Re: [R] ANCOVA with defined error terms
Could someone or Richard explain to me what he meant by This also shows a singular Error(). We look at the data and see that plot is identical to the three-way veget:fruit:block interaction. It seems to me that I just needed to recoded the plots, in order to get rid of the Error message. If that is true, then why I can't go back to the original model proposed by Richard track.aov - aov(mice ~ coon + block * veget * fruit * time - block:veget:fruit:time + Error(block/Plot), data = track) and calculate F values myself? Thank you very much for the feedback. -- View this message in context: http://www.nabble.com/ANCOVA-with-defined-error-terms-tp25055311p25147303.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Thank you David : Re : Odp: Re : Odp: Re : table function
Thank you very very much David De : David Winsemius dwinsem...@comcast.net à : David Winsemius dwinsem...@comcast.net Envoyé le : Mardi, 25 Août 2009, 23h32mn 53s Objet : Re: [R] Re : Odp: Re : Odp: Re : table function On Aug 25, 2009, at 9:23 AM, David Winsemius wrote: On Aug 25, 2009, at 6:25 AM, Inchallah Yarab wrote: [[elided Yahoo spam]] [[elided Yahoo spam]] [[elided Yahoo spam]] ?table ?xtab I read this incorrectly. Try instead; by(z, z.c, sum) z.c: 0 - 1000 [1] 1010 z.c: 1000 - 3000 [1] 6400 z.c: 3000 [1] 14200 De : Petr PIKAL petr.pi...@precheza.cz Cc : r-help@r-project.org Envoyé le : Mardi, 25 Août 2009, 11h53mn 21s Objet : Odp: [R] Re : Odp: Re : table function Hi r-help-boun...@r-project.org napsal dne 25.08.2009 11:28:31: à Thank you Peter, in my vector Z i have missing value NA and i want to count its number in the vector i had did alla this i know the difference between a numeirc and a factor OK. If you know difference between factor and numeric you probably have seen ?factor where is note about how to make NA an extra level #Here is z z-c(10,100, 1000, 1200, 2000, 2200, 3000, 3200, 5000, 6000) #let's put some NA values into it z[c(2,5)] -NA #let's make a cut z.c-cut(z, breaks = c(-Inf, 1000, 3000, Inf), labels = c(0 - 1000, 1000 - 3000, 3000)) #as you see there are NA values but they are not extra level z.c [1] 0 - 1000  NA    0 - 1000  1000 - 3000 NA [6] 1000 - 3000 1000 - 3000 3000    3000    3000 Levels: 0 - 1000 1000 - 3000 3000 is.na(z.c) [1] FALSE TRUE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE # so let's try what help page says factor(z.c, exclude=NULL) [1] 0 - 1000  NA    0 - 1000  1000 - 3000 NA [6] 1000 - 3000 1000 - 3000 3000    3000    3000 Levels: 0 - 1000 1000 - 3000 3000 NA # wow we have NA as extra level, let's do table table(factor(z.c, exclude=NULL))   0 - 1000 1000 - 3000    3000    NA      2      3      3      2 Regards Petr the goal of this exercice that to count the number of missing value, number betwwen 0-1000 , 1000-3000, 3000. Thank you again for your help De : Petr PIKAL petr.pi...@precheza.cz Cc : r-help@r-project.org EnvoyÄ© le : Mardi, 25 AoÄ»t 2009, 11h15mn 23s Objetà : Odp: [R] Re : table function Hi r-help-boun...@r-project.org napsal dne 25.08.2009 10:08:36: Hi Mark, Thank you for your answer !! it works but if i have NA in the vector z what i shoud do to count its number in Z? You do not have NA in z, you manage to convert it somehow to factor. Please try to read about data types and its behaviour. Start with this chapter 2.8 Other types of objects in R intro manual which I suppose you have in doc folder of R program directory. You possibly can convert it back to numeric by e.g. DF$z - as.numeric(as.character(DF$z)) but I presume you need to check your original data maybe by str(your.data) what mode they are and why they are factor if you expect them numeric. Regards Petr xàààyààààz 1àà0ààà100 5àà1ààà1500 6àà1àààNA 2àà2ààà500 1àà1àààNA 5àà2àà2000 8àà5àà4500 i did the same but it gives me this erroràmessage: à[0 - 1000] [1000 - 3000]àààà3000 àààààà0àààààà0àààààà0 Warning message: In inherits(x, factor) : NAs introduced by coercion Thank you De : Marc Schwartz marc_schwa...@me.com Cc : r-help@r-project.org EnvoyÄ© le : Lundi, 24 Aoűt 2009, 18h33mn 52s Objet : Re: [R] table function On Aug 24, 2009, at 10:59 AM, Inchallah Yarab wrote: hi, i want to use the function table to build a table not of frequence (number of time the vareable is repeated in a list or a data frame!!) but in function of classes [[elided Yahoo spam]] example xàààyààààz 1àà0ààà100 5àà1ààà1500 6àà1ààà1200 2àà2ààà500 1àà1ààà3500 5àà2àà2000 8àà5àà4500 i want to do a table summerizing the number of variable where z is in [0-1000],],[1000-3000], [ 3000] thank you very much for your help See ?cut, which bins a continuous variable. DF àx yààz 1 1 0à100 2 5 1 1500 3 6 1 1200 4 2 2à500 5 1 1 3500 6 5 2 2000 7 8 5 4500
[R] Select top three values from data frame
Hi, I'm trying to find an easy way to do this. I want to select the top three values of a specific column in a subset of rows in a data.frame. I'll demonstrate. ABC x21 x41 x32 y15 y26 y38 I want the top 3 values of B from the data.frame where A=X and C 2 I could extract all the rows where C2, then sort by B, then take the first 3. But that seems like the wrong way around, and it also will get messy with real data of over 100 columns. Any suggestions? __ R-help@r-project.org 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] ptproc package
If you have read the posting guide and then tell us the website, the version, the R version, your OS, and so on, we might finally be able to help. Uwe Ligges oyelola.adegb...@student.uhasselt.be wrote: Dear All, Please can anyone assist on installing 'ptproc', I downloaded it on the contributors website and tried to install it manual but R wont unzip it. Thank you __ R-help@r-project.org 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] [Rd] Formulas in gam function of mgcv package
I am trying to understand the relationships between: y~s(x1)+s(x2)+s(x3)+s(x4) and y~s(x1,x2,x3,x4) Does the latter contain the former? what about the smoothers of all interaction terms? The first says that you want a model E(y) = f_1(x_1) + f_2(x_2) + f_3(x_3) + f_4(x_4) (1) where the f_j are smooth functions. The additive decomposition is quite a strong assumption, since it assumes that the effect of x_j is not dependent on x_k unless j=k. The second model is just E(y) = f(x_1,x_2,x_3,x4) (2) where f is a smooth function. This looks very general, but actually `s' terms assume isotropic smoothness, which is also quite a strong assumption. Now if I simply state that f and the f_j are `smooth functions', and leave it at that, then (2) would of course contain (1), but to actually estimate the models I need to state, mathematically, what I mean by `smooth'. Once I've done that I've pretty much determined the function spaces in which f and the f_j will lie, and in general (2) will no longer strictly contain (1). mgcv's `s' terms use a thin plate spline measure of smoothness for multivariate smooths, and this means that (1) will not be strictly nested within (2), since e.g. a 4D thin plate spline can not generally represent exactly what the sum of 4 1D splines can represent. If you want to acheive exact nesting then using tensor product smooths with something like y~te(x1)+te(x2)+te(x3)+te(x4) (3) y~te(x1,x2,x3,x4) (4) will do the trick (because the function space for (4) is built up from the function spaces used in (3)). As to where all the 2 and 3 way interactions have gone in (4)... it's just like ANOVA - if you put in a 4 way interaction then the lower order interactions are not identifiable, unless you choose to add constraints to make them so. `mgcv' will allow you add main effects and interactions, and will handle the constraints automatically, but if this sort of functional ANOVA is a major component of what you want to do, then it is probably worth checking out the gss package and Chong Gu's book on smoothing spline ANOVA. best, Simon -- Simon Wood, Mathematical Sciences, University of Bath, Bath, BA2 7AY UK +44 1225 386603 www.maths.bath.ac.uk/~sw283 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Select top three values from data frame
Hi r-help-boun...@r-project.org napsal dne 26.08.2009 10:36:22: Hi, I'm trying to find an easy way to do this. I want to select the top three values of a specific column in a subset of rows in a data.frame. I'll demonstrate. ABC x21 x41 x32 y15 y26 y38 I want the top 3 values of B from the data.frame where A=X and C 2 I could extract all the rows where C2, then sort by B, then take the first 3. But that seems like the wrong way around, and it also will get messy with real data of over 100 columns. One way is to use subset, order and head head(subset(your.data[order(your.data$B, decreasing=T),], subset = C2 A==x), 3) Regards Petr Any suggestions? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Select top three values from data frame
Noah Silverman a écrit : Hi, I'm trying to find an easy way to do this. I want to select the top three values of a specific column in a subset of rows in a data.frame. I'll demonstrate. Hi, did you try this? data[data$A=='x' data$C2,]$B # data = your data frame ABC x21 x41 x32 y15 y26 y38 I want the top 3 values of B from the data.frame where A=X and C 2 I could extract all the rows where C2, then sort by B, then take the first 3. But that seems like the wrong way around, and it also will get messy with real data of over 100 columns. Any suggestions? regards ML __ R-help@r-project.org 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. -- Mohamed Lajnef INSERM Unité 955. 40 rue de Mesly. 94000 Créteil. Courriel : mohamed.laj...@inserm.fr tel. : 01 49 81 31 31 (poste 18470) Sec : 01 49 81 32 90 fax : 01 49 81 30 99 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: draw n-1 lines with n X and n Y
Hi r-help-boun...@r-project.org napsal dne 25.08.2009 18:15:25: I want to draw lines using a matrix with the X-axe on the first column, and the the Y-axe on the second column. These lines have to link the n points of a graph, so they are n-1, but the resout using these two commands: X - matrix(scan(MaxInterEdges.R,0), ncol=2); lines(X[,1], X[,2], type = l, col = red); is wrong, becose the lines are many more. What can I do? Well, without reproducible code could help to identify a your problem. here is a guess X[,1] is not ordered and the lines are connecting points as they appear in your data. You could order X by first column and then try to plot lines. X.ord-X[order(X[,1]),] lines(X.ord[,1], X.ord[,2], col = red) Regards Petr -- View this message in context: http://www.nabble.com/draw-n-1-lines-with-n-X- and-n-Y-tp25137514p25137514.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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@r-project.org 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] Select top three values from data frame
df.mydata[df.mydata$A==X AND df.mydata$C 2, ] will do the job ? 8rino Noah Silverman ha scritto: Hi, I'm trying to find an easy way to do this. I want to select the top three values of a specific column in a subset of rows in a data.frame. I'll demonstrate. ABC x21 x41 x32 y15 y26 y38 I want the top 3 values of B from the data.frame where A=X and C 2 I could extract all the rows where C2, then sort by B, then take the first 3. But that seems like the wrong way around, and it also will get messy with real data of over 100 columns. Any suggestions? __ R-help@r-project.org 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] Select top three values from data frame
I only have a few values in my example, but the real data set might have 20-100 rows with A=X. So how do I pick just the three highest ones? -N On 8/26/09 2:46 AM, Ottorino-Luca Pantani wrote: df.mydata[df.mydata$A==X AND df.mydata$C 2, ] will do the job ? 8rino Noah Silverman ha scritto: Hi, I'm trying to find an easy way to do this. I want to select the top three values of a specific column in a subset of rows in a data.frame. I'll demonstrate. ABC x21 x41 x32 y15 y26 y38 I want the top 3 values of B from the data.frame where A=X and C 2 I could extract all the rows where C2, then sort by B, then take the first 3. But that seems like the wrong way around, and it also will get messy with real data of over 100 columns. Any suggestions? [[alternative HTML version deleted]] __ R-help@r-project.org 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] Issues with factors with duplicate (empty) levels
Hello! I imported a DJI survey[1] from an SPSS file. When looking at some of the variables, I noticed problems with the `table` function and similar. It seems to be caused by duplicate levels which are generated from the value labels. Not all values have labels, so those who don’t get an empty string as the level, which leads to duplicates. I hope the code and output below illustrates the problem. Is it possible to prevent this? I’d still like to use the labels, so using numeric vectors instead of factors is not the best solution. Regards, Frederik library(foreign) Data - read.spss(js2003_16_29_db.sav, to.data.frame=TRUE, reencode=latin1) table(Data$J203_A) überhaupt nicht wichtig 352256 0 0 0 0 sehr wichtig Mehrfachnennung 4660 0 table(as.numeric(Data$J203_A)) 1234567 35 39 84 227 626 1280 4660 is.factor(Data$J203_A) [1] TRUE levels(Data$J203_A) [1] überhaupt nicht wichtig [3] [5] [7] sehr wichtigMehrfachnennung [1] http://213.133.108.158/surveys/index.php?m=msw,0sID=54 signature.asc Description: Dies ist ein digital signierter Nachrichtenteil __ R-help@r-project.org 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] Select top three values from data frame
Colin, Unless I missed something, the head function doesn't sort. So if I have 1000 values that match, head just gives me the first three, not the HIGHEST three. [df.mydata$A==X df.mydata$C 2, On 8/26/09 3:01 AM, Colin Millar wrote: Hi, This should work - head is quite a usefull summary function head(df.mydata[df.mydata$A==X df.mydata$C 2, ],3) Colin. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Noah Silverman Sent: 26 August 2009 10:54 To: ottorino-luca.pant...@unifi.it Cc: r help Subject: Re: [R] Select top three values from data frame I only have a few values in my example, but the real data set might have 20-100 rows with A=X. So how do I pick just the three highest ones? -N On 8/26/09 2:46 AM, Ottorino-Luca Pantani wrote: df.mydata[df.mydata$A==X AND df.mydata$C 2, ] will do the job ? 8rino Noah Silverman ha scritto: Hi, I'm trying to find an easy way to do this. I want to select the top three values of a specific column in a subset of rows in a data.frame. I'll demonstrate. ABC x21 x41 x32 y15 y26 y38 I want the top 3 values of B from the data.frame where A=X and C2 I could extract all the rows where C2, then sort by B, then take the first 3. But that seems like the wrong way around, and it also will get messy with real data of over 100 columns. Any suggestions? [[alternative HTML version deleted]] __ R-help@r-project.org 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. __ This email has been scanned by the MessageLabs Email Security System. For more information please visit http://www.messagelabs.com/email __ [[alternative HTML version deleted]] __ R-help@r-project.org 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] Select top three values from data frame
Noah Silverman a écrit : I only have a few values in my example, but the real data set might have 20-100 rows with A=X. So how do I pick just the three highest ones? -N Hi, and now? df.mydata$B[order(df.mydata[df.mydata$A==X AND df.mydata$C 2, ]$B)][length(df.mydata$B)-3:length(df.mydata$B)] cheers, ML On 8/26/09 2:46 AM, Ottorino-Luca Pantani wrote: df.mydata[df.mydata$A==X AND df.mydata$C 2, ] will do the job ? 8rino Noah Silverman ha scritto: Hi, I'm trying to find an easy way to do this. I want to select the top three values of a specific column in a subset of rows in a data.frame. I'll demonstrate. ABC x21 x41 x32 y15 y26 y38 I want the top 3 values of B from the data.frame where A=X and C 2 I could extract all the rows where C2, then sort by B, then take the first 3. But that seems like the wrong way around, and it also will get messy with real data of over 100 columns. Any suggestions? [[alternative HTML version deleted]] __ R-help@r-project.org 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. -- Mohamed Lajnef INSERM Unité 955. 40 rue de Mesly. 94000 Créteil. Courriel : mohamed.laj...@inserm.fr tel. : 01 49 81 31 31 (poste 18470) Sec : 01 49 81 32 90 fax : 01 49 81 30 99 __ R-help@r-project.org 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] [Rd] Formulas in gam function of mgcv package
This will not work... 2) y~s(x1, ,x36) Estimating a 36 dimensional functions reasonably well would require a tremendous quantity of data, but in any case the 36 dimensional TPS smoothnes measure will involve such high order derivatives that it will no longer be practically useful: in fact you will not have enough data to estimate the unpenalized coefficients of the smoother (and if you did R would run out of memory first). In such a high dimensional situation, I think that GAMs are really only useful if you have some prior knowledge of which variables are likely to interact (and it's not too many of them). If there's no prior information saying roughly what sort of smooth additive structure might be useful then, I'm not sure that GAMs are the right way to go, and some sort of machine learning approach might be better. Then again, the real problem with y~s(x1, ,x36) is that the data just won't contain enough information to estimate s, if all you can say is that s is smooth, but this also means that it's very unlikely that you really need to estimate s(x1, ,x36) in order to predict well. In that case, starting from y ~ s(x1) + + s(x36) and building the model up might result in something that does a reasonable predictive job. On the subject of tensor product smoothing vs isotropic smoothing. Isotropic smooths are really only reasonable if you think that the smooth should display approximately the same amount of wiggliness in all directions. If this is not the case then tensor product smoothing is a better bet. Centering and scaling alone is not enough to ensure that isotropy is reasonable (although in particular cases it may help, of course). best, Simon I am trying to build a predictive model. Since the the variables are centred and scaled, I think I need an isotropic smooth. I am also interested in having the interactions between the variables included, that is not a purely additive model. It is not clear to me when should I give preference to tensor smooths, possibly because I have not understood well how they work. I am reading Wood(2003) as recommended and I have also read rather extensively Simon N. Wood. Generalized Additive Models: An Introduction, 2006, but still I am stuck. Any additional suggestion or reading recommendation would be greatly appreciated. I have also some difficulties in understanding the values you have chosen for k in the first example (why 60?). Thanks Best, On Monday 24 August 2009 17:33:55 Gavin Simpson wrote: [Note R-Devel is the wrong list for such questions. R-Help is where this should have been directed - redirected there now] On Mon, 2009-08-24 at 17:02 +0100, Corrado wrote: Dear R-experts, I have a question on the formulas used in the gam function of the mgcv package. I am trying to understand the relationships between: y~s(x1)+s(x2)+s(x3)+s(x4) and y~s(x1,x2,x3,x4) Does the latter contain the former? what about the smoothers of all interaction terms? I'm not 100% certain how this scales to smooths of more than 2 variables, but Sections 4.10.2 and 5.2.2 of Simon Wood's book GAM: An Introduction with R (2006, Chapman Hall/CRC) discuss this for smooths of 2 variables. Strictly y ~ s(x1) + s(x2) is not nested in y ~ s(x1, x2) as the bases used to produce the smoothers in the two models may not be the same in both models. One option to ensure nestedness is to fit the more complicated model as something like this: ## if simpler model were: y ~ s(x1, k=20) + s(x2, k = 20) y ~ s(x1, k=20) + s(x2, k = 20) + s(x1, x2, k = 60) ^ where the last term (^^^ above) has the same k as used in s(x1, x2) Note that these are isotropic smooths; are x1 and x2 measured in the same units etc.? Tensor product smooths may be more appropriate if not, and if we specify the bases when fitting models s(x1) + s(x2) *is* strictly nested in te(x1, x2), eg. y ~ s(x1, bs = cr, k = 10) + s(x2, bs = cr, k = 10) is strictly nested within y ~ te(x1, x2, k = 10) ## is the same as y ~ te(x1, x2, bs = cr, k = 10) [Note that bs = cr is the default basis in te() smooths, hence we don't need to specify it, and k = 10 refers to each individual smooth in the te().] HTH G I have (tried to) read the manual pages of gam, formula.gam, smooth.terms, linear.functional.terms but could not understand properly. Regards -- Simon Wood, Mathematical Sciences, University of Bath, Bath, BA2 7AY UK +44 1225 386603 www.maths.bath.ac.uk/~sw283 __ R-help@r-project.org 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] [Rd] Formulas in gam function of mgcv package
