Re: [R] cor() function, method=spearman
[EMAIL PROTECTED] writes: Hello, Does anyone know if the cor function, when method = spearman, returns a correlation coefficient corrected for any ties in the ranks of the data? I have data with quite a few ties and am thinking that I should use a calculation of the coefficient corrected for ties, but before I try and code this calculation myself, I thought I should check whether or not cor() automatically does this. Thank you! Joanna McGarvie The tie-corrected spearman correlation is the correlation of the tie-corrected ranks, so yes, it does do that. Previous versions of R had an issue with p-values in cor.test, where it used the exact distribution of the test for non-tied data, even when ties were present. This was straightened out in 2.2.0, where it now uses an asymptotic formula instead. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ 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
Re: [R] Converting PROC NLMIXED code to NLME
Singh, Jatinder [EMAIL PROTECTED] writes:
Hi,
I am trying to convert the following NLMIXED code to NLME, but am
running into problems concerning 'Singularity in backsolve'. As I am new
to R/S-Plus, I thought I may be missing something in the NLME code.
Which version of R and NLME? R 2.2.0 ships with a version where the
internal optimizer is changed to nlminb(). As I understand it, this
was in response to reports where code that worked in S-PLUS refused to
work in R.
NLMIXED
***
proc nlmixed data=kidney.kidney;
parms delta=0.03 gamma=1.1 b1=-0.003 b2=-1.2 b3=0.09 b4=0.35 b5=-1.43
varu=0.5;
eta=b1*age+b2*sex+b3*gn+b4*an+b5*pkn+u;
hazard=eta+log(delta)+log(gamma)+(gamma-1)*log(rtime);
survivor=(-exp(eta))*delta*(rtime**gamma);
ll=(event*hazard)+survivor;
model rtime ~ general(ll);
random u~normal(0,varu) subject=patient out=frailty;
run;
NLME
kidney.nlme-nlme(model=rtime~
(event*
((b1*age+b2*sex+b3*gn+b4*an+b5*pkn+u)+log(delta)+log(gamma)+(gamma-1)*lo
g(rtime))
+((-exp(b1*age+b2*sex+b3*gn+b4*an+b5*pkn+u))*delta*(rtime**gamma))
),
fixed=list(delta~1,gamma~1,b1~1,b2~1,b3~1,b4~1,b5~1),
random=u~1|patient,
start=c(delta=0.03,gamma=1.1,b1=-0.003,b2=-1.2,b3=0.09,b4=0.35,b5=-1.43)
,
data=(kidney),
method=ML,
na.action=na.include
)
Error in NLME
*
traceback()
11: eval(action, sys.parent())
10: doErrorAction(Problem in .C(\fit_nlme\,: Singularity in
backsolve, 1000)
9: .C(fit_nlme,
8: nlme(model = rtime ~ (event * ((b1 * age + b2 * sex + b3 * gn + b4 *
an + b5 *
7: NULL
6: nlme.formula(model = rtime ~ (event * ((b1 * age + b2 * sex + b3 * gn
+ b4 *
5: eval(i, local)
4: source(auto.print = auto.print, exprs = substitute(exprs.literal))
3: script.run(exprs.literal = {
2: eval(expression(script.run(exprs.literal = {
1:
Message: Problem in .C(fit_nlme,: Singularity in backsolve
I am actually trying to fit a parametric model to the kidney catheter
data and compare NLMIXED with NLME. I am aware that COXPH and SURVREG
are also available with a frailty element added in, but wanted to fit
the likelihood model as above for a direct comparison.
Cheers,
Jindi
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--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [R] missing values in step procedure
At 11:11 7/10/2005, you wrote: Hi, I have the problem that for the step procedure stops due to missing values. There are no options in Step or stepAIC to handle missing values. Is there any way to run stepwise modelselection in R in an automated way in this case? Here is the last step before it stops. Hope someone knows. Best regards, Andreas Step: AIC= 1999.16 EF ~ SF120_KS + SF120_PS + HADA0 + SOZU0 + LVEDD + logPROBNP + ALTER + SD0_01 + ASE_UK + DS140POS + RSQSICH0 + SD0_01:ASE_UK + SD0_01:DS140POS + SD0_01:RSQSICH0 + ASE_UK:DS140POS + ASE_UK:RSQSICH0 + DS140POS:RSQSICH0 + SD0_01:ASE_UK:RSQSICH0 + SD0_01:DS140POS:RSQSICH0 + ASE_UK:DS140POS:RSQSICH0 Df Sum of Sq RSS AIC - SOZU0 1 3.0 25356.0 1997.2 - HADA0 1 7.6 25360.6 1997.3 - ALTER 1 13.0 25365.9 1997.4 - SF120_PS 1 14.7 25367.6 1997.5 - ASE_UK:DS140POS:RSQSICH0 1 20.1 25373.1 1997.6 - SD0_01:DS140POS:RSQSICH0 1 44.8 25397.7 1998.0 - SD0_01:ASE_UK:RSQSICH01 54.4 25407.4 1998.2 none 25352.9 1999.2 - LVEDD 1 382.2 25735.1 2004.6 - SF120_KS 1 476.4 25829.3 2006.4 - logPROBNP 1 891.9 26244.9 2014.4 Error in step(mod2, direction = back) : number of rows in use has changed: remove missing values? Andreas Try data-na.omit(original database) before you run step() or stepAIC() Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil -- __ 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
Re: [R] Converting PROC NLMIXED code to NLME
At 20:40 7/10/2005, Singh, Jatinder wrote: NLME kidney.nlme-nlme(model=rtime~ (event* ((b1*age+b2*sex+b3*gn+b4*an+b5*pkn+u)+log(delta)+log(gamma)+(gamma-1)*lo g(rtime)) +((-exp(b1*age+b2*sex+b3*gn+b4*an+b5*pkn+u))*delta*(rtime**gamma)) ), fixed=list(delta~1,gamma~1,b1~1,b2~1,b3~1,b4~1,b5~1), random=u~1|patient, start=c(delta=0.03,gamma=1.1,b1=-0.003,b2=-1.2,b3=0.09,b4=0.35,b5=-1.43) , data=(kidney), method=ML, na.action=na.include ) Hi! Try change na.action=na.include for na.action=na.omit The singularity occurs because the fixed effects matrix is not of full rank due to the unused factor levels. Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil -- __ 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
[R] color for points
Hi,
I have the following code to
randomly generate the
points:
csr -function(n=60){
x=runif(n)
y=runif(n)
f=cbind(x,y)
}
plot(csr())
I wonder how to code to make the first twenty points to be BLUE; second twenty
points to be RED; the last twenty points to be GREEN?
Thanks,
Sam
-
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Re: [R] Assign references
[EMAIL PROTECTED] wrote:
Patrick,
I'll have to check your S poetry, it's not clear why 'changing things
invisibly' is a bad thing in R, but is OK in C. Perhaps the answer lies
therein.
I think Tony's reply has the main part of the answer. R is
designed as a functional language -- meaning, essentially,
that side effects don't happen except for assignment. The
- operator (and 'assign') are compromises from that
ideal.
In C it is standard practice to pass by reference and that is
a great deal of its power. In R it is standard practice not to
have to worry about objects being changed behind your
back -- that freedom is a part of the power of R.
On the surface this looks contradictory -- one of the two
approaches must be the better way. But the two languages
have different uses. C computes extremely efficiently in terms
of having operations that closely match the machine. R computes
efficiently by having operations that closely match how people
think.
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and A Guide for the Unwilling S User)
I'd considered returning a sequence, but it seems more straight forward
to be able to pass the arguments by reference. The difficulty in doing
this in R points to a philosophical point which thus far has eluded me.
Certainly more thinking is in order.
Thanks for your help on this,
cur
--
Curt Seeliger, Data Ranger
CSC, EPA/WED contractor
541/754-4638
[EMAIL PROTECTED]
Patrick Burns [EMAIL PROTECTED] wrote on 10/07/2005 01:23:07
PM:
Because the function is using 'foo' and 'bar' as the global
variables, not 'x' and 'y'. What you might have missed from
Thomas's statement (if I can take some liberties) is that you
almost surely don't want to do that -- it is bad style because
it changes things invisibly. S Poetry has more on that.
