decided to use R though it seems a bit daunting to learn. I have read data in from a CSV the structure is : Species1 Species2 Species3 Site1 4 4 7 Site2 3 1 0 Site3 0 99 6 Site4 75 3 33 There are many more species than shown above this is just an example. Here are the questions. How do I read one row of data so as to load site2 into a variable called site2? Once I plot a graph using ordiplot how do I extract it from R so that I can put it into a Word for Windows document? Once I have the data in varables I hope to use designdist and Sørensen to discover diversity indices. I had a crack at this once but because I had sites as the columns it didn't work. Now that I think I have the data correct I can proceed. x Input data (this will be the whole data set that I read into my variable 'allSites' from a CSV.)? The variables for Sørensen will contain terms J for shared quantity, A and B for totals, N for the number of rows (sites) and P for the number of columns (species) and 'Binary' as the term. How do I get the shared number of species for each row? Probably very beginner type questions but I want to get on an haven't yet found the answers in my trawl throgh the help. Is there a book that I can buy to learn R? All the best, Richard Price MSc student University Birmingham. [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odp: Very new - beginners questions
Petr PIKAL [EMAIL PROTECTED] 13/08/2007 14:10:40 How do I read one row of data so as to load site2 into a variable called site2? ?read.table and other read. commands. Don't forget scan(). But Petr is right, you probably do not need to read one row at a time. Once I plot a graph using ordiplot how do I extract it from R so that I can put it into a Word for Windows document? 1. save a graph from menu Use windows metafile format for Word compatibility. 2. use ?png, ?jpeg and others to see how to output graph to files (devices) In particular, see ?savePlot and note the emf type for windows compatibility. *** This email contains information which may be confidential and/or privileged, and is intended only for the individual(s) or organisation(s) named above. If you are not the intended recipient, then please note that any disclosure, copying, distribution or use of the contents of this email is prohibited. Internet communications are not 100% secure and therefore we ask that you acknowledge this. If you have received this email in error, please notify the sender or contact +44(0)20 8943 7000 or [EMAIL PROTECTED] immediately, and delete this email and any attachments and copies from your system. Thank you. LGC Limited. Registered in England 2991879. Registered office: Queens Road, Teddington, Middlesex TW11 0LY, UK __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Artifacts in pdf() of image() (w/o comments)
Dear Friends,
Thanks for your input.
FYI: Preview doesn't show PDF aliasing in the image I produced if I
uncheck the Anti-alias text and line art box under the PDF tab in
Preferences. So I'm not yet ready to drop Preview from my toolbox.
MK
On Aug 13, 2007, at 12:02 AM, Mark Wardle wrote:
It may be worth outputting postscript and converting to PDF from
there. Although Preview can do this, it may be worth looking at
Ghostscript which *may* not have simila problems.
I have also had PDFs which have displayed well in Preview and open
source tools and have been garbled in Adobe Acrobat, so the problems
aren't limited to Preview.
Best wishes,
Mark
On 13/08/07, Duncan Murdoch [EMAIL PROTECTED] wrote:
Michael Kubovy wrote:
On Aug 12, 2007, at 6:24 AM, Duncan Murdoch wrote:
Michael Kubovy wrote:
Dear r-helpers,
In my previous message there were comments in the code that may
have made cutting and pasting awkward. Here it is w/o them.
I have two questions:
(1) The following produces a pdf with artifacts. How do I prevent
them?
What artifacts do you see? It looks like a smoothly varying field
when produced by R 2.5.1 and viewed in Acrobat Reader 6.0 on
Windows.
Duncan Murdoch
require(grDevices)
imSize - 200
lambda - 10
theta - 15
sigma - 40
x - 1:imSize
x0 - x / imSize -.5
freq = imSize/lambda
xf = x0 * freq * 2 * pi
f - function(x, y){r - -((x^2 + y^2)/(sigma ^2)); exp(r)}
z - outer(xf, xf, f)
f1 - function(x, y){cos(.1 * x)}
z1 - outer(xf, xf, f1)
pdf('gabor.pdf')
image(xf, xf, z * z1, col = gray(250:1000/1000),
xlab = '', ylab = '', bty = 'n', axes = FALSE, asp = 1)
dev.off()
I'm working on a Mac. You're right, Acrobat 6.05 renders the figure
nicely, but when it's included in a LaTeX-produced pdf or viewed
with
the Mac Preview program, a grid of fine white lines is superimposed
on the figure. So I believe that it's a matter of aliasing, which I
might be able to prevent by adjusting the parameters of the figures.
I just don't know enough to figure this out, and would appreciate
guidance.
I see the artifacts in Preview on a Mac too. So it looks to me
like a
Mac bug.
Preview is actually pretty poor at graphics display; see
http://www.geuz.org/pipermail/gl2ps/2007/000223.html.
My only suggestion is not to use Preview.
Duncan Murdoch
--
Dr. Mark Wardle
Clinical research fellow and specialist registrar, Neurology
Cardiff, UK
_
Professor Michael Kubovy
University of Virginia
Department of Psychology
USPS: P.O.Box 400400Charlottesville, VA 22904-4400
Parcels:Room 102Gilmer Hall
McCormick RoadCharlottesville, VA 22903
Office:B011+1-434-982-4729
Lab:B019+1-434-982-4751
Fax:+1-434-982-4766
WWW:http://www.people.virginia.edu/~mk9y/
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Artifacts in pdf() of image() (w/o comments)
On 8/13/2007 11:07 AM, Michael Kubovy wrote: Dear Friends, Thanks for your input. FYI: Preview doesn't show PDF aliasing in the image I produced if I uncheck the Anti-alias text and line art box under the PDF tab in Preferences. So I'm not yet ready to drop Preview from my toolbox. An alternative to dropping Preview is to report the bug in it to Apple. Apple has an online bug reporting web page somewhere; I haven't found them as helpful as R-help, but your mileage may vary. Duncan Murdoch __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Convert factor to numeric vector of labels
This is one of R's rather _endearing_ little idiosyncrasies. I ran into it a while ago. http://finzi.psych.upenn.edu/R/Rhelp02a/archive/98090.html For some reason, possibly historical, the option stringAsFactors is set to TRUE. As Prof Ripley says FAQ 7.10 will tell you as.numeric(as.character(f)) # for a one-off conversion From Gabor Grothendieck A one-off solution for a complete data.frame DF - data.frame(let = letters[1:3], num = 1:3, stringsAsFactors = FALSE) str(DF) # to see what has happened. You can reset the option globally, see below. However you might want to read Gabor Grothendieck's comment about this in the thread referenced above since it could cause problems if you transfer files alot. Personally I went with the global option since I don't tend to transfer programs to other people and I was getting tired of tracking down errors in my programs caused by numeric and character variables suddenly deciding to become factors. From Steven Tucker: You can also this option globally with options(stringsAsFactors = TRUE) # in \library\base\R\Rprofile --- Falk Lieder [EMAIL PROTECTED] wrote: Hi, I have imported a data file to R. Unfortunately R has interpreted some numeric variables as factors. Therefore I want to reconvert these to numeric vectors whose values are the factor levels' labels. I tried as.numeric(factor), but it returns a vector of factor levels (i.e. 1,2,3,...) instead of labels (i.e. 0.71, 1.34, 2.61, ). What can I do instead? Best wishes, Falk __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Artifacts in pdf() of image() (w/o comments)
But is it a bug? Can a program anti-alias text and line drawings and not bitmaps? On Aug 13, 2007, at 9:30 AM, Duncan Murdoch wrote: On 8/13/2007 11:07 AM, Michael Kubovy wrote: Dear Friends, Thanks for your input. FYI: Preview doesn't show PDF aliasing in the image I produced if I uncheck the Anti-alias text and line art box under the PDF tab in Preferences. So I'm not yet ready to drop Preview from my toolbox. An alternative to dropping Preview is to report the bug in it to Apple. Apple has an online bug reporting web page somewhere; I haven't found them as helpful as R-help, but your mileage may vary. Duncan Murdoch _ Professor Michael Kubovy University of Virginia Department of Psychology USPS: P.O.Box 400400Charlottesville, VA 22904-4400 Parcels:Room 102Gilmer Hall McCormick RoadCharlottesville, VA 22903 Office:B011+1-434-982-4729 Lab:B019+1-434-982-4751 Fax:+1-434-982-4766 WWW:http://www.people.virginia.edu/~mk9y/ __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] feature's contribution into classification
Hi, there:
The following question is more of statistics:
assume i have 5 features in a classification, and I am wondering which
methodology can help me identify which feature contributes the most
to classify a specific sample?
I knew some simple modeling like logistic regression probably can do
it since it provides an explicit formulae for that. Any others?
I tried to use lda{MASS} and posted the question last week but I did
not get any response. So again, I re-phrase my question.
Thanks.
WEIWEI
My previous questions are also attached for reference.
hi,
maybe I should re-phrase my question a bit:
is there a way to get explicit formulae like Y ~ sum of CiXi from the
model build by lda{MASS} to calculate $x (value) ?
I assume scaling is the coeff and Xi is from test data and Y is $x
called LD1. But I want to confirm this.
##
hi,
assume
val is the test data while m is lda model value by using CV=F
x = predict(m, val)
val2 = val[, 1:(ncol(val)-1)] # the last column is class label
# col is sample, row is variable
then I am wondering if
x$x == (apply(val2*m$scaling), 2, sum)
i.e., the scaling (is it coeff vector?) times val data and sum is the
discrimant result $x?