Dear Simon, thanks for your answer. I am running the model with both s and te smoothing, to compare. A few questions on your email: 1) Isotropic smoothness: my variables are centred and scaled. I assumed an isotropic smoother (that is, a smoother that treats all the variables in the same way) was good. What do you think? Is my understanding of isotropic smoothing wrong? 2) s(x1,, xn): it does not contains (1), but I thought it was true that it does improve on (1) by being free of including some interaction, albeit not explicitly is my interpretation wrong? 3) te: I am confused! What does it mean that the function space for (4) is built up from the function spaces used in (3)? Does it mean that te(xi,,xn) is an expansion on the te(xi), including all the terms te(x1)*te(x2)**te(xj)**te(xn) of the different orders? Example: in the case of 4 variables, including te(x1)*te(x2), te(x2)*te(x3), te(x1)*te(x2)*te(x3) to te(x1)*te(x2)*te(x3)*te(x4) . Sorry for being particularly daft Regards On Wednesday 26 August 2009 09:56:13 you wrote: I am trying to understand the relationships between: y~s(x1)+s(x2)+s(x3)+s(x4) and y~s(x1,x2,x3,x4) Does the latter contain the former? what about the smoothers of all interaction terms? The first says that you want a model E(y) = f_1(x_1) + f_2(x_2) + f_3(x_3) + f_4(x_4) (1) where the f_j are smooth functions. The additive decomposition is quite a strong assumption, since it assumes that the effect of x_j is not dependent on x_k unless j=k. The second model is just E(y) = f(x_1,x_2,x_3,x4) (2) where f is a smooth function. This looks very general, but actually `s' terms assume isotropic smoothness, which is also quite a strong assumption. Now if I simply state that f and the f_j are `smooth functions', and leave it at that, then (2) would of course contain (1), but to actually estimate the models I need to state, mathematically, what I mean by `smooth'. Once I've done that I've pretty much determined the function spaces in which f and the f_j will lie, and in general (2) will no longer strictly contain (1). mgcv's `s' terms use a thin plate spline measure of smoothness for multivariate smooths, and this means that (1) will not be strictly nested within (2), since e.g. a 4D thin plate spline can not generally represent exactly what the sum of 4 1D splines can represent. If you want to acheive exact nesting then using tensor product smooths with something like y~te(x1)+te(x2)+te(x3)+te(x4) (3) y~te(x1,x2,x3,x4) (4) will do the trick (because the function space for (4) is built up from the function spaces used in (3)). As to where all the 2 and 3 way interactions have gone in (4)... it's just like ANOVA - if you put in a 4 way interaction then the lower order interactions are not identifiable, unless you choose to add constraints to make them so. `mgcv' will allow you add main effects and interactions, and will handle the constraints automatically, but if this sort of functional ANOVA is a major component of what you want to do, then it is probably worth checking out the gss package and Chong Gu's book on smoothing spline ANOVA. best, Simon -- Corrado Topi Global Climate Change Biodiversity Indicators Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org 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] Select top three values from data frame
Noah Silverman ha scritto: I only have a few values in my example, but the real data set might have 20-100 rows with A=X. So how do I pick just the three highest ones? -N On 8/26/09 2:46 AM, Ottorino-Luca Pantani wrote: df.mydata[df.mydata$A==X AND df.mydata$C 2, ] will do the job ? 8rino Noah Silverman ha scritto: Hi, I'm trying to find an easy way to do this. I want to select the top three values of a specific column in a subset of rows in a data.frame. I'll demonstrate. ABC x21 x41 x32 y15 y26 y38 I want the top 3 values of B from the data.frame where A=X and C 2 I could extract all the rows where C2, then sort by B, then take the first 3. But that seems like the wrong way around, and it also will get messy with real data of over 100 columns. Any suggestions? my.data - cbind.data.frame(expand.grid(A = c(X, Y), B = 1:100), C = rnorm(100)) myA.data - my.data[my.data$A == X, ] myA.sorted.data - myA.data[order(myA.data$C, decreasing=TRUE), ][1:3.] Do this solve your problem ? -- Ottorino-Luca Pantani, Università di Firenze Dip. Scienza del Suolo e Nutrizione della Pianta P.zle Cascine 28 50144 Firenze Italia Tel 39 055 3288 202 (348 lab) Fax 39 055 333 273 olpant...@unifi.it http://www4.unifi.it/dssnp/ __ R-help@r-project.org 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] simple graph question: manipulating variable names
Donald Braman wrote:
This is a simple problem that has stumped me: I'm trying to loop through a
few dozen variable names in graphs. I've tried various approaches like
this:
attach(mydata)
ivs - c(oneiv, anotheriv, yetanotheriv)
dvs - c(onedv, anotherdv, yetanotherdv)
for (iv in ivs) {
for (dv in dvs) {
graphname - paste(iv, dv, .png, sep = )
png(file=graphname, width=300, height=300)
plot(dv ~ iv, pch=.)
lines(loess.smooth(iv, dv), lty=1)
dev.off()
}
}
Clearly that doesn't work. I'm not sure how to make R see the iv and dv
strings as variables. Advice?
Hi Donald,
I think the problem is that you are trying to plot the strings that you
are using for your filename rather than the elements of mydata. Try this:
for(ivindex in 1:3) {
for(dvindex in 1:3) {
graphname-paste(iv[ivindex],dv[dvindex],.png,sep=)
png(graphname,width=300,height=300)
plot(mydata[,3*iv-dv-1],mydata[,3*iv-dv],pch=.)
lines(loess.smooth(mydata[,3*iv-dv-1],mydata[,3*iv-dv],lty=1)
dev.off()
}
}
remembering that I have made up the indexing of mydata out of thin
air. You will have to work out how to index the columns or rows of
mydata to get the right iv and dv for each pass of the loops.
Jim
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[R] how to set crontab for updating the repositories?
Hi, I have downloaded around 60GB package repositories of bioconductor to use it locally and to set up mirror at my university site. I have installed the mirror with rsync command and able to access also. Now I have to set a cron job for its daily updating from bioconductor website. How should I do it? I know rsync have to be used but I don't know the proper syntax. I request to send proper syntax. Thanks, Sukhbir Singh Rattan. [[alternative HTML version deleted]] __ R-help@r-project.org 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] Select top three values from data frame
Or perhaps use a temporary vector might be neater? tmp - with(df.mydata, B[A==X C 2]) df.mydata[order(tmp) %in% 1:3,] # gives df with highest three values of B or head(df.mydata[order(tmp),],3) # gives first 3 rows of df sorted by B Colin. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Mohamed Lajnef Sent: 26 August 2009 11:25 To: Noah Silverman Cc: r help Subject: Re: [R] Select top three values from data frame Noah Silverman a écrit : I only have a few values in my example, but the real data set might have 20-100 rows with A=X. So how do I pick just the three highest ones? -N Hi, and now? df.mydata$B[order(df.mydata[df.mydata$A==X AND df.mydata$C 2, ]$B)][length(df.mydata$B)-3:length(df.mydata$B)] cheers, ML On 8/26/09 2:46 AM, Ottorino-Luca Pantani wrote: df.mydata[df.mydata$A==X AND df.mydata$C 2, ] will do the job ? 8rino Noah Silverman ha scritto: Hi, I'm trying to find an easy way to do this. I want to select the top three values of a specific column in a subset of rows in a data.frame. I'll demonstrate. ABC x21 x41 x32 y15 y26 y38 I want the top 3 values of B from the data.frame where A=X and C 2 I could extract all the rows where C2, then sort by B, then take the first 3. But that seems like the wrong way around, and it also will get messy with real data of over 100 columns. Any suggestions? [[alternative HTML version deleted]] __ R-help@r-project.org 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. -- Mohamed Lajnef INSERM Unité 955. 40 rue de Mesly. 94000 Créteil. Courriel : mohamed.laj...@inserm.fr tel. : 01 49 81 31 31 (poste 18470) Sec : 01 49 81 32 90 fax : 01 49 81 30 99 __ R-help@r-project.org 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. __ This email has been scanned by the MessageLabs Email Security System. For more information please visit http://www.messagelabs.com/email __ R-help@r-project.org 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] Re : error matrix, cross table,
Hi see ?table i hope that helps!! inchallah De : dalposso gustavodalpo...@hotmail.com À : r-help@r-project.org Envoyé le : Mercredi, 26 Août 2009, 1h20mn 31s Objet : [R] error matrix, cross table, how to create a cross table to quantify the classes of two thematic maps? -- View this message in context: http://www.nabble.com/error-matrix%2C-cross-table%2C-tp25143926p25143926.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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@r-project.org 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] changing the Date format.
Hi everyone, i have a data frame called 'sample'. in that 'sample' data frame variable called 'starts' is there like this starts 37987 37988 37989 37990 37991 now i want change that variable into '-mm-dd' format. can any body help in this aspect. Thanks in Advance. -- View this message in context: http://www.nabble.com/changing-the-Date-format.-tp25150494p25150494.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] changing the Date format.
On Wed, 26-Aug-2009 at 03:48AM -0700, rajclinasia wrote:
|
| Hi everyone,
|
| i have a data frame called 'sample'. in that 'sample' data frame variable
| called 'starts' is there like this
|
| starts
| 37987
| 37988
| 37989
| 37990
| 37991
|
| now i want change that variable into '-mm-dd' format. can any body help
| in this aspect.
Check out the as.Date function.
?as.Date
HTH
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
(:_~*~_:) Small minds discuss people
(_)-(_) . Eleanor Roosevelt
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
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Re: [R] Select top three values from data frame
Hi, This should work - head is quite a usefull summary function head(df.mydata[df.mydata$A==X df.mydata$C 2, ],3) Colin. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Noah Silverman Sent: 26 August 2009 10:54 To: ottorino-luca.pant...@unifi.it Cc: r help Subject: Re: [R] Select top three values from data frame I only have a few values in my example, but the real data set might have 20-100 rows with A=X. So how do I pick just the three highest ones? -N On 8/26/09 2:46 AM, Ottorino-Luca Pantani wrote: df.mydata[df.mydata$A==X AND df.mydata$C 2, ] will do the job ? 8rino Noah Silverman ha scritto: Hi, I'm trying to find an easy way to do this. I want to select the top three values of a specific column in a subset of rows in a data.frame. I'll demonstrate. ABC x21 x41 x32 y15 y26 y38 I want the top 3 values of B from the data.frame where A=X and C 2 I could extract all the rows where C2, then sort by B, then take the first 3. But that seems like the wrong way around, and it also will get messy with real data of over 100 columns. Any suggestions? [[alternative HTML version deleted]] __ R-help@r-project.org 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. __ This email has been scanned by the MessageLabs Email Security System. For more information please visit http://www.messagelabs.com/email __ R-help@r-project.org 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] GLMs
Hi, I am starting to work with R. I need to performe a General linear model and a Generalized mixed model, what are the package I have to use for? what is the difference between them? thanks letizia _ [[alternative HTML version deleted]] __ R-help@r-project.org 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] Filtering matrices
Tena koe Ben -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of bwgoudey Sent: Wednesday, 26 August 2009 2:55 p.m. To: r-help@r-project.org Subject: Re: [R] Filtering matrices r-rcorr(d[[1]]) #d is matrix containing observation r[[1]] #r values age sex BMI age 1.000 -0.30010322 -0.13702263 sex -0.3001032 1. 0.06300528 BMI -0.1370226 0.06300528 1. r[[2]] #Number of obervations age sex BMI age 100 100 100 sex 100 100 100 BMI 100 100 100 r[[3]] #P values age sex BMI age NA 0.002416954 0.1740134 sex 0.002416954 NA 0.5334484 BMI 0.174013354 0.533448366NA If I wanted to return a matrix containing all points where correlation was above 0.75 and P-value was below 0.05, how would I do this? Cheers Ben It is not clear to me what 'matrix' you wish to have returned since you'll be keeping an unknown number of values. However, if you are happy just to replace the unwanted values with a missing value then something like r[[1]][r[[1]]0.75 | (r[[3]]0.05 !is.na(r[[3]]))] - NA might work. I say 'might' because I am unfamiliar with rcorr() and thus am not sure whether the components of r are matrices or dataframes, and the above is untested. You might be interested in abs(r[[1]]). HTH Peter Alspach __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help in plotting a legend
Ashutosh Nandeshwar wrote: Hello, List, I am a new user of the R project, and I need some help in plotting a legend. I am using the PBSmapping library to plot map of Ohio and heat color it with the count of employees in each county. As a guide, I am using Data Mashups in R. I am able to plot the map with the colors; however, I would like to put a legend with a single box full of these colors, but only show the max and the min value on the left corner and the right corner respectively. Here's a part of the code I am using: #empdata has the employee data with the latitude and longitude for each employee library(PBSmapping) addressEvents-as.EventData(empdataRC,projection=NA) #myShapeFile is a shape file of Ohio imported using importShapefile addressPolys-findPolys(addressEvents,myShapeFile) myTrtFC- table(factor(addressPolys$PID,levels=levels(as.factor(myShapeFile$PID log(myTrtFC)-lTrt mapColors-heat.colors(max(lTrt)+1,alpha=.6)[max(lTrt)-lTrt+1] mapColors[ is.na(mapColors) ] - white pdf(RC-Employees.pdf,version=1.4) plotPolys(myShapeFile,axes=FALSE,bg=white,main= Employees ,xlab=,ylab=,col=mapColors) #here is the current legend, but as you can see, the colors and the box count doesn't match #i would like to see only one box with the gradients of the colors that I used, and the max and the min value on top. I even tried a rectangle but I could not plot it Hi Ashutosh, Have a look at color.legend in the plotrix package. Jim __ R-help@r-project.org 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] faulty formatting of toLatex(sessionInfo())
Dear all I am writing an Sweave document and have encountered formatting issues with the locale part of toLatex(sessionInfo()). The fact that there is no spaces between the various locale variables means that LaTeX cannot easily find an appropriate place to break the lines, and some will get printed off screen. Below is the text output, and this .pdf document [1] shows the (faulty) tex result. Could anyone suggest how to get around this issue? Thank you Liviu [1] http://s000.tinyupload.com/index.php?file_id=06259544336960829228 sessionInfo() R version 2.9.1 (2009-06-26) x86_64-pc-linux-gnu locale: LC_CTYPE=en_GB.UTF-8;LC_NUMERIC=C;LC_TIME=en_GB.UTF-8;LC_COLLATE=en_GB.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_GB.UTF-8;LC_PAPER=en_GB.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_GB.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] grDevices utils datasets stats graphics methods base other attached packages: [1] fortunes_1.3-6 hints_1.0.1-1 boot_1.2-38relimp_1.0-1 xtable_1.5-5 [6] Hmisc_3.6-1 loaded via a namespace (and not attached): [1] cluster_1.12.0 grid_2.9.1 lattice_0.17-25 tcltk_2.9.1 -- Do you know how to read? http://www.alienetworks.com/srtest.cfm Do you know how to write? http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail __ R-help@r-project.org 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] Number of CPU's
For windows you can do: Sys.getenv(NUMBER_OF_PROCESSORS) On Mon, Aug 24, 2009 at 3:16 PM, Håvard Rue havard@math.ntnu.no wrote: Any way to get access to the number of CPU's, optionally their type, from within R? In linux I can just read /proc/cpuinfo but for win/mac ? Thanks! Håvard -- Håvard Rue Department of Mathematical Sciences Norwegian University of Science and Technology N-7491 Trondheim, Norway Voice: +47-7359-3533URL : http://www.math.ntnu.no/~hruehttp://www.math.ntnu.no/%7Ehrue Fax : +47-7359-3524Email: havard@math.ntnu.no This message was created in a Microsoft-free computing environment. __ R-help@r-project.org 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org 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] faulty formatting of toLatex(sessionInfo())
Dear Liviu, On Wed, Aug 26, 2009 at 7:35 AM, Liviu Androniclandronim...@gmail.com wrote: Dear all I am writing an Sweave document and have encountered formatting issues with the locale part of toLatex(sessionInfo()). The fact that there is no spaces between the various locale variables means that LaTeX cannot easily find an appropriate place to break the lines, and some will get printed off screen. Below is the text output, and this .pdf document [1] shows the (faulty) tex result. Could anyone suggest how to get around this issue? Thank you Liviu There was a closely related discussion last April: https://stat.ethz.ch/pipermail/r-devel/2009-April/053094.html and IIRC this was fixed for R version 2.10. Hope this helps, Jay -- *** G. Jay Kerns, Ph.D. Associate Professor Department of Mathematics Statistics Youngstown State University Youngstown, OH 44555-0002 USA Office: 1035 Cushwa Hall Phone: (330) 941-3310 Office (voice mail) -3302 Department -3170 FAX VoIP: gjke...@ekiga.net E-mail: gke...@ysu.edu http://www.cc.ysu.edu/~gjkerns/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to set crontab for updating the repositories?