Better is:
myFunk - function(a, b) {
x - a + b
y - a * b
list(x=x, y=y)
}
xy - myFunk(4, 5)
x - xy$x
y - xy$y
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and A Guide for the Unwilling S User)
[EMAIL PROTECTED] wrote:
Folks,
I've run into trouble while writing functions that I hope will create
and modify a dataframe or two. To that end I've written a toy
function
that simply sets a couple of variables (well, tries but fails).
Searching the archives, Thomas Lumley recently explained the -
operator, showing that it was necessary for x and y to exist prior to
the function call, but I haven't the faintest why this isn't working:
myFunk-function(a,b,foo,bar) {foo-a+b; bar-a*b;}
x-0; y-0;
myFunk(4,5,x,y)
x-0; y-0;
myFunk(4,5,x,y)
x
[1] 0
y
[1] 0
What (no doubt simple) reason is there for x and y not changing?
Thank you,
cur
--
Curt Seeliger, Data Ranger
CSC, EPA/WED contractor
541/754-4638
[EMAIL PROTECTED]
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[R] square and points on the same figure
Hi,
I have the following code to
randomly generate the
points:
csr -function(n=60){
x=runif(n)
y=runif(n)
f=cbind(x,y)
}
plot(csr())
I wonder how to code to draw a square centering on 0.5,0.5; with the length of
each side to be 0.3 on the same figure with these plotted points?
Thanks,
Sam
-
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Re: [R] color for points
Sam R. Smith a écrit :
Hi,
I have the following code to randomly generate the points:
csr -function(n=60){
x=runif(n)
y=runif(n)
f=cbind(x,y)
}
plot(csr())
I wonder how to code to make the first twenty points to be BLUE; second
twenty points to be RED; the last twenty points to be GREEN?
mynewfct = function(n=60)
{
x=runif(n)
y=runif(n)
f=cbind(x,y)
plot(f[1:20] , col='blue');
par(new=T);
plot(f[21:40] , col='red');
par(new=T);
plot(f[41:60] , col='green');
}
hih
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Re: [R] R version 2.01.1, Crimson Editor and the one from nowhere
Hello Olivier, The problem prob ably comes from TpR.exe. It first sends Alt-w 1 before the command, to make sure the console is the active window. You should try to switch R in SDI mode, which is definitely the best mode for running R side-by-side with a text editor. Tell me if it solves the problem. Best, Philippe Grosjean ..°})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Pentagone (3D08) ( ( ( ( (Academie Universitaire Wallonie-Bruxelles ) ) ) ) ) 8, av du Champ de Mars, 7000 Mons, Belgium ( ( ( ( ( ) ) ) ) ) phone: + 32.65.37.34.97, fax: + 32.65.37.30.54 ( ( ( ( (email: [EMAIL PROTECTED] ) ) ) ) ) ( ( ( ( (web: http://www.umh.ac.be/~econum ) ) ) ) ) http://www.sciviews.org ( ( ( ( ( .. Olivier ETERRADOSSI wrote: Dear List sorry to bother you R-gurus with such an unstatistical question... but I face a problem using Crimson Editor with R 2.01.1 that I never had using R 2.00.1. I already posted on the Crimson Editor forum but it seems to be VERY few R-users there I successfully used R v2.00.1until now (under Windows XP professionnal, version 2002, Service Pack 2, P4 processor CPU 1.8 GHz), together with Crimson Editor. This editor is linked to R using three files (TpR.exe, R.SPC and R.KEYS). I recently upgraded to R 2.01.1. I kept using my old TpR.exe, R.SPC and R.KEYS, because I did not find any new files on the Crimson Editor Release web page. When I now launch a script, instead of getting my old, well known prompt : source(C:/Program Files/R/fooscript.txt) I get : 1source(C:/Program Files/R/fooscript.txt) with a 1 in front of the line and of course R greets me with a syntax error message. Then I have to remove the 1 by hand (pretty prehistoric, ...and does not work if my script is meant to launch other scripts during the night) I cannot figure where this 1 comes from !! Did some of you already encountered this problem, and how did you get rid of it ? Thanks a lot, have a nice week-end. Olivier __ 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
Re: [R] square and points on the same figure
?rect (and don't forget par(new=T)) hih __ 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
Re: [R] R version 2.01.1, Crimson Editor and the one from nowhere
... OK, I have spot the problem: TpR.exe expects RGui running in English. Shortcut for the Windows menu is Alt-w, which is what it sends to R. Then, it sends 1, meaning, activate first window (the console). You have probably RGui running in French, or in another language. In French the menu is called Fenêtres, with the corresponding shortcut being Alt-n. Consequently, the menu is not triggered and the 1 is send to the command line. Two solutions to continue using TpR.exe with R 2.1.1 or more: 1) Switch R in SDI mode, 2) Use RGui in MDI mode, but in English. The third solution is to patch TpR.exe, which I will not do, because I need to program a separate command for each different language of R! Best, Philippe Grosjean ..°})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Pentagone (3D08) ( ( ( ( (Academie Universitaire Wallonie-Bruxelles ) ) ) ) ) 8, av du Champ de Mars, 7000 Mons, Belgium ( ( ( ( ( ) ) ) ) ) phone: + 32.65.37.34.97, fax: + 32.65.37.30.54 ( ( ( ( (email: [EMAIL PROTECTED] ) ) ) ) ) ( ( ( ( (web: http://www.umh.ac.be/~econum ) ) ) ) ) http://www.sciviews.org ( ( ( ( ( .. Olivier ETERRADOSSI wrote: Dear List sorry to bother you R-gurus with such an unstatistical question... but I face a problem using Crimson Editor with R 2.01.1 that I never had using R 2.00.1. I already posted on the Crimson Editor forum but it seems to be VERY few R-users there I successfully used R v2.00.1until now (under Windows XP professionnal, version 2002, Service Pack 2, P4 processor CPU 1.8 GHz), together with Crimson Editor. This editor is linked to R using three files (TpR.exe, R.SPC and R.KEYS). I recently upgraded to R 2.01.1. I kept using my old TpR.exe, R.SPC and R.KEYS, because I did not find any new files on the Crimson Editor Release web page. When I now launch a script, instead of getting my old, well known prompt : source(C:/Program Files/R/fooscript.txt) I get : 1source(C:/Program Files/R/fooscript.txt) with a 1 in front of the line and of course R greets me with a syntax error message. Then I have to remove the 1 by hand (pretty prehistoric, ...and does not work if my script is meant to launch other scripts during the night) I cannot figure where this 1 comes from !! Did some of you already encountered this problem, and how did you get rid of it ? Thanks a lot, have a nice week-end. Olivier __ 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
[R] keeping interaction terms
Hello,?xml:namespace prefix = o ns = urn:schemas-microsoft-com:office:office /o:p/o:p while doing my thesis in habitat modelling I´ve come across a problem with interaction terms. My question concerns the usage of interaction terms for linear regression modelling with R. If an interaction-term (predictor) is chosen for a multiple model, then, according to ?xml:namespace prefix = st1 ns = urn:schemas-microsoft-com:office:smarttags /st1:place w:st=onCrawley/st1:place its single term has to be added to the multiple model: lrm(N~a*b+a+b).o:p/o:p This nearly always leads to high correlation rates between the interaction term a*b and its single term a or b. With regards to the law of colinearity modelling should not include correlated variables with an Spearman index 0,7. Does this mean that the interaction term has to be discarded or can the variables stay within the model when correlated? I do not necessarily want to do a PCA on this issue.o:p/o:p Thanks for helpingo:p/o:p Christiano:p/o:p Verschicken Sie romantische, coole und witzige Bilder per SMS! Jetzt bei WEB.DE FreeMail: http://f.web.de/?mc=021193 [[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
[R] square and points on the same figure
Hi,
I have the following code to
randomly generate the
points:
csr -function(n=60){
x=runif(n)
y=runif(n)
f=cbind(x,y)
}
plot(csr())
I wonder how to code to draw a square centering on 0.5,0.5; with the length of
each side to be 0.3 on the same figure with these plotted points?
