Re: [R] How to open a file with a name changed?
On May 25, 2012, at 02:38 , Rolf Turner wrote: > ?paste or ?sprintf, which might lead to more readable code. Also > format(today, "C:/Bonos/%Y%m%d.csv") [1] "C:/Bonos/20120525.csv" > On 25/05/12 11:45, Minerva Mora wrote: >> Hi, I apologize for my english. >> >> I´m trying to read a file, but the name of this file changes every day, for >> example: today is May 24, 2012 >> >> bonos<- read.table("C:/Bonos/*20120524*.csv", header=TRUE, sep="\t") >> >> So, tomorrow I want to read the file again, but i don´t want to put the date >> by myself, i want this automatically. I know that if a put day() this >> instruccion gives me the date. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.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] R does not recognise columns and rows as they are supposed to be
Yes I exactly followed what you all suggested: X<-(82:92) ; Y<-(364:369) # for sellected region > extract <- double(365) > setwd("C:\\Users\\aalyaari\\Desktop\\New folder (10)\\") > listfile<-dir() > for (i in 1:365) { + conne <- file(listfile[i], "rb") + file1<- readBin(conne, double(), n=360*720) + file2<-matrix(data=file1,ncol=720,nrow=360) + extract[i]<-mean(file2[X,Y],na.rm=TRUE) + close(conne) } write.table(extract,"C:\\Users\\aalyaari\\Desktop\\New folder (10)\\samregion1.txt") But I am still getting(negative values) all values like: -3.75E+306 -1.30E+54 -1.22E+58 and the right ones should be like: 22.25 22.76 33.25 -- View this message in context: http://r.789695.n4.nabble.com/R-does-not-recognise-columns-and-rows-as-they-are-supposed-to-be-tp4631217p4631279.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] Filling NA with cumprod?
On Thu, May 24, 2012 at 08:24:38PM -0700, igorre25 wrote: > Hello, > > I need to build certain interpolation logic using R. Unfortunately, I just > started using R, and I'm not familiar with lots of advanced or just > convenient features of the language to make this simpler. So I struggled > for few days and pretty much reduced the whole exercise to the following > problem, which I cannot resolve: > > Assume we have a vector of some values with NA: > a <- c(1, 2, 3, NA, NA, 6, 7, NA, NA, 10) > > and some coefficients as a vector of the same length: > > f <- c(0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1) > > I need to come up with function to get the following output > > o[1] = a[1] > o[2] = a[2] > o[3] = a[3] > o[4] = o[3]*[f3] # Because a[3] is NA > o[5] = o[4]*[f4] # Because a[4] is NA; This looks like recursive > calculations; If the rest of the elements we NA, I would use a * c(rep(1, > 3), cumprod(f[3:9])), but that's not the case > o[6] = a[6] # Not NA anymore > o[7] = a[7] > o[8] = o[7]*f[7] # Again a[8] is NA > o[9] = o[8]*f[8] > o[10] = a[10] # Not NA > > Even though my explanation may seems complex, in reality the requirement is > pretty simple and in Excel is achieved with a very short formula. > > The need to use R is to demonstrate capabilities of the language and then to > expand to more complex problems. Hello: How is the output defined, if a[1] is NA? I think, you are not asking for a loop solution. However, in this case, it can be a reasonable option. For example a <- c(1, 2, 3, NA, NA, 6, 7, NA, NA, 10) f <- c(0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1) n <- length(a) o <- rep(NA, times=n) prev <- 1 for (i in 1:n) { if (is.na(a[i])) { o[i] <- f[i]*prev } else { o[i] <- a[i] } prev <- o[i] } A more straightforward translation of the Excel formulas is getCell <- function(i) { if (i == 0) return(1) if (is.na(a[i])) { return(f[i]*getCell(i-1)) } else { return(a[i]) } } x <- rep(NA, times=n) for (i in 1:n) { x[i] <- getCell(i) } identical(o, x) # [1] TRUE Petr Savicky. __ 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] package metafor: specify weights?
Dear Wolfgang, Thank you very much for the quick reply. I already assumed that it might get too complicated, so I will just stick to using a moderator (as you also suggested). Best, Anke On 24.05.2012 17:05, Viechtbauer Wolfgang (STAT) wrote: At the moment, there is no possibility of specifying the weights with the rma() function. While the main model fitting part could be easily adapted to incorporate user-specified weights, the problem comes in with all the additional statistics that can be computed based on a fitted model. How should the predict() function now work? What would be the definition of I^2 now? How would one generalize the influence and outlier statistics to that case? Just to give some examples. Of course, I could leave out such things when the user has specified the weights, but then things also get confusing for the user. For example, it is already less than ideal that you can only use the trim and fill method with models that do not incorporate moderators. Nobody has (as of yet) generalized the trim and fill method to that case. But when there are too many "special cases", the package becomes unusable. I did consider user-specified weights at one point, but it opened up so many cans of worms that I preferred to quickly put the lid pack on those cans. That item is written down in my to-do list, but to be honest, it is somewhere at the very end of that list. If you are hesitant to combine the results from those two types of studies, what about simply using a moderator to distinguish the two groups? Best, Wolfgang -- Wolfgang Viechtbauer, Ph.D., Statistician Department of Psychiatry and Psychology School for Mental Health and Neuroscience Faculty of Health, Medicine, and Life Sciences Maastricht University, P.O. Box 616 (VIJV1) 6200 MD Maastricht, The Netherlands +31 (43) 388-4170 | http://www.wvbauer.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Anke Stein Sent: Thursday, May 24, 2012 15:59 To: r-help@r-project.org Subject: [R] package metafor: specify weights? Dear R-experts, Dear Wolfgang, Weighted model fitting in metafor uses the inverse of the study specific variances as weights. I am wondering if it is possible to specify different weights. In my meta-analysis, there are two types of studies with (intrinsic) differences in their range of sample sizes (which are used to calculate the variances of Fisher's z). I would like to try normalizing the sample sizes within each set of the two study types and use these normalized sample sizes as weights. Would that be possible with rma()? So far, I only found the option "weighted = TRUE/FALSE", but no possibility to specify which weights should be used. Many thanks in advance, Anke -- __ Anke Stein (Dipl.-Biol.) Biodiversity, Macroecology& Conservation Biogeography Head Prof. Dr. Holger Kreft Georg-August University of Göttingen Büsgenweg 2 | 37077 Göttingen | Germany phone +49(0)551-39-13761 fax +49(0)551-39-3618 ast...@uni-goettingen.de http://www.uni-goettingen.de/biodiversity __ 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] Filling NA with cumprod?
