Re: [R] fixed effects
There is a very similar example worked through in section 7.2 of `S
Programming'.
On Mon, 27 Mar 2006, Liaw, Andy wrote:
> Hmm... didn't read the whole post before I hit `send'...
>
> I think you basically will have to fit the model `by hand', which is not
> that hard, given the simple structure of the model(s). The formulae for the
> quantities of interests are quite straightforward and easy to code in R
> (similar to the rsq function I showed). Adding a continuous variable to the
> model simply requires regressing the residuals; i.e., ave(y, x, function(x)
> x - mean(x)), on the new variable z (easy for lm()), and take one more
> degree of freedom from SS(total).
>
> Andy
>
> From: Liaw, Andy
>>
>> I guess you meant X has 20,000 levels (and 40 observations
>> each)? In that case lm() will attempt to create a design
>> matrix that's 8e5 by 2e4, and that's unlikely to fit in the RAM.
>>
>> It is very easy to compute by hand. I'm using a smaller data
>> size first to check the result against summary(lm()), then
>> the size you specified (hopefully with more correct
>> arithmetics...) to show that it's quite doable even on a
>> modest laptop:
>>
>>> set.seed(1)
>>> y <- rnorm(80)
>>> x <- factor(rep(paste("x", 1:20, sep=""), 4))
>>> summary(lm(y ~ x))$r.squared
>> [1] 0.2555885
>>> rsq <- function(x, y) {
>> + sse <- sum(ave(y, x, FUN=function(x) x - mean(x))^2)
>> + sstotal <- var(y) * (length(y) - 1)
>> + 1 - sse / sstotal
>> + }
>>> rsq(x, y)
>> [1] 0.2555885
>>> set.seed(1)
>>> y <- rnorm(8e5)
>>> x <- factor(rep(paste("x", 1:2e4, sep=""), 40))
>> system.time(rsq(x, y))
>> [1] 1.99 0.03 2.06 NA NA
>>
>> Andy
>>
>>
>> From: ivo welch
>>>
>>> dear R wizards:
>>>
>>> X is factor with 20,000*20=800,000 observations of 20,000 factors.
>>> I.e., each factor has 20 observations. y is 800,000 normally
>>> distributed data points. I want to see how much R^2 the X
>>> factors can provide. Easy, right?
>>>
lm ( y ~ X)
>>> and
aov( y ~ X)
>>> Error: cannot allocate vector of size 3125000 Kb
>>>
>>> is this computationally infeasible? (I am not an expert, but
>>> off-hand, I thought this can work as long as the X's are just
>>> factors = fitted means.)
>>>
help.search("fixed.effects");
>>>
>>> fixed.effects(nlme) Extract Fixed Effects
>>> fixed.effects.lmList(nlme) Extract lmList Fixed Effects
>>> lme(nlme) Linear Mixed-Effects Models
>>> lmeStruct(nlme) Linear Mixed-Effects Structure
>>> nlme(nlme) Nonlinear Mixed-Effects Models
>>> nlmeStruct(nlme)Nonlinear Mixed-Effects
>> Structure
>>>
>>> Type 'help(FOO, package = PKG)' to inspect entry 'FOO(PKG) TITLE'.
>>>
>>> ok---I want to read the fixed.effects help. could the help
>>> system tell me how to inspect these entries?
>>>
help("fixed.effects")
>>> wrong
help.search("fixed.effects")
>>> two entries, as above. nothing new.
?fixed.effects
>>> wrong
>>>
>>> eventually, it dawned on me that nlme in parens was not a
>>> function argument, but the name of the package within which
>>> fixed.effects lives. Suggestion: maybe a different notation
>>> to name packages would be more intuitive in the help system.
>>> yes, I know it now, but other novices may not. even a colon
>>> instead of a () may be more intuitive in this context.
>>>
library(nlme); ?lme
>>> and then
lme(y ~ X)
>>> Error in getGroups.data.frame(dataMix, groups) :
>>> Invalid formula for groups
>>>
>>>
>>> now I have to beg for some help. ok, blatant free-riding.
>>> the lme docs tells me it does the Laird and Ware model, but I
>>> do not know this model. the only two examples given at the
>>> end of the lme help file seem to be similar to what I just
>>> specified. so, how do I execute a simple fixed-effects
>>> model? (later, I may want to add a variable Z that is a
>>> continuous random variable.) could someone please give me
>>> one quick example? help is, as always, highly appreciated.
>>>
>>> sincerely,
>>>
>>> /ivo welch
>>>
>>> __
>>> 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-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 confidentia
Re: [R] Having trouble with tsdiag function on a time series
tsdiag is for diagnostic plots from time-series fits, not time series. On Tue, 28 Mar 2006, Rafael Algara wrote: > Hello, > > I'm getting the following error message when I try to run 'tsdiag' on what > seems to be a valid time series: > >> tsdiag(small) > > returns: > > [Error in tsdiag(small) : no applicable method for "tsdiag"] > > where small is a little test series where I have isolated this problem (the > original has 30-years worth of daily data) > > When I print it (small), it looks like the following: > > Time Series: > Start = c(1990, 1) > End = c(1990, 31) > Frequency = 365 > [1] 0.0 16.0 10.0 0.0 0.0 0.0 0.0 0.0 2.0 2.2 0.0 0.0 0.0 0.0 > 0.0 0.0 0.0 2.5 0.1 > [20] 2.4 0.0 0.0 0.0 0.0 9.6 0.0 0.0 0.0 0.0 25.2 7.0 > > I've found some postings in the r-bugs list which suggest that "no > applicable method" can stem from a conflict across namespaces. I've > tried qualifying the call: Not relevant. Try > methods(tsdiag) [1] tsdiag.Arima*tsdiag.arima0* tsdiag.StructTS* Non-visible functions are asterisked > >> stats::tsdiag(small) > > with the same result. > > Many thanks in advance for any help. > > Rafael > [[alternative HTML version deleted]] > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html PLEASE do, and do not send HTML mail as we ask. -- 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] Default lag.max in ACF
The default is taken from S-PLUS, so the reference is the S-PLUS manual. It is pretty similar to the recommendation of Brockwell & Davis. On Mon, 27 Mar 2006, Spencer Graves wrote: > I don't know why the default lag.max is 10*log10(N/m) for m series. > The "acf" help page includes the following: > > Author(s): > > Original: Paul Gilbert, Martyn Plummer. Extensive modifications > and univariate case of 'pacf' by B.D. Ripley. > > I've copied these three contributors on this email. With luck, one > of them will enlighten us both. > > Best Wishes, > spencer graves > > Rafal Stankiewicz wrote: > >> Hi, >> >> The default value for lag.max in ACF implementation is 10*log10(N) >> >> There several publications recommending setting lag.max to: >> - N/4 (Box and Jenkins, 1970; Chatfield, 1975; Anderson, 1976; Pankratz, >> 1983; Davis, 1986; etc.) >> - sqrt(N)+10 (Cryer, 1986) >> - 20<=N<=40 (Brockwell and Davis) >> >> Why R uses 10*log10(N) as a default? >> >> Please, give me a reference to a book or article where the recommendation >> for using lag.max=10*log10(N) is proposed and explained. >> >> Thanks >> Rafal -- 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
[R] Having trouble with tsdiag function on a time series
Hello, I'm getting the following error message when I try to run 'tsdiag' on what seems to be a valid time series: > tsdiag(small) returns: [Error in tsdiag(small) : no applicable method for "tsdiag"] where small is a little test series where I have isolated this problem (the original has 30-years worth of daily data) When I print it (small), it looks like the following: Time Series: Start = c(1990, 1) End = c(1990, 31) Frequency = 365 [1] 0.0 16.0 10.0 0.0 0.0 0.0 0.0 0.0 2.0 2.2 0.0 0.0 0.0 0.0 0.0 0.0 0.0 2.5 0.1 [20] 2.4 0.0 0.0 0.0 0.0 9.6 0.0 0.0 0.0 0.0 25.2 7.0 I've found some postings in the r-bugs list which suggest that "no applicable method" can stem from a conflict across namespaces. I've tried qualifying the call: > stats::tsdiag(small) with the same result. Many thanks in advance for any help. Rafael [[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] fixed effects
Have you tried the following:
lme(y~1, random=~1|X, data=DF)
where DF = a data.frame with columns y and X.
The authoritative reference on library(nlme) is Pinheiro and Bates
(2000) Mixed-Effects Models in S and S-Plus (Springer). I've learned a
lot from Bates and from this book in particular.
hope this helps,
spencer graves
ivo welch wrote:
> dear R wizards:
>
> X is factor with 20,000*20=800,000 observations of 20,000 factors.
> I.e., each factor has 20 observations. y is 800,000 normally
> distributed data points. I want to see how much R^2 the X factors can
> provide. Easy, right?
>
>
>>lm ( y ~ X)
>
> and
>
>> aov( y ~ X)
>
> Error: cannot allocate vector of size 3125000 Kb
>
> is this computationally infeasible? (I am not an expert, but
> off-hand, I thought this can work as long as the X's are just factors
> = fitted means.)
>
>
>>help.search("fixed.effects");
>
>
> fixed.effects(nlme) Extract Fixed Effects
> fixed.effects.lmList(nlme) Extract lmList Fixed Effects
> lme(nlme) Linear Mixed-Effects Models
> lmeStruct(nlme) Linear Mixed-Effects Structure
> nlme(nlme) Nonlinear Mixed-Effects Models
> nlmeStruct(nlme)Nonlinear Mixed-Effects Structure
>
> Type 'help(FOO, package = PKG)' to inspect entry 'FOO(PKG) TITLE'.
>
> ok---I want to read the fixed.effects help. could the help system
> tell me how to inspect these entries?
>
>
>>help("fixed.effects")
>
> wrong
>
>>help.search("fixed.effects")
>
> two entries, as above. nothing new.
>
>>?fixed.effects
>
> wrong
>
> eventually, it dawned on me that nlme in parens was not a function
> argument, but the name of the package within which fixed.effects
> lives. Suggestion: maybe a different notation to name packages would
> be more intuitive in the help system. yes, I know it now, but other
> novices may not. even a colon instead of a () may be more intuitive
> in this context.
>
>
>>library(nlme); ?lme
>
> and then
>
>>lme(y ~ X)
>
> Error in getGroups.data.frame(dataMix, groups) :
> Invalid formula for groups
>
>
> now I have to beg for some help. ok, blatant free-riding. the lme
> docs tells me it does the Laird and Ware model, but I do not know this
> model. the only two examples given at the end of the lme help file
> seem to be similar to what I just specified. so, how do I execute a
> simple fixed-effects model? (later, I may want to add a variable Z
> that is a continuous random variable.) could someone please give me
> one quick example? help is, as always, highly appreciated.