Thanks,
Sam
-
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Re: [R] color for points
If you know explicitly that there are just 60 points, you can use:
plot(csr(),col=c(rep('blue',20), rep('red',20), rep('green',20)))
On 10/8/05, Sam R. Smith [EMAIL PROTECTED] wrote:
Hi,
I have the following code to
randomly generate the
points:
csr -function(n=60){
x=runif(n)
y=runif(n)
f=cbind(x,y)
}
plot(csr())
I wonder how to code to make the first twenty points to be BLUE; second
twenty points to be RED; the last twenty points to be GREEN?
Thanks,
Sam
-
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Cincinnati, OH
+1 513 247 0281
What the problem you are trying to solve?
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Re: [R] square and points on the same figure
Sam R. Smith wrote:
Hi,
I have the following code to
randomly generate the
points:
csr -function(n=60){
x=runif(n)
y=runif(n)
f=cbind(x,y)
}
plot(csr())
I wonder how to code to draw a square centering on 0.5,0.5; with the length
of each side to be 0.3 on the same figure with these plotted points?
?symbols
symbols(x=0.5, y=0.5, squares=c(0.3), add=TRUE)
Thanks,
Sam
-
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--
Chuck Cleland, Ph.D.
NDRI, Inc.
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 452-1424 (M, W, F)
fax: (917) 438-0894
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Re: [R] square and points on the same figure
after typing
symbols(x=0.5, y=0.5, squares=c(0.3), add=TRUE)
it works.
I want to add one more using different length such as
symbols(x=0.5, y=0.5, squares=c(0.15), add=TRUE)
but this square did not show.
Chuck Cleland [EMAIL PROTECTED] wrote:
Sam R. Smith wrote:
Hi,
I have the following code to
randomly generate the
points:
csr -function(n=60){
x=runif(n)
y=runif(n)
f=cbind(x,y)
}
plot(csr())
I wonder how to code to draw a square centering on 0.5,0.5; with the length
of each side to be 0.3 on the same figure with these plotted points?
?symbols
symbols(x=0.5, y=0.5, squares=c(0.3), add=TRUE)
Thanks,
Sam
-
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--
Chuck Cleland, Ph.D.
NDRI, Inc.
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 452-1424 (M, W, F)
fax: (917) 438-0894
-
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Re: [R] square and points on the same figure
Sam R. Smith wrote:
after typing
symbols(x=0.5, y=0.5, squares=c(0.3), add=TRUE)
it works.
I want to add one more using different length such as
symbols(x=0.5, y=0.5, squares=c(0.15), add=TRUE)
but this square did not show.
Specify both (all) squares in the same call to symbols:
csr -function(n=60){
x=runif(n)
y=runif(n)
f=cbind(x,y)
}
plot(csr())
symbols(x=c(0.5, 0.5), y=c(0.5, 0.5), squares=c(0.30,0.15), add=TRUE)
Chuck Cleland [EMAIL PROTECTED] wrote:
Sam R. Smith wrote:
Hi,
I have the following code to
randomly generate the
points:
csr -function(n=60){
x=runif(n)
y=runif(n)
f=cbind(x,y)
}
plot(csr())
I wonder how to code to draw a square centering on 0.5,0.5; with the length
of each side to be 0.3 on the same figure with these plotted points?
?symbols
symbols(x=0.5, y=0.5, squares=c(0.3), add=TRUE)
Thanks,
Sam
-
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NDRI, Inc.
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tel: (212) 845-4495 (Tu, Th)
tel: (732) 452-1424 (M, W, F)
fax: (917) 438-0894
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[R] Two-factor ANOVA Help
Hello, I am trying to perform a two-factor ANOVA analysis using a blocking design with Vol as the response variable. My intent is to have Rater treated as the treatment variable and the Pipe treated as the blocking variable. I am reading and preparing my dataset using the following three lines of code: values - read.table(filename, header=TRUE) attach(values) values = data.frame(values) The dataset is the following: Pipe Rater Volume 1A 5.129 1B 5.296 1C 4.679 1D 4.776 2A 8.519 2B 8.482 2C 7.659 2D 7.798 3A 13.769 3B 14.621 3C 12.418 3D 13.189 Below there are 2 versions which I've used to run my analysis. Option #1: g - lm(Volume ~ factor(Rater) + factor(Pipe), values) print(anova(g)) Option #2: Rater - as.factor(Rater) Pipe - as.factor(Pipe) g - lm(Volume ~ Rater + Pipe, values) print(anova(g)) A couple of questions I have are: 1. I thought that option #1 and option #2 would have given me the same results and they don't appear to. The only difference (to me) is how I have specified the factors used in the model. However, there appears to be something else I am missing and I was just wondering if anyone has any insight as to which is the correct way to code this analysis? 2. Unless otherwise specified is there a particular reference level that R uses by default - for example in this case, the second treatment level (Rater B)? 3. Is there a good reference someone can point me to for more insight on the two-factor ANOVA analysis with R? Thank you kindly in advance, Jim [[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
Re: [R] keeping interaction terms
Your note is formatted strangely. You seem to be using Microsoft - please tell your software to send plain text e-mails - Microsoft doesn't own plain ASCII text format, at least not yet (have they applied for a patent for it?). Christian Jones wrote: Hello,?xml:namespace prefix = o ns = urn:schemas-microsoft-com:office:office /o:p/o:p while doing my thesis in habitat modelling I´ve come across a problem with interaction terms. My question concerns the usage of interaction terms for linear regression modelling with R. If an interaction-term (predictor) is chosen for a multiple model, then, according to ?xml:namespace prefix = st1 ns = urn:schemas-microsoft-com:office:smarttags /st1:place w:st=onCrawley/st1:place its single term has to be added to the multiple model: lrm(N~a*b+a+b).o:p/o:p This nearly always leads to high correlation rates between the interaction term a*b and its single term a or b. With regards to the law of colinearity modelling should not include correlated variables with an Spearman index 0,7. Does this mean that the interaction term has to be discarded or can the variables stay within the model when correlated? I do not necessarily want to do a PCA on this issue.o:p/o:p Thanks for helpingo:p/o:p Christiano:p/o:p Your query opens up many issues. First, the statement that a main effect has to be added if an interaction term is chosen assumes that an interaction has meaning without adjustment for main effects. This is not the case. The hierarchy principle needs to be executed in a forward manner. Second, you are implying that you are not fitting a single pre-specified model but are doing variable selection based on p-values. This creates a host of problems. Third, you imply that correlations between main effects and interactions are not to be tolerated. Again this is not the case. It is a fact of life that we must accomodate. [Some people like to center main effects to reduce this correlation but that is an artificial and not helpful approach.] Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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
[R] obfuscated identity
Globe == Globe Trotter [EMAIL PROTECTED] on Fri, 7 Oct 2005 16:39:10 -0700 (PDT) writes: . Globe I use this Yahoo! e-mail address as a junk e-mail Globe address since the archives contain the complete Globe e-mail address (no anti-spam measures taken) so I do Globe not include my real name. [there are many different archives and some do some anti-spam measures.. to strip or obfuscate e-mail addresses before mail is sent out, is really not an option!] Globe It would be nice if the R-archivers stripped e-mail Globe addresses from posters (or messed it up) so that Globe spider programs would not be able to get it. (yes, but *all* of them would have to do it; and do it so smartly that smart address collecting software cannot regenerate it; ... ) Globe Many thanks and best wishes, GT and you still don't tell us who you are! Bert's remark was very much on point: Using a yahoo or gmail or ... e-mail address is one thing; but then we'd expect at least a full name (and often affiliation) as signature. It not only netiquette, it's real life etiquette to not send anonymous letters Martin Maechler, ETH Zurich __ 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
[R] Warning: condition has length 1
Hi -
I was wondering if anyone came across a problem such as mine! I was trying
to create a user-defined function, however, the results from calling the
function are unexpected!