Hello, I need to build certain interpolation logic using R. Unfortunately, I just started using R, and I'm not familiar with lots of advanced or just convenient features of the language to make this simpler. So I struggled for few days and pretty much reduced the whole exercise to the following problem, which I cannot resolve: Assume we have a vector of some values with NA: a <- c(1, 2, 3, NA, NA, 6, 7, NA, NA, 10) and some coefficients as a vector of the same length: f <- c(0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1) I need to come up with function to get the following output o[1] = a[1] o[2] = a[2] o[3] = a[3] o[4] = o[3]*[f3] # Because a[3] is NA o[5] = o[4]*[f4] # Because a[4] is NA; This looks like recursive calculations; If the rest of the elements we NA, I would use a * c(rep(1, 3), cumprod(f[3:9])), but that's not the case o[6] = a[6] # Not NA anymore o[7] = a[7] o[8] = o[7]*f[7] # Again a[8] is NA o[9] = o[8]*f[8] o[10] = a[10] # Not NA Even though my explanation may seems complex, in reality the requirement is pretty simple and in Excel is achieved with a very short formula. The need to use R is to demonstrate capabilities of the language and then to expand to more complex problems. Any idea? Thank you. -- View this message in context: http://r.789695.n4.nabble.com/Filling-NA-with-cumprod-tp4631269.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] How to open a file with a name changed?
?paste On 25/05/12 11:45, Minerva Mora wrote: Hi, I apologize for my english. I´m trying to read a file, but the name of this file changes every day, for example: today is May 24, 2012 bonos<- read.table("C:/Bonos/*20120524*.csv", header=TRUE, sep="\t") So, tomorrow I want to read the file again, but i don´t want to put the date by myself, i want this automatically. I know that if a put day() this instruccion gives me the date. My problem is, I don´t know how to concatenate these: -"C:/Bonos/ -the date -.csv" I tried this: bonos<- read.table(cat(cit,dia,extension), header=TRUE, sep="\t") where --cit is *"\"C:/Bonos/"* --dia is *dia<-format(today, "%Y%m%d")* --extension is *extension<-".csv\""* but i have this problem: "C:/Bonos/ 20120524 .csv"*Error in read.table(cat(cit, dia, extension), header = TRUE, sep = "\t") : * * 'file' must be a character string or connection* What can i do? Thanks! -- View this message in context: http://r.789695.n4.nabble.com/How-to-open-a-file-with-a-name-changed-tp4631261.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.
[R] levels of comma separated data
I have a data set that has some comma separated strings in each row. I'd like to create a vector consisting of all distinct strings that occur. The number of strings in each row may vary. Thanks for any help. __ 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] Manually modifying an hclust dendrogram to remove singletons
On Thu, May 24, 2012 at 9:31 AM, wrote: > Dear R-Help, > > I have a clustering problem with hclust that I hope someone can help > me with. Consider the classic hclust example: > > hc <- hclust(dist(USArrests), "ave") > plot(hc) > > I would like to cut the tree up in such a way so as to avoid small > clusters, so that we get a minimum number of items in each cluster, > and therefore avoid singletons. e.g. in this example, you can see that > Hawaii is split off onto its own at quite a high level. I would like > to avoid having a single item clustered on its own like this. How can > I achieve this? > > I have tried manually modifying the tree using dendrapply but have not > been able to produce a valid solution thus far.. > > Suggestions are welcome. > > Best wishes, > > Mark Hi Mark, I'm not sure how you want to handle the singletons if you don't want them in a separate cluster. The package WGCNA (I'm the maintainer) and its dependency dynamicTreeCut contain a few ways of avoiding singletons as separate clusters. One way is to remove them from the resulting clusters. To this end, use function cutreeStatic, specify the cut height and the minimum number of elements in the cluster. For example, clusters1 = cutreeStatic(hc, cutHeight = 35, minSize = 3); This way all branches that have size below 3 are labeled 0. To see what you get, use the function plotDendroAndColors like this: plotDendroAndColors(hc, clusters1, rowText = clusters1 ); Each color corresponds to a cluster, and the cluster label is shown by the numbers (each number is at the start of the corresponding cluster). If you'd like to assign everything but want to avoid cluster that are too small, use the dynamic tree cut approach (http://www.genetics.ucla.edu/labs/horvath/CoexpressionNetwork/BranchCutting/). For example: clusters2 = cutreeDynamic(hc, distM = as.matrix(dist(USArrests)), minClusterSize = 3, deepSplit = 2) To show the clusters: plotDendroAndColors(hc, clusters2, rowText = clusters2 ); If you think the clusters are too big, try setting deepSplit=3 in the cutreeDynamic call. The dynamic tree cut basically assigns all singletons and branches with size less than minClusterSize to the nearest existing cluster (notice Hawai and the Florida/North Carolina branch), thus basically combining hierarchical clustering and a PAM-like step Whether that's a good approach for your research goal is a question you need to answer. HTH, Peter > > __ > 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] Manually modifying an hclust dendrogram to remove singletons
Can't put my finger on it but something about your idea rubs me the wrong way. Maybe it's that the tree depends on the hierarchical clustering algorithm and the choice on how to trim it should be based on something more defensible than "avoid singletons". In this example Hawaii is really different than New Hampshire, why would you want them clustered together ? But, it's your work, field of study, whatever. If you are going to do it anyway, one way would be to loop over cut heights: hc <- hclust(dist(USArrests), "ave") plot(hc) hr <- range(hc$height) tol<- diff(hr)/100# set tolerance level for(i in seq(1e-4+hr[1],hr[2],tol)){ hcc <- rect.hclust(hc,h=i) if(all(sapply(hcc,length)>1)) break } str(hcc) # or if you prefer dendrogram dend1<- as.dendrogram(hc) for(i in seq(1e-4+hr[1],hr[2],tol)){ dend2 <- cut(dend1,h=i) if(all(sapply(dend2$lower,function(x) attr(x,'members'))>1)) break } dend2 Cheers On Thu, May 24, 2012 at 10:31 AM, wrote: > Dear R-Help, > > I have a clustering problem with hclust that I hope someone can help > me with. Consider the classic hclust example: > > hc <- hclust(dist(USArrests), "ave") > plot(hc) > > I would like to cut the tree up in such a way so as to avoid small > clusters, so that we get a minimum number of items in each cluster, > and therefore avoid singletons. e.g. in this example, you can see that > Hawaii is split off onto its own at quite a high level. I would like > to avoid having a single item clustered on its own like this. How can > I achieve this? > > I have tried manually modifying the tree using dendrapply but have not > been able to produce a valid solution thus far.. > > Suggestions are welcome. > > Best wishes, > > Mark > > __ > 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] How to open a file with a name changed?
