Re: [R] lattice panel.lmline problem
Thanks Frede
I didn't know about the r type.
Ross Darnell
-Original Message-
From: Frede Aakmann Tøgersen [mailto:[EMAIL PROTECTED]
Sent: Mon 10-Sep-07 4:45 PM
To: Ross Darnell; r-help@stat.math.ethz.ch
Subject: SV: [R] lattice panel.lmline problem
Why not use the more simple
xyplot(total.fat~x|variable,groups=Group,
data=tmp1,type=c(p,r))
???
See ?panel.xyplot and especially the type argument of that panel function.
Best regards
Frede Aakmann Tøgersen
Scientist
UNIVERSITY OF AARHUS
Faculty of Agricultural Sciences
Dept. of Genetics and Biotechnology
Blichers Allé 20, P.O. BOX 50
DK-8830 Tjele
Phone: +45 8999 1900
Direct: +45 8999 1878
E-mail: [EMAIL PROTECTED]
Web: http://www.agrsci.org
This email may contain information that is confidential.
Any use or publication of this email without written permission from Faculty of
Agricultural Sciences is not allowed.
If you are not the intended recipient, please notify Faculty of Agricultural
Sciences immediately and delete this email.
-Oprindelig meddelelse-
Fra: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] På vegne af Ross Darnell
Sendt: 10. september 2007 07:55
Til: r-help@stat.math.ethz.ch
Emne: [R] lattice panel.lmline problem
I am wanting to generate panels showing scatterplots with the
linear fitted line for two groups within each panel superimposed.
I have two conditioning factors, variable and Group and I
want separate panels for each level of variable
with different symbols and lmlines for each level of
Group. However all observations for a group are missing for
some variables so I would still like the points and lmline
for the observed group plotted for that variable(panel).
My attempt is
print(xyplot(total.fat~x|variable,data=tmp1,subscripts=TRUE,
scales=list(x=list(relation=free)),xlab=,groups=Group,
panel=function(x,y,subscripts,...){
panel.superpose(x,y,subscripts,...)
if(length(x[subscripts])!=0)
{panel.superpose(x,y,panel.groups=panel.lmline,subscripts,...)}}))
Which gives an error
Error in lm.fit(x, y, offset = offset, singular.ok =
singular.ok, ...) :
0 (non-NA) cases
Which occurs when the first panel with all values for one
group are missing.
The same error is returned if I replace the last line without the if
statement which obviously means it's ignored.
Of course I may be taking the wrong tack completely to get
the result I need. Any advice would be appreciated
Ross Darnell
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Re: [R] install packages automatically
Try ?update.packages Ross Darnell -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Wensui Liu Sent: Tuesday, 11 September 2007 11:45 AM To: r-help@stat.math.ethz.ch Subject: [R] install packages automatically Dear Listers, I am a little tired of installing all packages I want every time when I instill a new version of R. Say, if I have a list of packages I need to use, is it possible to tell R to install them all for me automatically rather than I install them one by one? Thx. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lattice panel.lmline problem
I am wanting to generate panels showing scatterplots with the linear
fitted line for two groups within each panel superimposed.
I have two conditioning factors, variable and Group and I want
separate panels for each level of variable
with different symbols and lmlines for each level of Group. However
all observations for a group are missing for some variables so I would
still like the points and lmline for the observed group plotted for that
variable(panel).
My attempt is
print(xyplot(total.fat~x|variable,data=tmp1,subscripts=TRUE,
scales=list(x=list(relation=free)),xlab=,groups=Group,
panel=function(x,y,subscripts,...){
panel.superpose(x,y,subscripts,...)
if(length(x[subscripts])!=0)
{panel.superpose(x,y,panel.groups=panel.lmline,subscripts,...)}}))
Which gives an error
Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
0 (non-NA) cases
Which occurs when the first panel with all values for one group are
missing.
The same error is returned if I replace the last line without the if
statement which obviously means it's ignored.
Of course I may be taking the wrong tack completely to get the result I
need. Any advice would be appreciated
Ross Darnell
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question on graphs and simple if statements
The answer to your first question is
curve(ifelse((x-1)(x1),x^2+1,x+2),from=-10,to=10)
Sorry I cannot understand the next part of your query.
Ross Darnell
On Sun, 2007-09-02 at 00:31 -0700, JHawk3 wrote:
Hi all,
I'm a relatively new user to the R interface, and am trying to do some basic
operations. So far I've used the curve() command to plot graphs. This has a
couple of limitations, namely trying to do piecewise graphs. Basically, I'm
looking for how to do if and while loops for making a graph.
in simple code, something like:
if x is on the interval [-1,1], x=x^2+1
otherwise, x=x^2
something along those lines. I also am wondering if it's possible to graph
something like this:
while(n20)
{
curve(1/n^2, -1,1)
n++
}
Along these same lines, is there a way to graph finite series?