--
Weiwei Shi, Ph.D
Research Scientist
GeneGO, Inc.
Did you always know?
No, I did not. But I believed...
---Matrix III
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Artifacts in pdf() of image() (w/o comments)
On 8/13/2007 11:43 AM, Michael Kubovy wrote: But is it a bug? Can a program anti-alias text and line drawings and not bitmaps? Anti-aliasing is the removal of artifacts caused by displaying an image on a low-resolution bitmapped display. Introducing artifacts is a bug. If it wasn't a bug, why did it bother you so much? And why do you think it's reasonable to complain about it on R-help, but not to complain about it to Apple, who are clearly responsible for it? Duncan Murdoch On Aug 13, 2007, at 9:30 AM, Duncan Murdoch wrote: On 8/13/2007 11:07 AM, Michael Kubovy wrote: Dear Friends, Thanks for your input. FYI: Preview doesn't show PDF aliasing in the image I produced if I uncheck the Anti-alias text and line art box under the PDF tab in Preferences. So I'm not yet ready to drop Preview from my toolbox. An alternative to dropping Preview is to report the bug in it to Apple. Apple has an online bug reporting web page somewhere; I haven't found them as helpful as R-help, but your mileage may vary. Duncan Murdoch _ Professor Michael Kubovy University of Virginia Department of Psychology USPS: P.O.Box 400400Charlottesville, VA 22904-4400 Parcels:Room 102Gilmer Hall McCormick RoadCharlottesville, VA 22903 Office:B011+1-434-982-4729 Lab:B019+1-434-982-4751 Fax:+1-434-982-4766 WWW:http://www.people.virginia.edu/~mk9y/ __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] xyplot() with segments() superposed?
On 8/11/07, Yuelin Li [EMAIL PROTECTED] wrote:
In the hypothetical example below, how do I add two segments() into
the two panels, respectively? Say segments(x0=5, y0=10, x1=5, y1=20)
on the left and segments(x0=15, y0=-10, x1=15, y1=-2) on the right?
Many thanks in advance,
Yuelin Li.
ps. part of the code came from a solution given by Deepayan Sarkar.
---
library(lattice)
set.seed(12345)
x - 0:20
y.male.obs - - 1.2 * x + 22 + rnorm(length(x), sd = 3)
y.male.prd - - 1.2 * x + 22
y.fema.obs - - 2.2 * x + 30 + rnorm(length(x), sd = 2)
y.fema.prd - - 2.2 * x + 30
tdat - data.frame(x = rep(x, 8),
y = rep(c(y.male.obs, y.male.prd, y.fema.obs, y.fema.prd), 2),
sex = rep(rep(c(m, f), each = 2*length(x)), 2),
cohort = rep(c(1970, 1980), each = 4*length(x)),
source = rep(rep(c(obs, prd), each = length(x)), 4) )
xyplot(y ~ x | as.factor(cohort), data = tdat,
groups = interaction(sex, source),
type = c(p, p, l, l), distribute.type = TRUE)
If this is a one-off requirement, the simplest solution is:
xyplot(y ~ x | as.factor(cohort), data = tdat,
groups = interaction(sex, source),
type = c(p, p, l, l), distribute.type = TRUE,
panel = function(...) {
panel.xyplot(...)
switch(panel.number(),
panel.segments(x0=5, y0=10, x1=5, y1=20),
panel.segments(x0=15, y0=-10, x1=15, y1=-2))
})
This is not generalizable, but you haven't told us your general use case.
-Deepayan
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Re: [R] Systematically biased count data regression model
Dear all, Thank you for your helpful comments. I have acquired and read Steyerberg (2001). I may try some of the methods from that paper. This did prompt a thorough examination of the data. Sure enough, my worst outlier, where little diversity was observed and the NDVI measurment was very high, had been stripped of vegetation between the time of the NDVI data and the arthropod sampling by an avalanche. Sincerely, Matthew Bowser On 8/9/07, Steven McKinney [EMAIL PROTECTED] wrote: Hi Matthew, You may be experiencing the classic 'regression towards the mean' phenomenon, in which case shrinkage estimation may help with prediction (extremely low and high values need to be shrunk back towards the mean) Here's a reference that discusses the issue in a manner somewhat related to your situation, and it has plenty of good references Application of Shrinkage Techniques in Logistic Regression Analysis: A Case Study E. W. Steyerberg Statistica Neerlandica, 2001, vol. 55, issue 1, pages 76-88 Steven McKinney Statistician Molecular Oncology and Breast Cancer Program British Columbia Cancer Research Centre email: smckinney +at+ bccrc +dot+ ca tel: 604-675-8000 x7561 BCCRC Molecular Oncology 675 West 10th Ave, Floor 4 Vancouver B.C. V5Z 1L3 Canada -Original Message- From: [EMAIL PROTECTED] on behalf of Matthew and Kim Bowser Sent: Thu 8/9/2007 8:43 AM To: r-help@stat.math.ethz.ch Subject: [R] Systematically biased count data regression model Dear all, I am attempting to explain patterns of arthropod family richness (count data) using a regression model. It seems to be able to do a pretty good job as an explanatory model (i.e. demonstrating relationships between dependent and independent variables), but it has systematic problems as a predictive model: It is biased high at low observed values of family richness and biased low at high observed values of family richness (see attached pdf). I have tried diverse kinds of reasonable regression models mostly as in Zeileis, et al. (2007), as well as transforming my variables, both with only small improvements. Do you have suggestions for making a model that would perform better as a predictive model? Thank you for your time. Sincerely, Matthew Bowser STEP student USFWS Kenai National Wildlife Refuge Soldotna, Alaska, USA M.Sc. student University of Alaska Fairbanks Fairbankse, Alaska, USA Reference Zeileis, A., C. Kleiber, and S. Jackman, 2007. Regression models for count data in R. Technical Report 53, Department of Statistics and Mathematics, Wirtschaftsuniversität Wien, Wien, Austria. URL http://cran.r-project.org/doc/vignettes/pscl/countreg.pdf. Code `data` - structure(list(D = c(4, 5, 12, 4, 9, 15, 4, 8, 3, 9, 6, 17, 4, 9, 6, 9, 3, 9, 7, 11, 17, 3, 10, 8, 9, 6, 7, 9, 7, 5, 15, 15, 12, 9, 10, 4, 4, 15, 7, 7, 12, 7, 12, 7, 7, 7, 5, 14, 7, 13, 1, 9, 2, 13, 6, 8, 2, 10, 5, 14, 4, 13, 5, 17, 12, 13, 7, 12, 5, 6, 10, 6, 6, 10, 4, 4, 12, 10, 3, 4, 4, 6, 7, 15, 1, 8, 8, 5, 12, 0, 5, 7, 4, 9, 6, 10, 5, 7, 7, 14, 3, 8, 15, 14, 7, 8, 7, 8, 8, 10, 9, 2, 7, 8, 2, 6, 7, 9, 3, 20, 10, 10, 4, 2, 8, 10, 10, 8, 8, 12, 8, 6, 16, 10, 5, 1, 1, 5, 3, 11, 4, 9, 16, 3, 1, 6, 5, 5, 7, 11, 11, 5, 7, 5, 3, 2, 3, 0, 3, 0, 4, 1, 12, 16, 9, 0, 7, 0, 11, 7, 9, 4, 16, 9, 10, 0, 1, 9, 15, 6, 8, 6, 4, 6, 7, 5, 7, 14, 16, 5, 8, 1, 8, 2, 10, 9, 6, 11, 3, 16, 3, 6, 8, 12, 5, 1, 1, 3, 3, 1, 5, 15, 4, 2, 2, 6, 5, 0, 0, 0, 3, 0, 16, 0, 9, 0, 0, 8, 1, 2, 2, 3, 4, 17, 4, 1, 4, 6, 4, 3, 15, 2, 2, 13, 1, 9, 7, 7, 13, 10, 11, 2, 15, 7), Day = c(159, 159, 159, 159, 166, 175, 161, 168, 161, 166, 161, 166, 161, 161, 161, 175, 161, 175, 161, 165, 176, 161, 163, 161, 168, 161, 161, 161, 161, 161, 165, 176, 175, 176, 163, 175, 163, 168, 163, 176, 176, 165, 176, 175, 161, 163, 163, 168, 163, 175, 167, 176, 167, 165, 165, 169, 165, 169, 165, 161, 165, 175, 165, 176, 175, 167, 167, 175, 167, 164, 167, 164, 181, 164, 167, 164, 176, 164, 167, 164, 167, 164, 167, 175, 167, 173, 176, 173, 178, 167, 173, 172, 173, 178, 178, 172, 181, 182, 173, 162, 162, 173, 178, 173, 172, 162, 173, 162, 173, 162, 173, 170, 178, 166, 166, 162, 166, 177, 166, 170, 166, 172, 172, 166, 172, 166, 174, 162, 164, 162, 170, 164, 170, 164, 170, 164, 177, 164, 164, 174, 174, 162, 170, 162, 172, 162, 165, 162, 165, 177, 172, 162, 170, 162, 170, 174, 165, 174, 166, 172, 174, 172, 174, 170, 170, 165, 170, 174, 174, 172, 174, 172, 174, 165, 170, 165, 170, 174, 172, 174, 172, 175, 175, 170, 171, 174, 174, 174, 172, 175, 171, 175, 174, 174, 174, 175, 172, 171, 171, 174, 160, 175, 160, 171, 170, 175, 170, 170, 160, 160, 160, 171, 171, 171, 171, 160, 160, 160, 171, 171, 176, 171, 176, 176, 171, 176, 171, 176, 176, 176, 176, 159, 166, 159, 159, 166, 168, 169, 159, 168, 169, 166, 163, 180, 163, 165, 164, 180, 166, 166, 164, 164, 177, 166), NDVI = c(0.187, 0.2, 0.379, 0.253, 0.356, 0.341, 0.268, 0.431, 0.282, 0.181, 0.243,
Re: [R] Unexpected behavior in PBSmapping package
Hi David, Thanks for the message. Essentially, the argument 'col' is treated like a vector which remains independent from (X,Y) and is simply repeated as need for the number of points. If you specify col=c(red,blue,green,yellow) as below, and there are only 2 points, only red and blue will be used (hence the perceived mismatch). To keep your colour column matched with X,Y, there is an argument called polyProps that automatically looks for the column col. Unfortunately, polyProps doesn't appear to be functional with EventData (bug, I think) but will work with PolyData. Take a look at the R-code below. I've redefined your EventData to be PolyData which can serve double duty as EventData and PolyData, i.e., you can pass events to plotPoints() as well as using polyProps=events. I've also added the label field to events to simplify things, and used PBSmapping's addLabels() function. The plotPoints() code should have been able to do this with EventData also. We will have to revisit this code. Thanks David for taking the time to show us user-related problems. It helps us to re-evaluate functionalities. Undoubtedly, the package has many little imperfections but we try to iron them out over time. Cheers, Rowan ### Begin Example ### library( PBSmapping ) # Define some EventData events - as.PolyData( read.table( textConnection( 'PID X Y col 1 494 1494 red 2 497 1497 blue 3 500 1500 green 4 503 1503 yellow' ), header=TRUE, strings=FALSE ), projection='UTM', zone=10 ) events$label - toupper(substring(events$col,1,1)); par( mfrow=c(3,1) ) # Plot the events with plot limits large enough to show # the full extent of all the symbols