On May 24, 2012, at 7:45 PM, Minerva Mora wrote: Hi, I apologize for my english. I´m trying to read a file, but the name of this file changes every day, for example: today is May 24, 2012 bonos<- read.table("C:/Bonos/*20120524*.csv", header=TRUE, sep="\t") So, tomorrow I want to read the file again, but i don´t want to put the date by myself, i want this automatically. I know that if a put day() this instruccion gives me the date. My problem is, I don´t know how to concatenate these: -"C:/Bonos/ -the date -.csv" or even : paste( gsub("-", "", Sys.Date() ) ) [1] "20120524" filname <- paste("C:/Bonos/", gsub("-", "", Sys.date() ) , sep="") -- David. I tried this: bonos<- read.table(cat(cit,dia,extension), header=TRUE, sep="\t") where --cit is *"\"C:/Bonos/"* --dia is *dia<-format(today, "%Y%m%d")* --extension is *extension<-".csv\""* but i have this problem: "C:/Bonos/ 20120524 .csv"*Error in read.table(cat(cit, dia, extension), header = TRUE, sep = "\t") : * * 'file' must be a character string or connection* What can i do? Thanks! -- View this message in context: http://r.789695.n4.nabble.com/How-to-open-a-file-with-a-name-changed-tp4631261.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. David Winsemius, MD 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] How to open a file with a name changed?
On May 24, 2012, at 7:45 PM, Minerva Mora wrote: Hi, I apologize for my english. I´m trying to read a file, but the name of this file changes every day, for example: today is May 24, 2012 bonos<- read.table(*20120524*.csv", header=TRUE, sep="\t") yr = "2012" mo = "01 dt = "25" filnam =paste("C:/Bonos/" , yr, mo, dt,sep="") bonos <- read.table(filnam, header=TRUE, sep="\t") So, tomorrow I want to read the file again, but i don´t want to put the date by myself, i want this automatically. I know that if a put day() this instruccion gives me the date. My problem is, I don´t know how to concatenate these: -"C:/Bonos/ -the date -.csv" I tried this: bonos<- read.table(cat(cit,dia,extension), header=TRUE, sep="\t") where --cit is *"\"C:/Bonos/"* --dia is *dia<-format(today, "%Y%m%d")* --extension is *extension<-".csv\""* but i have this problem: "C:/Bonos/ 20120524 .csv"*Error in read.table(cat(cit, dia, extension), header = TRUE, sep = "\t") : * * 'file' must be a character string or connection* What can i do? Thanks! -- View this message in context: http://r.789695.n4.nabble.com/How-to-open-a-file-with-a-name-changed-tp4631261.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. David Winsemius, MD 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] R Error: System is computationally singular
Thank you both, I will try using a zero inflated negative binomial as suggested. I had success with negative binomial on previous runs but only when I had fewer covariates and only ran a portion (10%) of the data. I may also try to reduce the number of covariates in the model (i.e., combine some of my landcover [LCOVER] classifications). I have considered using logistic regression and may end up trying that. I appreciate both of your input and will let you know (i.e., post) the results of these suggestions and what ends up working for the benefit of you and others. Thank you for your time and quick responses!! Nate Nathan Svoboda Graduate Research Assistant Carnivore Ecology Lab Mississippi State University From: Marc Schwartz [mailto:marc_schwa...@me.com] Sent: Thu 5/24/2012 3:09 PM To: Nathan Svoboda Cc: David Winsemius; r-help@r-project.org Subject: Re: [R] R Error: System is computationally singular Nathan, This does help, as in the first cut you provided, there was no variability in LOCS for LCOVER >= 5 and you have very few values of LOCS > 0 (you still do, relative to the scale of the total). Have you tried using a zero inflated negative binomial model (dist = "negbin") rather than poisson? I am not sure that the assumption of a zero inflated poisson distribution is reasonable with your data. Also, at least in this cut of the data, you have no 4's in LOCS and no 8's in LCOVER (same as before). If my math is correct only 0.006% of your LOCS values are > 0. I am also not convinced that you have enough data to differentiate between 1 and >=1 of whatever it is you are counting in LOCS. If that is the case, you might want to consider using logistic regression with a dichotomous response variable of LOCS == 0 versus LOCS >= 1. You seem to be in the general realm of very rare events given the distribution of LOCS in your data. Regards, Marc Schwartz On May 24, 2012, at 2:41 PM, Nathan Svoboda wrote: > Hi David, > > My apologies, I am not sure if this makes a big difference in your assessment > of the problem, but the results I just sent were only from a portion (1/15) > of the data. The dataset is rather large and the computer I am currently > using to set up the models is limited in its capabilities to analyze large > datasets. When I run the code you provided on a larger portion of the data > (1/2) this is the output I receive: > > LCOVER > LOCS 1 2 3 4 5 6 7 9 > 0 1692196 630659 550623 6140352 180896 255512 785929 63756 > 1 141 30 48 279 9 14 36 1 > 2 17 4 5 14 3 3 4 1 > 3 0 0 0 3 0 0 1 0 > 5 2 0 0 0 0 0 0 0 > > Thanks again for your time and assistance, > > Nate > > Nathan Svoboda > Graduate Research Assistant > Mississippi State University > > > On May 24, 2012, at 1:57 PM, Nathan Svoboda wrote: > >> Greetings, >> >> I am trying to fit a zero-inflated Poisson model using zeroinfl() >> from the >> pscl library. I have 5 covariates (4 continuous, 1 categorical); the >> categorical variable has 7 levels. I have had success fitting >> models that >> contain only the continuous covariates; however, when I add the >> categorical >> variable to any of the models (or if I run it by itself) I get the >> following >> error: >> >> Error in solve.default(as.matrix(fit$hessian)) : >> >> system is computationally singular: reciprocal condition number = >> 3.46934e-20 >> >> The code I am using is: >> >> library(pscl) >> f1 <- formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE + >> D_GRASS) >> ZIP1 <- zeroinfl(f1, dist="poisson", link = "logit", data = FAWNS) >> >> There is no correlation between my covariates. Also, I tried >> reducing my >> categorical covariate to 3 levels and still receive the same error. >> Can >> anyone suggest why I may be getting this error when I add the >> categorical >> covariate? >> > > What does this show: > > with( FAWNS, table(LOCS, LCOVER) ) > > -- > David Winsemius, MD > West Hartford, CT [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to open a file with a name changed?