Thank you for your time and your replies.
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Re: [R] customizing the color and point shape for each line drawn using lattice's xyplot
Does this help common.without.Method4 - subset(common, Method!=4) xyplot(MeanBxg ~ PercentVarExplained | bdg.f * bdx.f, data=common.without.Method4, groups=Method.f, type=l, auto.key=T) Ross Darnell -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Gen Sent: Thursday, 30 August 2007 2:51 PM To: r-help@stat.math.ethz.ch Subject: [R] customizing the color and point shape for each line drawn using lattice's xyplot Description of what I am trying to do: I am using the xyplot code below to plot the variable MeanBxg against the variable PercentVarExplained for all 9 possible combinations of variables bdg and bdx. Within each of these 9 scenarios I am plotting a separate line for each of up to 9 different methods that I used to estimate the variable MeanBxg. These methods are identified by the numeric variable called Method. Giving me one plot with 9 figures and each of the figures contains 9 lines. My problem arises because I would like to repeat the creation of this plot 8 times, in each instance only a subset (eg 6) of the 9 methods are used (a different subset each time). What I can't figure out: I would like to learn how to specify the exact line color that corresponds to each method such that Method==1 will always be represented by the same color (in every plot that it appears in). Where two methods that I used were of the same family of methods (say method==1 and method==2 made the same assumptions about the data) I would like to, if possible, represent the two methods using the same color and distinguish them by the symbol used to represent points on the line. My code as it currently stands: xyplot(MeanBxg ~ PercentVarExplained | bdg.f * bdx.f, data=common, groups=common$Method.f, type=l, subset= Method!=4, auto.key=T) As the code is, the default colors assigned are repeated causing different methods to be represented by the same color with no way to distinguish them (I have not succeeded in plotting lines and points simultaneously). Side question: When I subset the data to particular methods, is there a way to remove the excluded methods from the key as well? (in my code Method is a numeric variable, and Method.f corresponds to the lengthy descriptions of each method for the purpose of the key) Thank you very much for your help. Genevieve -- View this message in context: http://www.nabble.com/customizing-the-color-and-point-shape-for-each-lin e-drawn-using-lattice%27s-xyplot-tf4351934.html#a12400517 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] customizing the color and point shape for each line drawn using lattice's xyplot
Then you can regenerate the factor by common.without.Method4 - subset(common, Method!=4) common.without.Method4 - transform(common.without.Method4, Method = factor(Method)) Ross Darnell -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Gen Sent: Friday, 31 August 2007 1:42 AM To: r-help@stat.math.ethz.ch Subject: Re: [R] customizing the color and point shape for each line drawn using lattice's xyplot Thank you for your suggestion. Unfortunately it does not appear to matter if the data is subsetted before the xyplot command or within it. The problem remains. I was able to remove method 4 entirely from the key by using your suggestion to subset the dataset common, and then creating a new variable to assign labels for just this subset common.without.Method4$method.not4.f The resulting code was: xyplot(MeanBxg ~ PercentVarExplained | bdg.f * bdx.f, data=common.without.Method4, groups=method.not4.f, type=l, auto.key=T) This however leads to lack of continuity between plots, as when I exclude methods 4 and 6 in my second plot, using the same approach, the color that had corresponds to methods 7, 8 and 9 in plot one is automatically changed in plot two. The subsetting may work, so long as I can dictate line color and symbol type in each plot. Ross Darnell wrote: Does this help common.without.Method4 - subset(common, Method!=4) xyplot(MeanBxg ~ PercentVarExplained | bdg.f * bdx.f, data=common.without.Method4, groups=Method.f, type=l, auto.key=T) Ross Darnell -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Gen Sent: Thursday, 30 August 2007 2:51 PM To: r-help@stat.math.ethz.ch Subject: [R] customizing the color and point shape for each line drawn using lattice's xyplot Description of what I am trying to do: I am using the xyplot code below to plot the variable MeanBxg against the variable PercentVarExplained for all 9 possible combinations of variables bdg and bdx. Within each of these 9 scenarios I am plotting a separate line for each of up to 9 different methods that I used to estimate the variable MeanBxg. These methods are identified by the numeric variable called Method. Giving me one plot with 9 figures and each of the figures contains 9 lines. My problem arises because I would like to repeat the creation of this plot 8 times, in each instance only a subset (eg 6) of the 9 methods are used (a different subset each time). What I can't figure out: I would like to learn how to specify the exact line color that corresponds to each method such that Method==1 will always be represented by the same color (in every plot that it appears in). Where two methods that I used were of the same family of methods (say method==1 and method==2 made the same assumptions about the data) I would like to, if possible, represent the two methods using the same color and distinguish them by the symbol used to represent points on the line. My code as it currently stands: xyplot(MeanBxg ~ PercentVarExplained | bdg.f * bdx.f, data=common, groups=common$Method.f, type=l, subset= Method!