When passing X parameter as a single value variable (x-c(3)), everything
works fine. However, when passing a parameter as a vector with multiple
values (as the case in my code), the 'if condition' is only executed once!
Here is my code:
x
[1] 0.3 1.0 0.7 22.0
myfunction-function(x)
+ {
+ if (x=1) 0
+ else x^2
+ }
myfunction(x)
[1] 0.09 1.00 0.49 484.00
Warning message:
the condition has length 1 and only the first element will be used in: if
(x = 1) 0 else x^2
Is there a way to overcome this issue? Can you please explain what
modifications to the code I need to accommodate to make it work.
Thanks,
Nawaf
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Re: [R] keeping interaction terms
Adding a bit to Frank Harrell's good comments. 1. Regarding HTML infection: I rolled up my sleeves, washed my hands carefully, took a fine sharp knife, cut it all out, and then sowed up the incisions. 2. For the rest, see below. On 08-Oct-05 Christian Jones wrote: Hello, while doing my thesis in habitat modelling I´ve come across a problem with interaction terms. My question concerns the usage of interaction terms for linear regression modelling with R. If an interaction-term (predictor) is chosen for a multiple model, then, according to Crawley its single term has to be added to the multiple model: lrm(N~a*b+a+b). This nearly always leads to high correlation rates between the interaction term a*b and its single term a or b. With regards to the law of colinearity modelling should not include correlated variables with an Spearman index 0,7. Does this mean that the interaction term has to be discarded or can the variables stay within the model when correlated? I do not necessarily want to do a PCA on this issue. There's more than a suggestion in your statements that you tend to be drawn along by people's prescriptions. Instead, try to think simply about it. If, after fitting a+b, you make a significant difference by further including a:b, then the interaction between a and b matters, even if you observe high correlations. The latter should not lead you to ignore the former. How much it matters is of course another question. You could examine this, in R, by comparing the predicted values from the a+b model with the predicted values from the a*b model. Though they will be different, you will have to judge whether the amount of difference is large enough to be of real importance in your application. (It is possible to get highly significant results, i.e. small P-values, from small effects). Even if it does matter, in real terms, you are left with the fundamental difficulty, indicated by Frank, that interpreting interaction between variables a and b is simple only when the variables a and b are orthogonal in the data (either by accident or by design). If they are non-orthogonal, then you have to think carefully about how to interpret it, and this does depend on what it all means. Maybe we could help more with this if we knew more about your investigation (perhaps off-list, if you prefer). Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 08-Oct-05 Time: 14:14:48 -- XFMail -- __ 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
Re: [R] Warning: condition has length 1
nawaf b wrote:
Hi -
I was wondering if anyone came across a problem such as mine! I was trying
to create a user-defined function, however, the results from calling the
function are unexpected!
When passing X parameter as a single value variable (x-c(3)), everything
works fine. However, when passing a parameter as a vector with multiple
values (as the case in my code), the 'if condition' is only executed once!
Here is my code:
x
[1] 0.3 1.0 0.7 22.0
myfunction-function(x)
+ {
+ if (x=1) 0
+ else x^2
+ }
?ifelse
myfunction - function(x){ifelse(x = 1, 0, x^2)}
myfunction(c(.3, 1, .7, 22))
[1] 0.09 0.00 0.49 0.00
myfunction(x)
[1] 0.09 1.00 0.49 484.00
Warning message:
the condition has length 1 and only the first element will be used in: if
(x = 1) 0 else x^2
Is there a way to overcome this issue? Can you please explain what
modifications to the code I need to accommodate to make it work.
Thanks,
Nawaf
__
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
--
Chuck Cleland, Ph.D.
NDRI, Inc.
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 452-1424 (M, W, F)
fax: (917) 438-0894
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Re: [R] Warning: condition has length 1
ifelse(x = 1, 0, x^2)
or
(x = 1) * x^2
Also, you might need x (1 - e) where e is some small number since your
numbers may not be exact.
On 10/8/05, nawaf b [EMAIL PROTECTED] wrote:
Hi -
I was wondering if anyone came across a problem such as mine! I was trying
to create a user-defined function, however, the results from calling the
function are unexpected!
When passing X parameter as a single value variable (x-c(3)), everything
works fine. However, when passing a parameter as a vector with multiple
values (as the case in my code), the 'if condition' is only executed once!
Here is my code:
x
[1] 0.3 1.0 0.7 22.0
myfunction-function(x)
+ {
+ if (x=1) 0
+ else x^2
+ }
myfunction(x)
[1] 0.09 1.00 0.49 484.00
Warning message:
the condition has length 1 and only the first element will be used in: if
(x = 1) 0 else x^2
Is there a way to overcome this issue? Can you please explain what
modifications to the code I need to accommodate to make it work.
Thanks,
Nawaf
__
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
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Re: [R] Warning: condition has length 1
On Sat, 8 Oct 2005, nawaf b wrote:
Hi -
I was wondering if anyone came across a problem such as mine! I was trying
to create a user-defined function, however, the results from calling the
function are unexpected!
When passing X parameter as a single value variable (x-c(3)), everything
works fine. However, when passing a parameter as a vector with multiple
values (as the case in my code), the 'if condition' is only executed once!
Here is my code:
x
[1] 0.3 1.0 0.7 22.0
myfunction-function(x)
+ {
+ if (x=1) 0
+ else x^2
+ }
myfunction(x)
[1] 0.09 1.00 0.49 484.00
Warning message:
the condition has length 1 and only the first element will be used in: if
(x = 1) 0 else x^2
Is there a way to overcome this issue? Can you please explain what
modifications to the code I need to accommodate to make it work.
myfunction-function(x)
+ ifelse(x=1, 0, x^2)
x - c(0.3, 1.0, 0.7, 22.0)
myfunction(x)
[1] 0.09 0.00 0.49 0.00
ifelse() is vectorised, but if() only looks at the first element in x
before deciding what to do - so if the condition is longer, the rest gets
ignored.
Thanks,
Nawaf
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Roger Bivand
Economic Geography Section, Department of Economics, Norwegian School of
Economics and Business Administration, Helleveien 30, N-5045 Bergen,
Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43
e-mail: [EMAIL PROTECTED]
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Re: [R] Warning: condition has length 1
On 10/8/05, Gabor Grothendieck [EMAIL PROTECTED] wrote:
ifelse(x = 1, 0, x^2)
or
(x = 1) * x^2
Sorry, that should have been (x 1) * x^2
Also, you might need x (1 - e) where e is some small number since your
numbers may not be exact.
On 10/8/05, nawaf b [EMAIL PROTECTED] wrote:
Hi -
I was wondering if anyone came across a problem such as mine! I was trying
to create a user-defined function, however, the results from calling the
function are unexpected!
When passing X parameter as a single value variable (x-c(3)), everything
works fine. However, when passing a parameter as a vector with multiple
values (as the case in my code), the 'if condition' is only executed once!
Here is my code:
x
[1] 0.3 1.0 0.7 22.0
myfunction-function(x)
+ {
+ if (x=1) 0
+ else x^2
+ }
myfunction(x)
[1] 0.09 1.00 0.49 484.00
Warning message:
the condition has length 1 and only the first element will be used in: if
(x = 1) 0 else x^2
Is there a way to overcome this issue? Can you please explain what
modifications to the code I need to accommodate to make it work.