You probably want to use paste or paste0 instead of cat. paste0(cit,dia,extension) Also, this might work: tail(dir("C:/Bonos/",patt="csv$"),1) -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Minerva Mora Sent: Friday, 25 May 2012 11:45a To: r-help@r-project.org Subject: [R] How to open a file with a name changed? Hi, I apologize for my english. I´m trying to read a file, but the name of this file changes every day, for example: today is May 24, 2012 bonos<- read.table("C:/Bonos/*20120524*.csv", header=TRUE, sep="\t") So, tomorrow I want to read the file again, but i don´t want to put the date by myself, i want this automatically. I know that if a put day() this instruccion gives me the date. My problem is, I don´t know how to concatenate these: -"C:/Bonos/ -the date -.csv" I tried this: bonos<- read.table(cat(cit,dia,extension), header=TRUE, sep="\t") where --cit is *"\"C:/Bonos/"* --dia is *dia<-format(today, "%Y%m%d")* --extension is *extension<-".csv\""* but i have this problem: "C:/Bonos/ 20120524 .csv"*Error in read.table(cat(cit, dia, extension), header = TRUE, sep = "\t") : * * 'file' must be a character string or connection* What can i do? Thanks! -- View this message in context: http://r.789695.n4.nabble.com/How-to-open-a-file-with-a-name-changed-tp4631261.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] applying cbind (or any function) across all components in a list
On May 24, 2012, at 6:37 PM, Hans Thompson wrote: The function I am giving for context is cbind. Are you asking how I would like to apply the answer to my question? I am trying to take the results of a Fluidigm SNP microarray, organized by assay into a list (each component is the results of one assay), find coordinate midpoints ([1,] and [2,] of my XX, XY, and YY clusters (these are genotypes) and is represented by l1. l2 is the midpoint between XX/ XY and XY/YY although I did not give this in my example for simplicity, and I am now trying to find the midpoint between these new midpoints and their closest genotype clusters. This is represented as cbind( l1[[1]][,1]+l2[[1]][,1])/2, (l1[[1]][,2]+l2[[1]][,1])/2, (l1[[1]][,2]+l2[[1]][,2])/2, (l1[[1]][,3]+l2[[1]][,2])/2 ) cbind would not be much different than c() in this case. but only works for one assay in the list of 96. It's not really a function, but rather an expression that will be evaluated. If you had a function it might look like : bindfn <- function(l1, l1) { c( l1[[1]][,1]+l2[[1]][,1])/2, (l1[[1]][, 2]+l2[[1]][,1])/2, (l1[[1]][,2]+l2[[1]][,2])/2, (l1[[1]][,3]+l2[[1]][,2])/2 ) } And you could perhaps (depending on whether l1 and l2 were constructed) properly execute: do.call(bindfn, list(l1,l2)) perhaps: mapply(bindfn, l1, l2) You posted an example higher up in the thread but my wife is calling me to dinner. Will return later. I want to apply this to the entire list. My entire code so far is: ## OPEN .CSV and ORGANIZE BY ASSAY file="" { rawdata <- read.csv(file, skip = 15) OrgAssay <- split(rawdata, rawdata$Assay) ## RETURN MIDPOINTS FOR EACH CLUSTER WITHOUT NO CALLS #for loop ClustMidPts <-list() for(locus in 1:length(names(OrgAssay))){ ClustMidPts[[locus]]<-t(cbind(tapply(OrgAssay[[locus]][,"Allele.X. 1"], OrgAssay[[locus]][,"Final"], mean,na.rm=T), tapply(OrgAssay[[locus]][,"Allele.Y.1"], OrgAssay[[locus]][,"Final"], mean,na.rm=T)))} names(ClustMidPts)=names(OrgAssay) ## CREATE CLUSTER-CLUSTER MIDPOINT #for loop ClustClustMidPts <- list() for(locus in 1:length(names(ClustMidPts))){ ClustClustMidPts[[locus]] <- cbind(XXYX=(ClustMidPts[[locus]][,"XX"]+ClustMidPts[[locus]][,"YX"])/ 2, YXYY=(ClustMidPts[[locus]][,"YX"]+ClustMidPts[[locus]][,"YY"])/2) } names(ClustClustMidPts)=names(ClustMidPts) Please also let me know how I messed up the formatting because it shows up fine in gmail even when I post on Nabble. How did I assume you were using Nabble? Because you included no context. We cannot see your earlier postings. Expecting us to page out to Nabble or refer back to other item s in the thread is considered presumptuous on your part. Is this topic included in the posting guide? YES. It is. And you were asked to read the Posting Guide about 10 times by now. -- View this message in context: http://r.789695.n4.nabble.com/applying-cbind-or-any-function-across-all-components-in-a-list-tp4631128p4631260.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. David Winsemius, MD 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] R Error: System is computationally singular
On May 24, 2012, at 3:41 PM, Nathan Svoboda wrote: Re: [R] R Error: System is computationally singular Hi David, My apologies, I am not sure if this makes a big difference in your assessment of the problem, but the results I just sent were only from a portion (1/15) of the data. The dataset is rather large and the computer I am currently using to set up the models is limited in its capabilities to analyze large datasets. When I run the code you provided on a larger portion of the data (1/2) this is the output I receive: LCOVER LOCS 1 2 3 4 5 6 7 9 0 1692196 630659 550623 6140352 180896 255512 785929 63756 1 141 30 48 279 9 14 36 1 2 17 4 5 14 3 3 4 1 3 0 0 0 3 0 0 1 0 5 2 0 0 0 0 0 0 0 I do not see linear dependence (aka computational singularity) in that data, but if there are no LOCS values of 4, an missing levels has been reported as a show-stopper with zinf models with pscl in the past. There could also easily emerge linear dependence if you tabulated the entire data set. If level 4 had 3 at level 4 of LCOVER or 1 at level 7 then there would be linear dependence. Marc Schwartz, a smarter guy than I, has already suggested to you that your Poisson error structure might not be a good description of the data. Thanks again for your time and assistance, Nate Nathan Svoboda Graduate Research Assistant Mississippi State University From: David Winsemius [mailto:dwinsem...@comcast.net] Sent: Thu 5/24/2012 1:54 PM To: Nathan Svoboda Cc: r-help@r-project.org Subject: Re: [R] R Error: System is computationally singular On May 24, 2012, at 1:57 PM, Nathan Svoboda wrote: > Greetings, > > I am trying to fit a zero-inflated Poisson model using zeroinfl() > from the > pscl library. I have 5 covariates (4 continuous, 1 categorical); the > categorical variable has 7 levels. I have had success fitting > models that > contain only the continuous covariates; however, when I add the > categorical > variable to any of the models (or if I run it by itself) I get the > following > error: > > Error in solve.default(as.matrix(fit$hessian)) : > > system is computationally singular: reciprocal condition number = > 3.46934e-20 > > The code I am using is: > > library(pscl) > f1 <- formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE + > D_GRASS) > ZIP1 <- zeroinfl(f1, dist="poisson", link = "logit", data = FAWNS) > > There is no correlation between my covariates. Also, I tried > reducing my > categorical covariate to 3 levels and still receive the same error. > Can > anyone suggest why I may be getting this error when I add the > categorical > covariate? > What does this show: with( FAWNS, table(LOCS, LCOVER) ) -- David Winsemius, MD West Hartford, CT David Winsemius, MD 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.
[R] How to open a file with a name changed?