=4, auto.key=T) As the code is, the default colors assigned are repeated causing different methods to be represented by the same color with no way to distinguish them (I have not succeeded in plotting lines and points simultaneously). Side question: When I subset the data to particular methods, is there a way to remove the excluded methods from the key as well? (in my code Method is a numeric variable, and Method.f corresponds to the lengthy descriptions of each method for the purpose of the key) Thank you very much for your help. Genevieve -- View this message in context: http://www.nabble.com/customizing-the-color-and-point-shape-for-each-lin e-drawn-using-lattice%27s-xyplot-tf4351934.html#a12400517 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/customizing-the-color-and-point-shape-for-each-lin e-drawn-using-lattice%27s-xyplot-tf4351934.html#a12409939 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide
Re: [R] Conditional subsetting
Assuming *A* animal can only be in one location at any one time, I don't understand how once you have selected the location nearest 12pm how you can select based on the type of location? Do you mean to select on location before timei.e. from the best location visited that day which was closest to 12pm? Perhaps for condition 2 for a animal which.min(time-1200) (replace 1200 with properly defined timestamp representing 12:00pm) Ross Darnell -Original Message- From: [EMAIL PROTECTED] on behalf of Tim Sippel Sent: Wed 08-Aug-07 9:37 AM To: r-help@stat.math.ethz.ch Subject: [R] Conditional subsetting Hello- Upon searching the email archives and reading documentation I haven't found what I'm looking for. I have a dataset with a time/date stamp on a series of geographic locations (lat/lon) associated with the movements of animals. On some days there are numerous locations and some days there is only one location. Associated with each location is an indication that the quality of the location is either good, poor, or unknown. For each animal, I want to extract only one location for each day. And I want the location extracted to be nearest to 12pm on any given day, and highest quality possible. So the order of priority for my extraction will be: 1. only one location per animal each day; 2. the location selected be as close to 12pm as possible; 3. the selected location comes from the best locations available (ie. take from pool of good locations first, or select from poor locations if a good location isn't available, or unknown if nothing else is available). I think aspect of this task that has me particularly stumped is how to select only one location for each day, and for that location to be a close to 12pm as possible. An example of my dataset follows: DeployID Date.Time LocationQuality Latitude Longitude STM05-1 28/02/2005 17:35 Good -35.562 177.158 STM05-1 28/02/2005 19:44 Good -35.487 177.129 STM05-1 28/02/2005 23:01 Unknown -35.399 177.064 STM05-1 01/03/2005 07:28 Unknown -34.978 177.268 STM05-1 01/03/2005 18:06 Poor -34.799 177.027 STM05-1 01/03/2005 18:47 Poor -34.85 177.059 STM05-2 28/02/2005 12:49 Good -35.928 177.328 STM05-2 28/02/2005 21:23 Poor -35.926 177.314 Many thanks for your input. I'm using R 2.5.1 on Windows XP. Cheers, Tim Sippel (MSc) School of Biological Sciences Auckland University Private Bag 92019 Auckland 1020 New Zealand +64-9-373-7599 ext. 84589 (work) +64-9-373-7668 (Fax) +64-21-593-001 (mobile) [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with reading data files with different numbers of lines to skips
You could read the file and test each line (try readline()) but you need some
rule to distinguish between text lines and data lines. You could then reread
the file with the number of skip lines defined.
Ross Darnell
-Original Message-
From: [EMAIL PROTECTED] on behalf of Tom Cohen
Sent: Fri 03-Aug-07 7:14 AM
To: r-help@stat.math.ethz.ch
Subject: [R] problem with reading data files with different numbers of lines
to skips
Dear List,
I have 30 data files with different numbers of lines (31 and 33) that I want
to skip before reading the files. If I use the skip option I can only choose
either to skip 31 or 33 lines. The data files with 31 lines have no blank rows
between the lines and the header row. How can I read the files without manually
checking which files have 31 respectively 33 lines ? The only text line I want
to keep is the header.
Thamks for your help,
Tom
for (i in 1:num.files) {
a-read.table(file=data[i],
,header=T,skip=31,sep='\t',na.strings=NA)
}
-
Går det långsamt? Skaffa dig en snabbare bredbandsuppkoppling.
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Re: [R] Convert string to list?