plotPoints( events, pch=16, cex=5, polyProps=events, xlim=c(490,508), ylim=c(1490,1508),proj=UTM) addLabels(events, polyProps=events, font=2, cex=2, col=1) # Normal plot extents; partial symbols cut off by edges # of plotting region (as expected) plotPoints( events, pch=16, cex=5, polyProps=events,proj=UTM) addLabels(events, polyProps=events, font=2, cex=2, col=1) ## Now use more-restrictive plot limits plotPoints( events, pch=16, cex=5, polyProps=events, xlim=c(499,505), ylim=c(1499,1505),proj=UTM) addLabels(events, polyProps=events, font=2, cex=2, col=1) ### End example ### P.S. Once you've defined your EventData/PolyData/PolySet with attribute projection=UTM, plotMap() will attempt to use this but plotLines() and plotPoints() will simply tell you if it matched the specified projection in the function call. When you specified plotPoints(...,proj=TRUE), you are probably overriding the earlier specified UTM projection (of events) with proj=1. In this case, the 2 projections (UTM, 1) are the same. If your Poly dataset was set to projection=LL, they would not be. == -Original Message- From: Schnute, Jon Sent: Monday, August 13, 2007 8:19 AM To: Haigh, Rowan Subject: FW: Unexpected behavior in PBSmapping package Hi Rowan - Could you please check into this when you get a chance? Thanks - Jon -Original Message- From: D. Dailey [mailto:[EMAIL PROTECTED] Sent: Thursday, August 09, 2007 11:38 PM To: r-help@stat.math.ethz.ch; Schnute, Jon Subject: Unexpected behavior in PBSmapping package Using R 2.5.1 on Windows XP Professional, and PBSmapping package version 2.51, I have encountered some behavior which puzzles me. I am including the package's listed maintainer on this email but also seek the thoughts of the R-help community. I have a set of EventData, which I want to plot as points, and to color the points according to some criterion. It turns out that some of my points fall outside my desired plotting region. It looks like this causes the PBSmapping functions plotPoints and addPoints to incorrectly deal with the color assignments. Consider the following toy example: ### Begin Example ### library( PBSmapping ) # Define some EventData events - as.EventData( read.table( textConnection( 'EID X Y Color 1 494 1494 red 2 497 1497 blue 3 500 1500 green 4 503 1503 yellow' ), header=TRUE, strings=FALSE ), proj='UTM', zone=10 ) par( mfrow=c(3,1) ) # Plot the events with plot limits large enough to show # the full extent of all the symbols plotPoints( events, pch=16, cex=5, col=events$Color, xlim=c(490,508), ylim=c(1490,1508), proj=TRUE ) with( events, text( X, Y, toupper( substr( Color, 1, 1 ) ), font=2, cex=2 ) ) # Normal plot extents; partial symbols cut off by edges # of plotting region (as expected) plotPoints( events, pch=16, cex=5, col=events$Color, proj=TRUE ) with( events, text( X, Y, toupper( substr( Color, 1, 1 ) ), font=2, cex=2 ) ) ## Now use more-restrictive plot limits plotPoints( events, pch=16, cex=5, col=events$Color, xlim=c(499,505), ylim=c(1499,1505), proj=TRUE ) with( events, text( X, Y, toupper( substr( Color, 1, 1 ) ), font=2, cex=2 ) ) # Note that symbols are plotted in the right places (note text labels) # but colors are not as expected ### End example
Re: [R] Artifacts in pdf() of image() (w/o comments)
Hi Duncan, I was trying to learn to remove the artifacts by setting the parameters of the image so that anti-aliasing wouldn't produce them (analogous to making sure that the screens of two halftone screen process images are in register before combining them so as to avoid Moiré patterns). So, I wasn't complaining; I was looking for instruction. Since the image was computed with R, it's reasonable for me to ask R experts to help me out of what I thought was caused by my ineptitude. (Regarding your question, If it wasn't a bug, why did it bother you so much?---there are many things that bother me that are not bugs.) Moreover, I'm not opposed to complaining to Apple, once I have been assured that I'm not reporting a bug where there's none. On Aug 13, 2007, at 9:56 AM, Duncan Murdoch wrote: On 8/13/2007 11:43 AM, Michael Kubovy wrote: But is it a bug? Can a program anti-alias text and line drawings and not bitmaps? Anti-aliasing is the removal of artifacts caused by displaying an image on a low-resolution bitmapped display. Introducing artifacts is a bug. If it wasn't a bug, why did it bother you so much? And why do you think it's reasonable to complain about it on R-help, but not to complain about it to Apple, who are clearly responsible for it? Duncan Murdoch On Aug 13, 2007, at 9:30 AM, Duncan Murdoch wrote: On 8/13/2007 11:07 AM, Michael Kubovy wrote: Dear Friends, Thanks for your input. FYI: Preview doesn't show PDF aliasing in the image I produced if I uncheck the Anti-alias text and line art box under the PDF tab in Preferences. So I'm not yet ready to drop Preview from my toolbox. An alternative to dropping Preview is to report the bug in it to Apple. Apple has an online bug reporting web page somewhere; I haven't found them as helpful as R-help, but your mileage may vary. Duncan Murdoch _ Professor Michael Kubovy University of Virginia Department of Psychology USPS: P.O.Box 400400Charlottesville, VA 22904-4400 Parcels:Room 102Gilmer Hall McCormick RoadCharlottesville, VA 22903 Office:B011+1-434-982-4729 Lab:B019+1-434-982-4751 Fax:+1-434-982-4766 WWW:http://www.people.virginia.edu/~mk9y/ _ Professor Michael Kubovy University of Virginia Department of Psychology USPS: P.O.Box 400400Charlottesville, VA 22904-4400 Parcels:Room 102Gilmer Hall McCormick RoadCharlottesville, VA 22903 Office:B011+1-434-982-4729 Lab:B019+1-434-982-4751 Fax:+1-434-982-4766 WWW:http://www.people.virginia.edu/~mk9y/ __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Artifacts in pdf() of image() (w/o comments)
On 8/13/2007 1:33 PM, Michael Kubovy wrote: Hi Duncan, I was trying to learn to remove the artifacts by setting the parameters of the image so that anti-aliasing wouldn't produce them (analogous to making sure that the screens of two halftone screen process images are in register before combining them so as to avoid Moiré patterns). So, I wasn't complaining; I was looking for instruction. Since the image was computed with R, it's reasonable for me to ask R experts to help me out of what I thought was caused by my ineptitude. (Regarding your question, If it wasn't a bug, why did it bother you so much?---there are many things that bother me that are not bugs.) Moreover, I'm not opposed to complaining to Apple, once I have been assured that I'm not reporting a bug where there's none. Let me explain my point of view a bit more: - You saw ugly artifacts in a graph you produced. That's a bug. The question is, what caused it? - You rightly asked on R-help. It might have been a bug in the way you were doing things, or a bug in R, or a bug in the way you were viewing the image. - It was demonstrated that this only appears in Preview, not in Acrobat Reader, so it's probably a bug there. It's only at the last point that I object to what you did: your question quoted below makes it look as though you think Apple's time is more valuable than ours. I remind you that you paid for Preview, and you didn't pay for R. Apple has a stronger obligation to help you than any of us do. My experience with them (and with most other commercial software vendors) is that you'll get much worse help from them than from us --- but that doesn't mean you shouldn't try. Perhaps when anyone searching for Mac OSX Preview on Google shows up a page full of unhandled bug reports they'll actually do something. Duncan Murdoch On Aug 13, 2007, at 9:56 AM, Duncan Murdoch wrote: On 8/13/2007 11:43 AM, Michael Kubovy wrote: But is it a bug? Can a program anti-alias text and line drawings and not bitmaps? ... [deletions] ... An alternative to dropping Preview is to report the bug in it to Apple. Apple has an online bug reporting web page somewhere; I haven't found them as helpful as R-help, but your mileage may vary. Duncan Murdoch ... [the rest deleted] ... __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Extract part of vector
Dear R-users, How do I extract numbers between asp?P= and VID from my txt vector? I have tried grep function with no luck. txt - c( http://www.mysite.com/system/empty.asp?P=2VID=defaultSID=421384237289476S=1C=18631;, http://www.mysite.com/system/empty.asp?P=123VID=defaultSID=421384237289476S=1C=18643;, http://www.mysite.com/system/empty.asp?P=342VID=defaultSID=421384237289476S=1C=18634 , http://www.mysite.com/system/empty.asp?P=232VID=defaultSID=421384237289476S=1C=18645;, http://www.mysite.com/system/empty.asp?P=2345VID=defaultSID=421384237289476S=1C=18254;, http://www.mysite.com/system/empty.asp?P=257654VID=defaultSID=421384237289476S=1C=18732;, http://www.mysite.com/system/empty.asp?P=22VID=defaultSID=421384237289476S=1C=18637;, http://www.mysite.com/system/empty.asp?P=2463VID=defaultSID=421384237289476S=1C=18575 ) The result should be like 2 123 342 232 2345 257654 22 2463 Thanks, Lauri [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] SEM for categorical data
Hi I am looking for a structural equation modeling package in R which can be used for categorical data. Is anyone aware of the existence of such a package? Would appreciate any help on this. Thank you Upasna -- - Upasna Sharma Research Scholar Shailesh J. Mehta School of Management, Indian Institute of Technology, Bombay Powai, Mumbai - 400076, India - Homepage: http://www.som.iitb.ac.in/people/upasna/ - The past is a history, the future is a mystery, and this moment is a gift. That is why this moment is called 'the present'. Anonymous __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Q: how to extract coefisients from one glm and implement them in to an other glm?