Hi, I apologize for my english. I´m trying to read a file, but the name of this file changes every day, for example: today is May 24, 2012 bonos<- read.table("C:/Bonos/*20120524*.csv", header=TRUE, sep="\t") So, tomorrow I want to read the file again, but i don´t want to put the date by myself, i want this automatically. I know that if a put day() this instruccion gives me the date. My problem is, I don´t know how to concatenate these: -"C:/Bonos/ -the date -.csv" I tried this: bonos<- read.table(cat(cit,dia,extension), header=TRUE, sep="\t") where --cit is *"\"C:/Bonos/"* --dia is *dia<-format(today, "%Y%m%d")* --extension is *extension<-".csv\""* but i have this problem: "C:/Bonos/ 20120524 .csv"*Error in read.table(cat(cit, dia, extension), header = TRUE, sep = "\t") : * * 'file' must be a character string or connection* What can i do? Thanks! -- View this message in context: http://r.789695.n4.nabble.com/How-to-open-a-file-with-a-name-changed-tp4631261.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] applying cbind (or any function) across all components in a list
The function I am giving for context is cbind. Are you asking how I would like to apply the answer to my question? I am trying to take the results of a Fluidigm SNP microarray, organized by assay into a list (each component is the results of one assay), find coordinate midpoints ([1,] and [2,] of my XX, XY, and YY clusters (these are genotypes) and is represented by l1. l2 is the midpoint between XX/XY and XY/YY although I did not give this in my example for simplicity, and I am now trying to find the midpoint between these new midpoints and their closest genotype clusters. This is represented as cbind((l1[[1]][,1]+l2[[1]][,1])/2, (l1[[1]][,2]+l2[[1]][,1])/2, (l1[[1]][,2]+l2[[1]][,2])/2, (l1[[1]][,3]+l2[[1]][,2])/2) but only works for one assay in the list of 96. I want to apply this to the entire list. My entire code so far is: ## OPEN .CSV and ORGANIZE BY ASSAY file="" { rawdata <- read.csv(file, skip = 15) OrgAssay <- split(rawdata, rawdata$Assay) ## RETURN MIDPOINTS FOR EACH CLUSTER WITHOUT NO CALLS #for loop ClustMidPts <-list() for(locus in 1:length(names(OrgAssay))){ ClustMidPts[[locus]]<-t(cbind(tapply(OrgAssay[[locus]][,"Allele.X.1"], OrgAssay[[locus]][,"Final"], mean,na.rm=T), tapply(OrgAssay[[locus]][,"Allele.Y.1"], OrgAssay[[locus]][,"Final"], mean,na.rm=T)))} names(ClustMidPts)=names(OrgAssay) ## CREATE CLUSTER-CLUSTER MIDPOINT #for loop ClustClustMidPts <- list() for(locus in 1:length(names(ClustMidPts))){ ClustClustMidPts[[locus]] <- cbind(XXYX=(ClustMidPts[[locus]][,"XX"]+ClustMidPts[[locus]][,"YX"])/2, YXYY=(ClustMidPts[[locus]][,"YX"]+ClustMidPts[[locus]][,"YY"])/2) } names(ClustClustMidPts)=names(ClustMidPts) Please also let me know how I messed up the formatting because it shows up fine in gmail even when I post on Nabble. How did I assume you were using Nabble? Is this topic included in the posting guide? -- View this message in context: http://r.789695.n4.nabble.com/applying-cbind-or-any-function-across-all-components-in-a-list-tp4631128p4631260.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] applying cbind (or any function) across all components in a list
Yes. This gives me: [,1] [,2] [,3] [,4] [1,]1234 [2,]2345 [,1] [,2] [,3] [,4] [1,]6789 [2,]789 10 BUT, how can I have it still within components like [[1]] [,1] [,2] [,3] [,4] [1,]1234 [2,]2345 [[2]] [,1] [,2] [,3] [,4] [1,]6789 [2,]789 10 How should I have phrased my question to be specific to this result? Thanks. -- View this message in context: http://r.789695.n4.nabble.com/applying-cbind-or-any-function-across-all-components-in-a-list-tp4631128p4631258.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] R base package grid does not output raster image
I am running 64-bit R 2.15.0 on a Windows Server 2008 R2 Amazon EC2 instance. grid does not produce output. For example, the following code should print the R logo to the window() device: library(grid) library(png) img.path <- system.file("img", "Rlogo.png", package="png") bg <- readPNG(img.path) background <- rasterGrob(unclass(bg)) grid.draw(background) I, however, see nothing. Is there a flag or option I should be using on Windows Server 2008 R2 to enable grid output? For clarity, the following is how I am actually using images in R plots. This code works on my laptop (R 2.15.0, ggplot2 0.8.9): library(ggplot2) library(grid) library(maps) library(mapproj) library(png) library(RgoogleMaps) counties <- map_data("county", region="virginia") states <- map_data("state") tmp <- tempfile(fileext=".png") bg <- GetMap.bbox(range(counties$long), range(counties$lat), destfile=tmp, maptype="satellite", format="png32") background <- readPNG(tmp) background <- rasterGrob(unclass(background)) ggplot(counties, aes(long, lat)) + coord_map(xlim=c(bg$BBOX$ll[2], bg$BBOX$ur[2]), ylim=c(bg$BBOX$ll[1], bg$BBOX$ur[1])) + geom_path(aes(group=group), color="darkgrey") + geom_path(data=states, aes(group=group), color="white", size=1) + opts(axis.line=theme_blank(), axis.text.x=theme_blank(), axis.text.y=theme_blank(), axis.ticks=theme_blank(), axis.title.x=theme_blank(), axis.title.y=theme_blank(), axis.ticks.length=unit(0, "lines"), axis.ticks.margin=unit(0, "lines"), panel.border=theme_blank(), panel.background=function(...)background, panel.grid.major=theme_blank(), panel.grid.minor=theme_blank(), panel.margin=unit(0, "lines"), legend.position="none", legend.title=theme_blank(), legend.background=theme_blank(), plot.margin=unit(0*c(-1.5, -1.5, -1.5, -1.5), "lines")) Thank you, Patrick __ 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] svychisq using two frames
On Fri, May 25, 2012 at 4:10 AM, lasciel wrote: > > So, I could use ‘rbind’ to stack the two years of data together with their > appropriate weights into a single data frame. > svychisq(~MyVar+MyVar, BothYears, statistic=”Chisq”) > > But now I have a problem that the variable name is the same across years, so > how do I differentiate the different time periods in the syntax for the > formula? Do I need to also create two new variables per: > MyVar.1 [year = 1] <- MyVar > MyVar.2 [year = 2] <- MyVar > svychisq(~MyVar.1+MyVar.2, BothYears, statistic=”Chisq”, na.rm= TRUE) > > > (I feel like I may be overthinking this and the answer is much simpler) > > You don't need to create two new variables; you just need a year variable svychisq(~MyVar+Year, BothYears, statistic=”Chisq”, na.rm= TRUE) tests whether MyVar is independent of Year. -thomas -- Thomas Lumley Professor of Biostatistics University of Auckland __ 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] R Error: System is computationally singular