Manuel
Jim's may be what you want-- a list of numerics with names P, T and Q or
a list of character strings?
str - P = 0.0, T = 0.0, Q = 0.0
str(as.vector(unlist(strsplit(str,,)),mode=list))
List of 3
$ : chr P = 0.0
$ : chr T = 0.0
$ : chr Q = 0.0
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of jim holtman
Sent: Friday, 27 July 2007 11:20 AM
To: Manuel Morales
Cc: r-help
Subject: Re: [R] Convert string to list?
Is this what you want:
str - P = 0.0, T = 0.0, Q = 0.0
x - eval(parse(text=paste('list(', str, ')')))
str(x)
List of 3
$ P: num 0
$ T: num 0
$ Q: num 0
On 7/26/07, Manuel Morales [EMAIL PROTECTED] wrote:
Let's say I have the following string:
str - P = 0.0, T = 0.0, Q = 0.0
I'd like to find a function that generates the following object from
'str'.
list(P = 0.0, T = 0.0, Q = 0.0)
Thanks!
--
http://mutualism.williams.edu
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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Re: [R] Convert string to list?
Is this what your want? as.vector(unlist(strsplit(str,,)),mode=list) Ross Darnell -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Manuel Morales Sent: Friday, 27 July 2007 10:39 AM To: r-help Subject: [R] Convert string to list? Let's say I have the following string: str - P = 0.0, T = 0.0, Q = 0.0 I'd like to find a function that generates the following object from 'str'. list(P = 0.0, T = 0.0, Q = 0.0) Thanks! -- http://mutualism.williams.edu __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] questions on lme function
Ana You are estimating a random coefficient model on 5 individuals (mean and variance). Are you sure this is wise? Ross Darnell - Original Message - From: Ana Conesa [EMAIL PROTECTED] Date: Thursday, July 5, 2007 1:21 am Subject: [R] questions on lme function Dear list, I am using the lme funcion to fit a mixed model for the time response of a number of physiological variables. The random variable would be the individual on which physiological variables are measured at different time points. I have 4 time points, 5 individuals and 3 replicates per condition (time/individual), and I would like to fit a quadratic model on time. The model I am using is mm - lme(myvar ~ time + time2, random= ~ time|individual, data=clinical) being time2 = time*time I have a number of questions 1) I am not very sure the random effect is correctly modeled. Would I need to include the time2 variable aswell? 2) I would like to extract the F statistics of the model, and I do not find a function for this. Is this possible? 3) depending of the variable I take, I frequently obtain a convergence error as a result of the lme funcion. Any ideas on what to do to improve convergence? Thank you Ana Conesa, PhD Centro de Investigacion Principe Felipe Avda. Autopista Saler 16 46013 Valencia http://bioinfo.cipf.es __ 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.htmland provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Log-likelihood function
Alternatively generate the log-likelihood using the sum(dpois(y, fitted(model), log = TRUE)) Regards Ross Darnell Doxastic wrote: You're right. I do need to learn more. I never learned null/residual deviance. I know the deviance is equivalent to an anova decompostion. But I've never dealt with it seperated like this. I understand deviance as the difference between two model's log-likelihood difference between them and the most complex. I want to compare two models that are not the most complex. That is why I wanted the log-likelihood. I am using the poisson distribution because my response is count data, so the link is the log. If the deviance in R is computed by comparing the fitted model against the most saturated (which would make sense). Then yes, I can use that. I just picked the log-likelihood because I'm comparing two models. And that's the best way. But, it's equivalent if R compares the fitted to the most complex. I assumed the deviance print out tested the fitted model against the least complex. This tests whether the current model parameters can be dropped (that's what I thought the null deviance meant). I'm not sure what the residaul deviance means though. My main concern is why the likelihood functions differed between SAS and R. If anyone has encountered this or understands why, I would appreciate some help. Prof Brian Ripley wrote: I think you need to learn about deviances, which R does print. Log-likelihoods are only defined up to additive constants. In this case the conventional constant differs if you view this as a Poisson or as a product-multinomial log-linear model, and R gives you the log-likelihood for a Poisson log-linear model (assuming you specified family=poisson). However, deviances and differences in log-likelihoods do not depend on which. More details and worked examples can be found in MASS (the book, see the FAQ), including other ways to fit log-linear models in R. On Tue, 1 May 2007, someone ashamed of his real name wrote: I've computed a loglinear model on a categorical dataset. I would like to test whether an interaction can be dropped by comparing the log-likelihoods from two models(the model with the interaction vs. the model without). Since R does not immediately print the log-likelihood when I use the glm function, I used SAS initially. After searching for an extracting function, I found one in R. But, the log-likelihood given by SAS is different from the one given by R. I'm not sure if the logLik function in R is giving me something I don't want. Or if I'm misinterpreting the SAS output. Can anyone help? -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Log-likelihood-function-tf3678882.html#a10283755 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Lattice plots of the form xyplot(y1+y2~x)