Dear R ussers, I'm still working with the categorical models, and i can not do cross validation on them because odf their nature. not cv works perfectly on other models so i wonderd if it is possible to extract the beta's from a categorical model, then use their value in a 0/1 glm. Their formula is the same, so their assumptions should still hold. Is this possible in R. kind regards, Tom W. Disclaimer: click here [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] rates and rate ratios for survey-weighted data
Hello, I am looking at data on child survival from a survey with a stratified two-stage design. I have been trying out the excellent survey package, with which I have done a descriptive analysis and Cox models. I need to calculate rates (with a child-year denominator) and various rate ratios plus confidence intervals (basically, something like the STATA commands strate and stmh used after svyset and stset) but haven't been able to find a convenient way of doing this in R despite much searching. Is this possible, either with the survey package or any other way? Many thanks for any advice. Paul -- Dr. Paul Cleary, Specialist Registrar in Public Health, Chester Health Protection Unit. [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] xyplot() with segments() superposed?
Thanks for the quick response. The general use is like the following.
Suppose I have another variable, tkmark, which for cohort 1970 is
coded 1.0 at x = c(7, 19) and 0 otherwise. For cohort 1980, tkmark
is coded 1.0 at x = c(2, 11, 12, 19) and 0 otherwise. For each
cohort, I want to mark with a short vertical bar the observed Y at
(x[tkmark == 1 source == obs], y[tkmark == 1 source == obs]).
I may have up to 65 different cohorts. Different cohorts have
different numbers of tkmark positions (some have none).
The real data come from a computerized cigarette smoking reduction
program. I want to plot each smoker's tapering of cigarettes (y) over
days (x), and mark the quit attempts. Some participants try to quit
on days 3, 7, and 17 but resume smoking the next day, some follow the
programmed tapering exactly with no interim quit attempts, and some
ignored the tapering program. I want to plot each participant's
**scheduled** and **observed** smoking pattern over days and mark the
quit attempts. Hope this is clear.
Yuelin.
-- Deepayan Sarkar wrote --|Mon (Aug/13/2007)[09:35]|--:
On 8/11/07, Yuelin Li [EMAIL PROTECTED] wrote:
In the hypothetical example below, how do I add two segments() into
the two panels, respectively? Say segments(x0=5, y0=10, x1=5, y1=20)
on the left and segments(x0=15, y0=-10, x1=15, y1=-2) on the right?
Many thanks in advance,
Yuelin Li.
ps. part of the code came from a solution given by Deepayan Sarkar.
---
library(lattice)
set.seed(12345)
x - 0:20
y.male.obs - - 1.2 * x + 22 + rnorm(length(x), sd = 3)
y.male.prd - - 1.2 * x + 22
y.fema.obs - - 2.2 * x + 30 + rnorm(length(x), sd = 2)
y.fema.prd - - 2.2 * x + 30
tdat - data.frame(x = rep(x, 8),
y = rep(c(y.male.obs, y.male.prd, y.fema.obs, y.fema.prd), 2),
sex = rep(rep(c(m, f), each = 2*length(x)), 2),
cohort = rep(c(1970, 1980), each = 4*length(x)),
source = rep(rep(c(obs, prd), each = length(x)), 4) )
xyplot(y ~ x | as.factor(cohort), data = tdat,
groups = interaction(sex, source),
type = c(p, p, l, l), distribute.type = TRUE)
If this is a one-off requirement, the simplest solution is:
xyplot(y ~ x | as.factor(cohort), data = tdat,
groups = interaction(sex, source),
type = c(p, p, l, l), distribute.type = TRUE,
panel = function(...) {
panel.xyplot(...)
switch(panel.number(),
panel.segments(x0=5, y0=10, x1=5, y1=20),
panel.segments(x0=15, y0=-10, x1=15, y1=-2))
})
This is not generalizable, but you haven't told us your general use case.
-Deepayan
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[R] Problem with fisher.g.test
I am trying to analyse a mouse dataset with 45000 genes and 24 or 48 timpoints to find cycling genes using GeneCycle. However the function fisher.g.test never runs. I am pasting the input/output summary below. I would be grateful if someone could provide any insight. I 've run GeneCycle successfully in the past on similar data, but I had to reinstall R and the packages again, so this version may be different from the previous one. My R version is Release Version 1.20 (4534) - supplied with R 2.5.1 and GeneCycle was downloaded and applied only 3-4 days ago. Thanks in advance, S [Workspace restored from /Users/howler/.RData] ctraw-read.csv(ct.csv, header = TRUE, sep = ,, quote=\) ct-t(ctraw) ftraw-read.csv(ft.csv, header = TRUE, sep = ,, quote=\) ft-t(ftraw) library(GeneCycle) Loading required package: longitudinal Loading required package: corpcor Loading required package: fdrtool Loading required package: locfdr Loading required package: splines ctgtest-fisher.g.test(ctanalysis) Error in r[i1] - r[-length(r):-(length(r) - lag + 1)] : non-numeric argument to binary operator [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] help with scatterplot3d
Hello, I am having a bit of trouble with scatterplot3d(). I was able to plot a 3d cloud of points using the following code: my.3dplot-scatterplot3d(my.coords, pch=19, zlim=c(0,1), scale.y=0.5, angle=30, box=FALSE) where my.coords is a data frame that contains x, y, and z coordinates for grid points whose elevation we sampled. The problem occurs when I try to add points using points3d. I tried to follow the code in the examples of the package pdf. First, I tried the following code to add all of the points that I wanted: my.3dplot$points3d(seq(400,600,0.19), seq(600,400,0.295), seq(800,500,0.24), seq(1000,1400,0.22), seq(1200,600,0.24), seq(1200,1500,0.28), seq(1300,1400,0.205), seq(1700,500,0.26), seq(1700,600,0.21), seq(1900,1400,0.255), seq(2300,1400,0.275), seq(2600,1300,0.225), seq(2700,400,0.235), seq(2700,1300,0.265), seq(3100,1000,0.135), col=blue, type=h, pch=16) and got the following error: Error in seq.default(600, 400, 0.295) : wrong sign in 'by' argument So I just tried to add the first point using the following code: my.3dplot$points3d(seq(400,600,0.19), col=blue, type=h, pch=16) and got the following error message: Error in xyz.coords(x, y, z) : 'x', 'y' and 'z' lengths differ. Does anyone know how I can use this function in the scatterplot3d package? Thanks so much! Ryan ~~ Ryan D. Briscoe Runquist Population Biology Graduate Group University of California, Davis [EMAIL PROTECTED] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] rfImpute
I seem to recall that rfImpute() can sometimes come up with NAs at some
point in the iterations. Could you please send me a (small) set of
data/code that reproduces the problem?
Andy
From: Eric Turkheimer
I am having trouble with the rfImpute function in the
randomForest package.