Nathan, This does help, as in the first cut you provided, there was no variability in LOCS for LCOVER >= 5 and you have very few values of LOCS > 0 (you still do, relative to the scale of the total). Have you tried using a zero inflated negative binomial model (dist = "negbin") rather than poisson? I am not sure that the assumption of a zero inflated poisson distribution is reasonable with your data. Also, at least in this cut of the data, you have no 4's in LOCS and no 8's in LCOVER (same as before). If my math is correct only 0.006% of your LOCS values are > 0. I am also not convinced that you have enough data to differentiate between 1 and >=1 of whatever it is you are counting in LOCS. If that is the case, you might want to consider using logistic regression with a dichotomous response variable of LOCS == 0 versus LOCS >= 1. You seem to be in the general realm of very rare events given the distribution of LOCS in your data. Regards, Marc Schwartz On May 24, 2012, at 2:41 PM, Nathan Svoboda wrote: > Hi David, > > My apologies, I am not sure if this makes a big difference in your assessment > of the problem, but the results I just sent were only from a portion (1/15) > of the data. The dataset is rather large and the computer I am currently > using to set up the models is limited in its capabilities to analyze large > datasets. When I run the code you provided on a larger portion of the data > (1/2) this is the output I receive: > > LCOVER > LOCS 1 2 3 4 5 6 7 9 > 0 1692196 630659 550623 6140352 180896 255512 785929 63756 > 1 141 30 48 279 9 14 36 1 > 2 17 4 5 14 3 3 4 1 > 3 0 0 0 3 0 0 1 0 > 5 2 0 0 0 0 0 0 0 > > Thanks again for your time and assistance, > > Nate > > Nathan Svoboda > Graduate Research Assistant > Mississippi State University > > > On May 24, 2012, at 1:57 PM, Nathan Svoboda wrote: > >> Greetings, >> >> I am trying to fit a zero-inflated Poisson model using zeroinfl() >> from the >> pscl library. I have 5 covariates (4 continuous, 1 categorical); the >> categorical variable has 7 levels. I have had success fitting >> models that >> contain only the continuous covariates; however, when I add the >> categorical >> variable to any of the models (or if I run it by itself) I get the >> following >> error: >> >> Error in solve.default(as.matrix(fit$hessian)) : >> >> system is computationally singular: reciprocal condition number = >> 3.46934e-20 >> >> The code I am using is: >> >> library(pscl) >> f1 <- formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE + >> D_GRASS) >> ZIP1 <- zeroinfl(f1, dist="poisson", link = "logit", data = FAWNS) >> >> There is no correlation between my covariates. Also, I tried >> reducing my >> categorical covariate to 3 levels and still receive the same error. >> Can >> anyone suggest why I may be getting this error when I add the >> categorical >> covariate? >> > > What does this show: > > with( FAWNS, table(LOCS, LCOVER) ) > > -- > David Winsemius, MD > 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] R Error: System is computationally singular
Hi David, My apologies, I am not sure if this makes a big difference in your assessment of the problem, but the results I just sent were only from a portion (1/15) of the data. The dataset is rather large and the computer I am currently using to set up the models is limited in its capabilities to analyze large datasets. When I run the code you provided on a larger portion of the data (1/2) this is the output I receive: LCOVER LOCS 1 2 3 4 5 6 7 9 0 1692196 630659 550623 6140352 180896 255512 785929 63756 1 141 30 48 279 9 14 36 1 2 17 4 5 14 3 3 4 1 3 0 0 0 3 0 0 1 0 5 2 0 0 0 0 0 0 0 Thanks again for your time and assistance, Nate Nathan Svoboda Graduate Research Assistant Mississippi State University From: David Winsemius [mailto:dwinsem...@comcast.net] Sent: Thu 5/24/2012 1:54 PM To: Nathan Svoboda Cc: r-help@r-project.org Subject: Re: [R] R Error: System is computationally singular On May 24, 2012, at 1:57 PM, Nathan Svoboda wrote: > Greetings, > > I am trying to fit a zero-inflated Poisson model using zeroinfl() > from the > pscl library. I have 5 covariates (4 continuous, 1 categorical); the > categorical variable has 7 levels. I have had success fitting > models that > contain only the continuous covariates; however, when I add the > categorical > variable to any of the models (or if I run it by itself) I get the > following > error: > > Error in solve.default(as.matrix(fit$hessian)) : > > system is computationally singular: reciprocal condition number = > 3.46934e-20 > > The code I am using is: > > library(pscl) > f1 <- formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE + > D_GRASS) > ZIP1 <- zeroinfl(f1, dist="poisson", link = "logit", data = FAWNS) > > There is no correlation between my covariates. Also, I tried > reducing my > categorical covariate to 3 levels and still receive the same error. > Can > anyone suggest why I may be getting this error when I add the > categorical > covariate? > What does this show: with( FAWNS, table(LOCS, LCOVER) ) -- David Winsemius, MD 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] R Error: System is computationally singular
Thank you for your quick reply, When I run the code you provide I get this output: LCOVER LOCS 1 2 3 4 5 6 7 9 0 214507 79939 69803 778359 22932 32391 99630 8082 1 15 7 1 32 0 0 0 0 2 2 1 0 0 0 0 0 0 3 0 0 0 1 0 0 0 0 Nate -- View this message in context: http://r.789695.n4.nabble.com/R-Error-System-is-computationally-singular-tp4631242p4631251.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] R Error: System is computationally singular
Thank you for your quick reply! When I run the code you provide I get this output: LCOVER LOCS 1 2 3 4 5 6 7 9 0 214507 79939 69803 778359 22932 32391 99630 8082 1 15 7 1 32 0 0 0 0 2 2 1 0 0 0 0 0 0 3 0 0 0 1 0 0 0 0 Nate From: David Winsemius [mailto:dwinsem...@comcast.net] Sent: Thu 5/24/2012 1:54 PM To: Nathan Svoboda Cc: r-help@r-project.org Subject: Re: [R] R Error: System is computationally singular On May 24, 2012, at 1:57 PM, Nathan Svoboda wrote: > Greetings, > > I am trying to fit a zero-inflated Poisson model using zeroinfl() > from the > pscl library. I have 5 covariates (4 continuous, 1 categorical); the > categorical variable has 7 levels. I have had success fitting > models that > contain only the continuous covariates; however, when I add the > categorical > variable to any of the models (or if I run it by itself) I get the > following > error: > > Error in solve.default(as.matrix(fit$hessian)) : > > system is computationally singular: reciprocal condition number = > 3.46934e-20 > > The code I am using is: > > library(pscl) > f1 <- formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE + > D_GRASS) > ZIP1 <- zeroinfl(f1, dist="poisson", link = "logit", data = FAWNS) > > There is no correlation between my covariates. Also, I tried > reducing my > categorical covariate to 3 levels and still receive the same error. > Can > anyone suggest why I may be getting this error when I add the > categorical > covariate? > What does this show: with( FAWNS, table(LOCS, LCOVER) ) -- David Winsemius, MD 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] R Error: System is computationally singular