I would appreciate help trying to set different plot types in a lattice plot with multiple responses e.g. xyplot(y+fval~x) but have y as points and fval as a line. I have tried xyplot(y+fval~x,type=c(p,l)) but this results in both plots having both types. Thanks Ross Darnell University of Queensland, AUSTRALIA __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] MASS4 page 360 cumulative hazard plot
I cannot reproduce Fig. 13.2 in MASS4. plot(gehan.surv,fun='cloglog') Warning message: 2 x values = 0 omitted from logarithmic plot in: xy.coords(x, y, xlabel, ylabel, log) and the x-axis is badly scaled. I was wondering if someone can help Regards Ross Darnell __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sweave and the [ function
Hi Vincent This would seem logical but in this case doesn't work. It doesn't seem to be a Sweave problem (feature) at all but within R as Hadley stated. Within R try quote(women[1]) women[1] now try quote([(women,1)) women[1] So it's parsed and normalised (there's a familiar term); in this case to women[1] before its quoted. Curiously in the R Language Definition Guide in 10.4.3 it states R has three indexing constructs, two of which are syntactically similar although with somewhat different semantics: object [ arg1, .. , argn ] object [[ arg1, .. , argn ]] The object can formally be any valid expression, but it is understood to denote or evaluate to a subsettable object. The arguments generally evaluate to numerical or character indices, but other kinds of arguments are possible (notably drop = FALSE). Internally, these index constructs are stored as function calls with function name [ respectively [[. So I'm lost now. Can some one hand me a map and compass? Vincent Goulet wrote: Le Mardi 5 Septembre 2006 0:03, hadley wickham a écrit : = str(women) women$height women[,1] [(women,1) @ to show the equivalence of three methods of extracting an element from a data.frame. However Sweave returns the last of these as women[1] in the S input chunk How can I force it not to do this and return [(women,1) I don't think you can. Sweave parses your R code and from then on uses the internal R representation. R normalises the parse tree in certain ways (eg. strips comments, formats source code, and clearly normalises some function calls). Since sweave uses this, and not the original text, I don't think there is anyway to get around this, unless there is some trick during parsing. (And don't forget women[[1]]) Hadley So here's a workaround (untested): echo=TRUE, eval=TRUE= str(women) women$height women[,1] @ echo=TRUE, eval=FALSE= [(women,1) @ echo=FALSE, eval=TRUE= [(women,1) @ I often end up doing similar things. HTHVincent __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Sweave and the [ function
I am wanting to use the [ operator in an S-chunk, e.g. = str(women) women$height women[,1] [(women,1) @ to show the equivalence of three methods of extracting an element from a data.frame. However Sweave returns the last of these as women[1] in the S input chunk How can I force it not to do this and return [(women,1) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Lattice plot of multiply-nested data
I would like to replicate Figure 3.12 from Pinheiro and Bates's Mixed Effects Models in S and S-PLUS but I would like to plot both the average line as well as the individual sites lines on the same panel. I did try modifying the panel function but I cannot seem to get it as described. I would appreciate any suggestions Ross Darnell University of Queensland __ 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] Coefficients for aliased models
I find the reporting of aliased terms in class lm objects as little inconsistent. The best way to show this is to give an example. X - mvrnorm(10,c(0,0),Sigma=matrix(c(1,0.5,0.5,1),nrow=2)) Xnew - as.data.frame(cbind(X,X[,2])) # make copy of X[,2] names(Xnew) - c(y,X,Xcopy) model - lm(y~X+Xcopy,data=Xnew) coef(model) # reports NA for Xcopy estimate summary(model) # ditto coef(summary(model)) # drops NA for Xcopy estimate Is this intentional? Is there a way of extracting the coefficient table from the summary output with the aliased term included? Many thanks Ross Darnell __ 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] Calculation of r.squared for linear model with offset
R^2 for a model is usually defined as 1-RSS/TSS where TSS is the SS about the mean and RSS is the residual SS from the model. Consider the model in R z - runif(20) y - z+rnorm(20) my.model - lm(y~offset(z)) summary(my.model)$r.squared Here the RSS is equivalent to the TSS and gives 0 when it should (IMHO and a few others perhaps) be 1 - RSS/TSS(corrected for the mean only) RSS = sum(resid(my.model)^2) TSS = sum((y-mean(y))^2) Have I missed this somewhere in the FAQ's or elsewhere? Regards Ross Darnell __ 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] Sweave: How can I include S input in paragraph mode
Sweavers As the title suggests I would appreciate any help to include S code in ordinary paragraph mode. I can use the textsl font but is isn't the same. Thanks Ross Darnell __ 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] Translating lme model call to lme4
I would appreciate help translating the following lme model to an lmer function. lme(lognrms ~ Group*Rotation*muscle*side*support*arms, random=~1|Subject/Stratum2/rep, data=Data) Many thanks Ross Darnell [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] anova.lmlist output change
R-colleagues I have adapted the anova.lmlist function to use the model object name as the first column in the output instead of the string Model n. If there is general agreement can the change be implemented into the stats package? Regards Ross Darnell -- University of Queensland, Brisbane QLD 4072 AUSTRALIA Email: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] A suggestion for predict function(s)
Liaw, Andy wrote:
I must respectfully disagree. Why carry extra copies of data arround? This
is probably OK for small to medium sized data, but definitely not for large
data.