Thanks,
Nawaf
__
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
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[R] how to control ticks in plots with yasp or xasp
Hi, A few times I tried to control the number and position of tick marks in plots with the yasp or xasp parameters. For example, a y axis was drawn by default with tick marks at 0, 20, 40, 80 and 100. I tried to get tick marks every 10 by adding yasp=(0, 100, 10) but this had no effect at all. I know I can draw the axis and tick marks manually, but often this simple option would suffice if I could understand how to make it work. Thanks in advance, Denis Chabot __ 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
Re: [R] Two-factor ANOVA Help
On 10/8/05, Jim Brindle [EMAIL PROTECTED] wrote: Hello, I am trying to perform a two-factor ANOVA analysis using a blocking design with Vol as the response variable. My intent is to have Rater treated as the treatment variable and the Pipe treated as the blocking variable. I am reading and preparing my dataset using the following three lines of code: values - read.table(filename, header=TRUE) attach(values) values = data.frame(values) The dataset is the following: Pipe Rater Volume 1A 5.129 1B 5.296 1C 4.679 1D 4.776 2A 8.519 2B 8.482 2C 7.659 2D 7.798 3A 13.769 3B 14.621 3C 12.418 3D 13.189 Below there are 2 versions which I've used to run my analysis. Option #1: g - lm(Volume ~ factor(Rater) + factor(Pipe), values) print(anova(g)) Option #2: Rater - as.factor(Rater) Pipe - as.factor(Pipe) g - lm(Volume ~ Rater + Pipe, values) print(anova(g)) A couple of questions I have are: 1. I thought that option #1 and option #2 would have given me the same results and they don't appear to. The only difference (to me) is how I have specified the factors used in the model. However, there appears to be something else I am missing and I was just wondering if anyone has any insight as to which is the correct way to code this analysis? Note that values, as returned from read.table, is already a data frame and Rater is already a factor so you only need to convert Pipe to a factor: values - read.table(filename.dat, header = TRUE) # shows classes of columns among other things # note that Rater is already a factor and values is already a data frame str(g) # convert Pipe to a factor values$Pipe - factor(values$Pipe) g - lm(Volume ~., values) g 2. Unless otherwise specified is there a particular reference level that R uses by default - for example in this case, the second treatment level (Rater B)? By default R uses treatment contrasts and uses the first level as the baseline. You can change this using contrasts and contr.treatment. e.g. To use treatment effects on Pipe with level 2 as the baseline: contrasts(values$Pipe) - contr.treatment(3, base = 2) g2 - lm(Volume ~., values) g2 3. Is there a good reference someone can point me to for more insight on the two-factor ANOVA analysis with R? See ?read.table, ?contrasts, ?contr.treatment __ 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
Re: [R] Two-factor ANOVA Help
On 10/8/05, Gabor Grothendieck [EMAIL PROTECTED] wrote: On 10/8/05, Jim Brindle [EMAIL PROTECTED] wrote: Hello, I am trying to perform a two-factor ANOVA analysis using a blocking design with Vol as the response variable. My intent is to have Rater treated as the treatment variable and the Pipe treated as the blocking variable. I am reading and preparing my dataset using the following three lines of code: values - read.table(filename, header=TRUE) attach(values) values = data.frame(values) The dataset is the following: Pipe Rater Volume 1A 5.129 1B 5.296 1C 4.679 1D 4.776 2A 8.519 2B 8.482 2C 7.659 2D 7.798 3A 13.769 3B 14.621 3C 12.418 3D 13.189 Below there are 2 versions which I've used to run my analysis. Option #1: g - lm(Volume ~ factor(Rater) + factor(Pipe), values) print(anova(g)) Option #2: Rater - as.factor(Rater) Pipe - as.factor(Pipe) g - lm(Volume ~ Rater + Pipe, values) print(anova(g)) A couple of questions I have are: 1. I thought that option #1 and option #2 would have given me the same results and they don't appear to. The only difference (to me) is how I have specified the factors used in the model. However, there appears to be something else I am missing and I was just wondering if anyone has any insight as to which is the correct way to code this analysis? Note that values, as returned from read.table, is already a data frame and Rater is already a factor so you only need to convert Pipe to a factor: values - read.table(filename.dat, header = TRUE) # shows classes of columns among other things # note that Rater is already a factor and values is already a data frame str(g) # convert Pipe to a factor values$Pipe - factor(values$Pipe) g - lm(Volume ~., values) g 2. Unless otherwise specified is there a particular reference level that R uses by default - for example in this case, the second treatment level (Rater B)? By default R uses treatment contrasts and uses the first level as the baseline. You can change this using contrasts and contr.treatment. e.g. To use treatment effects on Pipe with level 2 as the baseline: contrasts(values$Pipe) - contr.treatment(3, base = 2) g2 - lm(Volume ~., values) g2 3. Is there a good reference someone can point me to for more insight on the two-factor ANOVA analysis with R? See ?read.table, ?contrasts, ?contr.treatment One more point. In your #2 the second argument to lm is specified as values so that's where lm will look for the variables. The Rater and Pipe defined immediately above that line will never be found since those variables will already have been found in the values data frame. __ 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
Re: [R] how to control ticks in plots with yasp or xasp
On Sat, 2005-10-08 at 09:28 -0400, Denis Chabot wrote: Hi, A few times I tried to control the number and position of tick marks in plots with the yasp or xasp parameters. For example, a y axis was drawn by default with tick marks at 0, 20, 40, 80 and 100. I tried to get tick marks every 10 by adding yasp=(0, 100, 10) but this had no effect at all. I know I can draw the axis and tick marks manually, but often this simple option would suffice if I could understand how to make it work. Thanks in advance, Denis Chabot I suspect that one problem you are having is that there is no par(xasp) or par(yasp)unless these are typos and you are trying to use par(xaxp) and par(yaxp)? There is an 'asp' argument to some of the plot functions (ie. plot.default), but this has a different intention. par(xaxp) and par(yaxp) are not listed as read only pars in ?par, however, I cannot recall an instance where R does not overwrite the user settings during the calculation of the axes, whether passed as arguments to a plot function or set a priori via a par() call. If you want explicit control over the tick marks, you will need to use axis(), perhaps in combination with axTicks(), after using 'xaxt = n' and/or 'yaxt = n' in the plot call, depending upon the circumstances. HTH, Marc Schwartz __ 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
Re: [R] Matrix calculations in R--erroneous?
Hi Spencer: Thanks! This gives me a number of other ways of thinking about this problem. My one concern is that these approaches would also run into some difficulties with how long it takes to calculate. I'm interested not in a single value but a matrix of over 300k values that has to be recomputed more than a million times. If I had to apply ifelse or floor to each of these, it might take too long. I'll have to see. Peter On 10/7/05 5:55 PM, Spencer Graves [EMAIL PROTECTED] wrote: Rather than adding 1e-15 to all numbers, I suggest you simply make that the floor. (Or use .Machine$double.eps or 2*.Machine$double.eps in place of 1e-15.) Another alternative that may or may not apply in your case is to develop an asymptotic expansion for the log(likelihood) for the small numbers. I've had good success with this kind of method. For example, consider the Box-Cox transformation: bc(y, b) = (y^b-1)/b What do we do with b = 0? We can test for b = 0 and replace those cases by the limit log(y). However, it is numerically more stable to use the following: bc(y, b) = ifelse(abs(b*log(y)).Machine$double.eps, (expm1(b*log(y))/b, log(y)). I don't have time to study your example to see if I could see anything like this that could be done, but I think there should be a good chance of finding something like this. Of course, if there are only very few 0's, then it hardly matters. However, if there are quite a few, then you need something like this. hope this helps. spencer graves __ 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
[R] add leading 0s to %d from png() {was Automatic creation of file names}
Dear useRs,
Is there a way to 'properly' format %d when plotting more than one
page on png()? 'Properly' means to me with leading 0s, so that the
PNGs become easy to navigate in a file/image browser. Lacking a better
solution I ended up using the code below, but would much prefer
something like
png(test_%d.png,bg=white,width=1000,height=700)
where %d could be formatted like
formatC(%d,digits=0,wid=3,flag=0,mode=integer)
Thank you,
b.