On May 24, 2012, at 1:57 PM, Nathan Svoboda wrote: Greetings, I am trying to fit a zero-inflated Poisson model using zeroinfl() from the pscl library. I have 5 covariates (4 continuous, 1 categorical); the categorical variable has 7 levels. I have had success fitting models that contain only the continuous covariates; however, when I add the categorical variable to any of the models (or if I run it by itself) I get the following error: Error in solve.default(as.matrix(fit$hessian)) : system is computationally singular: reciprocal condition number = 3.46934e-20 The code I am using is: library(pscl) f1 <- formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE + D_GRASS) ZIP1 <- zeroinfl(f1, dist="poisson", link = "logit", data = FAWNS) There is no correlation between my covariates. Also, I tried reducing my categorical covariate to 3 levels and still receive the same error. Can anyone suggest why I may be getting this error when I add the categorical covariate? What does this show: with( FAWNS, table(LOCS, LCOVER) ) -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R Error: System is computationally singular
Greetings, I am trying to fit a zero-inflated Poisson model using zeroinfl() from the pscl library. I have 5 covariates (4 continuous, 1 categorical); the categorical variable has 7 levels. I have had success fitting models that contain only the continuous covariates; however, when I add the categorical variable to any of the models (or if I run it by itself) I get the following error: Error in solve.default(as.matrix(fit$hessian)) : system is computationally singular: reciprocal condition number = 3.46934e-20 The code I am using is: library(pscl) f1 <- formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE + D_GRASS) ZIP1 <- zeroinfl(f1, dist="poisson", link = "logit", data = FAWNS) There is no correlation between my covariates. Also, I tried reducing my categorical covariate to 3 levels and still receive the same error. Can anyone suggest why I may be getting this error when I add the categorical covariate? I appreciate your time and input. Thank you, Nate Nathan Svoboda Graduate Research Assistant Carnivore Ecology Lab Mississippi State University -- View this message in context: http://r.789695.n4.nabble.com/R-Error-System-is-computationally-singular-tp4631242.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] R does not recognise columns and rows as they are supposed to be
> > X<-c(364:369) ; Y<-c(82:92) # for sellected region > > ... > > file2<-matrix(data=file,ncol=720,nrow=360) > > extract[i]<-mean(file2[X,Y],na.rm=TRUE) Note that 2-dimensional subscripts are given as mat[ROWS, COLUMNS] and you are using the reverse order, hence you get the error message that you are asking for rows 364:369 of a 360-row matrix. > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On > Behalf > Of Rui Barradas > Sent: Thursday, May 24, 2012 9:48 AM > To: r-help@r-project.org > Subject: Re: [R] R does not recognise columns and rows as they are supposed > to be > > Hello, > > There are several things with your code, most of which are not errors. > > 1. X<-c(364:369) ; Y<-c(82:92 and later c(1:365) > Expressions of the form m:n are integer sequences, the c() is not needed. > > 2. extract<-vector() > First you create the vector, then keep extending it throughout the loop. > This is much slower. > If you know the final length (and type), create it like this: > extract <- numeric(365) # or double(365) > > 3. 'dir1' is not used. > > 4. Now the readBin instruction: > 4.1. 'file' is the name of a function, use, say, 'file1' instead. > 4.2. Option 'signed' defaults to TRUE but it's only valid for integer types. > > 5. And, maybe this is the problem, in R, doubles are 8 bytes, not 4. Get rid > of option 'size', like the help page says, > > "If size is specified and not the natural size of the object, each element > of the vector is coerced to an appropriate type before being written or as > it is read." > > You are NOT reading doubles. The file may well contain single precision numbers and using what="double", size=4 will read them and convert them to doubles so other functions in R can use them. (Likewise, readBin can read files containing 1- or 2-byte signed or unsigned integers and store them as the 4-byte signed integers that the rest of R can deal with.) If the file was created on another sort of machine, then you may have worry about the byte order - try adding endian="big" or endian="little". > > As for the rest, it seems ok to me. (Keep option 'n' in readBin.) > > Hope this helps, > > Rui Barradas > > Jonsson wrote > > > > Dear All, > > The code given bellow is to extract values of one region and write that > > to a text file(there are 365 binary files in the directory). > > The problem which I am facing is that all my files are binary with size of > > 360 rows and 720 columns. > > I specified that in this line: > > file2<-matrix(data=file,ncol=720,nrow=360) but I got an error : Error in > > mean(file2[X, Y], na.rm = TRUE) : subscript out of bounds. > > and then I rewrote the above line as > > :file2<-matrix(data=file,ncol=360,nrow=720.I put ncol=360 and nrows =720 > > which is not right.But that worked and I didn't get any error.however,the > > results were not correct. > > Any help please > > X<-c(364:369) ; Y<-c(82:92) # for sellected region > > extract<-vector() > > dir1<- list.files("C:\\Users\\aalyaari\\Desktop\\New folder > > (10)\\Climate_Rad_f_GAMMA_%d.img", full.names = TRUE) > > listfile<-dir() > > for (i in c(1:365)) { > > conne <- file(listfile[i], "rb") > > file<- readBin(conne, double(), size=4, n=720*360, signed=T) > > file2<-matrix(data=file,ncol=720,nrow=360) > > extract[i]<-mean(file2[X,Y],na.rm=TRUE) > > close(conne) > > write.table(extract,"C:\\Users\\aalyaari\\Desktop\\New folder > > (10)\\sam.txt")} > > > > > -- > View this message in context: > http://r.789695.n4.nabble.com/R-does-not-recognise- > columns-and-rows-as-they-are-supposed-to-be-tp4631217p4631227.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] applying cbind (or any function) across all components in a list
On May 24, 2012, at 12:49 PM, Hans Thompson wrote: I'm confused why I haven't made clear what I am asking for help with. I have two different lists with two (or many) components, [[1]] and [[2]]. One of the list has components with dim=c(2,3) and the other has dim=c(2,2). I want to create a new list with components dim=c(2,4) by binding together the averages of the columns using cbind((l1[[1]][,1]+l2[[1]][,1])/2, (l1[[1]][,2]+l2[[1]][,1])/2, (l1[[1]][,2]+l2[[1]][,2])/2, (l1[[1]][,3]+l2[[1]][,2])/2 ) which should put out: [,1] [,2] [,3] [,4] [1,]1234 [2,]2345 my problem is finding out how I can apply this function to all the components within the list in the same function. Well, it's not clear what "function" you want applied to "what". In fact, it's even less clear in this posting than it was in your first one (since you assume, incorrectly, that we are looking at this on Nabble. Readers of this list expect you to include context.) I could not figure out in your first posting how we should be considering the fourth expression {(l1[[1]][,3]+l2[[1]][,2])/2 } ... to be an "average" of anything since its indices do not match? Perhaps equivalently, why should a 2 x 3 structure merged to a 2 x 2 structure yield a 2 x 4 structure? -- David Winsemius, MD 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] plot(summary) quantreg - Not all outputs needed