Besides, in your example, it may do different things depending on whether
newdata is supplied: model.matrix is not necessarily the same as the
original data frame. You need a bit more work to get the right model.matrix
that correspond to the newdata. It's not clear to me whether you want to
return model matrix or model frame, but in either case it's not sufficient
to just use `newdata'.
Andy
From: Ross Darnell
Maybe a useful addition to the predict functions would be to
return the
values of the predictor variables. It just (unless there are
problems)
requires an extra line. I have inserted an example below.
predict.glm -
function (object, newdata = NULL, type = c(link, response,
terms), se.fit = FALSE,
dispersion = NULL, terms = NULL,
na.action = na.pass, ...)
{
type - match.arg(type)
na.act - object$na.action
object$na.action - NULL
if (!se.fit) {
if (missing(newdata)) {
pred - switch(type, link = object$linear.predictors,
response = object$fitted, terms =
predict.lm(object,
se.fit =
se.fit, scale
= 1, type = terms,
terms = terms))
if (!is.null(na.act))
pred - napredict(na.act, pred)
}
else {
pred - predict.lm(object, newdata, se.fit, scale = 1,
type = ifelse(type == link,
response, type),
terms = terms, na.action = na.action)
switch(type, response = {
pred - family(object)$linkinv(pred)
}, link = , terms = )
}
}
else {
if (inherits(object, survreg))
dispersion - 1
if (is.null(dispersion) || dispersion == 0)
dispersion - summary(object, dispersion =
dispersion)$dispersion
residual.scale - as.vector(sqrt(dispersion))
pred - predict.lm(object, newdata, se.fit, scale =
residual.scale,
type = ifelse(type == link,
response, type),
terms = terms, na.action = na.action)
fit - pred$fit
se.fit - pred$se.fit
switch(type, response = {
se.fit - se.fit * abs(family(object)$mu.eta(fit))
fit - family(object)$linkinv(fit)
}, link = , terms = )
if (missing(newdata) !is.null(na.act)) {
fit - napredict(na.act, fit)
se.fit - napredict(na.act, se.fit)
}
predictors - if (missing(newdata)) model.matrix(object)
else newdata
pred - list(predictors=predictors,
fit = fit, se.fit = se.fit,
residual.scale = residual.scale)
}
pred
#__ end of R code
Ross Darnell
--
School of Health and Rehabilitation Sciences
University of Queensland, Brisbane QLD 4072 AUSTRALIA
Email: [EMAIL PROTECTED]
Phone: +61 7 3365 6087 Fax: +61 7 3365 4754 Room:822,
Therapies Bldg.
http://www.shrs.uq.edu.au/shrs/school_staff/ross_darnell.html
__
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PLEASE do read the posting guide!
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Notice: This e-mail message, together with any attachment...{{dropped}}
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] A suggestion for predict function(s)
Maybe a useful addition to the predict functions would be to return the
values of the predictor variables. It just (unless there are problems)
requires an extra line. I have inserted an example below.
predict.glm -
function (object, newdata = NULL, type = c(link, response,
terms), se.fit = FALSE,
dispersion = NULL, terms = NULL,
na.action = na.pass, ...)