>
> sincerely,
>
> /ivo welch
>
> __
> 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-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] fixed effects
Hmm... didn't read the whole post before I hit `send'...
I think you basically will have to fit the model `by hand', which is not
that hard, given the simple structure of the model(s). The formulae for the
quantities of interests are quite straightforward and easy to code in R
(similar to the rsq function I showed). Adding a continuous variable to the
model simply requires regressing the residuals; i.e., ave(y, x, function(x)
x - mean(x)), on the new variable z (easy for lm()), and take one more
degree of freedom from SS(total).
Andy
From: Liaw, Andy
>
> I guess you meant X has 20,000 levels (and 40 observations
> each)? In that case lm() will attempt to create a design
> matrix that's 8e5 by 2e4, and that's unlikely to fit in the RAM.
>
> It is very easy to compute by hand. I'm using a smaller data
> size first to check the result against summary(lm()), then
> the size you specified (hopefully with more correct
> arithmetics...) to show that it's quite doable even on a
> modest laptop:
>
> > set.seed(1)
> > y <- rnorm(80)
> > x <- factor(rep(paste("x", 1:20, sep=""), 4))
> > summary(lm(y ~ x))$r.squared
> [1] 0.2555885
> > rsq <- function(x, y) {
> + sse <- sum(ave(y, x, FUN=function(x) x - mean(x))^2)
> + sstotal <- var(y) * (length(y) - 1)
> + 1 - sse / sstotal
> + }
> > rsq(x, y)
> [1] 0.2555885
> > set.seed(1)
> > y <- rnorm(8e5)
> > x <- factor(rep(paste("x", 1:2e4, sep=""), 40))
> system.time(rsq(x, y))
> [1] 1.99 0.03 2.06 NA NA
>
> Andy
>
>
> From: ivo welch
> >
> > dear R wizards:
> >
> > X is factor with 20,000*20=800,000 observations of 20,000 factors.
> > I.e., each factor has 20 observations. y is 800,000 normally
> > distributed data points. I want to see how much R^2 the X
> > factors can provide. Easy, right?
> >
> > > lm ( y ~ X)
> > and
> > > aov( y ~ X)
> > Error: cannot allocate vector of size 3125000 Kb
> >
> > is this computationally infeasible? (I am not an expert, but
> > off-hand, I thought this can work as long as the X's are just
> > factors = fitted means.)
> >
> > > help.search("fixed.effects");
> >
> > fixed.effects(nlme) Extract Fixed Effects
> > fixed.effects.lmList(nlme) Extract lmList Fixed Effects
> > lme(nlme) Linear Mixed-Effects Models
> > lmeStruct(nlme) Linear Mixed-Effects Structure
> > nlme(nlme) Nonlinear Mixed-Effects Models
> > nlmeStruct(nlme)Nonlinear Mixed-Effects
> Structure
> >
> > Type 'help(FOO, package = PKG)' to inspect entry 'FOO(PKG) TITLE'.
> >
> > ok---I want to read the fixed.effects help. could the help
> > system tell me how to inspect these entries?
> >
> > > help("fixed.effects")
> > wrong
> > > help.search("fixed.effects")
> > two entries, as above. nothing new.
> > > ?fixed.effects
> > wrong
> >
> > eventually, it dawned on me that nlme in parens was not a
> > function argument, but the name of the package within which
> > fixed.effects lives. Suggestion: maybe a different notation
> > to name packages would be more intuitive in the help system.
> > yes, I know it now, but other novices may not. even a colon
> > instead of a () may be more intuitive in this context.
> >
> > > library(nlme); ?lme
> > and then
> > > lme(y ~ X)
> > Error in getGroups.data.frame(dataMix, groups) :
> > Invalid formula for groups
> >
> >
> > now I have to beg for some help. ok, blatant free-riding.
> > the lme docs tells me it does the Laird and Ware model, but I
> > do not know this model. the only two examples given at the
> > end of the lme help file seem to be similar to what I just
> > specified. so, how do I execute a simple fixed-effects
> > model? (later, I may want to add a variable Z that is a
> > continuous random variable.) could someone please give me
> > one quick example? help is, as always, highly appreciated.
> >
> > sincerely,
> >
> > /ivo welch
> >
> > __
> > 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-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 n
Re: [R] fixed effects
I guess you meant X has 20,000 levels (and 40 observations each)? In that
case lm() will attempt to create a design matrix that's 8e5 by 2e4, and
that's unlikely to fit in the RAM.
It is very easy to compute by hand. I'm using a smaller data size first to
check the result against summary(lm()), then the size you specified
(hopefully with more correct arithmetics...) to show that it's quite doable
even on a modest laptop:
> set.seed(1)
> y <- rnorm(80)
> x <- factor(rep(paste("x", 1:20, sep=""), 4))
> summary(lm(y ~ x))$r.squared
[1] 0.2555885
> rsq <- function(x, y) {
+ sse <- sum(ave(y, x, FUN=function(x) x - mean(x))^2)
+ sstotal <- var(y) * (length(y) - 1)
+ 1 - sse / sstotal
+ }
> rsq(x, y)
[1] 0.2555885
> set.seed(1)
> y <- rnorm(8e5)
> x <- factor(rep(paste("x", 1:2e4, sep=""), 40))
> system.time(rsq(x, y))
[1] 1.99 0.03 2.06 NA NA
Andy
From: ivo welch
>
> dear R wizards:
>
> X is factor with 20,000*20=800,000 observations of 20,000 factors.
> I.e., each factor has 20 observations. y is 800,000 normally
> distributed data points. I want to see how much R^2 the X
> factors can provide. Easy, right?
>
> > lm ( y ~ X)
> and
> > aov( y ~ X)
> Error: cannot allocate vector of size 3125000 Kb
>
> is this computationally infeasible? (I am not an expert, but
> off-hand, I thought this can work as long as the X's are just
> factors = fitted means.)
>
> > help.search("fixed.effects");
>
> fixed.effects(nlme) Extract Fixed Effects
> fixed.effects.lmList(nlme) Extract lmList Fixed Effects
> lme(nlme) Linear Mixed-Effects Models
> lmeStruct(nlme) Linear Mixed-Effects Structure
> nlme(nlme) Nonlinear Mixed-Effects Models
> nlmeStruct(nlme)Nonlinear Mixed-Effects Structure
>
> Type 'help(FOO, package = PKG)' to inspect entry 'FOO(PKG) TITLE'.
>
> ok---I want to read the fixed.effects help. could the help
> system tell me how to inspect these entries?
>
> > help("fixed.effects")
> wrong
> > help.search("fixed.effects")
> two entries, as above. nothing new.
> > ?fixed.effects
> wrong
>
> eventually, it dawned on me that nlme in parens was not a
> function argument, but the name of the package within which
> fixed.effects lives. Suggestion: maybe a different notation
> to name packages would be more intuitive in the help system.
> yes, I know it now, but other novices may not. even a colon
> instead of a () may be more intuitive in this context.
>
> > library(nlme); ?lme
> and then
> > lme(y ~ X)
> Error in getGroups.data.frame(dataMix, groups) :
> Invalid formula for groups
>
>
> now I have to beg for some help. ok, blatant free-riding.
> the lme docs tells me it does the Laird and Ware model, but I
> do not know this model. the only two examples given at the
> end of the lme help file seem to be similar to what I just
> specified. so, how do I execute a simple fixed-effects
> model? (later, I may want to add a variable Z that is a
> continuous random variable.) could someone please give me
> one quick example? help is, as always, highly appreciated.
>
> sincerely,
>
> /ivo welch
>
> __
> 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-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] Help understanding behavior of apply vs sapply
I'll give it a shot: apply() is only a wrapper around a for loop through the requested dimension(s). In this case it would run zls() once for each row in m; i.e., twice. The `Value' section of apply() explains what it does if the function being applied returns vector of length 0. sapply() is basically lapply() with nice post-processing if possible. It doesn't know about m being 2x2, just that it's a vector with four elements, so it's going to call the function four times, and return a list with the result of each of those calls. Andy From: Seth Falcon > > Hi, > > I was surprised that apply and sapply don't return the same > results in the example below. Can someone tell me what I'm missing? > > > > zls <- function(x) character(0) > > m <- matrix(0, nrow=2, ncol=2) > > apply(m, 1, zls) > character(0) > > sapply(m, zls) > [[1]] > character(0) > > [[2]] > character(0) > > [[3]] > character(0) > > [[4]] > character(0) > > > > > > > > R.version >_ > platform powerpc-apple-darwin8.5.0 > arch powerpc > os darwin8.5.0 > system powerpc, darwin8.5.0 > status alpha > major 2 > minor 3.0 > year 2006 > month 03 > day27 > svn rev37590 > language R > version.string Version 2.3.0 alpha (2006-03-27 r37590) > > __ > 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-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] fixed effects
dear R wizards:
X is factor with 20,000*20=800,000 observations of 20,000 factors.
I.e., each factor has 20 observations. y is 800,000 normally
distributed data points. I want to see how much R^2 the X factors can
provide. Easy, right?
> lm ( y ~ X)
and
> aov( y ~ X)
Error: cannot allocate vector of size 3125000 Kb
is this computationally infeasible? (I am not an expert, but
off-hand, I thought this can work as long as the X's are just factors
= fitted means.)
> help.search("fixed.effects");
fixed.effects(nlme) Extract Fixed Effects
fixed.effects.lmList(nlme) Extract lmList Fixed Effects
lme(nlme) Linear Mixed-Effects Models
lmeStruct(nlme) Linear Mixed-Effects Structure
nlme(nlme) Nonlinear Mixed-Effects Models
nlmeStruct(nlme)Nonlinear Mixed-Effects Structure
Type 'help(FOO, package = PKG)' to inspect entry 'FOO(PKG) TITLE'.
ok---I want to read the fixed.effects help. could the help system
tell me how to inspect these entries?
> help("fixed.effects")
wrong
> help.search("fixed.effects")
two entries, as above. nothing new.
> ?fixed.effects
wrong
eventually, it dawned on me that nlme in parens was not a function
argument, but the name of the package within which fixed.effects
lives. Suggestion: maybe a different notation to name packages would
be more intuitive in the help system. yes, I know it now, but other
novices may not. even a colon instead of a () may be more intuitive
in this context.