#---works, but is rather complicated---
pngno - 0 ; i - 1
for (w in 1:53) {
if (i %in% c(4*0:100+1)) {
pngno - pngno + 1
png(paste(test_,formatC(pngno,digits=0,wid=4,flag=0,mode=integer),
.png,sep=),bg=white,width=1000,height=750)
par(mfrow=c(2,2),mai=c(4,5,3,2)/10,omi=c(0.2,0,0,0),
cex.axis=1,cex.main=1.2)
}
plot(1:10,main=w)
if (i %in% c(4*1:100)) dev.off()
i - i+1
}
dev.off()
From: Mike Prager Mike.Prager at noaa.gov
Subject: Re: [R] Automatic creation of file names
Newsgroups: gmane.comp.lang.r.general
Date: 2005-09-22 14:51:54 GMT (2 weeks, 1 day, 23 hours and 55 minutes ago)
Walter --
P.S. The advantage of using formatC over pasting the digits (1:1000)
directly is that when one uses leading zeroes, as in the formatC example
shown, the resulting filenames will sort into proper order.
...MHP
You can use paste() with something like
formatC(number,digits=0,wid=3,flag=0)
(where number is your loop index) to generate the filenames.
on 9/22/2005 10:21 AM Leite,Walter said the following:
I have a question about how to save to the hard drive the one thousand
datasets I generated in a simulation. ://www.R-project.org/posting-guide.html
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[R] request
Dear All, Can someone please tell me if there is a provision in R to fit a random coefficient multinomial logistic regression. Thanks in advance for your help and suggestions. Regards Murthy.N.M __ 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
Re: [R] how to control ticks in plots with yasp or xasp
On Sat, 8 Oct 2005, Marc Schwartz wrote: On Sat, 2005-10-08 at 09:28 -0400, Denis Chabot wrote: Hi, A few times I tried to control the number and position of tick marks in plots with the yasp or xasp parameters. For example, a y axis was drawn by default with tick marks at 0, 20, 40, 80 and 100. I tried to get tick marks every 10 by adding yasp=(0, 100, 10) but this had no effect at all. I know I can draw the axis and tick marks manually, but often this simple option would suffice if I could understand how to make it work. Thanks in advance, Denis Chabot I suspect that one problem you are having is that there is no par(xasp) or par(yasp)unless these are typos and you are trying to use par(xaxp) and par(yaxp)? In any case, (0, 100, 10) is invalid syntax, and c(0, 100, 10) is needed. There is an 'asp' argument to some of the plot functions (ie. plot.default), but this has a different intention. par(xaxp) and par(yaxp) are not listed as read only pars in ?par, however, I cannot recall an instance where R does not overwrite the user settings during the calculation of the axes, whether passed as arguments to a plot function or set a priori via a par() call. Really? Try plot(1:100, xaxt=n) par(xaxp=c(0, 50, 5)) # the value is reset at each plot axis(1) for how it works (but not inline, which is probably a bug). If you want explicit control over the tick marks, you will need to use axis(), perhaps in combination with axTicks(), after using 'xaxt = n' and/or 'yaxt = n' in the plot call, depending upon the circumstances. That is usually as easy. -- 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
Re: [R] add leading 0s to %d from png() {was Automatic creation of file names}
The example on the png help page, Rplot%03d.png, _is_ what you want.
(More details are on ?postscript.)
formatC() is an S/R peculiarity: sprintf() is the cross-language way to do
this sort of thing.
On Sat, 8 Oct 2005, bogdan romocea wrote:
Dear useRs,
Is there a way to 'properly' format %d when plotting more than one
page on png()? 'Properly' means to me with leading 0s, so that the
PNGs become easy to navigate in a file/image browser. Lacking a better
solution I ended up using the code below, but would much prefer
something like
png(test_%d.png,bg=white,width=1000,height=700)
where %d could be formatted like
formatC(%d,digits=0,wid=3,flag=0,mode=integer)
Thank you,
b.
#---works, but is rather complicated---
pngno - 0 ; i - 1
for (w in 1:53) {
if (i %in% c(4*0:100+1)) {
pngno - pngno + 1
png(paste(test_,formatC(pngno,digits=0,wid=4,flag=0,mode=integer),
.png,sep=),bg=white,width=1000,height=750)
par(mfrow=c(2,2),mai=c(4,5,3,2)/10,omi=c(0.2,0,0,0),
cex.axis=1,cex.main=1.2)
}
plot(1:10,main=w)
if (i %in% c(4*1:100)) dev.off()
i - i+1
}
dev.off()
From: Mike Prager Mike.Prager at noaa.gov
Subject: Re: [R] Automatic creation of file names
Newsgroups: gmane.comp.lang.r.general
Date: 2005-09-22 14:51:54 GMT (2 weeks, 1 day, 23 hours and 55 minutes ago)
Walter --
P.S. The advantage of using formatC over pasting the digits (1:1000)
directly is that when one uses leading zeroes, as in the formatC example
shown, the resulting filenames will sort into proper order.
...MHP
You can use paste() with something like
formatC(number,digits=0,wid=3,flag=0)
(where number is your loop index) to generate the filenames.
on 9/22/2005 10:21 AM Leite,Walter said the following:
I have a question about how to save to the hard drive the one thousand
datasets I generated in a simulation.
://www.R-project.org/posting-guide.html
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University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] how to control ticks in plots with yasp or xasp
Denis Chabot [EMAIL PROTECTED] writes: Oops, Mark, bad typo indeed. But yaxp is what I had in my R program and it did not help. I did obtain control over my tick marks by drawing them myself, as you suggest. But I was curious as to how to use yaxp since the help on par gives it as a possible way of controlling ticks. Maybe it should be removed from the help file? plot(1) par(xaxp) par(xaxp=c(.6,1.4,8)) axis(3) So it does do something... Thanks for your help, Denis Le 05-10-08 à 11:16, Marc Schwartz a écrit : On Sat, 2005-10-08 at 09:28 -0400, Denis Chabot wrote: Hi, A few times I tried to control the number and position of tick marks in plots with the yasp or xasp parameters. For example, a y axis was drawn by default with tick marks at 0, 20, 40, 80 and 100. I tried to get tick marks every 10 by adding yasp=(0, 100, 10) but this had no effect at all. I know I can draw the axis and tick marks manually, but often this simple option would suffice if I could understand how to make it work. Thanks in advance, Denis Chabot I suspect that one problem you are having is that there is no par(xasp) or par(yasp)unless these are typos and you are trying to use par(xaxp) and par(yaxp)? There is an 'asp' argument to some of the plot functions (ie. plot.default), but this has a different intention. par(xaxp) and par(yaxp) are not listed as read only pars in ?par, however, I cannot recall an instance where R does not overwrite the user settings during the calculation of the axes, whether passed as arguments to a plot function or set a priori via a par() call. If you want explicit control over the tick marks, you will need to use axis(), perhaps in combination with axTicks(), after using 'xaxt = n' and/or 'yaxt = n' in the plot call, depending upon the circumstances. HTH, Marc Schwartz __ 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 -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ 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
Re: [R] how to control ticks in plots with yasp or xasp
On Sat, 2005-10-08 at 16:37 +0100, Prof Brian Ripley wrote: On Sat, 8 Oct 2005, Marc Schwartz wrote: On Sat, 2005-10-08 at 09:28 -0400, Denis Chabot wrote: Hi, A few times I tried to control the number and position of tick marks in plots with the yasp or xasp parameters. For example, a y axis was drawn by default with tick marks at 0, 20, 40, 80 and 100. I tried to get tick marks every 10 by adding yasp=(0, 100, 10) but this had no effect at all. I know I can draw the axis and tick marks manually, but often this simple option would suffice if I could understand how to make it work. Thanks in advance, Denis Chabot I suspect that one problem you are having is that there is no par(xasp) or par(yasp)unless these are typos and you are trying to use par(xaxp) and par(yaxp)? In any case, (0, 100, 10) is invalid syntax, and c(0, 100, 10) is needed. Indeed. There is an 'asp' argument to some of the plot functions (ie. plot.default), but this has a different intention. par(xaxp) and par(yaxp) are not listed as read only pars in ?par, however, I cannot recall an instance where R does not overwrite the user settings during the calculation of the axes, whether passed as arguments to a plot function or set a priori via a par() call. Really? Try plot(1:100, xaxt=n) par(xaxp=c(0, 50, 5)) # the value is reset at each plot axis(1) for how it works (but not inline, which is probably a bug). AhI had not thought about that 'par'ticular combination... ;-) Hence, not R.O. So it must be used _after_ a plot call, which makes sense. I had reached for my copy of Paul's book and on page 96 (last paragraph in section 3.4.5 on Axes), he suggests using the approach I elucidate below with axTicks(). I thought he might have some other ideas and that I was missing something. This is also referenced on page 70, third paragraph in section 3.2.5 on Axes. If you want explicit control over the tick marks, you will need to use axis(), perhaps in combination with axTicks(), after using 'xaxt = n' and/or 'yaxt = n' in the plot call, depending upon the circumstances. That is usually as easy. Agreed. Thanks, Marc __ 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
[R] Rpart -- using predict() when missing data is present?