Sukhbir Rattan wrote: Hi, I have downloaded around 60GB package repositories of bioconductor to use it locally and to set up mirror at my university site. I have installed the mirror with rsync command and able to access also. Now I have to set a cron job for its daily updating from bioconductor website. How should I do it? I know rsync have to be used but I don't know the proper syntax. I request to send proper syntax. Thanks, Sukhbir Singh Rattan. [[alternative HTML version deleted]] __ R-help@r-project.org 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. Hi, Typing: crontab -e Will open the crontab file for the current user, here you can add commands that are to executed at certain times. The crontab will look something like: # m h dom mon dow command 0 0 * * * /foo/bar where the command /foo/bar will be executed every day (dom, mon and dow are a *, meaning 'for all') at 12 pm. cheers, Paul -- Drs. Paul Hiemstra Department of Physical Geography Faculty of Geosciences University of Utrecht Heidelberglaan 2 P.O. Box 80.115 3508 TC Utrecht Phone: +3130 274 3113 Mon-Tue Phone: +3130 253 5773 Wed-Fri http://intamap.geo.uu.nl/~paul __ R-help@r-project.org 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] attach package
Hello, Is there a solution to attach a package without run the hook function .onAttach() Thanks Jérémy Mazet [[alternative HTML version deleted]] __ R-help@r-project.org 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] Within factor random factor
Hi, I am quite new to R and trying to analyze the following data. I have 28 controls and 25 patients. I measured X values of 4 different locations (A,B,C,D) in the brain image of each subject. And X ranges from 0 to 1. I think control or patient is a between subject factor and location is a within subject factor. So, controls: 28 patients: 25 (unbalanced data set) respone measure: X values (ranging 0 to 1) fixed factor: control vs. patient (between subject factor) random factor: location (level: A,B,C,D ;no order) (within subject factor) random factor: subjectID 1-53 My data looks like this; CorPX locationsubjectID control 0.708 A 1 control 0.648 A 2 patient 0.638 C 3 control 0.547 D 4 patient 0.632 B 5 control 0.723 C 6 ... I want to know (a) if there is a significant difference between controls and patients in X values. (b) where (A,B,C,D?) the difference is between controls and patients in X values. (There may be an interaction) I constructed linear mixed model with lme as followings; (1) model1 - lme(X ~ CorP*location, random= ~ 1| subjectID, mydata) (2) model2 - lme(X ~ CorP*location, random= ~ location| subjectID, mydata) I am not familiar with lme syntax. I'm just wondering which formula [(1) or (2)] is appropriate for my model to know answers of (a) and (b) questions. Or may be both of the formulas are wrong. I would appreciate it very much if somebody could help me. Sincerely, Kohkichi Hosoda __ R-help@r-project.org 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] counting subgroup sums within a data frame
Hi, I'm sure there's an easy approach to this issue, I'm just not seeing it. I have a data frame of the following form: Date classsubclass count 8/1/2009AX 1 8/1/2009BX 2 8/1/2009AY 9 8/1/2009BY 3 8/2/2009AX 1 8/2/2009BX 5 8/2/2009AY 4 8/2/2009BY 2 8/3/2009AX 6 8/3/2009BX 4 8/3/2009AY 3 8/3/2009BY 4 8/4/2009AX 1 8/4/2009BX 9 8/4/2009AY 3 8/4/2009BY 5 8/5/2009AX 3 8/5/2009BX 7 8/5/2009AY 2 8/5/2009BY 1 I would like to create a data frame of the sum the daily counts for, say, class 'A', like so: Date sum_of_counts 8/1/2009 10 8/2/20095 8/3/20099 8/4/20094 8/5/20095 I ultimately would like to do sum of counts on all classes and subclasses. It seems that this is equivalent to a GROUP BY query in SQL. I'm sure this is possible in R. Any suggestions? [[alternative HTML version deleted]] __ R-help@r-project.org 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] teaching R
Hello all, I am going to be running a small statistics workshop using R sometime in November. I am restricted to R because of the specific libraries I will be using - a good thing in my book - however the attendees are unfamiliar with R. I plan on giving as little R information as possible - just what is absolute necessary to run the statistics (the workshop is short, no time to spend hours teaching R). Has anyone done anything like this? Are there any public powerpoint/pdfs/etc. available for this type of application? Any advice / what works / what doesn't work is appreciated for those that have tried this before me. Sincerely, -Michael -- Michael A. Nestrud Cornell U. Sensory Science PhD Student m...@ataraxis.org All that you taste... all that you eat. __ R-help@r-project.org 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] Appending strings at the beginning of a text file
It works with paste, but I forgotten the close(fConn) after the writeLines.
On Tue, Aug 25, 2009 at 2:48 PM, Paul Smith phh...@gmail.com wrote:
Thanks, Henrique.
I think you mean the following:
fConn - file('test.txt', 'r+')
Lines - readLines(fConn)
writeLines(c(Text at beginning of file\n, Lines), con = fConn)
close(fConn)
(With paste(), it does not work.)
Paul
On Tue, Aug 25, 2009 at 5:02 PM, Henrique Dallazuannawww...@gmail.com
wrote:
Try this;
fConn - file('test.txt', 'r+')
Lines - readLines(fConn)
writeLines(paste(Text at beginning of file, Lines, sep = \n),
con = fConn)
On Tue, Aug 25, 2009 at 12:54 PM, Paul Smith phh...@gmail.com wrote:
Dear All,
I have a piece of text that I want to append to a text file at the
beginning of the text file.
I have thought about using cat() with the option 'append=T', but the
appending, in this case, is done at the bottom of the text file. Any
ideas?
Thanks in advance,
Paul
__
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--
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25° 25' 40 S 49° 16' 22 O
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25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
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Re: [R] counting subgroup sums within a data frame
Try this: with(d, tapply(count, list(Date, class), sum)) On Wed, Aug 26, 2009 at 10:07 AM, Shaun Grannis sgran...@regenstrief.orgwrote: Hi, I'm sure there's an easy approach to this issue, I'm just not seeing it. I have a data frame of the following form: Date classsubclass count 8/1/2009AX 1 8/1/2009BX 2 8/1/2009AY 9 8/1/2009BY 3 8/2/2009AX 1 8/2/2009BX 5 8/2/2009AY 4 8/2/2009BY 2 8/3/2009AX 6 8/3/2009BX 4 8/3/2009AY 3 8/3/2009BY 4 8/4/2009AX 1 8/4/2009BX 9 8/4/2009AY 3 8/4/2009BY 5 8/5/2009AX 3 8/5/2009BX 7 8/5/2009AY 2 8/5/2009BY 1 I would like to create a data frame of the sum the daily counts for, say, class 'A', like so: Date sum_of_counts 8/1/2009 10 8/2/20095 8/3/20099 8/4/20094 8/5/20095 I ultimately would like to do sum of counts on all classes and subclasses. It seems that this is equivalent to a GROUP BY query in SQL. I'm sure this is possible in R. Any suggestions? [[alternative HTML version deleted]] __ R-help@r-project.org 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org 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] teaching R
On Wed, Aug 26, 2009 at 2:08 PM, Michael Nestrudm...@ataraxis.org wrote: Hello all, I am going to be running a small statistics workshop using R sometime in November. I am restricted to R because of the specific libraries I will be using - a good thing in my book - however the attendees are unfamiliar with R. I plan on giving as little R information as possible - just what is absolute necessary to run the statistics (the workshop is short, no time to spend hours teaching R). Has anyone done anything like this? Are there any public powerpoint/pdfs/etc. available for this type of application? Any advice / what works / what doesn't work is appreciated for those that have tried this before me. You could try my 'R in one page' document (which will be two pages if you use a non-double-sided printer). Side 1 is a very simple introduction, and side 2 has some more examples for people to cut and paste. OpenOffice source and PDF here: http://www.maths.lancs.ac.uk/~rowlings/R/Simple/ Feel free to take these and edit to your needs. There's more user-contributed docs here: http://cran.r-project.org/other-docs.html although a 360 page PDF called An Introduction to R is probably a bit verbose for your needs. Barry __ R-help@r-project.org 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] counting subgroup sums within a data frame
Wow. That was fast -- and spot on! Thanks so much. Best Regards, Shaun On Aug 26, 2009, at 9:11 AM, Henrique Dallazuanna wrote: Try this: with(d, tapply(count, list(Date, class), sum)) On Wed, Aug 26, 2009 at 10:07 AM, Shaun Grannis sgran...@regenstrief.org wrote: Hi, I'm sure there's an easy approach to this issue, I'm just not seeing it. I have a data frame of the following form: Date classsubclass count 8/1/2009AX 1 8/1/2009BX 2 8/1/2009AY 9 8/1/2009BY 3 8/2/2009AX 1 8/2/2009BX 5 8/2/2009AY 4 8/2/2009BY 2 8/3/2009AX 6 8/3/2009BX 4 8/3/2009AY 3 8/3/2009BY 4 8/4/2009AX 1 8/4/2009BX 9 8/4/2009AY 3 8/4/2009BY 5 8/5/2009AX 3 8/5/2009BX 7 8/5/2009AY 2 8/5/2009BY 1 I would like to create a data frame of the sum the daily counts for, say, class 'A', like so: Date sum_of_counts 8/1/2009 10 8/2/20095 8/3/20099 8/4/20094 8/5/20095 I ultimately would like to do sum of counts on all classes and subclasses. It seems that this is equivalent to a GROUP BY query in SQL. I'm sure this is possible in R. Any suggestions? [[alternative HTML version deleted]] __ R-help@r-project.org 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org 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] Appending strings at the beginning of a text file
Henrique,
With paste(), it works only if the file has only one line; otherwise,
Text at beginning of file
is repeated after each line of the file.
Paul
On Wed, Aug 26, 2009 at 2:03 PM, Henrique Dallazuannawww...@gmail.com wrote:
It works with paste, but I forgotten the close(fConn) after the writeLines.
On Tue, Aug 25, 2009 at 2:48 PM, Paul Smith phh...@gmail.com wrote:
Thanks, Henrique.
I think you mean the following:
fConn - file('test.txt', 'r+')
Lines - readLines(fConn)
writeLines(c(Text at beginning of file\n, Lines), con = fConn)
close(fConn)
(With paste(), it does not work.)
Paul
On Tue, Aug 25, 2009 at 5:02 PM, Henrique Dallazuannawww...@gmail.com
wrote:
Try this;
fConn - file('test.txt', 'r+')
Lines - readLines(fConn)
writeLines(paste(Text at beginning of file, Lines, sep = \n),
con = fConn)
On Tue, Aug 25, 2009 at 12:54 PM, Paul Smith phh...@gmail.com wrote:
Dear All,
I have a piece of text that I want to append to a text file at the
beginning of the text file.
I have thought about using cat() with the option 'append=T', but the
appending, in this case, is done at the bottom of the text file. Any
ideas?
Thanks in advance,
Paul
__
R-help@r-project.org 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.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
__
R-help@r-project.org 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.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] counting subgroup sums within a data frame
Have a look at the reshape package.
Assuming that your data is in a data.frame called dataset.
cast(Date ~ ., data = dataset, value = count, fun = sum)
cast(Date ~ class, data = dataset, value = count, fun = sum)
cast(Date + class ~ ., data = dataset, value = count, fun = sum)
Or the plyr package
ddply(dataset, c(Date), function(x){c(Sum_of_counts = sum(x$count))})
ddply(dataset, c(Date, class), function(x){c(Sum_of_counts =
sum(x$count))})
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
thierry.onkel...@inbo.be
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Namens Shaun Grannis
Verzonden: woensdag 26 augustus 2009 15:07
Aan: r-help@r-project.org
Onderwerp: [R] counting subgroup sums within a data frame
Hi,
I'm sure there's an easy approach to this issue, I'm just not seeing it.
I have a data frame of the following form:
Date classsubclass count
8/1/2009AX 1
8/1/2009BX 2
8/1/2009AY 9
8/1/2009BY 3
8/2/2009AX 1
8/2/2009BX 5
8/2/2009AY 4
8/2/2009BY 2
8/3/2009AX 6
8/3/2009BX 4
8/3/2009AY 3
8/3/2009BY 4
8/4/2009AX 1
8/4/2009BX 9
8/4/2009AY 3
8/4/2009BY 5
8/5/2009AX 3
8/5/2009BX 7
8/5/2009AY 2
8/5/2009BY 1
I would like to create a data frame of the sum the daily counts for,
say, class 'A', like so:
Date sum_of_counts
8/1/2009 10
8/2/20095
8/3/20099
8/4/20094
8/5/20095
I ultimately would like to do sum of counts on all classes and
subclasses. It seems that this is equivalent to a GROUP BY query in
SQL.
I'm sure this is possible in R. Any suggestions?
[[alternative HTML version deleted]]
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Druk dit bericht a.u.b. niet onnodig af.
Please do not print this message unnecessarily.
Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer
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Re: [R] Filtering matrices
Hi, On Aug 25, 2009, at 10:54 PM, bwgoudey wrote: r-rcorr(d[[1]]) #d is matrix containing observation r[[1]] #r values age sex BMI age 1.000 -0.30010322 -0.13702263 sex -0.3001032 1. 0.06300528 BMI -0.1370226 0.06300528 1. r[[2]] #Number of obervations age sex BMI age 100 100 100 sex 100 100 100 BMI 100 100 100 r[[3]] #P values age sex BMI age NA 0.002416954 0.1740134 sex 0.002416954 NA 0.5334484 BMI 0.174013354 0.533448366NA Just a quick note: please provide data in an easy way for us to enter into our R session -- the way you provide the data requires more work on someone who is trying to help you in order to enter it in R, for instance this might have been better: rvals - matrix(c( 1.000, -0.30010322, -0.13702263, -0.3001032, 1., 0.06300528, -0.1370226, 0.06300528, 1.), byrow=TRUE, nrow=3) obs - matrix(c( 100, 100, 100, 100, 100, 100, 100, 100, 100), byrow=TRUE, nrow=3) pval - matrix(c( NA, 0.002416954, 0.1740134, 0.002416954, NA, 0.5334484, 0.174013354, 0.533448366,NA), byrow=TRUE, nrow=3) Since I can just copy and paste that into my R session and get right to answering your question. If I wanted to return a matrix containing all points where correlation was above 0.75 and P-value was below 0.05, how would I do this? Where is your points matrix that you want to return the data from? Assuming this is also a 3x3 matrix, you just build the indexing vectors using the data matrices of interest, and use those to pull the points out of your data/points matrix. Let's assume my data/points matrix is called my.data: #1 Get indices of points w/ correlation above 0.75 good.cor - rvals .75 #2 Get indices of points w/ p-value 0.05, since you have NA values # in your pval matrix, you have to *explicitly exclude* them from the query good.p - !is.na(pval) pval 0.05 # since you want their intersection, take an of the two vectors, # but this is empty in this case. Either way, this would get # the points you're after my.data[good.cor good.p] Does that make sense? -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org 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] [Rd] Formulas in gam function of mgcv package
Dear Simon, thanks again. Concerning the whole 36 variables well, I have run a principal components analysis, and I am only using part of them (I am running a test with the pc which cover the 95% of variance and then the 99%). :) so I will possibly end up with s(x1,,x8). I wonder if using isotropic smoothers on principal component is a good idea the variance diminishes from component to component, so theoretically also the wiggliness of the smoother should be less and less what do you think? am I saying something stupid? If that is the case, and if I want to enclose some interaction, then I have so include the interaction terms manually like s(x1,x2). Is that right? Sorry for the avalanche of questions, but I am trying to understand the principles underlying the working of gam in mgcv. It looks very powerful, particularly for exploring dependencies. I have run te() instead of s(), but the predictive power seems to be less than with s() in this particular situation. At the same time, does te() include the interaction? I did not understand well your previous point on interaction term in te(): is te(x1,,xn) build as an expansion from the t(x1), ,t(xn)? Then all the interaction terms should be included Finally, is it possible to incorporate both s() and te() terms in the formula? Machine learning: I am not too well versed in the area. Did you mean regression trees or maximum entropy models? Best, On Wednesday 26 August 2009 10:27:08 Simon Wood wrote: This will not work... 2) y~s(x1, ,x36) Estimating a 36 dimensional functions reasonably well would require a tremendous quantity of data, but in any case the 36 dimensional TPS smoothnes measure will involve such high order derivatives that it will no longer be practically useful: in fact you will not have enough data to estimate the unpenalized coefficients of the smoother (and if you did R would run out of memory first). In such a high dimensional situation, I think that GAMs are really only useful if you have some prior knowledge of which variables are likely to interact (and it's not too many of them). If there's no prior information saying roughly what sort of smooth additive structure might be useful then, I'm not sure that GAMs are the right way to go, and some sort of machine learning approach might be better. Then again, the real problem with y~s(x1, ,x36) is that the data just won't contain enough information to estimate s, if all you can say is that s is smooth, but this also means that it's very unlikely that you really need to estimate s(x1, ,x36) in order to predict well. In that case, starting from y ~ s(x1) + + s(x36) and building the model up might result in something that does a reasonable predictive job. On the subject of tensor product smoothing vs isotropic smoothing. Isotropic smooths are really only reasonable if you think that the smooth should display approximately the same amount of wiggliness in all directions. If this is not the case then tensor product smoothing is a better bet. Centering and scaling alone is not enough to ensure that isotropy is reasonable (although in particular cases it may help, of course). best, Simon I am trying to build a predictive model. Since the the variables are centred and scaled, I think I need an isotropic smooth. I am also interested in having the interactions between the variables included, that is not a purely additive model. It is not clear to me when should I give preference to tensor smooths, possibly because I have not understood well how they work. I am reading Wood(2003) as recommended and I have also read rather extensively Simon N. Wood. Generalized Additive Models: An Introduction, 2006, but still I am stuck. Any additional suggestion or reading recommendation would be greatly appreciated. I have also some difficulties in understanding the values you have chosen for k in the first example (why 60?). Thanks Best, On Monday 24 August 2009 17:33:55 Gavin Simpson wrote: [Note R-Devel is the wrong list for such questions. R-Help is where this should have been directed - redirected there now] On Mon, 2009-08-24 at 17:02 +0100, Corrado wrote: Dear R-experts, I have a question on the formulas used in the gam function of the mgcv package. I am trying to understand the relationships between: y~s(x1)+s(x2)+s(x3)+s(x4) and y~s(x1,x2,x3,x4) Does the latter contain the former? what about the smoothers of all interaction terms? I'm not 100% certain how this scales to smooths of more than 2 variables, but Sections 4.10.2 and 5.2.2 of Simon Wood's book GAM: An Introduction with R (2006, Chapman Hall/CRC) discuss this for smooths of 2 variables. Strictly y ~ s(x1) + s(x2) is not nested in y ~ s(x1, x2) as the bases used to produce the smoothers in
Re: [R] teaching R
Hello On 8/26/09, Michael Nestrud m...@ataraxis.org wrote: Any advice / what works / what doesn't work is appreciated for those that have tried this before me. It could prove helpful to forward similar questions to r-sig-teaching. Also, there was a recent discussion on the topic [1]. Liviu [1] https://stat.ethz.ch/pipermail/r-sig-teaching/2009q2/000157.html __ R-help@r-project.org 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] [Rd] Formulas in gam function of mgcv package
On Tue, 2009-08-25 at 10:00 +0100, Corrado wrote: Dear Gavin / Rlings, thanks for your kind answer and sorry for posting to the dev mailing list. Concerning the specific of your answer: I am working with 6 to 36 covariates, and they are all centred and scaled. I represented the problem with two variables to simplify the question. So ideally, the situation is: 1) y ~ s(x1) + + s(x36) vs. 2) y~s(x1, ,x36) I think you are pushing things a bit with such a complicated smooth. You're unlikely to be able to fit that either due to insufficient data and / or hardware limits on your machine. I see that Simon has responded to this as well, in a far more comprehensive and informed manner than I could manage So I'll leave it at that... I am trying to build a predictive model. Since the the variables are centred and scaled, I think I need an isotropic smooth. I am also interested in having the interactions between the variables included, that is not a purely additive model. That sounds a bit like data fishing; throw everything into the pot and see what comes out of it. snip / I have also some difficulties in understanding the values you have chosen for k in the first example (why 60?). Sorry, that was a complication on my part. The main point was to show that you need to try to get the same bases used in the s(x1) and s(x2) parts of the formula; So if you had this model y ~ s(x1, k = 20) + s(x2, k = 20) You need something like y ~ s(x1, k = 20) + s(x2, k = 20) + s(x2, x2) [If you wanted the bivariate smooth to be more complicated than the default in mgcv, then you might have done: y ~ s(x2, x2, k = 60) ## for example in which case you could fit that model as y ~ s(x1, k = 20) + s(x2, k = 20) + s(x2, x2, k = 60) ] That was where the k = 60 came from, but in simplifying my response I forgot to remove it. Simon has since provided a more thorough response (Thanks Simon). HTH G Thanks Best, On Monday 24 August 2009 17:33:55 Gavin Simpson wrote: [Note R-Devel is the wrong list for such questions. R-Help is where this should have been directed - redirected there now] On Mon, 2009-08-24 at 17:02 +0100, Corrado wrote: Dear R-experts, I have a question on the formulas used in the gam function of the mgcv package. I am trying to understand the relationships between: y~s(x1)+s(x2)+s(x3)+s(x4) and y~s(x1,x2,x3,x4) Does the latter contain the former? what about the smoothers of all interaction terms? I'm not 100% certain how this scales to smooths of more than 2 variables, but Sections 4.10.2 and 5.2.2 of Simon Wood's book GAM: An Introduction with R (2006, Chapman Hall/CRC) discuss this for smooths of 2 variables. Strictly y ~ s(x1) + s(x2) is not nested in y ~ s(x1, x2) as the bases used to produce the smoothers in the two models may not be the same in both models. One option to ensure nestedness is to fit the more complicated model as something like this: ## if simpler model were: y ~ s(x1, k=20) + s(x2, k = 20) y ~ s(x1, k=20) + s(x2, k = 20) + s(x1, x2, k = 60) ^ where the last term (^^^ above) has the same k as used in s(x1, x2) Note that these are isotropic smooths; are x1 and x2 measured in the same units etc.? Tensor product smooths may be more appropriate if not, and if we specify the bases when fitting models s(x1) + s(x2) *is* strictly nested in te(x1, x2), eg. y ~ s(x1, bs = cr, k = 10) + s(x2, bs = cr, k = 10) is strictly nested within y ~ te(x1, x2, k = 10) ## is the same as y ~ te(x1, x2, bs = cr, k = 10) [Note that bs = cr is the default basis in te() smooths, hence we don't need to specify it, and k = 10 refers to each individual smooth in the te().] HTH G I have (tried to) read the manual pages of gam, formula.gam, smooth.terms, linear.functional.terms but could not understand properly. Regards -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org 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] Statistical question about logistic regression simulation
Hi R help list
I'm simulating logistic regression data with a specified odds ratio
(beta) and have a problem/unexpected behaviour that occurs.
The datasets includes a lognormal exposure and diseased and healthy
subjects.
Here is my loop:
ors - vector()
for(i in 1:200){
# First, I create a vector with a lognormally distributed exposure:
n - 1 # number of study subjects
mean - 6
sd - 1
expo - rlnorm(n, mean, sd)
# Then I assign each study subject a probability of disease with a
# specified Odds ratio (or beta coefficient) according to a logistic
# model:
inter - 0.01 # intercept
or - log(1.5) # an odds ratio of 1.5 or a beta of ln(1.5)
p - exp(inter + or * expo)/(1 + exp(inter + or * expo))
# Then I use the probability to decide who is having the disease and who
# is not:
disease - rbinom(length(p), 1, p) # 1 = disease, 0 = healthy
# Then I calculate the logistic regression and extract the odds ratio
model - glm(disease ~ expo, family = binomial)
ors[i] - exp(summary(model)$coef[2]) # exponentiated beta = OR
}
Now to my questions:
1. I was expecting the mean of the odds ratios over all simulations to
be close to the specified one (1.5 in this case). This is not the case
if the mean of the lognormal distribution is, say 6.
If I reduce the mean of the exposure distribution to say 3, the mean of
the simulated ORs is very close to the specified one. So the simulation
seems to be quite sensitive to the parameters of the exposure distribution.
2. Is it somehow possible to stabilize the simulation so that it is
not that sensitive to the parameters of the lognormal exposure
distribution? I can't make up the parameters of the exposure
distribution, they are estimations from real data.
3. Are there general flaws or errors in my approach?
Thanks a lot for any help on this!
All the best,
Denis
--
Denis Aydin
Institute of Social and Preventive Medicine at Swiss Tropical Institute
Basel
Associated Institute of the University of Basel
Steinengraben 49 – 4051 Basel – Switzerland
Phone: +41 (0)61 270 22 04
Fax: +41 (0)61 270 22 25
denis.ay...@unibas.ch
www.ispm-unibasel.ch
__
R-help@r-project.org 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] lme: how to nest a random factor in a fixed factor?
Dear all, I have an experimental setup in which a random variable is nested within a fixed variable; however I have troubles specifying the correct LMM with lme. I have searched the lists but haven't been able to find an example like my setup, which I unfortunately need to get this stuff right. Pinheiro Bates is great but I still can't figure out how to do it. My experimental setup was as follows: 100 measurements per treatment plot 2 treatment plots per site 4 sites: 2 in one area and 2 in another area Both treatment and area are fixed factors,while site is random. I am interested in the significance of the fixed effects,less in the magnitude of the random effect. I have tried: mod1 - lme(response ~ area*treatment, data=data,random= ~1|site) but now site is not nested in area… mod2 - lme(response ~ area*treatment, data=data,random= ~1|area/site) but now area is both a fixed and a random variable, which doesn't seem to make sense, plus I run out of df for treatment mod3 - lme(response ~ area*treatment, data=data,random= ~1| plot) but here plots are not grouped according to site I hope someone would be willing to help me, thank you in advance! Robert Buitenwerf Ecologist South African Environmental Observation Network _ [[elided Hotmail spam]] __ R-help@r-project.org 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] teaching R
Hello all, I am going to be running a small statistics workshop using R sometime in November. I am restricted to R because of the specific libraries I will be using - a good thing in my book - however the attendees are unfamiliar with R. I plan on giving as little R information as possible - just what is absolute necessary to run the statistics (the workshop is short, no time to spend hours teaching R). Has anyone done anything like this? Are there any public powerpoint/pdfs/etc. available for this type of application? Any advice / what works / what doesn't work is appreciated for those that have tried this before me. R can be very intimidating to an audience that has very little experience with command line interfaces. Unlike a menu driven system there are no reminder clues, and it takes a little time to learn how to find help productively if you haven't lived in a command line world before. If your goal is to focus on the task at hand, my recommendation is to prepare a short 1-2 page handout that summarizes 1) how to get datasets from a standard format into a usable form for you application; 2) how to do each of the manipulations you will talk about; and 3) how to print/save the output (graphical and data) that you produce. An optional 4th section, if appropriate, might summarize how to learn more if they pursue the application on their own (i.e., use of appropriate help facilities). Your audience can then focus on the big picture of what you are talking about with a security blanket of being able to reproduce it later just as if they had a menu to help remind them of what you said. It may still be a little bit of an up hill battle with an inexperienced audience. Sincerely, -Michael -- Michael A. Nestrud Cornell U. Sensory Science PhD Student m...@ataraxis.org All that you taste... all that you eat. __ R-help@r-project.org 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@r-project.org 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] Grid lines in cloud plot for lattice
Hi all, I was wondering if there was a way to draw out grid lines in the cloud plots for lattice. I hunted all throughout the panel.cloud and panel. 3dscatter help pages and couldn't find anything to help me move forward with this. Do I have to get deeper into the drawing commands, or is there something more obvious I'm just not seeing? My goal would be to place grid lines on the bottom face of the box and make drop lines to the grid. #example dat set nirK.env.nmds$points = cbind(rnorm(30),rnorm(30),rnorm(30)) par.set - list(axis.line = list(col = transparent),clip = list(panel = off)) print(cloud(nirk.env.nmds$points[,3]~nirk.env.nmds$points[, 1]*nirk.env.nmds$points[,2], col=red,pch=16,cex=1.5,type=c(p,h), strip = strip.custom(strip.names = TRUE), scales=list(arrows=FALSE,distance=2), xlab=NMS 1, ylab=NMS 2, zlab=NMS 3), par.settings=par.set, split = c(1,1,2,1), more=TRUE) Many thanks!! Chris Jones Ph. D. Student, Department of Microbiology Swedish University of Agricultural Sciences chris.jo...@mikrob.slu.se [[alternative HTML version deleted]] __ R-help@r-project.org 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] Installing rJava RJDBC bad interpreter: Permission denied
Trying to install the above two packages via the install.packages(package_name) command and the R CMD INSTALL file.tar.gz. I receive the following error either way sh: ./configure: /bin/sh: bad interpreter: Permission denied. I have tried to chmod and chown permissions and also ran dos2unix in hopes that there is a CRLF in some of the tar.gz files, but that doesn't seem to fix the problem. I'm looking for any help I can get. Must I compile this to get this to work because of some incompatibility with the rpm packages?? * Installing *source* package ârJavaâ ... sh: ./configure: /bin/sh: bad interpreter: Permission denied ERROR: configuration failed for package ârJavaâ * Removing â/usr/lib64/R/library/rJavaâ * Installing *source* package âRJDBCâ ... ** R ** preparing package for lazy loading Error : package 'rJava' required by 'RJDBC' could not be found ERROR: lazy loading failed for package âRJDBCâ * Removing â/usr/lib64/R/library/RJDBCâ I'm using the rpm version of R found here http://cran.cnr.berkeley.edu/bin/linux/redhat/el5/x86_64 I ran the R CMD javareconf as root and set the JAVA_HOME environment variable. Here is my system configuration: OS: CentOS 5.2 x86_64 JDK: java version 1.6.0_12 Java(TM) SE Runtime Environment (build 1.6.0_12-b04) Java HotSpot(TM) 64-Bit Server VM (build 11.2-b01, mixed mode) R: 2.9.1 rJava: rJava_0.7-0.tar.gz RJDBC: RJDBC_0.1-5.tar.gz Thanks for your time and knowledge. Best, Matt __ R-help@r-project.org 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] mann whitney u
Dear Sir, I am comparing two samples using wilcox.test in R. Literature appears to describe mann whitney u test as the most appropriate test to use on my data. is the wilcox.test function equivalent to mann-whitney u? Is there a way to gain the U-value as apposed to the W-value in R? Thank you __ R-help@r-project.org 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] increasing significant digits in smooth.spline function
Hello All I have a very long vector of unique predictor values and 6 significant digits setting for the smooth.spline rounds them off. Is there any way of increasing the significant digits withour recompiling a lot if code (simple editing and tham sourcing of smooth.spline.r function does not work, probably due to presence of Fortan functional calls)? Thank you very much in advance Sergii R version 2.9.1 (2009-06-26) x86_64-redhat-linux-gnu locale: LC_CTYPE=en_GB.UTF-8;LC_NUMERIC=C;LC_TIME=en_GB.UTF-8;LC_COLLATE=en_GB.U TF-8;LC_MONETARY=C;LC_MESSAGES=en_GB.UTF-8;LC_PAPER=en_GB.UTF-8;LC_NAME= C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_GB.UTF-8;LC_IDENTIFICATI ON=C attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.9.1 [[alternative HTML version deleted]] __ R-help@r-project.org 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] Batch replacement, by factor, of values in a data frame
Dear List, I'm wondering if there is a better/cleaner/more efficient way of replacing 0 values in a variable with the minimum of the non-missing and non-zero values of that same variable, but doing it within the levels of a factor? Consider the dummy example data presented at the end of my message. Within each 'Site' there are some 0 values and possibly some NA's. I can compute the minimum of the non-missing and non-zero values by 'Site' as indicated below using aggregate for example. Save for looping over the 'Site's and replacing 0's with the relevant minimum is there a way of using a vectorised approach to do the replacement? Thanks in advance, G ## dummy data set.seed(123) D - data.frame(Site = factor(rep(LETTERS[1:5], times = 10)), Var = runif(5*10)) D - D[with(D, order(Site, Var)), ] ## simulate some 0's D[c(1,3,11,12,23,27,34,36,41,49), Var] - 0 ## just to complicate matters, some NA D[sample(NROW(D), 3), Var] - NA head(D) ## Compute minimums per Site aggregate(D$Var, by = list(Site = D$Site), FUN = function(x) min(x[x0], na.rm = TRUE)) ## How replace the appropriate 0's with the appropriate minimum? -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org 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] attach package
On 8/26/2009 8:17 AM, Jeremy MAZET wrote: Hello, Is there a solution to attach a package without run the hook function .onAttach() You could modify the source code to remove the hook, but why you'd want to do this, I don't know. Presumably the author of the package had a reason to put the hook there, and the package may not work properly without it. Duncan Murdoch __ R-help@r-project.org 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] Statistical question about logistic regression simulation
Your exposure variable has very large values, so all your probabilities are
1. You also get a bunch of NaN's because the `expit' (inverse logit)
function to calculate the probabilities cannot be evaluated. You need to use
values of exposure that will yield some 0's and 1's so that the binomial
model can be estimated.