> library(nlme); ?lme
and then
> lme(y ~ X)
Error in getGroups.data.frame(dataMix, groups) :
Invalid formula for groups
now I have to beg for some help. ok, blatant free-riding. the lme
docs tells me it does the Laird and Ware model, but I do not know this
model. the only two examples given at the end of the lme help file
seem to be similar to what I just specified. so, how do I execute a
simple fixed-effects model? (later, I may want to add a variable Z
that is a continuous random variable.) could someone please give me
one quick example? help is, as always, highly appreciated.
sincerely,
/ivo welch
__
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] Help understanding behavior of apply vs sapply
Hi, I was surprised that apply and sapply don't return the same results in the example below. Can someone tell me what I'm missing? > zls <- function(x) character(0) > m <- matrix(0, nrow=2, ncol=2) > apply(m, 1, zls) character(0) > sapply(m, zls) [[1]] character(0) [[2]] character(0) [[3]] character(0) [[4]] character(0) > R.version _ platform powerpc-apple-darwin8.5.0 arch powerpc os darwin8.5.0 system powerpc, darwin8.5.0 status alpha major 2 minor 3.0 year 2006 month 03 day27 svn rev37590 language R version.string Version 2.3.0 alpha (2006-03-27 r37590) __ 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] error message
Hi Does anyone know what this means: > glm.model = glm(formula = as.factor(nextDay) ~ ., family=binomial, data=spi[1:1000,]) > pred <- predict(glm.model, spi[1001:1250,-9], type="response") Warning message: prediction from a rank-deficient fit may be misleading in: predict.lm(object, newdata, se.fit, scale = 1, type = ifelse(type == 9 is my determinant and I still get this message even when I remove the 9. Stephen [[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] products and polynomials in formulae
Hi I can do this: formula = as.factor(outcome) ~ . in glm and other model building functions. I think there is a way to get the product of the determinants (that is d1 * d2, d1 * d3, etc) and also another way to get all the polynomials (that is like poly(d1,2) would produce for a single determinant). Can anyone tell me how you write them? Stephen [[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] why does lmer not give p valuses for quasibinomial family?
Dear All, I'm running a binomial model using the lmer() function, and I get p values for the parameter estimates only with family=binomial, but not with quasibinomial? Why is that so? I wanted to use quasibinomial family, because my data were overdispersed. Thanks, Istvan __ 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] graphing and scrolling
Does the following (or some simple modification of it) do what you
want?:
library(tkrplot)
y <- rnorm(1, 10, 2) + 5*sin( (1:1)/1000 )
tt <- tktoplevel()
left <- tclVar(1)
oldleft <- tclVar(1)
right <- tclVar(100)
f1 <- function(){
lleft <- as.numeric(tclvalue(left))
rright <- as.numeric(tclvalue(right))
x <- seq(lleft,rright,by=1)
plot(x,y[x], type='b',ylim=range(y))
}
img <- tkrplot(tt, f1)
f2 <- function(...){
ol <- as.numeric(tclvalue(oldleft))
tclvalue(oldleft) <- tclvalue(left)
r <- as.numeric(tclvalue(right))
tclvalue(right) <- as.character(r + as.numeric(...) - ol)
tkrreplot(img)
}
f3 <- function(...){
tkrreplot(img)
}
f4 <- function(...){
ol <- as.numeric(tclvalue(oldleft))
tclvalue(left) <- as.character(ol+100)
tclvalue(oldleft) <- as.character(ol+100)
r <- as.numeric(tclvalue(right))
tclvalue(right) <- as.character(r+100)
tkrreplot(img)
}
s1 <- tkscale(tt, command=f2, from=1, to=length(y),
variable=left, orient="horiz",label='left')
s2 <- tkscale(tt, command=f3, from=1, to=length(y),
variable=right, orient="horiz",label='right')
b1 <- tkbutton(tt, text='->', command=f4)
tkpack(img,s1,s2,b1)
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Fred J.
> Sent: Monday, March 27, 2006 1:05 PM
> To: r-help@stat.math.ethz.ch
> Subject: [R] graphing and scrolling
>
> Dear R users
>
> graphing with plot(x) seams to work for a small length(x),
> when length(x) is too large it seams to clutter the display,
> a solution would be to display subsets of x at a time, yet a
> better way which I hope R supports is to place a sliding bar
> on the display window to control length(x) and thus the
> resolution, which will involve auto scaling the y axis as
> well as automatically place a side-to-side scrolling bar at
> the bottom of the chart to move through and display the
> subsets of x according to the resolution. is there such an
> implementation?
>
> thank you
>
>
> -
>
> [[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
>
__
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[R] DSC 2007
Hi Following on from the "Distributed Statistical Computing" conferences in Vienna (1999, 2001, 2003) and the "Directions in Statistical Computing" conference in Seattle last year ... DSC 2007, a conference on systems and environments for statistical computing, will take place in Auckland, New Zealand on February 15 & 16, 2007. This is just an announcement of date and location so that interested parties living on the far side of the planet have time to book their travel; further details and a web site for the conference will be announced in due course. Paul -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 [EMAIL PROTECTED] http://www.stat.auckland.ac.nz/~paul/ ___ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-announce __ 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] graphing and scrolling
Dear R users graphing with plot(x) seams to work for a small length(x), when length(x) is too large it seams to clutter the display, a solution would be to display subsets of x at a time, yet a better way which I hope R supports is to place a sliding bar on the display window to control length(x) and thus the resolution, which will involve auto scaling the y axis as well as automatically place a side-to-side scrolling bar at the bottom of the chart to move through and display the subsets of x according to the resolution. is there such an implementation? thank you - [[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] reading in multi-dimensional data from .csv
How is the data organized? You could 'linearize' the data (e.g., column order) and supply the dimensions of the data that you would apply after the data was read in. If you had three dimensions and you separated each 2D page with a blank line, you again could reconstruct the data. If you have the ability to change the program doing the output, then look at the output of 'dput' for a multidimensional object and create a text string that looks like that and 'source' it in. On 3/27/06, Werner Wernersen <[EMAIL PROTECTED]> wrote: > > Hi, > > I would like to read in multi-dimensional data from a > text file, i.e. "tables" with more than 2 dimensions. > I have looked for a function which I can abuse for > that but haven't found anything. > > I would appreciate it a lot if somebody gave me a hint > if such functions already exist somewhere. > > Thanks, > Werner > > __ > 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 > -- Jim Holtman Cincinnati, OH +1 513 646 9390 (Cell) +1 513 247 0281 (Home) What the problem you are trying to solve? [[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] reading in multi-dimensional data from .csv
Hi, I would like to read in multi-dimensional data from a text file, i.e. "tables" with more than 2 dimensions. I have looked for a function which I can abuse for that but haven't found anything. I would appreciate it a lot if somebody gave me a hint if such functions already exist somewhere. Thanks, Werner __ 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] Arrays of functions or results of functions.
If I understand you correctly, I would say that this is a standard analysis
of covariance problem that you are approaching incorrectly. You should not
be testing for "equal variances," IMO. Instead,
1. Combine all your data into 3 columnsS x, y, and group= subset.
2. Model.1: y ~ x
3. Model.2: y ~ x*group. This gives a separate slope and intercept for each
group.
4. anova(Model.1, Model.2) to compare.
A better approach would be to use lmList() from the nlme package. Better yet
would be model the data as a mixed effect model via lme(), which assumes
that slopes and intercepts vary randomly among your subsets about their
corresponfding central values. This gets you shrinkage, which is highly
desirable. See Bates and Pinheiro "MIXED EFFECTS MODELS IN S AND S-PLUS" for
details.
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
"The business of the statistician is to catalyze the scientific learning
process." - George E. P. Box
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Gabor
> Grothendieck
> Sent: Monday, March 27, 2006 9:48 AM
> To: Gary Mallard
> Cc: r-help@stat.math.ethz.ch
> Subject: Re: [R] Arrays of functions or results of functions.
>
> Is your question how to run a regression with separate slopes
> and then with one slope and then to complare them? If that
> is it here is an example:
>
> # test data - ind is factor which defines the groups
> set.seed(1)
> y1 <- 10 + 20 * seq(100) + rnorm(100)
> y2 <- -200 + 35 * seq(100) + rnorm(100)
> yy <- pmax(y1, y2)
> ind <- factor(y1 > y2)
>
> # lm.N has N slopes
> lm.1 <- lm(yy ~ ind + seq(100) - 1)
> lm.2 <- lm(yy ~ ind/seq(100)-1)
> lm.1
> lm.2
> anova(lm.1, lm.2)
>
> On 3/27/06, Gary Mallard <[EMAIL PROTECTED]> wrote:
> > The general problem I am trying to solve is to determine if
> a series of
> > subsets of data can be described with a single regression
> slope. This
> > involves fitting the data to each subset, calculating a joint slope
> > followed by F tests to show that the variances are equal
> the final slope is
> > valid.
> > The data for is characterized by a parameter PC (for the 4
> - in this case)
> > sets and a dependent variable RI and an independent variable ROH.
> > The data are contained in a variable "joint".
> > The joint has been attached and has RI, ROH and PC for each element.
> > The following gives the initial results:
> > Mline<-lm(RI[PC==1]~ROH[PC==1])
> > Eline<-lm(RI[PC==2]~ROH[PC==2])
> > Iline<-lm(RI[PC==3]~ROH[PC==3])
> > Pline<-lm(RI[PC==4]~ROH[PC==4])
> >
> >
> > joint_reduced <- joint;
> > for(i in 1:4) {
> >
> joint_reduced$RI[joint_reduced$PC==i]<-joint$RI[joint$PC==i]-m
> ean(joint$RI[joint$PC==1]}
> > AllLine<-lm(joint_reduced$RI~joint_reduced$ROH);
> >
> > Now the statistics from AllLine can be compared with each
> of the individual
> > statistics.
> >
> > NOW THE QUESTION:
> > From a lot of point of view it would be useful to have a parameter
> > generated by
> > for (i in 1:4){ Xline[i]=lm(RI[PC==i]~ROH[PC==i])}
> > And now all of the work of comparison can be done with
> calls to Xline[i]
> > rather than having to work individually with Mline, Eline, etc.
> >
> > This appears to be impossible. The constructor for Xline[i] is not
> > automatic (as it is for Mline, etc) noted above. I cannot
> determine how to
> > construct the Xline[i] object so that this kind of process can be
> > generalized. Is it possible? Is there another way to set
> us such tests of
> > multiple line linearity that is already in a package?
> >
> > Comments or pointers would be appreciated.
> > Gary
> >
> > __
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide!