{
type - match.arg(type)
na.act - object$na.action
object$na.action - NULL
if (!se.fit) {
if (missing(newdata)) {
pred - switch(type, link = object$linear.predictors,
response = object$fitted, terms = predict.lm(object,
se.fit = se.fit, scale
= 1, type = terms,
terms = terms))
if (!is.null(na.act))
pred - napredict(na.act, pred)
}
else {
pred - predict.lm(object, newdata, se.fit, scale = 1,
type = ifelse(type == link, response, type),
terms = terms, na.action = na.action)
switch(type, response = {
pred - family(object)$linkinv(pred)
}, link = , terms = )
}
}
else {
if (inherits(object, survreg))
dispersion - 1
if (is.null(dispersion) || dispersion == 0)
dispersion - summary(object, dispersion = dispersion)$dispersion
residual.scale - as.vector(sqrt(dispersion))
pred - predict.lm(object, newdata, se.fit, scale = residual.scale,
type = ifelse(type == link, response, type),
terms = terms, na.action = na.action)
fit - pred$fit
se.fit - pred$se.fit
switch(type, response = {
se.fit - se.fit * abs(family(object)$mu.eta(fit))
fit - family(object)$linkinv(fit)
}, link = , terms = )
if (missing(newdata) !is.null(na.act)) {
fit - napredict(na.act, fit)
se.fit - napredict(na.act, se.fit)
}
predictors - if (missing(newdata)) model.matrix(object) else newdata
pred - list(predictors=predictors,
fit = fit, se.fit = se.fit,
residual.scale = residual.scale)
}
pred
#__ end of R code
Ross Darnell
--
School of Health and Rehabilitation Sciences
University of Queensland, Brisbane QLD 4072 AUSTRALIA
Email: [EMAIL PROTECTED]
Phone: +61 7 3365 6087 Fax: +61 7 3365 4754 Room:822, Therapies Bldg.
http://www.shrs.uq.edu.au/shrs/school_staff/ross_darnell.html
__
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Histogram without common borders
Is it possible to produce a histogram directly using the hist() function with the common borders removed? It can be done by plotting the histogram object using type 's'teps. my.hist - hist(x,plot=FALSE) plot(my.hist$breaks,c(0,my.hist$counts),type='s') I would appreciate help Ross Darnell -- University of Queensland, Brisbane QLD 4067 AUSTRALIA Email: [EMAIL PROTECTED] Phone: +61 7 3365 6087 Fax: +61 7 3365 4754 http://www.shrs.uq.edu.au/shrs/school_staff/ross_darnell.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] WinMenu's question
I am using the Windows menu functions below which will work on the
first pass, but if I repeat the same script I cannot get the
WinMenuAddItem to work. This is a problem if I change the menu
structure and reread the source code I am forced to quit and restart Rgui.
try.menu - function(){
OS - .Platform$OS.type
GUI - .Platform$GUI
if (!(OS == windows GUI == Rgui)) return(Sorry, you must be running R using
Rgui.exe on MS Windows)
try(winMenuDelItem('EMG/Graphics','Plot.trace'))
try(winMenuDel('EMG/Graphics'))
try(winMenuDel(EMG))
winMenuAdd(EMG)
winMenuAdd(EMG/Graphics)
winMenuAddItem(EMG/Graphics,Plot trace,plot.trace())
}
plot.trace - function(){
x - eval(parse(text=trace.dialog()))
plot(x,type=l)
invisible()
}
Help much appreciated
Ross Darnell
--
University of Queensland, Brisbane QLD 4067 AUSTRALIA
Email: [EMAIL PROTECTED]
Phone: +61 7 3365 6087 Fax: +61 7 3365 4754
http://www.shrs.uq.edu.au/shrs/school_staff/ross_darnell.html
__
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[R] Frequent crash printing graphics windows and wavethresh
I often have R (1.8.1) crash after I generate several graphics windows using windows() or X11(), print a graphics window and then rerun the same script. A windows message comes up saying Program error Rterm.exe .. (I can get the same problem using Rgui.) Xemacs tells me Process R trace trap at time and date The entry in the Dr Watson logs starts with Application exception occurred: App: (pid=134697032) When: 26/11/2002 @ 21:12:10.057 Exception number: c005 (access violation) The situation in which I can consistently produce this is when I am using the wavethresh3 dll and associated functions. I have tried to generate the same problem without using wavelet functions (just producing several graphics windows with plots of random numbers) but cannot reproduce this problem consistently from a new session. I was wondering if anyone else is experiencing this problem Regards Ross Darnell -- University of Queensland, Brisbane QLD 4067 AUSTRALIA Email: [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] help with nomogram function
Frank Harrell wrote On Wed, 05 Nov 2003 09:22:34 +1000 Ross Darnell [EMAIL PROTECTED] wrote: I have fitted a logistic regression model failed.lr2$call lrm(formula = failed ~ Age + task2 + Age:task2, data = time.long, na.action = na.omit) Use Age*task2 and omit na.action as na.omit is the default using the Design package functions and would like to generate a nomogram from this model. the datadist information is generated and stored in ddist time.long$Age time.long$task2 Low:effect 45NA Adjust to 56 both.foam High:effect68NA Low:prediction 21 both.foam High:prediction80 right Low20 both.foam High 80 right This looks most strange. You did not include the original code but I suspect you had $ in a term. $ should not appear in column headings above. Design wants you to use data= or attach, and avoid $ in terms. Values: time.long$task2 : both.foam left right The model fitted and then when I try the nomgram function nomogram(failed.lr2) Error in value.chk(at, i, NA, -nint, Limval, type.range = full) : variable Age does not have limits defined by datadist I get an error. The NA values in ddist seem to be the problem but I don't understand the datadist information. I tried using the data argument in datadist names(my.data) [1] subject Age failed task ddist - datadist(Age,task,data=my.data) Error in datadist(Age, task, data = my.data) : Object Age not found Strange. Thanks Ross Darnell -- __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] help with nomogram function