Here is a sample...
clunk.roughfix-na.roughfix(clunk)
clunk.impute-rfImpute(CONVERT~.,data=clunk)
ntree OOB 1 2
300: 26.80% 3.83% 85.37%
ntree OOB 1 2
300: 18.56% 5.74% 51.22%
Error in randomForest.default(xf, y, ntree = ntree, ...,
do.trace = ntree,
:
NA not permitted in predictors
So roughFix works, but rfImpute doesn't
Thanks,
Eric
ent3c *at* virginia.edu
--
Eric Turkheimer, PhD
Department of Psychology
University of Virginia
PO Box 400400
Charlottesville, VA 22904-4400
434-982-4732
434-982-4766 (FAX)
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Re: [R] xyplot() with segments() superposed?
On 8/13/07, Yuelin Li [EMAIL PROTECTED] wrote: Thanks for the quick response. The general use is like the following. Suppose I have another variable, tkmark, which for cohort 1970 is coded 1.0 at x = c(7, 19) and 0 otherwise. For cohort 1980, tkmark is coded 1.0 at x = c(2, 11, 12, 19) and 0 otherwise. For each cohort, I want to mark with a short vertical bar the observed Y at (x[tkmark == 1 source == obs], y[tkmark == 1 source == obs]). I may have up to 65 different cohorts. Different cohorts have different numbers of tkmark positions (some have none). The real data come from a computerized cigarette smoking reduction program. I want to plot each smoker's tapering of cigarettes (y) over days (x), and mark the quit attempts. Some participants try to quit on days 3, 7, and 17 but resume smoking the next day, some follow the programmed tapering exactly with no interim quit attempts, and some ignored the tapering program. I want to plot each participant's **scheduled** and **observed** smoking pattern over days and mark the quit attempts. Hope this is clear. Sounds like you need to pass your 'tkmarks' variable on to the panel function and then use 'subscripts' inside your panel function. Read the entry on 'panel' in ?xyplot. Also, there's a fairly long thread on a similar problem which you might find useful. It starts with https://stat.ethz.ch/pipermail/r-help/2007-July/136188.html -Deepayan __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with scatterplot3d
On 8/13/2007 3:03 PM, Ryan Briscoe Runquist wrote: Hello, I am having a bit of trouble with scatterplot3d(). I was able to plot a 3d cloud of points using the following code: my.3dplot-scatterplot3d(my.coords, pch=19, zlim=c(0,1), scale.y=0.5, angle=30, box=FALSE) where my.coords is a data frame that contains x, y, and z coordinates for grid points whose elevation we sampled. The problem occurs when I try to add points using points3d. I tried to follow the code in the examples of the package pdf. First, I tried the following code to add all of the points that I wanted: my.3dplot$points3d(seq(400,600,0.19), seq(600,400,0.295), seq(800,500,0.24), seq(1000,1400,0.22), seq(1200,600,0.24), seq(1200,1500,0.28), seq(1300,1400,0.205), seq(1700,500,0.26), seq(1700,600,0.21), seq(1900,1400,0.255), seq(2300,1400,0.275), seq(2600,1300,0.225), seq(2700,400,0.235), seq(2700,1300,0.265), seq(3100,1000,0.135), col=blue, type=h, pch=16) The header to the function (which you can see by evaluating my.3dplot$points3d) is function (x, y = NULL, z = NULL, type = p, ...) so you are setting x to seq(400,600,0.19), y to seq(600,400,0.295), etc. I think you probably want my.3dplot$points3d(x=c(400, 600, ...), y=c(600, 400, ...), z=c(0.19, 0.295, ...), ...) (where the ... is to be filled in by you.) and got the following error: Error in seq.default(600, 400, 0.295) : wrong sign in 'by' argument So I just tried to add the first point using the following code: my.3dplot$points3d(seq(400,600,0.19), col=blue, type=h, pch=16) and got the following error message: Error in xyz.coords(x, y, z) : 'x', 'y' and 'z' lengths differ. Does anyone know how I can use this function in the scatterplot3d package? try my.3dplot$points3d(x=400, y=600, z=0.19, col=blue, type=h, pch=16) Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extract part of vector
This should do it: txt [1] \nhttp://www.mysite.com/system/empty.asp?P=2VID=defaultSID=421384237289476S=1C=18631; [2] \nhttp://www.mysite.com/system/empty.asp?P=123VID=defaultSID=421384237289476S=1C=18643; [3] \nhttp://www.mysite.com/system/empty.asp?P=342VID=defaultSID=421384237289476S=1C=18634\n; [4] \nhttp://www.mysite.com/system/empty.asp?P=232VID=defaultSID=421384237289476S=1C=18645; [5] \nhttp://www.mysite.com/system/empty.asp?P=2345VID=defaultSID=421384237289476S=1C=18254; [6] \nhttp://www.mysite.com/system/empty.asp?P=257654VID=defaultSID=421384237289476S=1C=18732; [7] \nhttp://www.mysite.com/system/empty.asp?P=22VID=defaultSID=421384237289476S=1C=18637; [8] \nhttp://www.mysite.com/system/empty.asp?P=2463VID=defaultSID=421384237289476S=1C=18575\n; gsub(^.*asp.P=([[:digit:]]+).*$, '\\1', txt) [1] 2 1233422322345 257654 22 2463 On 8/13/07, Lauri Nikkinen [EMAIL PROTECTED] wrote: Dear R-users, How do I extract numbers between asp?P= and VID from my txt vector? I have tried grep function with no luck. txt - c( http://www.mysite.com/system/empty.asp?P=2VID=defaultSID=421384237289476S=1C=18631;, http://www.mysite.com/system/empty.asp?P=123VID=defaultSID=421384237289476S=1C=18643;, http://www.mysite.com/system/empty.asp?P=342VID=defaultSID=421384237289476S=1C=18634 , http://www.mysite.com/system/empty.asp?P=232VID=defaultSID=421384237289476S=1C=18645;, http://www.mysite.com/system/empty.asp?P=2345VID=defaultSID=421384237289476S=1C=18254;, http://www.mysite.com/system/empty.asp?P=257654VID=defaultSID=421384237289476S=1C=18732;, http://www.mysite.com/system/empty.asp?P=22VID=defaultSID=421384237289476S=1C=18637;, http://www.mysite.com/system/empty.asp?P=2463VID=defaultSID=421384237289476S=1C=18575 ) The result should be like 2 123 342 232 2345 257654 22 2463 Thanks, Lauri [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] 'From' and 'to' arguments in panel.abline
Dear R-users,
The help does not give much details on the use of the arguments 'from'
and 'to' in the panel.abline function. I have looked in the archives but
did not find how to implement them. My different tries failed miserably.
E.g, the following code doesn't seem to work, in a sense that the line
is not limited to the (0,10) range.
Do these arguments really apply to panel.abline? If so, how should they
be specified?
xy-data.frame(x-0.1:10,y-0.1:10)
xyplot(y~x,data=xy,
panel = function(x, y, ...){
panel.abline(a=0,b=1,from=0,to=10)
panel.xyplot(x,y)
})
Thanks you in advance for your help.
Sebastien
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[R] vertically oriented color key in heatmaps
Hi, I have some data which I was plotting using image(). I wanted to add a vertical color key to the plot and I found that heatmap.2 in gplots does let me add a color key. However, I was thinking of a vertical bar with the color range rather than the style that gplots provides. Is there any package (or code snippet) that would let me add a vertical color key to an image() or heatmap plot? Thanks, --- Rajarshi Guha [EMAIL PROTECTED] GPG Fingerprint: 0CCA 8EE2 2EEB 25E2 AB04 06F7 1BB9 E634 9B87 56EE --- Q: What do you get when you put a spinning flywheel in a casket and turn a corner? A: A funeral precession. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] GML with tweedie: AIC=NA
Dear Catarina, I prefer to leave the AIC value as NA for the tweedie GLM family because it takes extra time to compute and is only occasionally wanted. It's easy to compute the AIC yourself using the dtweedie() function of the tweedie package. Best wishes Gordon At 03:05 AM 14/08/2007, Catarina Miranda wrote: Dear Gordon; I have also sent this email to R help mailing list, so I apologize for duplicated mailing. I am modelling densities of some species of birds, and I have a problem with a great amount of zeros. I have decided to try GLMs with the tweedie family, but in all the models I have tried I got an NA for the AIC value. Just to check the problem I've compared the a glm using the Gaussian family with the identity link and a glm using the tweedie family with var.power=0 and link.power=1. These are equal, as expected, except the fact that the tweedie output gives me an NA for the AIC. Could you help me with this problem? Below you can find the two outputs I refer. Best Wishes; Catarina summary(glm(formula=ACIN~DIST_REF+DIST_H2O+DIST_OST+ COTA+H2O_SUP+vasa,family=gaussian(link=identity))) Call:glm(formula = ACIN ~ DIST_REF + DIST_H2O + DIST_OST + COTA + H2O_SUP + vasa, family = gaussian(link = identity)) Deviance Residuals: Min 1Q Median 3QMax -0.112792 -0.042860 -0.021113 -0.006311 1.551824 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) -6.625e-02 5.454e-02 -1.215 0.2256 DIST_REF 3.581e-06 1.336e-05 0.268 0.7889 DIST_H2O- 3.168e-05 1.527e-05 -2.074 0.0391 *DIST_OST-1.799e-05 1.953e-05 -0.921 0.3579 COTA 5.648e-04 2.470e-04 2.287 0.0230 *H2O_SUP -2.172e-04 3.994e-04 -0.544 0.5870 vasa 3.695e-02 4.573e-020.808 0.4199 ---Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for gaussian family taken to be 0.02151985) Null deviance: 5.6028 on 257 degrees of freedomResidual deviance: 5.4015 on 251 degrees of freedomAIC: -249.33 Number of Fisher Scoring iterations: 2 summary(glm(formula=ACIN~DIST_REF+DIST_H2O+DIST_OST+ COTA+H2O_SUP+vasa,control= glm.control(maxit=750),family=tweedie(var.power=0, link.power=1))) Call:glm(formula = ACIN ~ DIST_REF + DIST_H2O + DIST_OST + COTA + H2O_SUP + vasa, family = tweedie(var.power = 0, link.power = 1), control = glm.control (maxit = 750)) Deviance Residuals: Min 1Q Median 3QMax -0.112792 -0.042860 -0.021113 -0.006311 1.551824 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) - 6.625e-02 5.454e-02 -1.215 0.2256 DIST_REF 3.581e-06 1.336e-05 0.268 0.7889 DIST_H2O-3.168e-05 1.527e-05 -2.074 0.0391 *DIST_OST-1.799e-05 1.953e-05 -0.921 0.3579 COTA 5.648e-04 2.470e-042.287 0.0230 *H2O_SUP -2.172e-04 3.994e-04 -0.544 0.5870 vasa 3.695e-02 4.573e-02 0.808 0.4199 ---Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for Tweedie family taken to be 0.02151985) Null deviance: 5.6028 on 257 degrees of freedomResidual deviance: 5.4015 on 251 degrees of freedomAIC: NA Number of Fisher Scoring iterations: 2 __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] 'From' and 'to' arguments in panel.abline
On 8/13/07, Sébastien [EMAIL PROTECTED] wrote:
Dear R-users,
The help does not give much details on the use of the arguments 'from'
and 'to' in the panel.abline function.