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor, The Center on Aging and Health
Division of Geriatric Medicine and Gerontology
Johns Hopkins University
Ph: (410) 502-2619
Fax: (410) 614-9625
Email: rvarad...@jhmi.edu
Webpage:
http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h
tml
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Denis Aydin
Sent: Wednesday, August 26, 2009 10:18 AM
To: r-help@r-project.org
Subject: [R] Statistical question about logistic regression simulation
Hi R help list
I'm simulating logistic regression data with a specified odds ratio
(beta) and have a problem/unexpected behaviour that occurs.
The datasets includes a lognormal exposure and diseased and healthy
subjects.
Here is my loop:
ors - vector()
for(i in 1:200){
# First, I create a vector with a lognormally distributed exposure:
n - 1 # number of study subjects
mean - 6
sd - 1
expo - rlnorm(n, mean, sd)
# Then I assign each study subject a probability of disease with a
# specified Odds ratio (or beta coefficient) according to a logistic
# model:
inter - 0.01 # intercept
or - log(1.5) # an odds ratio of 1.5 or a beta of ln(1.5)
p - exp(inter + or * expo)/(1 + exp(inter + or * expo))
# Then I use the probability to decide who is having the disease and who
# is not:
disease - rbinom(length(p), 1, p) # 1 = disease, 0 = healthy
# Then I calculate the logistic regression and extract the odds ratio
model - glm(disease ~ expo, family = binomial)
ors[i] - exp(summary(model)$coef[2]) # exponentiated beta = OR
}
Now to my questions:
1. I was expecting the mean of the odds ratios over all simulations to
be close to the specified one (1.5 in this case). This is not the case
if the mean of the lognormal distribution is, say 6.
If I reduce the mean of the exposure distribution to say 3, the mean of
the simulated ORs is very close to the specified one. So the simulation
seems to be quite sensitive to the parameters of the exposure distribution.
2. Is it somehow possible to stabilize the simulation so that it is
not that sensitive to the parameters of the lognormal exposure
distribution? I can't make up the parameters of the exposure
distribution, they are estimations from real data.
3. Are there general flaws or errors in my approach?
Thanks a lot for any help on this!
All the best,
Denis
--
Denis Aydin
Institute of Social and Preventive Medicine at Swiss Tropical Institute
Basel
Associated Institute of the University of Basel
Steinengraben 49 - 4051 Basel - Switzerland
Phone: +41 (0)61 270 22 04
Fax: +41 (0)61 270 22 25
denis.ay...@unibas.ch
www.ispm-unibasel.ch
__
R-help@r-project.org 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@r-project.org 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] glmmPQL model selection
Sorry for the late reply. Just use the first 90% of your data to fit and then predict the last 10% and see which one is better. If the random effects are not good it will become very obvious. If the concern is with fixed effects then just use gls which puts the random effects in the error and select model as usual. Emmanuelle TASTARD wrote: Hi, Im sorry, I know that it is a recurrent question but I have not been able to find the response in the Rhelp archives. I think my data require the use of the glmmPQL function but I do not know how to make the model selection. Since the AIC and log-likelihood are apparently meaningless, how can we select the parameters for a model and compare the models to find which one fits best the data? Thanks a lot Emmanuelle Tastard Emmanuelle TASTARD UMR 5174 'Evolution et Diversité Biologique' Université Paul Sabatier Bat 4R3 31062 TOULOUSE CEDEX 9 France tel : 05 61 55 67 59 [[alternative HTML version deleted]] __ r-h...@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 -- View this message in context: http://www.nabble.com/glmmPQL-model-selection-tp3027224p25151417.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] lme: how to nest a random factor in a fixed factor?
Dear Robert, Since you have only 4 sites, a random effect is not so good. You would need at least 6 sites for a good estimate of the variance. You have enough data to treat site as a fixed effects. It only costs 2 extra degrees of freedom. Therefore I would model this like: lm(response ~ (area/site)*treatment, data = data) HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Robert Buitenwerf Verzonden: woensdag 26 augustus 2009 16:24 Aan: R Help Onderwerp: [R] lme: how to nest a random factor in a fixed factor? Dear all, I have an experimental setup in which a random variable is nested within a fixed variable; however I have troubles specifying the correct LMM with lme. I have searched the lists but haven't been able to find an example like my setup, which I unfortunately need to get this stuff right. Pinheiro Bates is great but I still can't figure out how to do it. My experimental setup was as follows: 100 measurements per treatment plot 2 treatment plots per site 4 sites: 2 in one area and 2 in another area Both treatment and area are fixed factors,while site is random. I am interested in the significance of the fixed effects,less in the magnitude of the random effect. I have tried: mod1 - lme(response ~ area*treatment, data=data,random= ~1|site) but now site is not nested in area... mod2 - lme(response ~ area*treatment, data=data,random= ~1|area/site) but now area is both a fixed and a random variable, which doesn't seem to make sense, plus I run out of df for treatment mod3 - lme(response ~ area*treatment, data=data,random= ~1| plot) but here plots are not grouped according to site I hope someone would be willing to help me, thank you in advance! Robert Buitenwerf Ecologist South African Environmental Observation Network _ [[elided Hotmail spam]] __ R-help@r-project.org 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. Druk dit bericht a.u.b. niet onnodig af. Please do not print this message unnecessarily. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ R-help@r-project.org 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] faulty formatting of toLatex(sessionInfo())
On 8/26/09, Frank E Harrell Jr f.harr...@vanderbilt.edu wrote:
G. Jay Kerns wrote:
There was a closely related discussion last April:
https://stat.ethz.ch/pipermail/r-devel/2009-April/053094.html
and IIRC this was fixed for R version 2.10.
If you still have trouble just do
s - toLatex(sessionInfo())
cat(s[-grep('Locale',s)], sep='\n')
This works great. Thank you both for the info.
Liviu
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Re: [R] mann whitney u
On Aug 26, 2009, at 7:18 AM, Mcdonald, Grant wrote: Dear Sir, I am comparing two samples using wilcox.test in R. Literature appears to describe mann whitney u test as the most appropriate test to use on my data. is the wilcox.test function equivalent to mann-whitney u? When used in its two-sample mode the answer is yes. Is there a way to gain the U-value as apposed to the W-value in R? Answering that question was as simple as going to : http://search.r-project.org/nmz.html typing in mann-whitney u ... and a few clicks. Try it yourself. -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help regarding frequency distribution Graphs
On Aug 26, 2009, at 7:32 AM, anupam sinha wrote: Hi all, I am trying to construct a frequency distribution graph i.e. suppose there is a variable *k* and it takes a range of values. What I want to do is to plot *P(k) *(probability/frequency of finding a specific value of *k*) vs *k*.I would like to get a smooth curve between P(k) vs k. I have already tried *hist* function in R but want something more specific. Can anyone help me out ? Thanks in advance. ?density David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org 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] Installing rJava RJDBC bad interpreter: Permission denied
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Matias Silva Sent: Wednesday, August 26, 2009 7:37 AM To: r-help@r-project.org Subject: [R] Installing rJava RJDBC bad interpreter: Permission denied Trying to install the above two packages via the install.packages(package_name) command and the R CMD INSTALL file.tar.gz. I receive the following error either way sh: ./configure: /bin/sh: bad interpreter: Permission denied. This may mean that your /bin/sh does not have execution permission. (It is similar to the message you get when you start a script with #!/bin/nosuchshell and run it with 'sh -c script'.) What does ls -l /bin/sh /bin/bash show? Bill Dunlap TIBCO Software Inc - Spotfire Division wdunlap tibco.com I have tried to chmod and chown permissions and also ran dos2unix in hopes that there is a CRLF in some of the tar.gz files, but that doesn't seem to fix the problem. I'm looking for any help I can get. Must I compile this to get this to work because of some incompatibility with the rpm packages?? * Installing *source* package ârJavaâ ... sh: ./configure: /bin/sh: bad interpreter: Permission denied ERROR: configuration failed for package ârJavaâ * Removing â/usr/lib64/R/library/rJavaâ * Installing *source* package âRJDBCâ ... ** R ** preparing package for lazy loading Error : package 'rJava' required by 'RJDBC' could not be found ERROR: lazy loading failed for package âRJDBCâ * Removing â/usr/lib64/R/library/RJDBCâ I'm using the rpm version of R found here http://cran.cnr.berkeley.edu/bin/linux/redhat/el5/x86_64 I ran the R CMD javareconf as root and set the JAVA_HOME environment variable. Here is my system configuration: OS: CentOS 5.2 x86_64 JDK: java version 1.6.0_12 Java(TM) SE Runtime Environment (build 1.6.0_12-b04) Java HotSpot(TM) 64-Bit Server VM (build 11.2-b01, mixed mode) R: 2.9.1 rJava: rJava_0.7-0.tar.gz RJDBC: RJDBC_0.1-5.tar.gz Thanks for your time and knowledge. Best, Matt __ R-help@r-project.org 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@r-project.org 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] increasing significant digits in smooth.spline function
On Aug 26, 2009, at 9:26 AM, Sergii, Ivakhno wrote: Hello All I have a very long vector of unique predictor values and 6 significant digits setting for the smooth.spline rounds them off. Is there any way of increasing the significant digits withour recompiling a lot if code (simple editing and tham sourcing of smooth.spline.r function does not work, probably due to presence of Fortan functional calls)? ?options options(digits=12) What you see on the screen is not necessarily what is happening inside. There is no rounding unless you force such. -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org 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] Statistical question about logistic regression simulatio
On 26-Aug-09 14:17:40, Denis Aydin wrote:
Hi R help list
I'm simulating logistic regression data with a specified odds ratio
(beta) and have a problem/unexpected behaviour that occurs.
The datasets includes a lognormal exposure and diseased and healthy
subjects.
Here is my loop:
ors - vector()
for(i in 1:200){
# First, I create a vector with a lognormally distributed exposure:
n - 1 # number of study subjects
mean - 6
sd - 1
expo - rlnorm(n, mean, sd)
# Then I assign each study subject a probability of disease with a
# specified Odds ratio (or beta coefficient) according to a logistic
# model:
inter - 0.01 # intercept
or - log(1.5) # an odds ratio of 1.5 or a beta of ln(1.5)
p - exp(inter + or * expo)/(1 + exp(inter + or * expo))
# Then I use the probability to decide who is having the disease and who
# is not:
disease - rbinom(length(p), 1, p) # 1 = disease, 0 = healthy
# Then I calculate the logistic regression and extract the odds ratio
model - glm(disease ~ expo, family = binomial)
ors[i] - exp(summary(model)$coef[2]) # exponentiated beta = OR
}
Now to my questions:
1. I was expecting the mean of the odds ratios over all simulations to
be close to the specified one (1.5 in this case). This is not the case
if the mean of the lognormal distribution is, say 6.
If I reduce the mean of the exposure distribution to say 3, the mean of
the simulated ORs is very close to the specified one. So the simulation
seems to be quite sensitive to the parameters of the exposure
distribution.
2. Is it somehow possible to stabilize the simulation so that it is
not that sensitive to the parameters of the lognormal exposure
distribution? I can't make up the parameters of the exposure
distribution, they are estimations from real data.
3. Are there general flaws or errors in my approach?
Thanks a lot for any help on this!
All the best,
Denis
--
Denis Aydin
You need to look at the probabilities 'p' being generated by your code.
Taking first your case mean - 6 (and sorting 'expo', and using
whatever seed my system had current at the time):
n - 1 # number of study subjects
mean - 6
sd - 1
expo - sort(rlnorm(n, mean, sd))
p - exp(inter + or * expo)/(1 + exp(inter + or * expo))
p[1:20]
# [1] 0.9763438 0.9918962 0.9924002 0.9965314 0.9980887 0.9984698
# [7] 0.9993116 0.9993167 0.9994007 0.9994243 0.9996288 0.9997037
# [13] 0.9998728 0.9998832 0.284 0.346 0.446 0.528
# [19] 0.561 0.645
so that almost all of your 'p's are very close to 1.0, which means
that almost all or even all) of your responses will be 1. Indeed,
continuiung from the above:
disease - rbinom(length(p), 1, p)
disease[1:20]
# [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
sum(disease)
# [1] NaN
sum(is.nan(disease))
# [1] 710
What has happened here is that the higher values of 'expo' are so
large (in the 1000s) that the calculation of 'p' gives NA, because
the value of exp(inter + or * expo) is +Inf, so the calculation of
'p' is in terms of (+Inf)/(+Inf), which is NA.
Now compare with what happens when mean - 3:
mean - 3
sd - 1
expo - sort(rlnorm(n, mean, sd))
p - exp(inter + or * expo)/(1 + exp(inter + or * expo))
p[1:20]
# [1] 0.5514112 0.5543155 0.5702318 0.5830025 0.5885994 0.5889078
# [7] 0.5908860 0.6004657 0.6029123 0.6042805 0.6048688 0.6122290
# [13] 0.6123407 0.6135233 0.6137499 0.6139299 0.6153900 0.6181017
# [19] 0.6184093 0.6203757
sum(is.na(p))
# [1] 0
max(expo)
# [1] 728.0519
So now no NAs (max(expo), though large, is now not large enough to
make the calculation of 'p' yield NA).
These smaller probabilities are now well away from 1.0, so a good mix
of 0 and 1 responses can be expected, although a good number of
the 'p's will still be very close to 1 or will be set equal to 1.
disease - rbinom(length(p), 1, p)
disease[1:20]
# [1] 0 1 0 0 1 0 1 1 1 0 1 0 0 1 0 1 1 0 0 1
(as expected), and
sum(disease)
# [1] 9740
As well as the problem with p = NA --- disease = NaN, when you
have all the probabiltiies close to 1 and (as above) get all the
'disease' outcomes = 1, the resulting attempt to fit the glm will
yield nonsense.
In summary: do not use silly paramater values for the model you
are simulating. It will almost always not work (for reasons
illustrated above), and even if it appreas to work the result
will be highly unreliable. If in doubt, have a look at what you
are getting, along the line, as illustrated above!
The above reasons almost certainly underlie your finding that the
mean of simulated OR estimates is markedly different from the value
which you set when you run the case mean - 6, and the much
better finding when you run the case mean - 3.
Hoping this helps,
Ted.
E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 26-Aug-09
Re: [R] mann whitney u
On Wed, 26 Aug 2009 12:18:53 +0100 Mcdonald, Grant grant.mcdonal...@imperial.ac.uk wrote: MG is the wilcox.test function equivalent to mann-whitney u? Is there Yes, the test is the same. It is also called wilcoxon mann whitney test because the authors created the test independently. MG a way to gain the U-value as apposed to the W-value in R? What would be the purpose of this when you already have the p-value given? As far as I know W and U are related. Look at http://en.wikipedia.org/wiki/Mann%E2%80%93Whitney_U and calculate the U1 U2 with the values of the example(wilcox.test). You have not described how your data look like. In case of tie's (equal values) use wilcox_test of the coin package. hth Stefan __ R-help@r-project.org 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] glmmPQL and variance structure
this is very late, but I saw this now as I am dealing with it now:
I think varPower should not be needed here. The family should be one of the
quasi families eg quasibinomial and that will automatically allow
variance/dispersion to become a function of the fit. This is a feature of
glm inherently, which does not exist in lme, so it is called with varPower
(in lme).