> http://www.R-project.org/posting-guide.html
> >
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
> http://www.R-project.org/posting-guide.html
>
__
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Glm poisson
Why are you trying to extract the values by calling a function with the name of the value? glm objects are stored as a list i.e. str(pAmeir_1) Hence, you can extract what you need by selecting the values on the list i.e. pAmeir_1$df.null pAmeir_1$null.deviance Cheers Francisco >From: "Guenther, Cameron" <[EMAIL PROTECTED]> >To: <[EMAIL PROTECTED]>, >Subject: [R] Glm poisson >Date: Mon, 27 Mar 2006 12:07:04 -0500 > >Hello, >I am using the glm model with a poisson distribution. The model runs >just fine but when I try to get the null deviance for the model of the >null degrees of freedom I get the following errors: > > > null.deviance(pAmeir_1) >Error: couldn't find function "null.deviance" > > > df.null(pAmeir_1) >Error: couldn't find function "df.null" > >When I do: > > > names(pAmeir_1) > [1] "coefficients" "residuals" "fitted.values" > [4] "effects" "R" "rank" > [7] "qr""family""linear.predictors" >[10] "deviance" "aic" "null.deviance" >[13] "iter" "weights" "prior.weights" >[16] "df.residual" "df.null" "y" >[19] "converged" "boundary" "model" >[22] "call" "formula" "terms" >[25] "data" "offset""control" >[28] "method""contrasts" "xlevels" > >It tells me that the values are there but I can not access them. > >Any comments would be greatly appreciated. > >Thank you > >Cameron Guenther, Ph.D. >Associate Research Scientist >FWC/FWRI, Marine Fisheries Research >100 8th Avenue S.E. >St. Petersburg, FL 33701 >(727)896-8626 Ext. 4305 >[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 __ 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] Arrays of functions or results of functions.
Is your question how to run a regression with separate slopes
and then with one slope and then to complare them? If that
is it here is an example:
# test data - ind is factor which defines the groups
set.seed(1)
y1 <- 10 + 20 * seq(100) + rnorm(100)
y2 <- -200 + 35 * seq(100) + rnorm(100)
yy <- pmax(y1, y2)
ind <- factor(y1 > y2)
# lm.N has N slopes
lm.1 <- lm(yy ~ ind + seq(100) - 1)
lm.2 <- lm(yy ~ ind/seq(100)-1)
lm.1
lm.2
anova(lm.1, lm.2)
On 3/27/06, Gary Mallard <[EMAIL PROTECTED]> wrote:
> The general problem I am trying to solve is to determine if a series of
> subsets of data can be described with a single regression slope. This
> involves fitting the data to each subset, calculating a joint slope
> followed by F tests to show that the variances are equal the final slope is
> valid.
> The data for is characterized by a parameter PC (for the 4 - in this case)
> sets and a dependent variable RI and an independent variable ROH.
> The data are contained in a variable "joint".
> The joint has been attached and has RI, ROH and PC for each element.
> The following gives the initial results:
> Mline<-lm(RI[PC==1]~ROH[PC==1])
> Eline<-lm(RI[PC==2]~ROH[PC==2])
> Iline<-lm(RI[PC==3]~ROH[PC==3])
> Pline<-lm(RI[PC==4]~ROH[PC==4])
>
>
> joint_reduced <- joint;
> for(i in 1:4) {
>
> joint_reduced$RI[joint_reduced$PC==i]<-joint$RI[joint$PC==i]-mean(joint$RI[joint$PC==1]}
> AllLine<-lm(joint_reduced$RI~joint_reduced$ROH);
>
> Now the statistics from AllLine can be compared with each of the individual
> statistics.
>
> NOW THE QUESTION:
> From a lot of point of view it would be useful to have a parameter
> generated by
> for (i in 1:4){ Xline[i]=lm(RI[PC==i]~ROH[PC==i])}
> And now all of the work of comparison can be done with calls to Xline[i]
> rather than having to work individually with Mline, Eline, etc.
>
> This appears to be impossible. The constructor for Xline[i] is not
> automatic (as it is for Mline, etc) noted above. I cannot determine how to
> construct the Xline[i] object so that this kind of process can be
> generalized. Is it possible? Is there another way to set us such tests of
> multiple line linearity that is already in a package?
>
> Comments or pointers would be appreciated.
> Gary
>
> __
> 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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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Differential Equations
Dominik, Adding (1) and (2) yields, A(t) + B(t) = constant = A(0) + B(0) = c So, plug in B = c - A in (1) and solve for A. This should be an easy solution. Hope this is helpful, Ravi. > -Original Message- > From: [EMAIL PROTECTED] [mailto:r-help- > [EMAIL PROTECTED] On Behalf Of Peter Dalgaard > Sent: Monday, March 27, 2006 11:37 AM > To: Dominik Heinzmann > Cc: r-help@stat.math.ethz.ch > Subject: Re: [R] Differential Equations > > Dominik Heinzmann <[EMAIL PROTECTED]> writes: > > > Dear R-community > > > > My ODE problems looks as follows: > > (1) dA/dt = u*A - v*B > > (2) dB/dt = v*B - u*A > > > > where u is a constant, and v=k*t (k=constant, t=time) > > > > Does anybody knows a good function/procedure of solving? Should one > > involve the equation (3) dv/dt = k? > > Thanks for your support. > > You probably need to look into the odesolve package. The v*B terms > make it nonlinear, so I wouldn't expect an analytic solution. > > -- >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 __ 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] Glm poisson
Hello, I am using the glm model with a poisson distribution. The model runs just fine but when I try to get the null deviance for the model of the null degrees of freedom I get the following errors: > null.deviance(pAmeir_1) Error: couldn't find function "null.deviance" > df.null(pAmeir_1) Error: couldn't find function "df.null" When I do: > names(pAmeir_1) [1] "coefficients" "residuals" "fitted.values" [4] "effects" "R" "rank" [7] "qr""family""linear.predictors" [10] "deviance" "aic" "null.deviance" [13] "iter" "weights" "prior.weights" [16] "df.residual" "df.null" "y" [19] "converged" "boundary" "model" [22] "call" "formula" "terms" [25] "data" "offset""control" [28] "method""contrasts" "xlevels" It tells me that the values are there but I can not access them. Any comments would be greatly appreciated. Thank you Cameron Guenther, Ph.D. Associate Research Scientist FWC/FWRI, Marine Fisheries Research 100 8th Avenue S.E. St. Petersburg, FL 33701 (727)896-8626 Ext. 4305 [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
[R] Arrays of functions or results of functions.
The general problem I am trying to solve is to determine if a series of
subsets of data can be described with a single regression slope. This
involves fitting the data to each subset, calculating a joint slope
followed by F tests to show that the variances are equal the final slope is
valid.
The data for is characterized by a parameter PC (for the 4 - in this case)
sets and a dependent variable RI and an independent variable ROH.
The data are contained in a variable "joint".
The joint has been attached and has RI, ROH and PC for each element.
The following gives the initial results:
Mline<-lm(RI[PC==1]~ROH[PC==1])
Eline<-lm(RI[PC==2]~ROH[PC==2])
Iline<-lm(RI[PC==3]~ROH[PC==3])
Pline<-lm(RI[PC==4]~ROH[PC==4])
joint_reduced <- joint;
for(i in 1:4) {
joint_reduced$RI[joint_reduced$PC==i]<-joint$RI[joint$PC==i]-mean(joint$RI[joint$PC==1]}
AllLine<-lm(joint_reduced$RI~joint_reduced$ROH);
Now the statistics from AllLine can be compared with each of the individual
statistics.
NOW THE QUESTION:
From a lot of point of view it would be useful to have a parameter
generated by
for (i in 1:4){ Xline[i]=lm(RI[PC==i]~ROH[PC==i])}
And now all of the work of comparison can be done with calls to Xline[i]
rather than having to work individually with Mline, Eline, etc.
This appears to be impossible. The constructor for Xline[i] is not
automatic (as it is for Mline, etc) noted above. I cannot determine how to
construct the Xline[i] object so that this kind of process can be
generalized. Is it possible? Is there another way to set us such tests of
multiple line linearity that is already in a package?
Comments or pointers would be appreciated.