I am doing library(rpart) m - rpart(y ~ x, D[insample,]) D[outsample,] y x 8 0.78391922 0.579025591 9 0.06629211 NA 10 NA 0.001593063 p - predict(m, newdata=D[9,]) Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid result from na.action How do I persuade him to give me NA since x is NA? I looked at ?predict.rpart but didn't find any mention about NAs. (In this problem, I can easily do it manually, but this is a part of something bigger where I want him to be able to gracefully handle prediction requests involving NA). -- Ajay Shah Consultant [EMAIL PROTECTED] Department of Economic Affairs http://www.mayin.org/ajayshah Ministry of Finance, New Delhi __ 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
Re: [R] how to control ticks in plots with yasp or xasp
Hi, sorry about the bad syntax, though the right syntax would not have worked either, according to your tests (Mark, Brian, Peter). Anyway it is too finicky, I will draw them myself. For instance, plot(1:100, xaxt=n) par(xaxp=c(0, 100, 10)) # the value is reset at each plot axis(1) Placed tick marks at intervals at 0, 10, ..., 100, as expected, but did not place a label under 100... Thanks for the answers, usufull as always, Denis Le 05-10-08 à 11:58, Marc Schwartz a écrit : On Sat, 2005-10-08 at 16:37 +0100, Prof Brian Ripley wrote: On Sat, 8 Oct 2005, Marc Schwartz wrote: On Sat, 2005-10-08 at 09:28 -0400, Denis Chabot wrote: Hi, A few times I tried to control the number and position of tick marks in plots with the yasp or xasp parameters. For example, a y axis was drawn by default with tick marks at 0, 20, 40, 80 and 100. I tried to get tick marks every 10 by adding yasp=(0, 100, 10) but this had no effect at all. I know I can draw the axis and tick marks manually, but often this simple option would suffice if I could understand how to make it work. Thanks in advance, Denis Chabot I suspect that one problem you are having is that there is no par(xasp) or par(yasp)unless these are typos and you are trying to use par(xaxp) and par(yaxp)? In any case, (0, 100, 10) is invalid syntax, and c(0, 100, 10) is needed. Indeed. There is an 'asp' argument to some of the plot functions (ie. plot.default), but this has a different intention. par(xaxp) and par(yaxp) are not listed as read only pars in ? par, however, I cannot recall an instance where R does not overwrite the user settings during the calculation of the axes, whether passed as arguments to a plot function or set a priori via a par() call. Really? Try plot(1:100, xaxt=n) par(xaxp=c(0, 50, 5)) # the value is reset at each plot axis(1) for how it works (but not inline, which is probably a bug). AhI had not thought about that 'par'ticular combination... ;-) Hence, not R.O. So it must be used _after_ a plot call, which makes sense. I had reached for my copy of Paul's book and on page 96 (last paragraph in section 3.4.5 on Axes), he suggests using the approach I elucidate below with axTicks(). I thought he might have some other ideas and that I was missing something. This is also referenced on page 70, third paragraph in section 3.2.5 on Axes. If you want explicit control over the tick marks, you will need to use axis(), perhaps in combination with axTicks(), after using 'xaxt = n' and/or 'yaxt = n' in the plot call, depending upon the circumstances. That is usually as easy. Agreed. Thanks, Marc __ 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
Re: [R] how to control ticks in plots with yasp or xasp
On Sat, 8 Oct 2005, Denis Chabot wrote: Hi, sorry about the bad syntax, though the right syntax would not have worked either, according to your tests (Mark, Brian, Peter). It DOES work according to my tests! (Do give us the credit for testing our advice: we would appreciate your showing equal care.) Anyway it is too finicky, I will draw them myself. For instance, plot(1:100, xaxt=n) par(xaxp=c(0, 100, 10)) # the value is reset at each plot axis(1) Placed tick marks at intervals at 0, 10, ..., 100, as expected, but did not place a label under 100... Does for me (2.2.0, Windows and X11 on Linux). Might the text be clipped on your unnamed device, so you need to set xpd? (If so, it will be clipped however you try to do this, and is an unrelated local problem.) -- 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
Re: [R] how to control ticks in plots with yasp or xasp
Hi Brian, Le 05-10-08 à 13:21, Prof Brian Ripley a écrit : On Sat, 8 Oct 2005, Denis Chabot wrote: Hi, sorry about the bad syntax, though the right syntax would not have worked either, according to your tests (Mark, Brian, Peter). It DOES work according to my tests! (Do give us the credit for testing our advice: we would appreciate your showing equal care.) I am very sorry, I did not express myself properly. I do appreciate your help, and I did take the time to test your solution. What I meant was that even if I had use the proper syntax, i.e. yaxp=c (10,100,10) instead of y=(10,100,10) [actually in my program I had used it but I was careless when I composed the message], it would not have worked because you demonstrated that it has to be used within a par statement, which I had not done. Therefore my attempt was doomed. I did not mean there was no way to make yaxp work, so I apologize if you think I take your advice lightly, and I cannot state strongly enough that it is not the case at all. Anyway it is too finicky, I will draw them myself. For instance, plot(1:100, xaxt=n) par(xaxp=c(0, 100, 10)) # the value is reset at each plot axis(1) Placed tick marks at intervals at 0, 10, ..., 100, as expected, but did not place a label under 100... Does for me (2.2.0, Windows and X11 on Linux). Might the text be clipped on your unnamed device, so you need to set xpd? (If so, it will be clipped however you try to do this, and is an unrelated local problem.) This is strange. I work on Mac OS X, so the default device is quartz. Label 100 is shown without clipping if plot without changing xaxp. But just in case I made more space at the right of the graph with par(mai=c(0.7, 0.7, 0.5, 0.5)) plot(1:100, xaxt=n) par(xaxp=c(0, 100, 10)) # the value is reset at each plot axis(1) and the 100 does not appear. If I run this again: plot(1:100, xaxt=n) axis(1) I'm back with R's default axis and the 100 is present. Anyone on a Mac wants to try? Denis -- 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
Re: [R] Rpart -- using predict() when missing data is present?
On Sat, 8 Oct 2005, Ajay Narottam Shah wrote: I am doing library(rpart) m - rpart(y ~ x, D[insample,]) D[outsample,] y x 8 0.78391922 0.579025591 9 0.06629211 NA 10 NA 0.001593063 p - predict(m, newdata=D[9,]) Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid result from na.action How do I persuade him to give me NA since x is NA? I think the point is to do something sensible! One x prediction problems are not what rpart is designed to do, and the default na.action (na.rpart) fails in that case. (The author forgot drop=F.) I looked at ?predict.rpart but didn't find any mention about NAs. How about ?rpart ? That does. (In this problem, I can easily do it manually, but this is a part of something bigger where I want him to be able to gracefully handle prediction requests involving NA). -- 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
Re: [R] quasi-random vector according to an independent graph
Hi
I was waiting for more advices and prepared to summarize to reduce the mail
traffic. I got two answers (I realize that my explanation was definitely
insufficient).
Thanks Reid Spencer ! The link http://www.r-project.org/gR/ was helpful,
although it does not directly answer my question.