If you see that, pls post whether it works as expected on the original
dataset.
regards
Stephen
Spencer Graves wrote:
Thanks for providing a partially reproducible example. I believe the
error message you cite came from lme. I say this, because I modified
your call to glmmPQL2 to call lme and got the following:
library(nlme)
fit.lme - lme(y ~ trt + I(week 2), random = ~ 1 | ID,
+ data = bacteria, weights=varPower(~1))
Error in unlist(x, recursive, use.names) :
argument not a list
I consulted Pinheiro and Bates (2000) Mixed-Effects Models in S and
S-Plus (Springer, sec. 5.2, p.211) to see that the syntax for varPower
appears to be correct. I removed ~ and it worked, mostly:
fit.lme - lme(y ~ trt + I(week 2), random = ~ 1 | ID,
+ data = bacteria, weights=varPower(1))
Warning message:
- not meaningful for factors in: Ops.factor(y[revOrder], Fitted)
I got an answer, though with a warning and not for the problem you
want to solve. However, I then made this modification to a call to my
own modification of Venables and Ripley's glmmPQL and it worked:
fit. - glmmPQL.(y ~ trt + I(week 2), random = ~ 1 | ID,
+ family = binomial, data = bacteria,
+ weights.lme=varPower(1))
iteration 1
iteration 2
iteration 3
fit.
Linear mixed-effects model fit by maximum likelihood
Data: bacteria
Log-likelihood: -541.0882
Fixed: y ~ trt + I(week 2)
(Intercept) trtdrugtrtdrug+ I(week 2)TRUE
2.7742329 -1.0852566 -0.5896635 -1.2682626
Random effects:
Formula: ~1 | ID
(Intercept) Residual
StdDev: 4.940885e-05 2.519018
Variance function:
Structure: Power of variance covariate
Formula: ~fitted(.)
Parameter estimates:
power
0.3926788
Number of Observations: 220
Number of Groups: 50
This function glmmPQL. adds an argument weights.lme to Venables
and Ripley's glmmPQL and uses that in place of
'quote(varFixed(~invwt))' when provided; see below.
hope this helps.
spencer graves
glmmPQL. -
function (fixed, random, family, data, correlation, weights,
weights.lme, control, niter = 10, verbose = TRUE, ...)
{
if (!require(nlme))
stop(package 'nlme' is essential)
if (is.character(family))
family - get(family)
if (is.function(family))
family - family()
if (is.null(family$family)) {
print(family)
stop('family' not recognized)
}
m - mcall - Call - match.call()
nm - names(m)[-1]
wts.lme - mcall$weights.lme
keep - is.element(nm, c(weights, data, subset, na.action))
for (i in nm[!keep]) m[[i]] - NULL
allvars - if (is.list(random))
allvars - c(all.vars(fixed), names(random),
unlist(lapply(random,
function(x) all.vars(formula(x)
else c(all.vars(fixed), all.vars(random))
Terms - if (missing(data))
terms(fixed)
else terms(fixed, data = data)
off - attr(Terms, offset)
if (length(off - attr(Terms, offset)))
allvars - c(allvars, as.character(attr(Terms, variables))[off
+
1])
m$formula - as.formula(paste(~, paste(allvars, collapse = +)))
environment(m$formula) - environment(fixed)
m$drop.unused.levels - TRUE
m[[1]] - as.name(model.frame)
mf - eval.parent(m)
off - model.offset(mf)
if (is.null(off))
off - 0
w - model.weights(mf)
if (is.null(w))
w - rep(1, nrow(mf))
mf$wts - w
fit0 - glm(formula = fixed, family = family, data = mf,
weights = wts, ...)
w - fit0$prior.weights
eta - fit0$linear.predictor
zz - eta + fit0$residuals - off
wz - fit0$weights
fam - family
nm - names(mcall)[-1]
keep - is.element(nm, c(fixed, random, data, subset,
na.action, control))
for (i in nm[!keep]) mcall[[i]] - NULL
fixed[[2]] - quote(zz)
mcall[[fixed]] - fixed
mcall[[1]] - as.name(lme)
mcall$random - random
mcall$method - ML
if (!missing(correlation))
mcall$correlation - correlation
# weights.lme
{
if(is.null(wts.lme))
mcall$weights - quote(varFixed(~invwt))
else
mcall$weights - wts.lme
}
mf$zz - zz
mf$invwt - 1/wz
mcall$data - mf
for (i in 1:niter) {
if (verbose)
cat(iteration, i, \n)
fit - eval(mcall)
etaold - eta
eta -
Re: [R] Batch replacement, by factor, of values in a data frame
The ave function is very handy for things like this: mins = ave(D$Var,D$Site,FUN=function(x)min(x[x0],na.rm=TRUE)) D$Var = ifelse(is.na(D$Var) | D$Var == 0,mins,D$Var) should do the required replacements. - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Wed, 26 Aug 2009, Gavin Simpson wrote: Dear List, I'm wondering if there is a better/cleaner/more efficient way of replacing 0 values in a variable with the minimum of the non-missing and non-zero values of that same variable, but doing it within the levels of a factor? Consider the dummy example data presented at the end of my message. Within each 'Site' there are some 0 values and possibly some NA's. I can compute the minimum of the non-missing and non-zero values by 'Site' as indicated below using aggregate for example. Save for looping over the 'Site's and replacing 0's with the relevant minimum is there a way of using a vectorised approach to do the replacement? Thanks in advance, G ## dummy data set.seed(123) D - data.frame(Site = factor(rep(LETTERS[1:5], times = 10)), Var = runif(5*10)) D - D[with(D, order(Site, Var)), ] ## simulate some 0's D[c(1,3,11,12,23,27,34,36,41,49), Var] - 0 ## just to complicate matters, some NA D[sample(NROW(D), 3), Var] - NA head(D) ## Compute minimums per Site aggregate(D$Var, by = list(Site = D$Site), FUN = function(x) min(x[x0], na.rm = TRUE)) ## How replace the appropriate 0's with the appropriate minimum? -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Select top three values from data frame
Do you want just the values (i.e., a vector), or do you also want the
corresponding rows of the data frame?
What if there is a tie, or do you know in advance that within any
particular subset the values of B are unique?
What if the subset that meets the constraints has fewer than 3 unique
values? (which I think is the case in your example)
tail( unique( sort( df$B[ df$A=='x' df$C 2 ] ) ) ,3 )
Should do it (but I haven't tested).
Why does it get messy with over 100 columns?
I'll pretend for the moment that you have exactly 100 columns:
1) you will be doing this many times, each time with a different
sets of 3 columns?
2) you want the three highest values in each of 98 columns based
on constraints on the other two?
3) you want the three highest values of B based on constraints on
all of the other 99 columns?
Depending on what changes when more columns are involved, you might
be able to loop over columns with syntax like,
for (nm in c('B','D','E') ) tail( unique( sort( df[[nm]][
df$A=='x' df$C 2 ] ) ) ,3 )
-Don
At 1:36 AM -0700 8/26/09, Noah Silverman wrote:
Hi,
I'm trying to find an easy way to do this.
I want to select the top three values of a specific column in a
subset of rows in a data.frame. I'll demonstrate.
ABC
x21
x41
x32
y15
y26
y38
I want the top 3 values of B from the data.frame where A=X and C 2
I could extract all the rows where C2, then sort by B, then take
the first 3. But that seems like the wrong way around, and it also
will get messy with real data of over 100 columns.
Any suggestions?
__
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--
--
Don MacQueen
Environmental Protection Department
Lawrence Livermore National Laboratory
Livermore, CA, USA
925-423-1062
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help regarding frequency distribution Graphs
Google R gallery :-) On Wed, Aug 26, 2009 at 7:32 AM, anupam sinha anupam.cont...@gmail.comwrote: Hi all, I am trying to construct a frequency distribution graph i.e. suppose there is a variable *k* and it takes a range of values. What I want to do is to plot *P(k) *(probability/frequency of finding a specific value of *k*) vs *k*.I would like to get a smooth curve between P(k) vs k. I have already tried *hist* function in R but want something more specific. Can anyone help me out ? Thanks in advance. Regards, Anupam Sinha [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org 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] GLMs
Hi, On Aug 26, 2009, at 6:53 AM, Letizia Campioni wrote: Hi, I am starting to work with R. I need to performe a General linear model and a Generalized mixed model, what are the package I have to use for? R RSiteSearch(general linear model) R RSiteSearch(Generalized mixed model) Will provide many answers. -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org 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 RHEL 5 repo and the latest R RPMs
Martyn Plummer wrote: It turns out that the default checksum for createrepo, the command that creates repository metadata, has changed from sha1 to sha256, whereas Enterprise Linux 5 still requires sha1. I have modified the scripts to use the older checksum for EL4 and EL5. Sorry for the slow reply. Lance Brown from Duke University already raised this problem with Bob Kinney and me. When I saw a query from Hugh Brown I was not paying enough attention to see that you were somebody else. This appears to have solved the problem. Thanks very much for your help! -- Hugh Brown, Systems Manager The Centre for High-Throughput Biology hbr...@chibi.ubc.ca __ R-help@r-project.org 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] Scripting - sort of
Dear R-ians: I'm running R2.9.2 on a 6 year old Windows XP DELL with 2 Gig RAM and a 3MHz Pentium 4 chip. I've written a package using the User Menus Under Windows commands (winMenuAdd, etc). It works very well. I have 6 test cases and running any one of them requires many selections form the menus and some keyboard entry, and it takes me hours to exercise them all to see that my changes to the R code produced no unexpected results. Is there a way to record my mouse movements and mouse clicks, as well as keyboard entry, in a script, so that I can just watch to see if all unfolds correctly? Many thanks for any suggestions. Charles Annis, P.E. mailto:charles.an...@statisticalengineering.com charles.an...@statisticalengineering.com phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com http://www.StatisticalEngineering.com [[alternative HTML version deleted]] __ R-help@r-project.org 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] tweedie and lmer
Hello all, I have count data with about 36% of observations being zeros. I found in some of the examples of the r-help mail archives that a tweedie family of distributions could be used to fit a model with random effects. Upon installing the tweedie package and attempting to fit the following model: lmer(SUS ~ 1 + (1| GRP),REML=FALSE,data=mydata,family=tweedie(var.power=1.55,link.power=0)) I get the following error: Error in famType(glmFit$family) : unknown GLM family: ‘Tweedie’ If it helps, im on a mac with R V 2.9.1, lme4 V.0.999375-31, Tweedie V2.0. Thanks, Mohammad AlMarzouq __ R-help@r-project.org 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] Scripting - sort of
autoit and autohotkey are two free Windows utilities based on the basic language that can be used for GUI scripting. On Wed, Aug 26, 2009 at 12:22 PM, Charles Annis, P.E.charles.an...@statisticalengineering.com wrote: Dear R-ians: I'm running R2.9.2 on a 6 year old Windows XP DELL with 2 Gig RAM and a 3MHz Pentium 4 chip. I've written a package using the User Menus Under Windows commands (winMenuAdd, etc). It works very well. I have 6 test cases and running any one of them requires many selections form the menus and some keyboard entry, and it takes me hours to exercise them all to see that my changes to the R code produced no unexpected results. Is there a way to record my mouse movements and mouse clicks, as well as keyboard entry, in a script, so that I can just watch to see if all unfolds correctly? Many thanks for any suggestions. Charles Annis, P.E. mailto:charles.an...@statisticalengineering.com charles.an...@statisticalengineering.com phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com http://www.StatisticalEngineering.com [[alternative HTML version deleted]] __ R-help@r-project.org 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@r-project.org 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] Applying do.call to a data.frame using function arguments
miller_2555 wrote: I'm trying to convert a data.frame to a series of strings (row-wise). There was a very good discussion awhile back (2002) entitled [R] string concatenate across rows of a matrix?? where Tony Plate recommended the following two alternatives (x2 is an R object of type data frame -- a matrix also works for solution #1): 1) apply(format(x2), 1, paste, collapse= ); 2) do.call(paste,x2) Nevermind. Stupid question. The solution is: do.call(paste,c(x2,sep=',')); Hope this helps somebody. -- View this message in context: http://www.nabble.com/Applying-do.call-to-a-data.frame-using-function-arguments-tp25151441p25151445.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] access to source code of a website ..?
hi, is it possible to read the source code of a website within R? __ R-help@r-project.org 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] Applying do.call to a data.frame using function arguments
On Wed, Aug 26, 2009 at 11:31 AM, miller_2555nabble.30.miller_2...@spamgourmet.com wrote: miller_2555 wrote: I'm trying to convert a data.frame to a series of strings (row-wise). There was a very good discussion awhile back (2002) entitled [R] string concatenate across rows of a matrix?? where Tony Plate recommended the following two alternatives (x2 is an R object of type data frame -- a matrix also works for solution #1): 1) apply(format(x2), 1, paste, collapse= ); 2) do.call(paste,x2) Nevermind. Stupid question. The solution is: do.call(paste,c(x2,sep=',')); I think you're missing some quotes: cat(do.call(paste,c(x2,sep=','))[1], \n) Hadley -- http://had.co.nz/ __ R-help@r-project.org 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] GLMs
Letizia Campioni wrote: Hi, I am starting to work with R. I need to performe a General linear model and a Generalized mixed model, what are the package I have to use for? what is the difference between them? General linear models (called GLM mostly by SAS users) are implemented by lm (no need to install or load additional packages). General*ized* linear models (called GLM by everyone else): use glm (still no need for extra packages). Linear mixed models (LMMs): nlme (library(nlme), no need to install extras. Generalized linear mixed models (GLMMS): lme4 -- install.packages(lme4); library(lme4), ?glmer This list is not generally for general statistics questions -- you're expected to know *what* you want to do, and we will help you with *how* to do it in R (provided you have read the posting guide and formulated a useful question). I would suggest a book -- check out the R books page ( http://www.r-project.org/doc/bib/R-books.html ) -- maybe Faraway? -- View this message in context: http://www.nabble.com/GLMs-tp25150823p25151455.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] changing equal values on matrix by same random number
Dear all, I have about 30,000 matrix (512x512), with values from 1 to N. Each value on a matrix represent a habitat patch on my matrix (i.e. my landscape). Non-habitat are stored as ZERO. No I need to change each 1-to-N values for the same random number. Just supose my matrix is: mymat-matrix(c(1,1,1,0,0,0,0,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 3,3,0,0,0,0,0,0,0,4,4, 3,3,0,0,0,0,0,0,0,0,0), nrow=5) I would like that all cells with 1 come to be runif(1,min=0.4, max=0.7), and cells with 2 be replace by another runif(...). I can do it using for(), but it is very time expensive. Any help are welcome. cheers milton [[alternative HTML version deleted]] __ R-help@r-project.org 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] changing equal values on matrix by same random number
You want to replace all 1s each with the same random nuumber from the uniform distribution, then all 2s each with the same random nuumber from the uniform distribution, and so forth? Are the arguments (i.e., the min and max of the distribution) to each call to runif identical? -- David - David Huffer, Ph.D. Senior Statistician CSOSA/Washington, DC david.huf...@csosa.gov - -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of milton ruser Sent: Wednesday, August 26, 2009 12:54 PM To: r-help@r-project.org Subject: [R] changing equal values on matrix by same random number Dear all, I have about 30,000 matrix (512x512), with values from 1 to N. Each value on a matrix represent a habitat patch on my matrix (i.e. my landscape). Non-habitat are stored as ZERO. No I need to change each 1-to-N values for the same random number. Just supose my matrix is: mymat-matrix(c(1,1,1,0,0,0,0,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 3,3,0,0,0,0,0,0,0,4,4, 3,3,0,0,0,0,0,0,0,0,0), nrow=5) I would like that all cells with 1 come to be runif(1,min=0.4, max=0.7), and cells with 2 be replace by another runif(...). I can do it using for(), but it is very time expensive. Any help are welcome. cheers milton [[alternative HTML version deleted]] __ R-help@r-project.org 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@r-project.org 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] changing equal values on matrix by same random number
On Aug 26, 2009, at 12:53 PM, milton ruser wrote: Dear all, I have about 30,000 matrix (512x512), with values from 1 to N. Each value on a matrix represent a habitat patch on my matrix (i.e. my landscape). Non-habitat are stored as ZERO. No I need to change each 1-to-N values for the same random number. Just supose my matrix is: mymat-matrix(c(1,1,1,0,0,0,0,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 3,3,0,0,0,0,0,0,0,4,4, 3,3,0,0,0,0,0,0,0,0,0), nrow=5) I would like that all cells with 1 come to be runif(1,min=0.4, max=0.7), and cells with 2 be replace by another runif(...). First the wrong way and then the right way: mymat[mymat==1] - runif(1,min=0.4,max=0.7) mymat [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [1,] 0.457316100200000 3 0 [2,] 0.457316100202000 0 0 [3,] 0.457316100202004 0 0 [4,] 0.00000002304 0 0 [5,] 0.00000000303 0 0 All the values are the same, clearly not what was desired. Put it back to your starting point: mymat-matrix(c(1,1,1,0,0,0,0,0,0,0,0, + 0,0,0,0,2,2,2,0,0,0,0, + 0,0,0,0,2,2,2,0,0,0,0, + 3,3,0,0,0,0,0,0,0,4,4, + 3,3,0,0,0,0,0,0,0,0,0), nrow=5) # So supply the proper number of random realizations: mymat[mymat==1] - runif(sum(mymat==1),min=0.4,max=0.7) mymat [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [1,] 0.574566500200000 3 0 [2,] 0.695641800202000 0 0 [3,] 0.693546600202004 0 0 [4,] 0.00000002304 0 0 [5,] 0.00000000303 0 0 If you want to supply a matrix of max and min values for the other integers there would probably be an *apply approach that could be used. I can do it using for(), but it is very time expensive. Any help are welcome. cheers David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org 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] changing equal values on matrix by same random number
Hi David, Thanks for the reply. This is what I need: mymat[mymat==1] - runif(1,min=0.4,max=0.7) mymat [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [1,] 0.457316100200000 3 0 [2,] 0.457316100202000 0 0 [3,] 0.457316100202004 0 0 [4,] 0.00000002304 0 0 [5,] 0.00000000303 0 0 But as my real landscapes have values from 1 to large number (~10,), so I think that if I put this on a for() looping it will be very time expensive, and as I have a lot of landscapes, I need to speed up it. Any suggestion? bests milton On Wed, Aug 26, 2009 at 1:12 PM, David Winsemius dwinsem...@comcast.netwrote: On Aug 26, 2009, at 12:53 PM, milton ruser wrote: Dear all, I have about 30,000 matrix (512x512), with values from 1 to N. Each value on a matrix represent a habitat patch on my matrix (i.e. my landscape). Non-habitat are stored as ZERO. No I need to change each 1-to-N values for the same random number. Just supose my matrix is: mymat-matrix(c(1,1,1,0,0,0,0,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 3,3,0,0,0,0,0,0,0,4,4, 3,3,0,0,0,0,0,0,0,0,0), nrow=5) I would like that all cells with 1 come to be runif(1,min=0.4, max=0.7), and cells with 2 be replace by another runif(...). First the wrong way and then the right way: mymat[mymat==1] - runif(1,min=0.4,max=0.7) mymat [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [1,] 0.457316100200000 3 0 [2,] 0.457316100202000 0 0 [3,] 0.457316100202004 0 0 [4,] 0.00000002304 0 0 [5,] 0.00000000303 0 0 All the values are the same, clearly not what was desired. Put it back to your starting point: mymat-matrix(c(1,1,1,0,0,0,0,0,0,0,0, + 0,0,0,0,2,2,2,0,0,0,0, + 0,0,0,0,2,2,2,0,0,0,0, + 3,3,0,0,0,0,0,0,0,4,4, + 3,3,0,0,0,0,0,0,0,0,0), nrow=5) # So supply the proper number of random realizations: mymat[mymat==1] - runif(sum(mymat==1),min=0.4,max=0.7) mymat [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [1,] 0.574566500200000 3 0 [2,] 0.695641800202000 0 0 [3,] 0.693546600202004 0 0 [4,] 0.00000002304 0 0 [5,] 0.00000000303 0 0 If you want to supply a matrix of max and min values for the other integers there would probably be an *apply approach that could be used. I can do it using for(), but it is very time expensive. Any help are welcome. cheers David Winsemius, MD Heritage Laboratories West Hartford, CT [[alternative HTML version deleted]] __ R-help@r-project.org 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] access to source code of a website ..?