Not surprising, since panel.abline doesn't actually have arguments
called 'from' and 'to'.
I have looked in the archives but
did not find how to implement them. My different tries failed miserably.
E.g, the following code doesn't seem to work, in a sense that the line
is not limited to the (0,10) range.
Do these arguments really apply to panel.abline?
No, they do not. It is common to have many functions documented in one
help page, and you need to look at the usage section (near the top) to
figure out which arguments are releant for which functions. The help
page has:
Usage:
panel.abline(a = NULL, b = 0,
h = NULL, v = NULL,
reg = NULL, coef = NULL,
col, col.line, lty, lwd, type,
...)
panel.curve(expr, from, to, n = 101,
curve.type = l,
col, lty, lwd, type,
...)
etc. If you want to limit the range, use panel.curve, e.g.
panel.curve(0 + 1 * x,from=0,to=10)
-Deepayan
If so, how should they
be specified?
xy-data.frame(x-0.1:10,y-0.1:10)
xyplot(y~x,data=xy,
panel = function(x, y, ...){
panel.abline(a=0,b=1,from=0,to=10)
panel.xyplot(x,y)
})
Thanks you in advance for your help.
Sebastien
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and provide commented, minimal, self-contained, reproducible code.
[R] R^2 for multilevel models
Hi there, In multiple regression one way to view R^2 is as (the square of) the correlation between original y's and the estimated y's. Suppose you fit a multilevel model with random intercept for each cluster. Would it be valid to compute an R^2 by using fixed effects plus the group intercepts to reduce the residuals? I suspect this has been done and, given its absence from the lmer output, refuted somewhere. A reference would be lovely. Code below. Many thanks, Andy # require(lme4) # First generate some data people = 100 obs = 4 # The random intercepts: sub_noise = data.frame(id = 1:people, rand_int = rnorm(people,0,60)) # Merge everything id = rep(1:people,obs) thedata = data.frame(id) thedata = merge(sub_noise, thedata) thedata$x = rep(1:obs,people) # Setup the relationship between x and y thedata$y = 23*thedata$x + 20 + thedata$rand_int + rnorm(people*obs, 0, 20) # plus residuals # Have a look plot(y~x, data = thedata) # Now fit a standard regression model lm1 = lm(y ~ x, data = thedata) summary(lm1) # Use the model to get the R^2 predicted = coef(lm1)[1] + thedata$x * coef(lm1)[2] plot(thedata$y, predicted) # It's a noisy mess cor(thedata$y, predicted)^2 # Now how about adjusting everyone's intercept towards the group # intercept? lmer1 = lmer(y ~ x + (1|id), data=thedata) summary(lmer1) # Get the random intercepts and stick them in a table ran_effects = data.frame(rownames(ranef(lmer1)$id), ranef(lmer1)$id[1]) names(ran_effects) = c(id, b) ran_effects_data = merge(thedata, ran_effects) # Now compute predicted.ml = fixef(lmer1)[1] + ran_effects_data$x * fixef(lmer1)[2] + ran_effects_data$b plot(thedata$y, predicted.ml) cor(thedata$y, predicted.ml)^2 # Looks much nicer __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] invert 160000x160000 matrix
Can R invert a 16x16 matrix with all positive numbers? Thanks a lot! __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] BDS test - results unclear to me
Hello, I would like to use the BDS test from the tseries package, but there is something I don't understand in the results of the test. Let's say, I want the BDS values for an embedding dimension equal to 2 : bds.test(c, m = 2, eps = seq(0.5 * sd(c), 2 * sd(c), length = 4),trace=FALSE); Here are the outputs: data: c Embedding dimension = 2 Epsilon for close points = 0.0097 0.0194 0.0291 0.0388 Standard Normal = [ 0.0097 ] [ 0.0194 ] [ 0.0291 ] [ 0.0388 ] [ 2 ]14.600613.900312.574511.4012 Now for dimension 3 (m=3), we obtain: Standard Normal = [ 0.0097 ] [ 0.0194 ] [ 0.0291 ] [ 0.0388 ] [ 2 ]14.554413.875812.554011.3897 [ 3 ]20.914918.764016.156214.2518 what I don't understand is why the values for embedding dimension 2 are not equal when BDS is computed with parameter m=2 and m=3, could someone please explain that to me ? In the documentation, it is said that m is an integer indicating that the BDS test statistic is computed for embedding dimensions 2, ..., m., so why don't we get the same result in both cases? Thank you for your help, Best regards, Nicolas __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] 'From' and 'to' arguments in panel.abline
Thanks Deepayan,
I actually looked at the upper section but couldn't not spot these
arguments. I thought they were optional arguments in '...'. I will try
to open my eyes next time ;)
Deepayan Sarkar a écrit :
On 8/13/07, Sébastien [EMAIL PROTECTED] wrote:
Dear R-users,
The help does not give much details on the use of the arguments 'from'
and 'to' in the panel.abline function.
Not surprising, since panel.abline doesn't actually have arguments
called 'from' and 'to'.
I have looked in the archives but
did not find how to implement them. My different tries failed miserably.
E.g, the following code doesn't seem to work, in a sense that the line
is not limited to the (0,10) range.
Do these arguments really apply to panel.abline?
No, they do not. It is common to have many functions documented in one
help page, and you need to look at the usage section (near the top) to
figure out which arguments are releant for which functions. The help
page has:
Usage:
panel.abline(a = NULL, b = 0,
h = NULL, v = NULL,
reg = NULL, coef = NULL,
col, col.line, lty, lwd, type,
...)
panel.curve(expr, from, to, n = 101,
curve.type = l,
col, lty, lwd, type,
...)
etc. If you want to limit the range, use panel.curve, e.g.
panel.curve(0 + 1 * x,from=0,to=10)
-Deepayan
If so, how should they
be specified?
xy-data.frame(x-0.1:10,y-0.1:10)
xyplot(y~x,data=xy,
panel = function(x, y, ...){
panel.abline(a=0,b=1,from=0,to=10)
panel.xyplot(x,y)
})
Thanks you in advance for your help.
Sebastien
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Re: [R] invert 160000x160000 matrix
You would need 200GB to store a since image, so if you have about 1TB of physical memory on your computer, it might be possible. On 8/13/07, Jiao Yang [EMAIL PROTECTED] wrote: Can R invert a 16x16 matrix with all positive numbers? Thanks a lot! __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Q: how to extract coefisients from one glm and implement them in to an other glm?
Part of the homework you were supposed to do before asking this question: help.search(coefficients) . . . coef(stats) Extract Model Coefficients . . . So, go read ?coef and PLEASE do read the posting guide http://www.R-project.org/posting-guide.html As your email says. Then go read ?predict which is probably more germane to your goal. On Mon, 13 Aug 2007, Tom Willems wrote: Dear R ussers, I'm still working with the categorical models, and i can not do cross validation on them because odf their nature. not cv works perfectly on other models so i wonderd if it is possible to extract the beta's from a categorical model, then use their value in a 0/1 glm. Their formula is the same, so their assumptions should still hold. Is this possible in R. kind regards, Tom W. Disclaimer: click here [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] invert 160000x160000 matrix
I don't think you can define a matrix this large in R, even if you have
the memory. Then, of course, inverting it there may be other programs
that have limitations.
Paul
Jiao Yang wrote:
Can R invert a 16x16 matrix with all positive numbers? Thanks a lot!