Gary
__
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] X11, fonts, R-2.0.1, R-2.2.1 and R-devel
The issue here is that you are using a UTF-8 locale (you sent this message in UTF-8), and you need appropriately encoded X11 fonts. R-2.0.1 did not support UTF-8, and so you got incorrect output for non-ASCII characters. It *is* an X11 installation/fontpath problem. On Thu, 23 Mar 2006, Xavier Fernández i Marín wrote: Hello, I am having some problems with the X11 display in a gentoo linux laptop with R compiled manually. (https://stat.ethz.ch/pipermail/r-help/2006-March/089701.html) Whether I can open the X11 device and use it when I am using 'ion' as a window manager, I can't open it using 'gnome', due to a problem related to fonts: -8<--- Error in X11() : could not find any X11 fonts Check that the Font Path is correct. -8<--- I have tried and compiled R-2.0.1 and _it works_. With the latest stable version of R it does not work. And with the latest development (22 march 06) it does not work, neither. It is not due to the Xorg installation, because the display is opened when using other window manager different from gnome (and other version of R) It is not something related to the compilation options, for the same reason. So my last option is that it seems to be a problem with R, X11 and gnome, specifically. Any ideas or suggestions? I have googled and somebody in the FreeBSD lists talked about more or less the same problems, but it seems that without success: http://www.archivum.info/[EMAIL PROTECTED]/2005-06/msg00313.html http://www.archivum.info/[EMAIL PROTECTED]/2005-06/msg00307.html I paste the sessions and the errors, if it helps: -8<--- R-2.0.1 $ ./bin/R R : Copyright 2004, The R Foundation for Statistical Computing Version 2.0.1 (2004-11-15), ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for a HTML browser interface to help. Type 'q()' to quit R. WARNING: UTF-8 locales are not currently supported X11() q() Save workspace image? [y/n/c]: n -8<--- Works. The display is opened. -8<--- R-2.2.1 $ ./bin/R R : Copyright 2005, The R Foundation for Statistical Computing Version 2.2.1 (2005-12-20 r36812) ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. X11() Error in X11() : could not find any X11 fonts Check that the Font Path is correct. q() -8<--- Doesn't work. -8<--- deu R-devel $ ./bin/R R : Copyright 2006, The R Foundation for Statistical Computing Version 2.3.0 Under development (unstable) (2006-03-22 r37566) ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. X11() Error in X11(display, width, height, pointsize, if (is.null(gamma)) 1 else gamma, : invalid 'width' or 'height' x <- rnorm(50) y <- rnorm(50) plot(x,y) Error in X11(display, width, height, pointsize, if (is.null(gamma)) 1 else gamma, : invalid 'width' or 'height' X11(width=200, height=200) Error in X11(width = 200, height = 200) : could not find any X11 fonts Check that the Font Path is correct. -8<--- -- 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
[R] useR! 2006 program online
Dear useRs, we are happy to inform you that the program for the 2nd R user conference useR! 2006 is now available online from the conference Web page at http://www.R-project.org/useR-2006/program.html We would like to thank the useR community for submitting so many interesting abstracts about an astonishing variety of R applications. We are looking forward to an exciting and diversified conference! The useR! organization team and program committee ___ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-announce __ 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] Differential Equations
Dominik Heinzmann <[EMAIL PROTECTED]> writes: > Dear R-community > > My ODE problems looks as follows: > (1) dA/dt = u*A - v*B > (2) dB/dt = v*B - u*A > > where u is a constant, and v=k*t (k=constant, t=time) > > Does anybody knows a good function/procedure of solving? Should one > involve the equation (3) dv/dt = k? > Thanks for your support. You probably need to look into the odesolve package. The v*B terms make it nonlinear, so I wouldn't expect an analytic solution. -- 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] comparing AIC values of models with transformed, untransformed, and weighted variables
Two comments: 1) The log-likelihood and hence AIC for a model for log X are not comparable with those of a model for X. You need to make an additive adjustment when you transform: it is quite easy to work out what from the definitions. 2) The AIC given by glm() for weighted models was wrong in R < 2.3.0 alpha. I am not sure why you are using a glm for what appears to be a least-squares fit: use lm() instead (or try 2.3.0 alpha). On Wed, 15 Mar 2006, Patrick Baker wrote: > Hi there, I have a question regarding model comparisons that seems simple > enough but to which I cannot find an answer. I am interested in developing a > predictive model relating some measure of a tree's stem to the total leaf > area (TLA) of the tree. Predictor variables might include, for example, the > total cross-sectional area of the tree (commonly referred to as basal area) > or the amount of sapwood area (SA) (which represents the amount of wood > involved in active transport of water up the tree to the leaves). A variety > of people have developed these models for a variety of tree species in a > variety of places around the world. Perhaps not surprisingly, different > studies have used different model forms in analyzing their data. I am > interested in comparing the range of models that have been previously used > (some of which are theoretically derived, others of which are empirically > driven) using a data set that I have collected (for yet another species in > yet another place). To compare the different model forms I had intended to > use the AIC. However, I have found, again perhaps not surprisingly, that when > I use log-transformed data, the AIC is substantially lower for a given > predictor variable. If I use a weighted glm the same issue arises. For > example, using BA vs TLA the (rounded) AIC values are 275 for a linear > model, 30 for a log-log model, and 8 for a glm weighted by 1/BA. I don't > believe that these vast differences reflect a major improvement in the model, > but rather the scaling of the variables by transformation or weighting. What > I'd like to get some advice or insight on is whether there is an appropriate > way to rescale the AIC values to permit comparisons across these models. Any > suggestions would be very welcome. Cheers, Patrick Baker > -- 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
[R] Graded Response Model Simulation (SAS code conversion)
I have used R a lot in the past, but never for simulation. I have a code in SAS for the Graded Response Model (GRM), also known as Samejima's model. This code simulates an ordinal response, provided item characteristics (A=item discrimination, BB(G) are thresholds between various categorical responses). It is a macro file. I am thinking that I can write this as a function, and call it up inside a simulation code. Here is the SAS code: %MACRO GRGEN; DO G=1 TO NCAT-1; Z=EXP(A*(THETA-BB(G))); PS(G)=Z/(1+Z); END; PP(1)=1-PS(1); PP(NCAT)=PS(NCAT-1); DO G=2 TO NCAT-1; PP(G)=PS(G-1)-PS(G); END; X=RANUNI(-1); SUMP=0; R(J)=1; DO K=1 TO NCAT-1; SUMP=SUMP+PP(K); IF X>SUMP THEN R(J)=K+1; END; %MEND GRGEN; Now, I am totally unfamiliar to simulation in R. So does anyone have a good reference I could go to convert this? Or have any suggestions for how to convert it to R? My biggest problem is all the loops inside this program. In particular, how to setup the updating of R(J). It seems if I built a function for this, I need the item parameters A and BB's (possibly the NCAT). Any suggestions? Thanks for any help or info. Keith Yang University of Tennessee - [[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] A plotting question - how to get error bars?
See plotCI in package gplots.
For dates you can make use of the builtin vector month.abb
plot(1:3, 11:13, xaxt = "n")
axis(1, 1:3, month.abb[1:3])
or use Date class:
xvals <- seq(as.Date("2006-01-01"), length = 3, by = "month")
plot(xvals, 1:3)
or with specific control over x axis:
xvals <- seq(as.Date("2006-01-01"), length = 3, by = "month")
plot(xvals, 1:3, xaxt = "n")
axis(1, xvals, format(xvals, "%b"))
More on dates is in the Help Desk article on R News 4/1.
As an aside, note that xlim=range(xvals) is a bit more compact..
On 3/27/06, Sean Davis <[EMAIL PROTECTED]> wrote:
>
>
>
> On 3/27/06 6:55 AM, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote:
>
> > Dear R list,
> >
> > Can anyone help with a plotting question? I'm trying to display some data
> > on a plot and I've almost got the format I need (see code below), but 2
> > things I can't get:
> >
> > 1. How to get "Jan","Feb","Mar" on the x=axis instead of 1:3?
>
> First, do your plot with (..., axes=F). Then, look at the help for axis()
> to put the axes on the plot.
>
> > 2. How to get "T"s on the end of my error bars like you have in standard
> > scientific plots?
>
> RSiteSearch('error bars') several answers that might be of interest.
>
> Sean
>
> __
> 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] Differential Equations
Dear R-community My ODE problems looks as follows: (1) dA/dt = u*A - v*B (2) dB/dt = v*B - u*A where u is a constant, and v=k*t (k=constant, t=time) Does anybody knows a good function/procedure of solving? Should one involve the equation (3) dv/dt = k? Thanks for your support. -- Dominik Heinzmann Master of Science in Mathematics, EPFL Ph.D. student in Biostatistics Institute of Mathematics University of 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
Re: [R] glmmML
yOn Mon, 27 Mar 2006, Szentirmai Istvan wrote: > Dear R Users, > > I'm looking for a similar function as step() or drop1() for glmmML models, > but couldn't yet find any. I would appreciate if anyone could help me find > such a function. If you read the help for those functions, you will see that 'all' you need is to specify an extractAIC() function for such models. However, I don't think even defining AIC for them is that easy (what do you do about zero variance parameters, for example?). -- 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] Date in dataframe manipulation
Thank you Marc and Don's help, especially Marc's. Output<- subset(FireDataAppling, select = c(STARTDATE, County, TOTAL, CAUSE)) Worked! STARTDATE IS a factor and I used the following command to get the -mm-dd format of the date Output$Date<- as.POSIXct(Output$STARTDATE) Thank you! Daniel Chan -Original Message- From: Marc Schwartz (via MN) [mailto:[EMAIL PROTECTED] Sent: Friday, March 24, 2006 9:22 PM To: Dan Chan Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Date in dataframe manipulation On Fri, 2006-03-24 at 15:29 -0500, Dan Chan wrote: > Hi, > > I have a dataframe with many columns, including date and I want to keep > only a few of the columns including date column. > > I used the following command: > with(FireDataAppling, cbind(STARTDATE, County, TOTAL, CAUSE) > > It works, but the date becomes days from Jan 1, 2001. > > FireDataAppling$STARTDATE[1] gives > [1] 2001-01-04 00:00:00 > 1703 Levels: . This output suggests that STARTDATE is a factor, rather than a Date related data type. Did you read this data in via one of the read.table() family of functions? If these values are quoted character fields in the imported text file, they will be converted to factors by default. > After the cbind command, the entry becomes a 4. > > I want to get 2001-01-04. What command should I use? > > Thank you. You might want to review the "Note" section in ?cbind, relative to the result of cbind()ing vectors of differing data types. By using with(), you are effectively taking the data frame columns as individual vectors and the resultant _matrix_ will be coerced to a single data type, in this case, presumably numeric. I am guessing that 'County' and 'CAUSE' are also factors, whereas 'TOTAL' is numeric. Using str(FireDataAppling) will give you some insight into the structure of your data frame. The '4' that you are getting is the factor level numeric code for the entry above, not the number of days since Jan 1, 2001, which is not a default 'origin' date in R. Jan 1, 1970 is. You might want to look at ?factor for more insight here. If you want to retain only a _subset_ of the columns in a data frame, use the subset() function: subset(FireDataAppling, select = c(STARTDATE, County, TOTAL, CAUSE)) This will return a data frame and retain the original data types. If you want to then perform actual Date based operations on those values, take a look at ?DateTimeClasses, paying attention to the "See Also" section relative to associated functions. 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] seq(2,5,-2) not an error but NULL
Hi Duncan et al > > I don't think seq() could reasonably be expected to handle "to" and > "by" arguments with complex values. Trying to divide the (to-from) > difference by (by) to find how many steps to take would usually > result in enough rounding error that the result wouldn't be real- > valued. It's enough of a miracle that it correctly handles > > seq(from=1, by=1+1i, len=4) > > Duncan Murdoch > well it depends on your definition of miracles, but I wouldn't say 1 + (0:3) * (1+1i) is particularly miraculous ;-\ best wishes rksh -- Robin Hankin Uncertainty Analyst National Oceanography Centre, Southampton European Way, Southampton SO14 3ZH, UK tel 023-8059-7743 __ 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] seq(2,5,-2) not an error but NULL
On 3/27/2006 8:28 AM, Robin Hankin wrote:
> Hi.
>
> seq() is a complex beast indeed. 'by' being the wrong
> sign is a special case of the behaviour seen in the
> following code snippets, the first of which is correctly
> rejected by seq(), the second of which should arguably
> return a three element complex vector.
>
>
> > seq(from=1,to=3,by=1+1i)
> Error in n < 0 : invalid comparison with complex values
>
> > seq(from=1,to=4+3i,by=1+1i)
> Error in n < 0 : invalid comparison with complex values
I don't think seq() could reasonably be expected to handle "to" and "by"
arguments with complex values. Trying to divide the (to-from)
difference by (by) to find how many steps to take would usually result
in enough rounding error that the result wouldn't be real-valued. It's
enough of a miracle that it correctly handles
seq(from=1, by=1+1i, len=4)
Duncan Murdoch
>
> best wishes
>
> Robin
>
>
>
> On 27 Mar 2006, at 13:23, Duncan Murdoch wrote:
>
>> On 3/27/2006 4:41 AM, Christian Hoffmann wrote:
>>> Hi,
>>>
>>> This may belong more to r-develop, but general discussion may be
>>> useful
>>> (for the how many-th time ?)