I think I have got some idea to do what I want. When I have a more or less
satisfactory solution I will report to the list to share (if I think it is
worthwhile).
Jinfang Wang
- Original Message -
From: Huntsinger, Reid [EMAIL PROTECTED]
To: 'Spencer Graves' [EMAIL PROTECTED]; Jinfang Wang
[EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Friday, September 30, 2005 2:00 AM
Subject: RE: [R] quasi-random vector according to an independent graph
Might the graphical models in R packages be of interest?
http://www.r-project.org/gR/
Reid Huntsinger
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Spencer Graves
Sent: Thursday, September 29, 2005 11:43 AM
To: Jinfang Wang
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] quasi-random vector according to an independent graph
Are you still interested in a reply to this post? I have not seen
any. If you are, it might help if you were more specific, e.g.,
following the posting guide www.R-project.org/posting-guide.html. I'm
not certain what you mean by a joint distribution defined by an
independent graph, and my efforts using RSiteSearch exposed several
things that might be useful but none that seemed to me to be obvious
answers to your question.
Sorry I could not be more helpful.
spencer graves
Jinfang Wang wrote:
Dear R-users,
Is anyone aware of any function/package for generating a random vector
from
a joint distribution defined by an independent graph? Or I have to work
it
out myself?
Thanks.
Jinfang
--
Jinfang Wang, Associate Professor
Chiba University, Japan
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Senior Development Engineer
PDF Solutions, Inc.
333 West San Carlos Street Suite 700
San Jose, CA 95110, USA
[EMAIL PROTECTED]
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Tel: 408-938-4420
Fax: 408-280-7915
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Re: [R] quasi-random vector according to an independent graph
Hi, Jihfang: Definitely, please report to the list when you think you have a more or less satisfactory solution. The archives are searchable, and your comments might help someone else. They might also provoke useful suggestions of other things you might try that might provide even better solutions. Also, if you get stuck in the process, please report to the list your progress and where you are stuck. Good Luck, spencer graves Jinfang Wang wrote: Hi I was waiting for more advices and prepared to summarize to reduce the mail traffic. I got two answers (I realize that my explanation was definitely insufficient). Thanks Reid Spencer ! The link http://www.r-project.org/gR/ was helpful, although it does not directly answer my question. I think I have got some idea to do what I want. When I have a more or less satisfactory solution I will report to the list to share (if I think it is worthwhile). Jinfang Wang - Original Message - From: Huntsinger, Reid [EMAIL PROTECTED] To: 'Spencer Graves' [EMAIL PROTECTED]; Jinfang Wang [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Friday, September 30, 2005 2:00 AM Subject: RE: [R] quasi-random vector according to an independent graph Might the graphical models in R packages be of interest? http://www.r-project.org/gR/ Reid Huntsinger -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Spencer Graves Sent: Thursday, September 29, 2005 11:43 AM To: Jinfang Wang Cc: r-help@stat.math.ethz.ch Subject: Re: [R] quasi-random vector according to an independent graph Are you still interested in a reply to this post? I have not seen any. If you are, it might help if you were more specific, e.g., following the posting guide www.R-project.org/posting-guide.html. I'm not certain what you mean by a joint distribution defined by an independent graph, and my efforts using RSiteSearch exposed several things that might be useful but none that seemed to me to be obvious answers to your question. Sorry I could not be more helpful. spencer graves Jinfang Wang wrote: Dear R-users, Is anyone aware of any function/package for generating a random vector from a joint distribution defined by an independent graph? Or I have to work it out myself? Thanks. Jinfang -- Jinfang Wang, Associate Professor Chiba University, Japan __ 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 -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ 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 -- Notice: This e-mail message, together with any attachments, contains information of Merck Co., Inc. (One Merck Drive, Whitehouse Station, New Jersey, USA 08889), and/or its affiliates (which may be known outside the United States as Merck Frosst, Merck Sharp Dohme or MSD and in Japan, as Banyu) that may be confidential, proprietary copyrighted and/or legally privileged. It is intended solely for the use of the individual or entity named on this message. If you are not the intended recipient, and have received this message in error, please notify us immediately by reply e-mail and then delete it from your system. -- -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ 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
[R] lm.ridge
Dear R People: I have a question about the lm.ridge function, please. In the example, there is one set of output values in the select function but another in the comment section. Am I missing something please? R Version 2.1.1 Windows Thanks, Sincerely, Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: [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
Re: [R] Warning: condition has length 1
nawaf b wrote:
Hi -
I was wondering if anyone came across a problem such as mine! I was trying
to create a user-defined function, however, the results from calling the
function are unexpected!
When passing X parameter as a single value variable (x-c(3)), everything
works fine. However, when passing a parameter as a vector with multiple
values (as the case in my code), the 'if condition' is only executed once!
Here is my code:
x
[1] 0.3 1.0 0.7 22.0
myfunction-function(x)
+ {
+ if (x=1) 0
+ else x^2
+ }
myfunction(x)
[1] 0.09 1.00 0.49 484.00
Warning message:
the condition has length 1 and only the first element will be used in: if
(x = 1) 0 else x^2
Is there a way to overcome this issue? Can you please explain what
modifications to the code I need to accommodate to make it work.
I am going to go out on a limb here (I don't know much about R) but I'm
not sure it has anything to do with that fact that you have defined a
function.
It seems to me that if (x 1) {...} expects a scalar and if x is not
a scalar, it will use the first value. To apply your function to each
element, check out the help for sapply (eg ?sapply). I am guessing the
solution is something like sapply(x, myfunction) but don't quote me.
paul sorenson
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Re: [R] R-2.2.0 compilation problem
On 07/10/2005, at 8:00 PM, Robin Hankin wrote: I tried to compile R-2.2.0 just now. configure worked fine, but compilation stopped with ld: Undefined symbols: restFP saveFP /usr/bin/libtool: internal link edit command failed make[3]: *** [libR.dylib] Error 1 make[2]: *** [R] Error 2 make[1]: *** [R] Error 1 make: *** [R] Error 1 octopus:~/scratch/R-2.2.0% This has been written up many times already. It would be nice if it was in the documentation. g77 can only be used with gcc-3.x and earlier. The default compiler on Tiger is gcc-4. All you need to do to build with g77 on Tiger is use the command sudo gcc_select 3.3 which will make gcc-3.3 the default compiler and it will all work. Bill Northcott __ 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
Re: [R] Warning: condition has length 1
ifelse() should work too.
Andrew
On Sun, Oct 09, 2005 at 03:30:42PM +1000, sosman wrote:
nawaf b wrote:
Hi -
I was wondering if anyone came across a problem such as mine! I was trying
to create a user-defined function, however, the results from calling the
function are unexpected!
When passing X parameter as a single value variable (x-c(3)), everything
works fine. However, when passing a parameter as a vector with multiple
values (as the case in my code), the 'if condition' is only executed once!
Here is my code:
x
[1] 0.3 1.0 0.7 22.0
myfunction-function(x)
+ {
+ if (x=1) 0
+ else x^2
+ }
myfunction(x)
[1] 0.09 1.00 0.49 484.00
Warning message:
the condition has length 1 and only the first element will be used in: if
(x = 1) 0 else x^2
Is there a way to overcome this issue? Can you please explain what
modifications to the code I need to accommodate to make it work.
I am going to go out on a limb here (I don't know much about R) but I'm
not sure it has anything to do with that fact that you have defined a
function.
It seems to me that if (x 1) {...} expects a scalar and if x is not
a scalar, it will use the first value. To apply your function to each
element, check out the help for sapply (eg ?sapply). I am guessing the
solution is something like sapply(x, myfunction) but don't quote me.
paul sorenson
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--
Andrew Robinson
Senior Lecturer in Statistics Tel: +61-3-8344-9763
Department of Mathematics and StatisticsFax: +61-3-8344-4599
University of Melbourne, VIC 3010 Australia
Email: [EMAIL PROTECTED]Website: http://www.ms.unimelb.edu.au
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