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La version française suit le texte anglais.
This email may contain privileged and/or confidential inform...{{dropped}}
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[R] simulate data from multivariate normal with pre-specified correlation matrix
For example, the correlation matrix is 3x3 and looks like
1 0.75 0 0 0
0.751 0 0 0
0 0 0 0 0
Can I write the code like this?
p- 3 # number of variables per observation
N- 10 # number of samples
# define population correlation matrix sigma
sigma-matrix(0,p,p) #creates a px p matrix of 0
rank-2
for (i in 1:rank){
for (j in 1:rank){
rho-0.75
sigma[i,j]-rho^abs(i-j)
sigma[i,i]-1
}
}
# create a data set of 10 observations from multivariate normalcovariance
matrix sigma
library(MASS)
Xprime-mvrnorm(N,rep(1,p), sigma )
Also, if I calculate
S- cov(Xprime)
will S be the sample correlation matrix?
Thank you very much for your time!
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[R] Error message when using zero-inflated count regression model in package zicounts
I have data on number of vines per tree for ~550 trees. Over half of the trees did not have any vines and the data is fairly skewed (median = 0, mean = 1.158, 3rd qu. = 1.000). I am attempting to investigate whether plot location (four sites), species (I'm using only the four most common species), or tree dbh has a significant influence on the number of vines per tree. When I attempted to use the zicounts function, R gave me the following error message: vines.zip-zicounts(resp=Total.vines~.,x=~Site+Species+DBH,z=~Site +Species+DBH,distrname=ZIP,data=sycamores.1) Error in ifelse(y == 0, 1, y/mu) : dim- : dims [product 12] do not match the length of object [549] In addition: Warning messages: 1: longer object length is not a multiple of shorter object length in: x[good, ] * w 2: longer object length is not a multiple of shorter object length in: eta + offset 3: longer object length is not a multiple of shorter object length in: y/mu I do not know enough about the calculations done in the function to interpret the error messages. Is there a glitch in my data and if yes, what is it? Thanks for your help. Jim Milks Graduate Student Environmental Sciences Ph.D. Program 136 Biological Sciences Wright State University 3640 Colonel Glenn Hwy Dayton, OH 45435 [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with scatterplot3d
Ryan Briscoe Runquist rdbriscoe at ucdavis.edu writes: Hello, I am having a bit of trouble with scatterplot3d(). I was able to plot a 3d cloud of points using the following code: my.3dplot-scatterplot3d(my.coords, pch=19, zlim=c(0,1), scale.y=0.5, angle=30, box=FALSE) where my.coords is a data frame that contains x, y, and z coordinates for grid points whose elevation we sampled. The problem occurs when I try to add points using points3d. I tried to follow the code in the examples of the package pdf. First, I tried the following code to add all of the points that I wanted: my.3dplot$points3d(seq(400,600,0.19), seq(600,400,0.295), seq(800,500,0.24), seq(1000,1400,0.22), seq(1200,600,0.24), seq(1200,1500,0.28), seq(1300,1400,0.205), seq(1700,500,0.26), seq(1700,600,0.21), seq(1900,1400,0.255), seq(2300,1400,0.275), seq(2600,1300,0.225), seq(2700,400,0.235), seq(2700,1300,0.265), seq(3100,1000,0.135), col=blue, type=h, pch=16) I think you probably want: xvals - c(400,600,800,1000,1200,1200,1300,1700,1700,1900, 2300,2600,2700,2700,3100) yvals - c(600,400,500,1400,600,1500,1400,500,600, 1400,1400,1300,400,1300,1000) zvals - c(0.19,0.295,0.24,0.22,0.24,0.28,0.205,0.26, 0.21,0.255,0.275,0.225,0.235,0.265,0.135) my.3dplot$points3d(x=xvals,y=yvals,z=zvals,col=blue,type=h,pch=16) Look carefully at ?seq and you may understand what you've done wrong ... Ben Bolker __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Reading polygon shapefiles in splancs
Dear List, I'm trying to import a multiple polygon shapefile in splancs. I've found the following command but it seems to work only for single polygons: Name_of_splancs_polygon-getPolygonCoordsSlot (getPolygonsPolygonsSlot(getSpPpolygonsSlot(name_of_sp_polygon) [[1]]) [[1]]) I've tried to change the numbers but it seams to choose polygons instead of importing them altogether. Can somebody help me? Many Thanks, Enrico - - [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Legend on graph
Hi Akki, Then you may need to increase y-axis scale by ylim=c(min,max) Cheers Nguyen On 8/12/07, akki [EMAIL PROTECTED] wrote: Hi, I have a problem when I want to put a legend on the graph. I do: legend(topright, names(o), cex=0.9, col=plot_colors,lty=1:5, bty=n) but the legend is writen into the graph (graphs' top but into the graph), because I have values on this position. How can I write the legend on top the graph without the legend writes on graph's values. Thanks. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] predict kernelmatrix
HI
I got a problem in the predict function of the kernlab.
I want to use ksvm and predict with kernelmatrix (S4 method for signature
'kernelMatrix')
#executing the following sentences
library(kernlab)
# identity kernel
k - function(x,y) {
n-length(x)
cont-0
for(i in 1:n){
if(x[i]==y[i]){
cont-cont+1
}
}
cont
}
class(k) - kernel
data(promotergene)
ind - sample(1:dim(promotergene)[1],20)
genetrain - promotergene[-ind, -1]
genetest - promotergene[ind,-1 ]
kx - kernelMatrix(k, as.matrix(genetrain))
#y-as.vector(promotergene[-ind,1 ])
y-as.factor(promotergene[-ind,1 ])
y
gene1 - ksvm(kx, y, type=C-svc)
gene1
genetype - predict(gene1,genetest)
Error en as.matrix(Z) : objeto Z no encontrado
#genetest1-as.matrix(genetest)
#genetype - predict(gene1,genetest1)
genetype
thank you,
nelsonhernandez
[EMAIL PROTECTED]
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Re: [R] deldir package - voronoi
Team, we figured this out, Roger at XLSOLUTIONS solved the puzzle for us..!! Okay you can easily change the color palette. see below: library(MASS) library(deldir) set.seed(1123) y - runif(20) x - runif(20) #run voronoi from deldir package ddd - deldir(x,y) ttt - tile.list(ddd) plot.deldir(ddd,wlines=tess) #play with colors *ccc - colors()[1:20] or ccc -heat.colors(20) or ccc -terrain.colors(20)* plot(ttt,polycol=ccc,close=T) Rolf Turner wrote: On 12/08/2007, at 1:22 PM, zubin wrote: Hello! I am using the deldir package to visualize my data, pretty neat. However, i need to fill the colors using polycol =, fill with colors like a heatmap - more of a gradient fill. The only colors i get are very blocky - how do i assign the correct colors for a gradient, even a grayscale? i tried the chart of R colors, using 200 numbers for grayscale but not getting them. The polycol = colors the cells in the tesselation with the value a specific vector for color. snip I'm sorry, but I can't help here. I've been struggling with colors, in a different context, recently myself, and I'm unclear as to how they work. There are a bunch of functions --- palette(), colorRamp(), colorRampPalette() that probably relate to what you want to do, but I'm not sure just *how* they relate. With a bit of luck, someone cleverer than I will come to your rescue. cheers, Rolf Turner ## Attention:This e-mail message is privileged and confidential. If you are not theintended recipient please delete the message and notify the sender.Any views or opinions presented are solely those of the author. This e-mail has been scanned and cleared by MailMarshalwww.marshalsoftware.com ## __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Artifacts in pdf() of image() (w/o comments)
Hi Duncan, Hi Michael, Duncan Murdoch murdoch at stats.uwo.ca writes: On 8/13/2007 1:33 PM, Michael Kubovy wrote: - You saw ugly artifacts in a graph you produced. That's a bug. The question is, what caused it? - You rightly asked on R-help. It might have been a bug in the way you were doing things, or a bug in R, or a bug in the way you were viewing the image. - It was demonstrated that this only appears in Preview, not in Acrobat Reader, so it's probably a bug there. It's only at the last point that I object to what you did: your question quoted below makes it look as though you think Apple's time is more valuable than ours. I remind you that you paid for Preview, and you didn't pay for R. Apple has a stronger obligation to help you than any of us do. My experience with them (and with most other commercial software vendors) is that you'll get much worse help from them than from us --- but that doesn't mean you shouldn't try. Perhaps when anyone searching for Mac OSX Preview on Google shows up a page full of unhandled bug reports they'll actually do something. Duncan Murdoch On Aug 13, 2007, at 9:56 AM, Duncan Murdoch wrote: On 8/13/2007 11:43 AM, Michael Kubovy wrote: But is it a bug? Can a program anti-alias text and line drawings and not bitmaps? ... [deletions] ... An alternative to dropping Preview is to report the bug in it to Apple. Apple has an online bug reporting web page somewhere; I haven't found them as helpful as R-help, but your mileage may vary. Duncan Murdoch I'm entering this late. I just wanted to point out that there are earlier threads on this issue on the R-Sig-Mac list that might be useful to follow through. https://stat.ethz.ch/pipermail/r-sig-mac/2005-March/001650.html https://stat.ethz.ch/pipermail/r-sig-mac/2006-February/002678.html Simon Urbanek suggested one patch on the later thread. My solution, at the time, was to capture the image from the acrobat rendition. best, ken -- Ken Knoblauch Inserm U846 Institut Cellule Souche et Cerveau Département Neurosciences Intégratives 18 avenue du Doyen Lépine 69500 Bron France tel: +33 (0)4 72 91 34 77 fax: +33 (0)4 72 91 34 61 portable: +33 (0)6 84 10 64 10 http://www.lyon.inserm.fr/846/english.html __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Legend on graph
You can get the legend outside the plot region by 1. First changing the clipping region via par(xpd = TRUE) ; (or xpd=NA). see ?par 2. Specifying x and y coodinates for legend placement outside the limits of the plot region. This allows you to include a legend without adding a bunch of useless whitespace to the plot region; or to add a grid to the plot without interfering with the legend. Bert Gunter Genentech Nonclinical Statistics -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Nguyen Dinh Nguyen Sent: Monday, August 13, 2007 3:42 PM To: [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Subject: [R] Legend on graph Hi Akki, Then you may need to increase y-axis scale by ylim=c(min,max) Cheers Nguyen On 8/12/07, akki [EMAIL PROTECTED] wrote: Hi, I have a problem when I want to put a legend on the graph. I do: legend(topright, names(o), cex=0.9, col=plot_colors,lty=1:5, bty=n) but the legend is writen into the graph (graphs' top but into the graph), because I have values on this position. How can I write the legend on top the graph without the legend writes on graph's values. Thanks. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] binomial simulation
As I understand this,
P(T+ | D-)=1-P(T+ | D+)=0.05
is the probability not to detect desease for a person
at ICU who has the desease. Correct?