>>>
>>> seq(2,5,-2)
>>> seq(5,2,2)
>>>
>>> both result in
>>>
>>> Error in seq.default(2, 5, -2) : wrong sign in 'by' argument
>>>
>>> But often, if not always, mathematicians and programmers want a
>>> behaviour e.g. in for loops, where this statement results in an empty
>>> statement, that is
>>>
>>> for (ii in seq(2,5,-2)) print(ii)
>>>
>>> were equivalent to
>>>
>>> for (ii in NULL) print(ii).
>>>
>>> The relevant part in seq.default is now
>>>
>>> if (n < 0)
>>> stop("wrong sign in 'by' argument")
>>>
>>> but could be changed by option to
>>>
>>>return(NULL)
>>>
>>> I think there should be an option to seq requiring this behaviour,
>>> or a
>>> specific function, may be even a special operator, e.g. %;%:
>>>
>>> 3;5 resulting in NULL.
>>>
>>> What do you think?
>>
>> If you want optional behaviour, the easiest way is to write your own
>> wrapper function. E.g.
>>
>> emptyseq <- function(from, to, by) {
>>if ((to-from)*by < 0) return(NULL)
>>else return(seq(from, to, by))
>> }
>>
>> I don't think this is a desirable default, though. We already have a
>> special way to handle the most common case, i.e.
>>
>> seq(1, length(x), 1)
>>
>> should be written as
>>
>> seq(along=x))
>>
>> to handle the length(x) == 0 case the way you're requesting.
>>
>> But I'm not so sure that seq(2,5,-2) should really be NULL; it looks
>> much more like an error to me. You say mathematicians and programmers
>> want this behaviour, but I really can't think of any examples other
>> than
>> the one above.
>>
>> As a general principle, I think it's better to throw an error on
>> ambiguous or apparently erroneous code rather than always returning an
>> answer. If the code can be made unambiguous, it should be. (R
>> doesn't
>> always follow this principle; for example, recycling of vectors of
>> lengths bigger than 1 is probably an error at least as often as it's
>> intended.)
>>
>> 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
>
> --
> Robin Hankin
> Uncertainty Analyst
> National Oceanography Centre, Southampton
> European Way, Southampton SO14 3ZH, UK
> tel 023-8059-7743
>
>
>
__
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] seq(2,5,-2) not an error but NULL
Hi.
seq() is a complex beast indeed. 'by' being the wrong
sign is a special case of the behaviour seen in the
following code snippets, the first of which is correctly
rejected by seq(), the second of which should arguably
return a three element complex vector.
> seq(from=1,to=3,by=1+1i)
Error in n < 0 : invalid comparison with complex values
> seq(from=1,to=4+3i,by=1+1i)
Error in n < 0 : invalid comparison with complex values
best wishes
Robin
On 27 Mar 2006, at 13:23, Duncan Murdoch wrote:
> On 3/27/2006 4:41 AM, Christian Hoffmann wrote:
>> Hi,
>>
>> This may belong more to r-develop, but general discussion may be
>> useful
>> (for the how many-th time ?)
>>
>> seq(2,5,-2)
>> seq(5,2,2)
>>
>> both result in
>>
>> Error in seq.default(2, 5, -2) : wrong sign in 'by' argument
>>
>> But often, if not always, mathematicians and programmers want a
>> behaviour e.g. in for loops, where this statement results in an empty
>> statement, that is
>>
>> for (ii in seq(2,5,-2)) print(ii)
>>
>> were equivalent to
>>
>> for (ii in NULL) print(ii).
>>
>> The relevant part in seq.default is now
>>
>> if (n < 0)
>> stop("wrong sign in 'by' argument")
>>
>> but could be changed by option to
>>
>>return(NULL)
>>
>> I think there should be an option to seq requiring this behaviour,
>> or a
>> specific function, may be even a special operator, e.g. %;%:
>>
>> 3;5 resulting in NULL.
>>
>> What do you think?
>
> If you want optional behaviour, the easiest way is to write your own
> wrapper function. E.g.
>
> emptyseq <- function(from, to, by) {
>if ((to-from)*by < 0) return(NULL)
>else return(seq(from, to, by))
> }
>
> I don't think this is a desirable default, though. We already have a
> special way to handle the most common case, i.e.
>
> seq(1, length(x), 1)
>
> should be written as
>
> seq(along=x))
>
> to handle the length(x) == 0 case the way you're requesting.
>
> But I'm not so sure that seq(2,5,-2) should really be NULL; it looks
> much more like an error to me. You say mathematicians and programmers
> want this behaviour, but I really can't think of any examples other
> than
> the one above.
>
> As a general principle, I think it's better to throw an error on
> ambiguous or apparently erroneous code rather than always returning an
> answer. If the code can be made unambiguous, it should be. (R
> doesn't
> always follow this principle; for example, recycling of vectors of
> lengths bigger than 1 is probably an error at least as often as it's
> intended.)
>
> 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
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
__
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to create a directoy with R
See ?dir.create, and take care at the 'recursive' argument in case you
have to create several subdir levels at once.
Best,
Philippe Grosjean
Sarah Goslee wrote:
> I think you need to use system("mkdir") or whatever is appropriate
> for your OS. Making directories is a function of the OS, not of R. If
> you need to make a truly cross-platform solution, you might need
> to check within your code what OS is being used, and call the
> appropriate system statement. (I think you can do this, but have
> never needed to.) That would be particularly important if you need to
> specify paths.
>
> Sarah
>
> On 3/27/06, pau carre <[EMAIL PROTECTED]> wrote:
>
>>Hello, I am trying to create directories with R. I would like R to
>>create directories because it is platform independent. I tried using
>>file() and searching in "R Data Import/Export" but I did not succeed.
>>I think it must be some function since exists the unlink to remove
>>directories (and files).
>>
>>Pau
>>
>>__
>>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
>>
>
>
>
>
> --
> Sarah Goslee
> http://www.stringpage.com
>
> [[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
>
>
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[R] step() for glmmML
Dear R Users,
I'm looking for a similar function as step() or drop1() for glmmML models,
but couldn't yet find any. I would appreciate if anyone could help me find
such a function.
Thanks,
Istvan
- Original Message -
From: <[EMAIL PROTECTED]>
To:
Sent: Monday, March 27, 2006 1:55 PM
Subject: [R] A plotting question - how to get error bars?
> Dear R list,
>
> Can anyone help with a plotting question? I'm trying to display some data
> on a plot and I've almost got the format I need (see code below), but 2
> things I can't get:
>
> 1. How to get "Jan","Feb","Mar" on the x=axis instead of 1:3?
> 2. How to get "T"s on the end of my error bars like you have in standard
> scientific plots?
>
> Any comments gratefully received!
>
> Thanks,
> Toby
>
> xvals=1:3 #couldn't get it to be "Jan, Feb, Mar" on the x-axis
> rgrT1=c(10,20,30)
> errbarsT1lo=c(0.5830952,0.3741657,0.8944272)
> errbarsT1up=errbarsT1lo
> rgrT2=c(25,30,35)
> errbarsT2lo=c(1.356466,3.535534,1.140175)
> errbarsT2up=errbarsT2lo
> minx=min(xvals);maxx=max(xvals)
> miny=min(rgrT1-errbarsT1lo,rgrT2-errbarsT2lo);maxy=max(rgrT1+errbarsT1up,rgrT2+errbarsT2up)
> plot(x=0,y=0,type="n",xlim=c(minx,maxx),ylim=c(miny,maxy),lab=c(2,20,0),bty="l",xlab="month",ylab="Relative
> Growth Rate")
> points(x=xvals,y=rgrT1,pch=21)
> symbols(x=xvals,y=rgrT1,boxplots=cbind(0,0,errbarsT1lo,errbarsT1up,0.5),inches=FALSE,add=TRUE)
> #symbols does the error bars, but without the "T"s at the end. The
> boxplot command does the Ts, but you can't have them without the box in
> the middle (and you can't have different symbols for points either)
> lines(x=xvals,y=rgrT1,lty=21)
> points(x=xvals,y=rgrT2,pch=24)
> symbols(x=xvals,y=rgrT2,boxplots=cbind(0,0,errbarsT2lo,errbarsT2up,0.5),inches=FALSE,add=TRUE)
> lines(x=xvals,y=rgrT2,lty=24)
> legend(x="right",c("Treatment 1","Treatment 2"),pch=c(21,24))
>
> __
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> PLEASE do read the posting guide!
> http://www.R-project.org/posting-guide.html
>
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Re: [R] A plotting question - how to get error bars?
On 3/27/06 6:55 AM, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote:
> Dear R list,
>
> Can anyone help with a plotting question? I'm trying to display some data
> on a plot and I've almost got the format I need (see code below), but 2
> things I can't get:
>
> 1. How to get "Jan","Feb","Mar" on the x=axis instead of 1:3?
First, do your plot with (..., axes=F). Then, look at the help for axis()
to put the axes on the plot.
> 2. How to get "T"s on the end of my error bars like you have in standard
> scientific plots?
RSiteSearch('error bars') several answers that might be of interest.
Sean
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Re: [R] Plotting with date
thanks a lot ! it runs well now Pierre Clauss Gabor Grothendieck <[EMAIL PROTECTED]> a écrit : Make sure your x axis variable really is of class Date: class(x) plot(Sys.Date() + 0:99, 1:100) See ?str ?class ?as.Date, ?axis.Date and the help desk article in R News 4/1 on dates for more info. On 3/24/06, pierre clauss wrote: > Hello ! > I need your help for plotting with date in the x-axis. > I do not manage to plot temporal curves with the label of Date (01/01/1990, > etc) in the x-axis. > Thanks a lot for your answer ! > Pierre Clauss. > > > > - > > [[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 > - [[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] apply(ing) to sum subset of a vector
apply(cbind(from,to), 1, function(x) sum(g[x[1]:x[2]])) Fred J. a écrit : >Dear R users > > I am trying to sum selective elements of a vector but my solution > is not cutting it. > > Example: > > g <- 1:5; > > > from <- 1:3; > > to <- 3:5; > from to > 1 3 > 2 4 > 3 5 > > so I expect 3 sums from g > 1+2+3 that is 1 to 3 of g > 2+3+4 that is 2 to 4 of g > 3+4+5 that is 3 to 5 of g > > my solution will not work. > sum.em <- function(g, c1, c2) sum(g[c1:c2]) > apply(g, 1, sum.em, ...) I don't think so because apply is not > aware of the from and to. and if I f <- list(g, from, to) that will not fit > with the second arg of apply. > > thank you > > >- > > [[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-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] glmmML
Dear R Users, I'm looking for a similar function as step() or drop1() for glmmML models, but couldn't yet find any. I would appreciate if anyone could help me find such a function. Thanks, Istvan __ 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 create a directoy with R
help.search("directory") would have given you:
R.home(base)Return the R Home Directory
files(base) File and Directory Manipulation
getwd(base) Get or Set Working Directory
list.files(base)List the Files in a Directory/Folder
unlink(base)Delete Files and Directories
and then ?files would have told you there's dir.create() and how to use it.