On 8/26/2009 12:44 PM, Martin Batholdy wrote: hi, is it possible to read the source code of a website within R? url(http://example.com;) creates a connection that you can do what you like with. For example, readLines(url(http://www.r-project.org;)) Duncan Murdoch __ R-help@r-project.org 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] Applying do.call to a data.frame using function arguments (nabble: message 8 of 20)
On Wed, Aug 26, 2009 at 12:47 PM, hadley wickham - I think you're missing some quotes: cat(do.call(paste,c(x2,sep=','))[1], \n) Thanks - the strings are actually substrings of larger strings (specifically, SQL statements), which will wrap with the leading and trailing quotes (though I should have pointed this out in my original post). __ R-help@r-project.org 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] changing equal values on matrix by same random number
You could try either of the two examples below. They both assume that min and
max are invariant:
mymat - matrix (
c (
1 , 1 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 2
, 2 , 2 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 2 , 2 , 2 , 0 , 0
, 0 , 0 , 3 , 3 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 4 , 4 , 3 , 3
, 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0
) , nrow = 5
)
## this way gives same random values for
## each integer between 0 and
## max (mymat):
milton - function ( x , min = 0.4 , max = 0.7 ) {
.rep - runif ( n = max ( x ) , min = min , max = max )
for ( i in 1:max ( x ) ) {
x[x == i] - .rep[i]
}
x
}
milton ( x = mymat )
## this way gives different random values
## for each integer between
## 0 and max (mymat):
milton - function ( x , min = 0.4 , max = 0.7 ) {
for ( i in 1:max ( x ) ) {
x [ x == i ] - runif (
sum ( x == i )
, min = min
, max = max
)
}
x
}
milton ( x = mymat )
--
David
-
David Huffer, Ph.D. Senior Statistician
CSOSA/Washington, DC david.huf...@csosa.gov
-
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of milton ruser
Sent: Wednesday, August 26, 2009 1:18 PM
To: David Winsemius
Cc: r-help@r-project.org
Subject: Re: [R] changing equal values on matrix by same random number
Hi David,
Thanks for the reply. This is what I need:
mymat[mymat==1] - runif(1,min=0.4,max=0.7)
mymat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,] 0.457316100200000 3 0
[2,] 0.457316100202000 0 0
[3,] 0.457316100202004 0 0
[4,] 0.00000002304 0 0
[5,] 0.00000000303 0 0
But as my real landscapes have values from 1 to large number (~10,),
so I think that if I put this on a for() looping it will be very time
expensive,
and as I have a lot of landscapes, I need to speed up it.
Any suggestion?
bests
milton
On Wed, Aug 26, 2009 at 1:12 PM, David Winsemius dwinsem...@comcast.netwrote:
On Aug 26, 2009, at 12:53 PM, milton ruser wrote:
Dear all,
I have about 30,000 matrix (512x512), with values from 1 to N.
Each value on a matrix represent a habitat patch on my
matrix (i.e. my landscape). Non-habitat are stored as ZERO.
No I need to change each 1-to-N values for the same random
number.
Just supose my matrix is:
mymat-matrix(c(1,1,1,0,0,0,0,0,0,0,0,
0,0,0,0,2,2,2,0,0,0,0,
0,0,0,0,2,2,2,0,0,0,0,
3,3,0,0,0,0,0,0,0,4,4,
3,3,0,0,0,0,0,0,0,0,0), nrow=5)
I would like that all cells with 1 come to be
runif(1,min=0.4, max=0.7), and cells with 2
be replace by another runif(...).
First the wrong way and then the right way:
mymat[mymat==1] - runif(1,min=0.4,max=0.7)
mymat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,] 0.457316100200000 3 0
[2,] 0.457316100202000 0 0
[3,] 0.457316100202004 0 0
[4,] 0.00000002304 0 0
[5,] 0.00000000303 0 0
All the values are the same, clearly not what was desired.
Put it back to your starting point:
mymat-matrix(c(1,1,1,0,0,0,0,0,0,0,0,
+ 0,0,0,0,2,2,2,0,0,0,0,
+ 0,0,0,0,2,2,2,0,0,0,0,
+ 3,3,0,0,0,0,0,0,0,4,4,
+ 3,3,0,0,0,0,0,0,0,0,0), nrow=5)
# So supply the proper number of random realizations:
mymat[mymat==1] - runif(sum(mymat==1),min=0.4,max=0.7)
mymat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,] 0.574566500200000 3 0
[2,] 0.695641800202000 0 0
[3,] 0.693546600202004 0 0
[4,] 0.00000002304 0 0
[5,] 0.00000000303 0 0
If you want to supply a matrix of max and min values for the other integers
there would probably be an *apply approach that could be used.
I can do it using for(), but it is very time expensive.
Any help are welcome.
cheers
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
[[alternative HTML version deleted]]
__
R-help@r-project.org 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@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
[R] rJava error for large XML object return in StatET plugin
Hi R List, I get this error using StatET R plugin in Eclipse. sessionInfo() R version 2.9.0 (2009-04-17) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 I am running Eclipse Platform Version: 3.4.1 Build id: M20080911-1700 on Windows XP My R process creates a XML object which when I cat, gives the error below. The XML object is large, containing 1000+ rows like the one below. data name=subset y=1.0121137732 x=0.8018022766 ypred=0.86 yloess=1.0472/ The XML object when created with less than 800 rows does get created. Does it have anything to do with the size of xml. org.eclipse.core.runtime.CoreException: Communication error at de.walware.statet.r.nico.impl.RjsController.rjsRunMainLoop(RjsController.java:191) at de.walware.statet.r.nico.impl.RjsController.doSubmit(RjsController.java:410) at de.walware.statet.nico.core.runtime.ToolController.submitToConsole(ToolController.java:1009) at de.walware.statet.nico.core.runtime.ToolController$ConsoleCommandRunnable.run(ToolController.java:124) at de.walware.statet.nico.core.runtime.ToolController.loopRunTask(ToolController.java:822) at de.walware.statet.nico.core.runtime.ToolController.loop(ToolController.java:767) at de.walware.statet.nico.core.runtime.ToolController.run(ToolController.java:293) at de.walware.statet.nico.core.runtime.ToolRunner.run(ToolRunner.java:37) at de.walware.statet.nico.core.runtime.ToolRunner.access$0(ToolRunner.java:35) at de.walware.statet.nico.core.runtime.ToolRunner$1.run(ToolRunner.java:47) Caused by: java.rmi.UnmarshalException: error unmarshalling return; nested exception is: java.io.UTFDataFormatException at sun.rmi.server.UnicastRef.invoke(Unknown Source) at java.rmi.server.RemoteObjectInvocationHandler.invokeRemoteMethod(Unknown Source) at java.rmi.server.RemoteObjectInvocationHandler.invoke(Unknown Source) at $Proxy0.runMainLoop(Unknown Source) at de.walware.statet.r.nico.impl.RjsController.rjsRunMainLoop(RjsController.java:172) ... 9 more Caused by: java.io.UTFDataFormatException at java.io.ObjectInputStream$BlockDataInputStream.readUTFSpan(Unknown Source) at java.io.ObjectInputStream$BlockDataInputStream.readUTFBody(Unknown Source) at java.io.ObjectInputStream$BlockDataInputStream.readUTF(Unknown Source) at java.io.ObjectInputStream.readUTF(Unknown Source) at de.walware.rj.server.ConsoleCmdItem.init(ConsoleCmdItem.java:58) at de.walware.rj.server.MainCmdList.readExternal(MainCmdList.java:70) at java.io.ObjectInputStream.readExternalData(Unknown Source) at java.io.ObjectInputStream.readOrdinaryObject(Unknown Source) at java.io.ObjectInputStream.readObject0(Unknown Source) at java.io.ObjectInputStream.readObject(Unknown Source) at sun.rmi.server.UnicastRef.unmarshalValue(Unknown Source) ... 14 more Thanks and regards, Harsh [[alternative HTML version deleted]] __ R-help@r-project.org 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] Batch replacement, by factor, of values in a data frame
On Wed, 2009-08-26 at 08:06 -0700, Phil Spector wrote: The ave function is very handy for things like this: mins = ave(D$Var,D$Site,FUN=function(x)min(x[x0],na.rm=TRUE)) D$Var = ifelse(is.na(D$Var) | D$Var == 0,mins,D$Var) should do the required replacements. Thanks Phil, that's great. I hadn't come across ave() before. Shortly after I received your email, it dawned on me to just rep the minimums the required number of times (by the number of non-missing 0's) for each site, and then insert this vector into the data frame whilst subsetting on whether it was zero or not. Cheers, G - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Wed, 26 Aug 2009, Gavin Simpson wrote: Dear List, I'm wondering if there is a better/cleaner/more efficient way of replacing 0 values in a variable with the minimum of the non-missing and non-zero values of that same variable, but doing it within the levels of a factor? Consider the dummy example data presented at the end of my message. Within each 'Site' there are some 0 values and possibly some NA's. I can compute the minimum of the non-missing and non-zero values by 'Site' as indicated below using aggregate for example. Save for looping over the 'Site's and replacing 0's with the relevant minimum is there a way of using a vectorised approach to do the replacement? Thanks in advance, G ## dummy data set.seed(123) D - data.frame(Site = factor(rep(LETTERS[1:5], times = 10)), Var = runif(5*10)) D - D[with(D, order(Site, Var)), ] ## simulate some 0's D[c(1,3,11,12,23,27,34,36,41,49), Var] - 0 ## just to complicate matters, some NA D[sample(NROW(D), 3), Var] - NA head(D) ## Compute minimums per Site aggregate(D$Var, by = list(Site = D$Site), FUN = function(x) min(x[x0], na.rm = TRUE)) ## How replace the appropriate 0's with the appropriate minimum? -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org 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. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org 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] changing equal values on matrix by same random number
Hi David,
Now is perfect!!
Thanks a lot
milton
On Wed, Aug 26, 2009 at 1:30 PM, David Huffer david.huf...@csosa.govwrote:
You could try either of the two examples below. They both assume that min
and max are invariant:
mymat - matrix (
c (
1 , 1 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 2
, 2 , 2 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 2 , 2 , 2 , 0 , 0
, 0 , 0 , 3 , 3 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 4 , 4 , 3 , 3
, 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0
) , nrow = 5
)
## this way gives same random values for
## each integer between 0 and
## max (mymat):
milton - function ( x , min = 0.4 , max = 0.7 ) {
.rep - runif ( n = max ( x ) , min = min , max = max )
for ( i in 1:max ( x ) ) {
x[x == i] - .rep[i]
}
x
}
milton ( x = mymat )
## this way gives different random values
## for each integer between
## 0 and max (mymat):
milton - function ( x , min = 0.4 , max = 0.7 ) {
for ( i in 1:max ( x ) ) {
x [ x == i ] - runif (
sum ( x == i )
, min = min
, max = max
)
}
x
}
milton ( x = mymat )
--
David
-
David Huffer, Ph.D. Senior Statistician
CSOSA/Washington, DC david.huf...@csosa.gov
-
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of milton ruser
Sent: Wednesday, August 26, 2009 1:18 PM
To: David Winsemius
Cc: r-help@r-project.org
Subject: Re: [R] changing equal values on matrix by same random number
Hi David,
Thanks for the reply. This is what I need:
mymat[mymat==1] - runif(1,min=0.4,max=0.7)
mymat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,] 0.457316100200000 3 0
[2,] 0.457316100202000 0 0
[3,] 0.457316100202004 0 0
[4,] 0.00000002304 0 0
[5,] 0.00000000303 0 0
But as my real landscapes have values from 1 to large number (~10,),
so I think that if I put this on a for() looping it will be very time
expensive,
and as I have a lot of landscapes, I need to speed up it.
Any suggestion?
bests
milton
On Wed, Aug 26, 2009 at 1:12 PM, David Winsemius dwinsem...@comcast.net
wrote:
On Aug 26, 2009, at 12:53 PM, milton ruser wrote:
Dear all,
I have about 30,000 matrix (512x512), with values from 1 to N.
Each value on a matrix represent a habitat patch on my
matrix (i.e. my landscape). Non-habitat are stored as ZERO.
No I need to change each 1-to-N values for the same random
number.
Just supose my matrix is:
mymat-matrix(c(1,1,1,0,0,0,0,0,0,0,0,
0,0,0,0,2,2,2,0,0,0,0,
0,0,0,0,2,2,2,0,0,0,0,
3,3,0,0,0,0,0,0,0,4,4,
3,3,0,0,0,0,0,0,0,0,0), nrow=5)
I would like that all cells with 1 come to be
runif(1,min=0.4, max=0.7), and cells with 2
be replace by another runif(...).
First the wrong way and then the right way:
mymat[mymat==1] - runif(1,min=0.4,max=0.7)
mymat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,] 0.457316100200000 3 0
[2,] 0.457316100202000 0 0
[3,] 0.457316100202004 0 0
[4,] 0.00000002304 0 0
[5,] 0.00000000303 0 0
All the values are the same, clearly not what was desired.
Put it back to your starting point:
mymat-matrix(c(1,1,1,0,0,0,0,0,0,0,0,
+ 0,0,0,0,2,2,2,0,0,0,0,
+ 0,0,0,0,2,2,2,0,0,0,0,
+ 3,3,0,0,0,0,0,0,0,4,4,
+ 3,3,0,0,0,0,0,0,0,0,0), nrow=5)
# So supply the proper number of random realizations:
mymat[mymat==1] - runif(sum(mymat==1),min=0.4,max=0.7)
mymat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,] 0.574566500200000 3 0
[2,] 0.695641800202000 0 0
[3,] 0.693546600202004 0 0
[4,] 0.00000002304 0 0
[5,] 0.00000000303 0 0
If you want to supply a matrix of max and min values for the other
integers
there would probably be an *apply approach that could be used.