What I asked was whether it is possible to mistakenly
detect the desease for a person who does not have it?
Assuming that this is impossible the formula is below:
If there are N patients, each has a probability p to
have the desease (p=0.6 in your case) and q is the
probability to detect the desease for a person who has
it (q = 0.95 for ICU and q = 0.8 for a regular unit),
then
P(k have the desease AND m are detected) =
P(k have the desease)*P(m are detected / k have the
desease) =
C(N,k)*p^k*(1-p)^(N-k)*C(k,m)*q^m*(1-q)^(k-m)
where C(a,b) is the Binomial coefficient a above b -
the number of ways to choose b items out of a (when
the order does not matter). You of course must assume
that N = k = m = 0 (otherwise the probability is
0).
To generate such pairs (k infected and m detected) you
can do the following:
k - rbinom(N,1,p)
m - rbinom(k,1,q)
Regards,
Moshe.
--- sigalit mangut-leiba [EMAIL PROTECTED] wrote:
Hi,
The probability of false detection is: P(T+ |
D-)=1-P(T+ | D+)=0.05.
and I want to find the joint probability
P(T+,D+)=P(T+|D+)*P(D+)
Thank you for your reply,
Sigalit.
On 8/13/07, Moshe Olshansky [EMAIL PROTECTED]
wrote:
Hi Sigalit,
Do you want to find the probability P(T+ = t AND
D+ =
d) for all the combinations of t and d (for ICU
and
Reg.)?
Is the probability of false detection (when there
is
no disease) always 0?
Regards,
Moshe.
--- sigalit mangut-leiba [EMAIL PROTECTED]
wrote:
hello,
I asked about this simulation a few days ago,
but
still i can't get what i
need.
I have 2 units: icu and regular. from icu I want
to
take 200 observations
from binomial distribution, when probability for
disease is: p=0.6.
from regular I want to take 300 observation with
the
same probability: p=0.6
.
the distribution to detect disease when disease
occurred- *for someone from
icu* - is: p(T+ | D+)=0.95.
the distribution to detect disease when disease
occurred- *for someone from
reg.unit* - is: p(T+ | D+)=0.8.
I want to compute the joint distribution for
each
unit: p(T+,D+) for icu,
and the same for reg.
I tried:
pdeti - 0
pdetr - 0
picu - pdeti*.6
preg - pdetr*.6
dept - c(icu,reg)
icu - rbinom(200, 1, .6)
reg - rbinom(300, 1, .6)
for(i in 1:300) {
if(dept==icu) pdeti==0.95
if (dept==reg) pdetr==0.80
}
print(picu)
print(preg)
and got 50 warnings:
the condition has length 1 and only the first
element will be used in: if
(dept == icu) pdeti == 0.95
the condition has length 1 and only the first
element will be used in: if
(dept == reg) pdetr == 0.8
I would appreciate any suggestions,
thank you,
Sigalit.
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Re: [R] invert 160000x160000 matrix
While inverting the matrix may be a problem, if you
need to solve an equation A*x = b you do not need to
invert A, there exist iterative methods which do need
A or inv(A) - all you need to provide is a function
that computes A*x for an arbitrary vector x.
For such a large matrix this may be slow but possible.
--- Paul Gilbert [EMAIL PROTECTED]
wrote:
I don't think you can define a matrix this large in
R, even if you have
the memory. Then, of course, inverting it there may
be other programs
that have limitations.
Paul
Jiao Yang wrote:
Can R invert a 16x16 matrix with all
positive numbers? Thanks a lot!
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[R] State Space Modelling
Hey all, I am trying to work under a State Space form, but I didn't get the help exactly. Have anyone eles used this functions? I was used to work with S-PLUS, but I have some codes I need to adpt. Thanks alot, Bernardo [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Fatal Error - R in Windows XP
I would greatly appreciate if someone can tell me how to solve the issue I have been facing. After creating a large sparse matrix (6300 by 8300 matrix), I saved image of it in my work space to do further computation. Since then whenever I start R (version 2.5.1 on Windows XP, SP2), I seem to get an error: Fatal error: unable to restore saved data in .RData and then R crashes. I reinstalled R, and same thing happens. Thanks in advance for suggestions. Sid Dalal - Pinpoint customers who are looking for what you sell. [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] vertically oriented color key in heatmaps
colorbar() from the package matlab. nightly.py wrote: Hi, I have some data which I was plotting using image(). I wanted to add a vertical color key to the plot and I found that heatmap.2 in gplots does let me add a color key. However, I was thinking of a vertical bar with the color range rather than the style that gplots provides. Is there any package (or code snippet) that would let me add a vertical color key to an image() or heatmap plot? -- View this message in context: http://www.nabble.com/vertically-oriented-color-key-in-heatmaps-tf4263271.html#a12138673 Sent from the R help mailing list archive at Nabble.com. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] State Space Modelling
try help.search(state space) you will get pointer where you want to go
cm.state(CreditMetrics) Computation of state space
dlmLL(dlm)Log likelihood evaluation for a
state space model
SS(dse1) State Space Models
balanceMittnik(dse1) Balance a state space model
estSSMittnik(dse1)Estimate a State Space Model
estSSfromVARX(dse1) Estimate a state space TSmodel
using VAR estimation
l.SS(dse1)Evaluate a state space TSmodel
nstates(dse1) State Dimension of a State Space
Model
toSS(dse1)Convert to State Space Model
toSSChol(dse1)Convert to Non-Innovation State
Space Model
toSSinnov(dse1) Convert to State Space
Innovations Model
generateSSmodel(dse2) Randomly generate a state space
model
ets(forecast) Exponential smoothing state
space model
SS(sspir) Representation of Gaussian State
Space Model
kfilter(sspir)Kalman filter for Gaussian state
space model
recursion(sspir) Simulate from a Gaussian state
space model
smoother(sspir) Kalman smoother for Gaussian
state space model
ssm(sspir)Define state-space model in a
glm-style call.
sspir(sspir) State Space Models in R
and
RSiteSearch(state space)
http://search.r-project.org/cgi-bin/namazu.cgi?query=state+spacemax=20result=normalsort=scoreidxname=Rhelp02aidxname=functionsidxname=docs
post your code of S Plus, i feel there are many who can help you in this
regards
HTH
Bernardo Ribeiro [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
08/14/2007 06:57 AM
To
r-help@stat.math.ethz.ch
cc
Subject
[R] State Space Modelling
Hey all,
I am trying to work under a State Space form, but I didn't get the help
exactly.
Have anyone eles used this functions?
I was used to work with S-PLUS, but I have some codes I need to adpt.
Thanks alot,
Bernardo
[[alternative HTML version deleted]]
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DISCLAIMER AND CONFIDENTIALITY CAUTION:\ \ This message and ...{{dropped}}
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[R] diffusing GIS data in maps
Hi- I am trying to find a way to diffuse GIS data on a European map. I have a dataset consisting of particular locations scattered across Europe, along with magnitude and value information. I can plot these as discrete points with something like the following: geocode is a dataframe with four columns: LAT; LONG; MAGNITUDE;VALUE. library(maps) library(mapdata) map(worldHires, regions=c(Germany, Belgium, Netherlands)) points(geocode$LONG, geocode$LAT, cex=geocode$MAGNITUDE / 2500, col=rainbow(length(geocode$VALUE), start=0, end=.4)[rank(geocode$VALUE)]) This gives me a map of Europe with my datapoints highlighted in two ways: magnitude is represented by the size of the point, and value is represented by the color. However, what I would really like is for there to be some sort of diffusion, such that instead of discrete points, the European map is covered in color so I can see more clearly whether there are regional patterns (something that will presumably look like this contour chart: http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=20 only on the European map). I have absolutely no idea where to start because I can't find a function that will allow me to diffuse the datapoints on a map. thank you for any help ldb __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