Andy
From: pau carre
>
> Hello, I am trying to create directories with R. I would like
> R to create directories because it is platform independent. I
> tried using
> file() and searching in "R Data Import/Export" but I did not
> succeed. I think it must be some function since exists the
> unlink to remove directories (and files).
>
> Pau
>
> __
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> PLEASE do read the posting guide!
> http://www.R-project.org/posting-guide.html
>
>
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Re: [R] How to create a directoy with R
On Mon, 27 Mar 2006, pau carre wrote:
> Hello, I am trying to create directories with R. I would like R to
> create directories because it is platform independent. I tried using
> file() and searching in "R Data Import/Export" but I did not succeed.
> I think it must be some function since exists the unlink to remove
> directories (and files).
?dir.create
[help.search("directory") leads to
files(base) File and Directory Manipulation
on which page it is documented.]
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] seq(2,5,-2) not an error but NULL
On Mon, 27 Mar 2006, Christian Hoffmann wrote:
> Hi,
>
> This may belong more to r-develop, but general discussion may be useful
> (for the how many-th time ?)
The place for general discussion of changes to R is the R-devel list.
There is almost no scope to change things like this, as there is so much
existing code which relies on it (and it is also compatible with S).
> seq(2,5,-2)
> seq(5,2,2)
>
> both result in
>
> Error in seq.default(2, 5, -2) : wrong sign in 'by' argument
>
> But often, if not always, mathematicians and programmers want a
> behaviour e.g. in for loops, where this statement results in an empty
> statement, that is
>
> for (ii in seq(2,5,-2)) print(ii)
>
> were equivalent to
>
> for (ii in NULL) print(ii).
>
> The relevant part in seq.default is now
>
> if (n < 0)
> stop("wrong sign in 'by' argument")
>
> but could be changed by option to
>
> return(NULL)
Why is NULL plausible? I would think integer(0) is more likely, but if
you think this should not be an error, then another plausible
interpretation is the intersection of {2 ... 5} and {2, 0, -2, ...}, that
is {2}. (The only language I can think of with a close analogue is
Fortran DO loops, where DO 10 I=2,5,-2 used to be {2} and was defined in
F77 to be empty, so I don't think the interpretation is unambiguous.)
[As an aside, one could argue that 'for (ii in NULL)' should be an error,
since the help page says
seq: An expression evaluating to a vector (including a list).
and NULL is not a vector.]
> I think there should be an option to seq requiring this behaviour, or a
> specific function, may be even a special operator, e.g. %;%:
>
> 3;5 resulting in NULL.
>
> What do you think?
It cannot be the default, and if you need it, why not write your own
function to do it? S has survived ca 18 years without it, so the need
cannot be overwhelming.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] seq(2,5,-2) not an error but NULL
You should be able to do it yourself; e.g.,
my.seq <- function(...) if((to - from) * by < 0) NULL else seq(...)
and use that instead when you want that behavior.
Andy
From: Christian Hoffmann
>
> Hi,
>
> This may belong more to r-develop, but general discussion may
> be useful
> (for the how many-th time ?)
>
> seq(2,5,-2)
> seq(5,2,2)
>
> both result in
>
> Error in seq.default(2, 5, -2) : wrong sign in 'by' argument
>
> But often, if not always, mathematicians and programmers want a
> behaviour e.g. in for loops, where this statement results in an empty
> statement, that is
>
> for (ii in seq(2,5,-2)) print(ii)
>
> were equivalent to
>
> for (ii in NULL) print(ii).
>
> The relevant part in seq.default is now
>
> if (n < 0)
> stop("wrong sign in 'by' argument")
>
> but could be changed by option to
>
>return(NULL)
>
> I think there should be an option to seq requiring this
> behaviour, or a
> specific function, may be even a special operator, e.g. %;%:
>
> 3;5 resulting in NULL.
>
> What do you think?
>
> Christian
> --
> Dr. Christian W. Hoffmann,
> Swiss Federal Research Institute WSL
> Mathematics + Statistical Computing
> Zuercherstrasse 111
> CH-8903 Birmensdorf, Switzerland
>
> Tel +41-44-7392-277 (office) -111(exchange)
> Fax +41-44-7392-215 (fax)
> [EMAIL PROTECTED] http://www.wsl.ch/staff/christian.hoffmann
>
> International Conference 5.-7.6.2006 Ekaterinburg Russia
> "Climate changes and their impact on boreal and temperate
> forests" http://ecoinf.uran.ru/conference/
>
> __
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> 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] seq(2,5,-2) not an error but NULL
On 3/27/2006 4:41 AM, Christian Hoffmann wrote:
> Hi,
>
> This may belong more to r-develop, but general discussion may be useful
> (for the how many-th time ?)
>
> seq(2,5,-2)
> seq(5,2,2)
>
> both result in
>
> Error in seq.default(2, 5, -2) : wrong sign in 'by' argument
>
> But often, if not always, mathematicians and programmers want a
> behaviour e.g. in for loops, where this statement results in an empty
> statement, that is
>
> for (ii in seq(2,5,-2)) print(ii)
>
> were equivalent to
>
> for (ii in NULL) print(ii).
>
> The relevant part in seq.default is now
>
> if (n < 0)
> stop("wrong sign in 'by' argument")
>
> but could be changed by option to
>
>return(NULL)
>
> I think there should be an option to seq requiring this behaviour, or a
> specific function, may be even a special operator, e.g. %;%:
>
> 3;5 resulting in NULL.
>
> What do you think?
If you want optional behaviour, the easiest way is to write your own
wrapper function. E.g.
emptyseq <- function(from, to, by) {
if ((to-from)*by < 0) return(NULL)
else return(seq(from, to, by))
}
I don't think this is a desirable default, though. We already have a
special way to handle the most common case, i.e.
seq(1, length(x), 1)
should be written as
seq(along=x))
to handle the length(x) == 0 case the way you're requesting.
But I'm not so sure that seq(2,5,-2) should really be NULL; it looks
much more like an error to me. You say mathematicians and programmers
want this behaviour, but I really can't think of any examples other than
the one above.
As a general principle, I think it's better to throw an error on
ambiguous or apparently erroneous code rather than always returning an
answer. If the code can be made unambiguous, it should be. (R doesn't
always follow this principle; for example, recycling of vectors of
lengths bigger than 1 is probably an error at least as often as it's
intended.)
Duncan Murdoch
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Re: [R] How to create a directoy with R
I think you need to use system("mkdir") or whatever is appropriate
for your OS. Making directories is a function of the OS, not of R. If
you need to make a truly cross-platform solution, you might need
to check within your code what OS is being used, and call the
appropriate system statement. (I think you can do this, but have
never needed to.) That would be particularly important if you need to
specify paths.
Sarah
On 3/27/06, pau carre <[EMAIL PROTECTED]> wrote:
>
> Hello, I am trying to create directories with R. I would like R to
> create directories because it is platform independent. I tried using
> file() and searching in "R Data Import/Export" but I did not succeed.
> I think it must be some function since exists the unlink to remove
> directories (and files).
>
> Pau
>
> __
> 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
>
--
Sarah Goslee
http://www.stringpage.com
[[alternative HTML version deleted]]
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[R] A plotting question - how to get error bars?
Dear R list,
Can anyone help with a plotting question? I'm trying to display some data
on a plot and I've almost got the format I need (see code below), but 2
things I can't get:
1. How to get "Jan","Feb","Mar" on the x=axis instead of 1:3?
2. How to get "T"s on the end of my error bars like you have in standard
scientific plots?
Any comments gratefully received!
Thanks,
Toby
xvals=1:3 #couldn't get it to be "Jan, Feb, Mar" on the x-axis
rgrT1=c(10,20,30)
errbarsT1lo=c(0.5830952,0.3741657,0.8944272)
errbarsT1up=errbarsT1lo
rgrT2=c(25,30,35)
errbarsT2lo=c(1.356466,3.535534,1.140175)
errbarsT2up=errbarsT2lo
minx=min(xvals);maxx=max(xvals)
miny=min(rgrT1-errbarsT1lo,rgrT2-errbarsT2lo);maxy=max(rgrT1+errbarsT1up,rgrT2+errbarsT2up)
plot(x=0,y=0,type="n",xlim=c(minx,maxx),ylim=c(miny,maxy),lab=c(2,20,0),bty="l",xlab="month",ylab="Relative
Growth Rate")
points(x=xvals,y=rgrT1,pch=21)
symbols(x=xvals,y=rgrT1,boxplots=cbind(0,0,errbarsT1lo,errbarsT1up,0.5),inches=FALSE,add=TRUE)
#symbols does the error bars, but without the "T"s at the end.
The
boxplot command does the Ts, but you can't have them without the box in
the middle (and you can't have different symbols for points either)
lines(x=xvals,y=rgrT1,lty=21)
points(x=xvals,y=rgrT2,pch=24)
symbols(x=xvals,y=rgrT2,boxplots=cbind(0,0,errbarsT2lo,errbarsT2up,0.5),inches=FALSE,add=TRUE)
lines(x=xvals,y=rgrT2,lty=24)
legend(x="right",c("Treatment 1","Treatment 2"),pch=c(21,24))
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Re: [R] How to create a directoy with R
?dir.create On Mon, 2006-03-27 at 13:07 +0200, pau carre wrote: > Hello, I am trying to create directories with R. I would like R to > create directories because it is platform independent. I tried using > file() and searching in "R Data Import/Export" but I did not succeed. > I think it must be some function since exists the unlink to remove > directories (and files). > > Pau > > __ > 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-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] Clustering question \ dist(datmat)
On 3/27/06 12:19 AM, "kumar zaman" <[EMAIL PROTECTED]> wrote:
> Dear Gabor and all ;
>
> I know this will work; but i already have a distance matrix calculated using
> my distance measure Dij = 0.5 * ( 1 - cos(theta_i - theta_j)), if i do
> hclust(as.dist(df)) then i am taking distance another time for a matrix " df "
> which is supposed to be a distance matrix, i hope i am clear ;
>
> ps: I just found out i can use " kmeans(df, 3, iter.max=100)" it will take
> df as calculated by Dij. I still need to use methods in hclust like " single,
> average, ward, median, mcquitty, ...etc"
>
> Thank u anyway.