I have fitted a logistic regression model failed.lr2$call lrm(formula = failed ~ Age + task2 + Age:task2, data = time.long, na.action = na.omit) using the Design package functions and would like to generate a nomogram from this model. the datadist information is generated and stored in ddist time.long$Age time.long$task2 Low:effect 45NA Adjust to 56 both.foam High:effect68NA Low:prediction 21 both.foam High:prediction80 right Low20 both.foam High 80 right Values: time.long$task2 : both.foam left right The model fitted and then when I try the nomgram function nomogram(failed.lr2) Error in value.chk(at, i, NA, -nint, Limval, type.range = full) : variable Age does not have limits defined by datadist I get an error. The NA values in ddist seem to be the problem but I don't understand the datadist information. Can anyone help explain why this is failing. Thanks Ross Darnell -- University of Queensland, Brisbane QLD 4067 AUSTRALIA Email: [EMAIL PROTECTED] Phone +61 7 3365 6087 http://www.shrs.uq.edu/shrs/school_staff/ross_darnell.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] R workshop, Qld Australia
Need some help with R? An R workshop, lead by John Maindonald, is planned for the 2nd October at the University of Southern Queensland, Toowoomba, Australia. The workshop is being organised by the Queensland branch of the Statistical Society of Australia as part of their conference. Anyone interested in attending can find out more by looking at http://www.sci.usq.edu.au/staff/dunn/qstatconf/workshops.html -- Ross Darnell Organising committee Email: [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Specifying an lme model
Thanks for the help. Sorry but now comes the second question. What we have now assumes equal within-subject variances. How can I fit separate (pooled) within-subject variances for each group (similar to an unequal variance t-test) and test if this is better than the existing constant within-subject variance. Many thanks Ross Douglas Bates [EMAIL PROTECTED] writes: Ross Darnell [EMAIL PROTECTED] writes: I would like some advice on how if possible, to test the following I have subjects each measured several times. The subjects are sampled from 3 subpopulations (groups). The question is Is the between-subject variance the same for the three groups? The null model is lme0 - lme(y~group,random=~1|subject) I did think that the model that defined a specific between-subject variance for each group was update(lme0,.~., weights=varIdent(form=~1|group)) but I am not sure. I think you have it right. You should then compare the two fitted models using the anova generic, which will provide a likelihood ratio test statistic and a p-value based on a chi-squared reference distribution. Regard the p-value as an approximation. -- Ross Darnell School of Health and Rehabilitation Sciences University of Queensland, Brisbane QLD 4067 AUSTRALIA Email: [EMAIL PROTECTED] Phone +61 7 3365 6087 http://www.uq.edu.au/~uqrdarne/ __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Specifying an lme model
I would like some advice on how if possible, to test the following I have subjects each measured several times. The subjects are sampled from 3 subpopulations (groups). The question is Is the between-subject variance the same for the three groups? The null model is lme0 - lme(y~group,random=~1|subject) I did think that the model that defined a specific between-subject variance for each group was update(lme0,.~., weights=varIdent(form=~1|group)) but I am not sure. I would appreciate any help. Ross Darnell __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] equivalence test
Hi, is it possible to do an equivalence test on paired quantitative datas in R? Is there a way to calculate sample size for such tests? I've tried to find some documentation on that subject but I was unsuccessfull. I'll be happy with any links on equivalence test. If such a test does'nt exist in R, i'll do it manually if I find a method to do so. Best regards Blaise The sample size for an equivalence trial with normally distributed outcomes can be calculated by n - 2 * s^2 / delta^2 * (qnorm(alpha/2) + qnorm(beta/2))^2 s = standard deviation delta = clinically important difference alpha = Type I error beta = Type II error -- Ross Darnell __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Re: Finding missing data patterns
The norm package of Joe Schafer can be used. It has a function called 'prelim' (or very similar) that does this. Regards Ross Darnell Email: [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list http://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Searching for glmmNQ
I cannot find the glmmNQ function in the MASS package (or anywhere else I have tried) mentioned on page 296 of MASS4. I would appreciate directions. Thanks -- Ross Darnell __ [EMAIL PROTECTED] mailing list http://www.stat.math.ethz.ch/mailman/listinfo/r-help