I can do it using for(), but it is very time expensive.
Any help are welcome.
cheers
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Re: [R] rJava error for large XML object return in StatET plugin
Hi, On Aug 26, 2009, at 1:31 PM, Harsh wrote: Hi R List, I get this error using StatET R plugin in Eclipse. You might have more luck asking the StatET user mailing list: https://lists.r-forge.r-project.org/cgi-bin/mailman/listinfo/statet-user -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org 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] specify a model in differential equations (nlme)
Dear all, I wonder if there is a way to specify a model in differential equations for nlme(). Or in other packages? Appreciate any comment. Thanks. Jun Shen [[alternative HTML version deleted]] __ R-help@r-project.org 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] ggplot2: geom_smooth and legend
Hi Benoit, You could turn the standard errors off with se = F. Then they'll be removed from the legend as well. Hadley On Tue, Aug 18, 2009 at 7:43 AM, Benoit Boulinguiezbenoit.boulingu...@ensc-rennes.fr wrote: Sorry I forgot the code that goes with **CODE desorb_plot-ggplot() + geom_smooth(data=DATA.B1_SA_N2, aes(Temp,DrTGA*100,colour=B1),span=0.1,size=1.6) + geom_smooth(data=DATA.FM30K_SA_N2, aes(Temp,DrTGA*100,colour=FM30K),span=0.2,size=1.6) + geom_smooth(data=DATA.NC60_SA_N2, aes(Temp,-DrTGA*100,colour=NC60),span=0.1,size=1.6) + geom_smooth(data=DATA.THC515_SA_N2, aes(Temp,DrTGA*100,colour=THC515),span=0.2,size=1.6) + scale_colour_hue(name=Adsorbent) + labs(x=Temp~(degree*C),y=Weight~Derivative~(%/*degree*C)) + opts(panel.grid.minor = theme_line(colour = grey94)) print(desorb_plot) Cordialement / Regards --- Benoit Boulinguiez Ecole de Chimie de Rennes (ENSCR) Bureau 1.20 Equipe CIP UMR CNRS 6226 Sciences Chimiques de Rennes Avenue du Général Leclerc CS 50837 35708 Rennes CEDEX 7 Tel 33 (0)2 23 23 80 83 Fax 33 (0)2 23 23 81 20 http://www.ensc-rennes.fr/ Quoting Benoit Boulinguiez benoit.boulingu...@ensc-rennes.fr: Hi all, Is that possible to remove the grey colour in the legend key that goes with the geom_smooth? In my case it doesn't ease the reading of the legend. http://www.4shared.com/file/125864977/e10644f8/desorb.html Cordialement / Regards --- Benoit Boulinguiez Ecole de Chimie de Rennes (ENSCR) Bureau 1.20 Equipe CIP UMR CNRS 6226 Sciences Chimiques de Rennes Avenue du Général Leclerc CS 50837 35708 Rennes CEDEX 7 Tel 33 (0)2 23 23 80 83 Fax 33 (0)2 23 23 81 20 http://www.ensc-rennes.fr/ __ R-help@r-project.org 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. -- http://had.co.nz/ __ R-help@r-project.org 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] changing equal values on matrix by same random number
Hi David all,
It is me again. When I try with a sample matrix (10x10) it appears to be
good.
But when I apply to a real 512x512 landscape, with values ranging from 1 to
10,000
It is still time expensive (please, see below):
mymat - matrix (
c (
1 , 1 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 2
, 2 , 2 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 2 , 2 , 2 , 0 , 0
, 0 , 0 , 3 , 3 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 4 , 4 , 3 , 3
, 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0
) , nrow = 5, byrow=T
)
## this way gives same random values for
## each integer between 0 and
## max (mymat):
milton - function ( x , min = 0.4 , max = 0.7 ) {
.rep - runif ( n = max ( x ) , min = min , max = max )
for ( i in 1:max ( x ) ) {
x[x == i] - .rep[i]
}
x
}
milton ( x = mymat )
mymat -matrix(sample(0:1,size=512*512, replace=T), ncol=512)
milton ( x = mymat )
May be using apply series it could be more efficient? I think I need to
avoid for() looping to speed up it.
cheers
milton
On Wed, Aug 26, 2009 at 1:30 PM, David Huffer david.huf...@csosa.govwrote:
You could try either of the two examples below. They both assume that min
and max are invariant:
mymat - matrix (
c (
1 , 1 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 2
, 2 , 2 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 2 , 2 , 2 , 0 , 0
, 0 , 0 , 3 , 3 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 4 , 4 , 3 , 3
, 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0
) , nrow = 5
)
## this way gives same random values for
## each integer between 0 and
## max (mymat):
milton - function ( x , min = 0.4 , max = 0.7 ) {
.rep - runif ( n = max ( x ) , min = min , max = max )
for ( i in 1:max ( x ) ) {
x[x == i] - .rep[i]
}
x
}
milton ( x = mymat )
## this way gives different random values
## for each integer between
## 0 and max (mymat):
milton - function ( x , min = 0.4 , max = 0.7 ) {
for ( i in 1:max ( x ) ) {
x [ x == i ] - runif (
sum ( x == i )
, min = min
, max = max
)
}
x
}
milton ( x = mymat )
--
David
-
David Huffer, Ph.D. Senior Statistician
CSOSA/Washington, DC david.huf...@csosa.gov
-
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of milton ruser
Sent: Wednesday, August 26, 2009 1:18 PM
To: David Winsemius
Cc: r-help@r-project.org
Subject: Re: [R] changing equal values on matrix by same random number
Hi David,
Thanks for the reply. This is what I need:
mymat[mymat==1] - runif(1,min=0.4,max=0.7)
mymat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,] 0.457316100200000 3 0
[2,] 0.457316100202000 0 0
[3,] 0.457316100202004 0 0
[4,] 0.00000002304 0 0
[5,] 0.00000000303 0 0
But as my real landscapes have values from 1 to large number (~10,),
so I think that if I put this on a for() looping it will be very time
expensive,
and as I have a lot of landscapes, I need to speed up it.
Any suggestion?
bests
milton
On Wed, Aug 26, 2009 at 1:12 PM, David Winsemius dwinsem...@comcast.net
wrote:
On Aug 26, 2009, at 12:53 PM, milton ruser wrote:
Dear all,
I have about 30,000 matrix (512x512), with values from 1 to N.
Each value on a matrix represent a habitat patch on my
matrix (i.e. my landscape). Non-habitat are stored as ZERO.
No I need to change each 1-to-N values for the same random
number.
Just supose my matrix is:
mymat-matrix(c(1,1,1,0,0,0,0,0,0,0,0,
0,0,0,0,2,2,2,0,0,0,0,
0,0,0,0,2,2,2,0,0,0,0,
3,3,0,0,0,0,0,0,0,4,4,
3,3,0,0,0,0,0,0,0,0,0), nrow=5)
I would like that all cells with 1 come to be
runif(1,min=0.4, max=0.7), and cells with 2
be replace by another runif(...).
First the wrong way and then the right way:
mymat[mymat==1] - runif(1,min=0.4,max=0.7)
mymat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,] 0.457316100200000 3 0
[2,] 0.457316100202000 0 0
[3,] 0.457316100202004 0 0
[4,] 0.00000002304 0 0
[5,] 0.00000000303 0 0
All the values are the same, clearly not what was desired.
Put it back to your starting point:
mymat-matrix(c(1,1,1,0,0,0,0,0,0,0,0,
+ 0,0,0,0,2,2,2,0,0,0,0,
+ 0,0,0,0,2,2,2,0,0,0,0,
+ 3,3,0,0,0,0,0,0,0,4,4,
+ 3,3,0,0,0,0,0,0,0,0,0), nrow=5)
# So supply the proper number of random realizations:
mymat[mymat==1] - runif(sum(mymat==1),min=0.4,max=0.7)
mymat
[,1] [,2] [,3] [,4] [,5]
Re: [R] specify a model in differential equations (nlme)
For nlme, no. However, take a look at the CRAN Task View for pharmokinetics and packages recommended there, especially nlmeODE . You might also try R's search capabilities: RSiteSearch(differential equations) ?RSiteSearch of other non-R search engines Bert Gunter Genentech Nonclinical Biostatisics -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jun Shen Sent: Wednesday, August 26, 2009 10:45 AM To: r-help@r-project.org Subject: [R] specify a model in differential equations (nlme) Dear all, I wonder if there is a way to specify a model in differential equations for nlme(). Or in other packages? Appreciate any comment. Thanks. Jun Shen [[alternative HTML version deleted]] __ R-help@r-project.org 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@r-project.org 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] Trying to make Nas 0
I have an lm object called mro A summary gives summary(mro) Call: lm(formula = REGRESSIONSTRING, data = wData) Residuals: Min 1Q Median 3Q Max -8.18077 -1.06867 -0.09387 1.03153 11.20201 Coefficients: (1 not defined because of singularities) Estimate Std. Error t value Pr(|t|) (Intercept)7.2096 1.0345 6.969 5.37e-11 *** log(H2403_P1) -0.3113 0.1305 -2.386 0.0180 * I(LEVDSCRT/100)3.4425 0.7818 4.403 1.79e-05 *** REG_UTIL_FLAG_P1 NA NA NA NA REALESTATE_ROI 0.2413 0.6211 0.389 0.6980 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 2.116 on 186 degrees of freedom Multiple R-squared: 0.1455, Adjusted R-squared: 0.1317 F-statistic: 10.56 on 3 and 186 DF, p-value: 1.917e-06 Now, I know my REG_UTIL_FLAG_P1 var is NA because it is missing/constant in the data frame. What I am trying to do is this tring(summary(mro)$coef[,3]) [1] 6.96900610837729, -2.38577388980627, 4.403441093625, 0.388556365853674 I'd really like the NA to not be omitted, but cast to 0 (I am going to load this into a DB) So I'd like to see [1] 6.96900610837729, -2.38577388980627, 4.403441093625, 0, 0.388556365853674 I was able to do that with the coefficients via mro$coef[is.na(mro$coef)]-0 mro$coef I can do this in a brute force way by by getting a vector of aliased columns and iterating thru it -- but was hoping for a more elegant solution. Thx! === Please access the attached hyperlink for an important electronic communications disclaimer: http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html __ R-help@r-project.org 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] Managing output
Hi,
Is there a way to build up a vector, item by item. In perl, we can
push an item onto an array. How can we can do this in R?
I have a loop that generates values as it goes. I want to end up with a
vector of all the loop results.
In perl it woud be:
for(item in list){
result - 2*item^2 (Or whatever formula, this is just a pseudo example)
Push(@result_list, result) (This is the step I can't do in R)
}
Thanks!
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Re: [R] Managing output
How about ?append, but R is vectorized, so why not just
result_list - 2*item^2 , or for more complicated tasks, the
apply/sapply/lapply/mapply family of functions?
In general, the for loop construct can be avoided so you don't have to think
about messy indexing. What exactly are you trying to do?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Noah Silverman
Sent: Wednesday, August 26, 2009 2:20 PM
To: r help
Subject: [R] Managing output
Hi,
Is there a way to build up a vector, item by item. In perl, we can
push an item onto an array. How can we can do this in R?
I have a loop that generates values as it goes. I want to end up with a
vector of all the loop results.
In perl it woud be:
for(item in list){
result - 2*item^2 (Or whatever formula, this is just a pseudo example)
Push(@result_list, result) (This is the step I can't do in R)
}
Thanks!
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Managing output
The actually process is REALLY complicate, I just gave a simple example
for the list.
I have a lot of steps to process the data before I get a final
score. (nested loops, conditional statements, etc.)
Right now, I'm just printing the scores to the screen. I'd like to
accumulate them in some kind of data structure so I can either write
them to disk or graph them.
-N
On 8/26/09 12:27 PM, Erik Iverson wrote:
How about ?append, but R is vectorized, so why not just
result_list- 2*item^2 , or for more complicated tasks, the
apply/sapply/lapply/mapply family of functions?
In general, the for loop construct can be avoided so you don't have to
think about messy indexing. What exactly are you trying to do?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Noah Silverman
Sent: Wednesday, August 26, 2009 2:20 PM
To: r help
Subject: [R] Managing output
Hi,
Is there a way to build up a vector, item by item. In perl, we can
push an item onto an array. How can we can do this in R?
I have a loop that generates values as it goes. I want to end up with a
vector of all the loop results.
In perl it woud be:
for(item in list){
result- 2*item^2 (Or whatever formula, this is just a pseudo example)
Push(@result_list, result) (This is the step I can't do in R)
}
Thanks!
__
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and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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Re: [R] changing equal values on matrix by same random number
Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of milton ruser
Sent: Wednesday, August 26, 2009 9:54 AM
To: r-help@r-project.org
Subject: [R] changing equal values on matrix by same random number
Dear all,
I have about 30,000 matrix (512x512), with values from 1 to N.
Each value on a matrix represent a habitat patch on my
matrix (i.e. my landscape). Non-habitat are stored as ZERO.
No I need to change each 1-to-N values for the same random
number.
Just supose my matrix is:
mymat-matrix(c(1,1,1,0,0,0,0,0,0,0,0,
0,0,0,0,2,2,2,0,0,0,0,
0,0,0,0,2,2,2,0,0,0,0,
3,3,0,0,0,0,0,0,0,4,4,
3,3,0,0,0,0,0,0,0,0,0), nrow=5)
I would like that all cells with 1 come to be
runif(1,min=0.4, max=0.7), and cells with 2
be replace by another runif(...).
I can do it using for(), but it is very time expensive.
Any help are welcome.
Is the following what you want? It uses the small
integers in mymat as indices into a vector of N random
numbers.
mymat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,]100200000 3 0
[2,]100202000 0 0
[3,]100202004 0 0
[4,]000002304 0 0
[5,]000000303 0 0
range(unique(mymat))
[1] 0 4
f-function(mat){
N-max(mat)
tmp - mat[mat0]
tmp - runif(N, 0.4, 0.7)[tmp]
mat[mat0] - tmp
mat
}
f(mymat)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[,9]
[1,] 0.536142700 0.67274910 0.000 0.0000
0.000
[2,] 0.536142700 0.67274910 0.6727491 0.0000
0.000
[3,] 0.536142700 0.67274910 0.6727491 0.0000
0.5357566
[4,] 0.00000 0.0000 0.6727491 0.64178320
0.5357566
[5,] 0.00000 0.0000 0.000 0.64178320
0.6417832
[,10] [,11]
[1,] 0.6417832 0
[2,] 0.000 0
[3,] 0.000 0
[4,] 0.000 0
[5,] 0.000 0
The 3 lines involving tmp could be collapsed into one, making
things more obscure
mat[mat0] - runif(N, 0.4, 0.7)[mat[mat0]]
Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com
cheers
milton
[[alternative HTML version deleted]]
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Re: [R] Managing output
On 26/08/2009 3:20 PM, Noah Silverman wrote:
Hi,
Is there a way to build up a vector, item by item. In perl, we can
push an item onto an array. How can we can do this in R?
I have a loop that generates values as it goes. I want to end up with a
vector of all the loop results.
In perl it woud be:
for(item in list){
result - 2*item^2 (Or whatever formula, this is just a pseudo example)
Push(@result_list, result) (This is the step I can't do in R)
}
Thanks!
Use the c() function, e.g.
result - numeric(0)
for (item in list) {
result - 2*item^2
result_list - c(result_list, result)
}
If the list is long, this is an extremely inefficient style of R
programming, but for a short list it should be fine.
Duncan Murdoch
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[R] contourLines() documentation
Hello, I have searched for documentation on the function contourLines's algorithm but cannot find a thing. I am about to submit a paper to a journal but cannot yet do so because I need to provide some reference for this function. Does anyone know what algorithm is used for this function? Thanks, Derek Lacoursiere -- View this message in context: http://www.nabble.com/contourLines%28%29-documentation-tp25151467p25151467.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Managing output
On 27/08/2009, at 8:00 AM, Duncan Murdoch wrote:
On 26/08/2009 3:20 PM, Noah Silverman wrote:
Hi,
Is there a way to build up a vector, item by item. In perl, we can
push an item onto an array. How can we can do this in R?
I have a loop that generates values as it goes. I want to end up
with a
vector of all the loop results.
In perl it woud be:
for(item in list){
result - 2*item^2 (Or whatever formula, this is just a
pseudo example)
Push(@result_list, result) (This is the step I can't do in R)
}
Thanks!
Use the c() function, e.g.
result - numeric(0)
That should be result_list - numeric(0)!!! :-)
cheers,
Rolf
for (item in list) {
result - 2*item^2
result_list - c(result_list, result)
}
If the list is long, this is an extremely inefficient style of R
programming, but for a short list it should be fine.
Duncan Murdoch
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guide.html
and provide commented, minimal, self-contained, reproducible code.
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[R] Plotting to stdout
Hello, is there a way to write the result of a plot to stdout? I mean even a binary thingy like a png-file, written to stdout?! I tried with the file-argument of png() and jpeg(), but did not get working results. Ciao, Oliver __ R-help@r-project.org 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.