Kumar,
If I understand Your point, you are misunderstanding what as.dist() does.
It does not compute a distance matrix. Instead, it simply makes a matrix
into a "dist" object, which is NOT just a matrix. However, the distances in
a matrix converted to a "dist" object are not altered. Therefore, you are
not "taking distance another time"; instead, you are simply converting the
distance matrix into a form that hclust can understand.
Hope that helps clarify.
Sean
> Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> A distance matrix must be of class "dist". Try
>
> hclust(as.dist(df))
>
>
> On 3/26/06, kumar zaman wrote:
>> Hello everybody. I am trying to cluster circular data (data points which are
>> angles), thus i can not use the "dist" function in "mclust" to generate my
>> distance matrix, I am using the function " Dij = 0.5*( 1 - cos(theta_i -
>> theta_j)). The thing is "hclust" will not accept this distance matrix, i
>> tried to put it in a data frame, but again i get an error message saying "
>> Error in if (n < 2) stop("must have n >= 2 objects to cluster") : argument is
>> of length zero". The distance matrix "dist" producing is a lower triangular
>> one, mine is a square matrix, which i think does not matter. My question how
>> to make "hclust" process my distance matrix, what i am doing wrong. I am sure
>> the problem is with the distance matrix format, Any suggestions are highly
>> apprciated, the code below shows what i have done.
>>
>> clust1<- as.vector(rvm(5,5,15))
>> clust2<- as.vector(rvm(5,10,15))
>> clust3<- as.vector(rvm(5,15,15))
>> clust4<- as.vector(rvm(5,20,15))
>> clust5<- as.vector(rvm(5,25,15))
>> data1<- rbind(clust1,clust2,clust3,clust4,clust5)
>> datmat<- matrix(data1,nrow=25,ncol=1,byrow=TRUE)
>> circ.plot(datmat)
>> df<- array(dim=c(25,25))
>> for (i in 1:25){
>> for (j in 1:25){
>> df[i,j]<- 0.5*(1 - cos(datmat[i] - datmat[j]))
>> }
>> }
>> hcA<-hclust(df,method="average")
>>
>> Ahmed
>> Florida
>>
>>
>> -
>>
>> [[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
>>
>
>
>
> Ahmed Albatineh,PhD
> Assistant Professor of Statistics
> Nova Southeastern University
> Fort Lauderdale, FL 33314
> U.S.A
>
> -
>
> [[alternative HTML version deleted]]
>
> __
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Re: [R] Missing Argument in optim()
On Mon, 27 Mar 2006, [EMAIL PROTECTED] wrote:
> Hello everybody,
>
> i already searched the archieves, but i still don't know what is wrong
> in my implementation, mybe anybody coud give me some advice
>
> ll1<-function(rho,theta,beta1,beta2,beta3,beta4,t,Szenariosw5,Testfaellew5,X1,X2)
> {
>n<-length(t)
>t<-cumsum(t)
> tn<-t[length(t)]
> Szenn<-Szenariosw5[length(Szenariosw5)]
> Testn<-Testfaellew5[length(Testfaellew5)]
> X1n<-X1[length(X1)]
>X2n<-X2[length(X2)]
>n/rho-(tn^theta)*(beta1*Szenn+beta2*Testn+beta3*X1n+beta4*X2n)
> }
>
> p0<-c(rho=1,theta=1,beta1=1,beta2=1,beta3=1,beta4=1)
>
> optim(p0,fn=ll1,t=t,Szenariosw5=Szenariosw5,Testfaellew5=Testfaellew5,X1=X1,X2=X2)
>
> i always got an error message: Argument "theta" is missing, with no default
> but i don't know what is wrong?
>From the help page:
fn: A function to be minimized (or maximized), with first
argument the vector of parameters over which minimization is
to take place. It should return a scalar result.
ll1 is not such a function, and I think you need to replace the first 6
arguments by a single 6-element vector.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] How to create a directoy with R
Hello, I am trying to create directories with R. I would like R to create directories because it is platform independent. I tried using file() and searching in "R Data Import/Export" but I did not succeed. I think it must be some function since exists the unlink to remove directories (and files). Pau __ 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] Missing Argument in optim()
Hello everybody,
i already searched the archieves, but i still don't know what is wrong
in my implementation, mybe anybody coud give me some advice
ll1<-function(rho,theta,beta1,beta2,beta3,beta4,t,Szenariosw5,Testfaellew5,X1,X2)
{
n<-length(t)
t<-cumsum(t)
tn<-t[length(t)]
Szenn<-Szenariosw5[length(Szenariosw5)]
Testn<-Testfaellew5[length(Testfaellew5)]
X1n<-X1[length(X1)]
X2n<-X2[length(X2)]
n/rho-(tn^theta)*(beta1*Szenn+beta2*Testn+beta3*X1n+beta4*X2n)
}
p0<-c(rho=1,theta=1,beta1=1,beta2=1,beta3=1,beta4=1)
optim(p0,fn=ll1,t=t,Szenariosw5=Szenariosw5,Testfaellew5=Testfaellew5,X1=X1,X2=X2)
i always got an error message: Argument "theta" is missing, with no default
but i don't know what is wrong?
thanks!
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Re: [R] load huge image
On 3/27/06, Martin Maechler <[EMAIL PROTECTED]> wrote:
> > "Gottfried" == Gottfried Gruber <[EMAIL PROTECTED]>
> > on Sun, 26 Mar 2006 10:27:35 +0200 writes:
>
> Gottfried> hello, i have run around 65000 regressions and
> Gottfried> stored them in a list. then i stored the session
> Gottfried> with save.image on my hard disk. the file is
> Gottfried> almost 1GB. when i now want to load the image it
> Gottfried> took tons of time. even after 12h of loading it
> Gottfried> was not done, although the saving was done fairly
> Gottfried> fast.
>
> I'm sure it takes so lang because you (i.e. R) run out of RAM
> and the machine starts to swap.
>
> Try to get access to a Linux (or other Unix-alike) machine with
> a 64-bit version of R and about 8 GB of RAM (maybe 4 GB is
> already sufficient). I guess then you should be able to read it
> much more quickly.
>
> For 65000 regressions, do you need more than the estimated
> coefficients or -- a bit more informatively -- the
>
> coef(summary( )) result ?
>
> If you had only saved these coefficient matrices, I'm sure you'd
> have need **much** less memory.
>
>
> Gottfried> i fear i have to run the regressions again and
> Gottfried> store them in a database ...
>
> or really store what you need instead of everything ...
>
> Gottfried> can i load this file? any suggestions?
Do you need to have all of them in memory at once? Instead of using
save.image() can't you use save()/load() on each of the regression
fits? You can name the files using sprintf("regression%05d.Rdata",
idx) or similar.
Also, as Martin says, "fit" objects contains a lot of information that
you might not need; remove these before saving by setting the elements
you don't want to NULL.
/Henrik
> Gottfried> thanks & bets regards, gg --
> Gottfried> ---
> Gottfried> Gottfried Gruber
> Gottfried> mailto:[EMAIL PROTECTED] www:
> Gottfried> http://gogo.sehrsupa.net
>
> Gottfried> __
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> Gottfried> PLEASE do read the posting guide!
> Gottfried> http://www.R-project.org/posting-guide.html
>
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>
>
--
Henrik Bengtsson
Mobile: +46 708 909208 (+2h UTC)
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Re: [R] apply(ing) to sum subset of a vector
create a matrix and then use apply: > g <- 1:5; > > from <- 1:3; > to <- 3:5; > > index <- cbind(from,to) > apply(index, 1, function(x) sum(g[x[1]:x[2]])) [1] 6 9 12 > On 3/27/06, Fred J. <[EMAIL PROTECTED]> wrote: > > Dear R users > > I am trying to sum selective elements of a vector but my solution > is not cutting it. > > Example: > > g <- 1:5; > > > from <- 1:3; > > to <- 3:5; > from to > 1 3 > 2 4 > 3 5 > > so I expect 3 sums from g > 1+2+3 that is 1 to 3 of g > 2+3+4 that is 2 to 4 of g > 3+4+5 that is 3 to 5 of g > > my solution will not work. > sum.em <- function(g, c1, c2) sum(g[c1:c2]) > apply(g, 1, sum.em, ...) I don't think so because apply is not > aware of the from and to. and if I f <- list(g, from, to) that will not > fit with the second arg of apply. > > thank you > > > - > >[[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 > -- Jim Holtman Cincinnati, OH +1 513 646 9390 (Cell) +1 513 247 0281 (Home) What the problem you are trying to solve? [[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] seq(2,5,-2) not an error but NULL
Hi,
This may belong more to r-develop, but general discussion may be useful
(for the how many-th time ?)
seq(2,5,-2)
seq(5,2,2)
both result in
Error in seq.default(2, 5, -2) : wrong sign in 'by' argument
But often, if not always, mathematicians and programmers want a
behaviour e.g. in for loops, where this statement results in an empty
statement, that is
for (ii in seq(2,5,-2)) print(ii)
were equivalent to
for (ii in NULL) print(ii).
The relevant part in seq.default is now
if (n < 0)
stop("wrong sign in 'by' argument")
but could be changed by option to
return(NULL)
I think there should be an option to seq requiring this behaviour, or a
specific function, may be even a special operator, e.g. %;%:
3;5 resulting in NULL.
What do you think?
Christian
--
Dr. Christian W. Hoffmann,
Swiss Federal Research Institute WSL
Mathematics + Statistical Computing
Zuercherstrasse 111
CH-8903 Birmensdorf, Switzerland
Tel +41-44-7392-277 (office) -111(exchange)
Fax +41-44-7392-215 (fax)
[EMAIL PROTECTED]
http://www.wsl.ch/staff/christian.hoffmann
International Conference 5.-7.6.2006 Ekaterinburg Russia
"Climate changes and their impact on boreal and temperate forests"
http://ecoinf.uran.ru/conference/
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[R] apply(ing) to sum subset of a vector
Dear R users I am trying to sum selective elements of a vector but my solution is not cutting it. Example: > g <- 1:5; > from <- 1:3; > to <- 3:5; from to 1 3 2 4 3 5 so I expect 3 sums from g 1+2+3 that is 1 to 3 of g 2+3+4 that is 2 to 4 of g 3+4+5 that is 3 to 5 of g my solution will not work. sum.em <- function(g, c1, c2) sum(g[c1:c2]) apply(g, 1, sum.em, ...) I don't think so because apply is not aware of the from and to. and if I f <- list(g, from, to) that will not fit with the second arg of apply. thank you - [[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