On 2012-05-24 03:26, stefan23 wrote: Hi Folks, I am currently trying to present some results I obtained by using the quantreg package developed by Roger Koenker. After calculating fit<-summary(rq(Y~X1+X2, tau=2:98/100) ) the function plot(fit) presents a really nice the results by showing the values for all "regressors" in the given interval tau. But in my case, I only need the output of a single variable, say X1 and I am not interested in plotting the others. Is there a way to hide the other graphics? Isn't this accomplished by specifying the argument 'parm'? plot(fit, parm = 2) etc. Peter Ehlers Thank you very much for your help, Cheers Stefan -- View this message in context: http://r.789695.n4.nabble.com/plot-summary-quantreg-Not-all-outputs-needed-tp4631184.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] applying cbind (or any function) across all components in a list
You don't make it clear if you want the two lists processed in parallel, but if so, this may be a step toward the function you want: Combine <- function(l1, l2) { pattern <- cbind(c(1, 2, 2, 3), c(1, 1, 2, 2)) for (i in 1:length(l1)) { print((l1[[i]][,pattern[,1]]+l2[[i]][,pattern[,2]])/2) } } Combine(l1, l2) This simply prints the results, but you didn't indicate how you wanted the output organized. -- David L Carlson Associate Professor of Anthropology Texas A&M University College Station, TX 77843-4352 > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- > project.org] On Behalf Of Greg Snow > Sent: Thursday, May 24, 2012 10:17 AM > To: Hans Thompson > Cc: r-help@r-project.org > Subject: Re: [R] applying cbind (or any function) across all components > in a list > > The "do.call" function may be what you want, but it is not completely > clear. If that does not solve your problem then try giving us an > example of what your list looks like (the dput function can help) and > what you want your end result to look like. > > On Thu, May 24, 2012 at 5:38 AM, Hans Thompson > wrote: > > The combination of column means in cbind is what I am trying to do. > I just > > don't know how to do it for every component (2 in this case) of the > list. > > I've been able to work with R to apply a function across all > components when > > there is just one list but I am having trouble working with multiple > lists > > at the same time. > > > > -- > > View this message in context: http://r.789695.n4.nabble.com/applying- > cbind-or-any-function-across-all-components-in-a-list- > tp4631128p4631187.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. > > > > -- > Gregory (Greg) L. Snow Ph.D. > 538...@gmail.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] R does not recognise columns and rows as they are supposed to be
Yes this helped a lot . I exactly followed what you suggested: X<-(82:92) ; Y<-(364:369) # for sellected region > extract <- double(365) > setwd("C:\\Users\\aalyaari\\Desktop\\New folder (10)\\") > listfile<-dir() > for (i in 1:365) { + conne <- file(listfile[i], "rb") + file1<- readBin(conne, double(), n=360*720) + file2<-matrix(data=file1,ncol=720,nrow=360) + extract[i]<-mean(file2[X,Y],na.rm=TRUE) + close(conne) + write.table(extract,"C:\\Users\\aalyaari\\Desktop\\New folder (10)\\samregion1.txt")} But I wonder why I got all values like -3.75E+306 -1.30E+54 -1.22E+58 and the right ones should be like: 22.25 22.76 33.25 -- View this message in context: http://r.789695.n4.nabble.com/R-does-not-recognise-columns-and-rows-as-they-are-supposed-to-be-tp4631217p4631241.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] inner_perc_table?
Thank you William, Does this mean I should use "" when inner_perc_table is called in the code then or am I supposed to define a full path? Regards, Charles On Thu, May 24, 2012 at 12:31 PM, William Dunlap wrote: > You need to supply a character string, e.g., "inner_perc_table", that names > the C entry point. If you supply a name, e.g., entry_point, or a more > general expression, > e.g, entry_points[j], then it will be evaluated with the usual R evaluation > rules and the result must be a character string. > > > getNativeSymbolInfo("inner_perc_table") > $name > [1] "inner_perc_table" > > $address > > attr(,"class") > [1] "NativeSymbol" > > $package > DLL name: nlme > Filename: c:/Program Files/R/R-2.15.0/library/nlme/libs/x64/nlme.dll > Dynamic lookup: FALSE > > $numParameters > [1] 6 > > attr(,"class") > [1] "CRoutine" "NativeSymbolInfo" > > or > > entry_point <- "inner_perc_table" > > entry_points <- c("outer_perc_table", "inner_perc_table") > > identical(getNativeSymbolInfo(entry_point), > getNativeSymbolInfo("inner_perc_table")) > [1] TRUE > > identical(getNativeSymbolInfo(entry_points[2]), > getNativeSymbolInfo("inner_perc_table")) > [1] TRUE > > Bill Dunlap > Spotfire, TIBCO Software > wdunlap tibco.com > > > > -Original Message- > > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] > On Behalf > > Of Charles Determan Jr > > Sent: Thursday, May 24, 2012 10:02 AM > > To: Duncan Murdoch > > Cc: r-help@r-project.org > > Subject: Re: [R] inner_perc_table? > > > > Thank you Duncan, > > > > I used the R function getNativeSymbolInfo(inner_perc_table) and the > object > > is not found: > > > > Error in is.vector(X) : object 'inner_perc_table' not found > > Done. > > > > I also tried the defining the directory to the nlme.dll file and it had > the > > same error. > > Is there something I missed while doing this? > > > > Thanks, > > Charles > > > > > > On Thu, May 24, 2012 at 11:51 AM, Duncan Murdoch > > wrote: > > > > > On 24/05/2012 11:35 AM, Charles Determan Jr wrote: > > > > > >> Hello, > > >> > > >> Does anyone on this list know what inner_perc_table is or where it is > > >> typically found? I am trying to modify some source code and it is > used > > >> with the .C() function. When I try and run it, it states that > > >> 'inner_perc_table is not found'. It is only called in such a way and > > >> isn't > > >> defined at any point in the previous code in which the function works. > > >> Should this be in some other directory of the source code files? I > have > > >> been scouring through them and have continue to come up empty. > > >> > > > > > > Look for the C function R_registerRoutines and/or the R function > > > getNativeSymbolInfo. > > > > > > Duncan Murdoch > > > > > > > [[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. > [[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] inner_perc_table?
You need to supply a character string, e.g., "inner_perc_table", that names the C entry point. If you supply a name, e.g., entry_point, or a more general expression, e.g, entry_points[j], then it will be evaluated with the usual R evaluation rules and the result must be a character string. > getNativeSymbolInfo("inner_perc_table") $name [1] "inner_perc_table" $address attr(,"class") [1] "NativeSymbol" $package DLL name: nlme Filename: c:/Program Files/R/R-2.15.0/library/nlme/libs/x64/nlme.dll Dynamic lookup: FALSE $numParameters [1] 6 attr(,"class") [1] "CRoutine" "NativeSymbolInfo" or > entry_point <- "inner_perc_table" > entry_points <- c("outer_perc_table", "inner_perc_table") > identical(getNativeSymbolInfo(entry_point), getNativeSymbolInfo("inner_perc_table")) [1] TRUE > identical(getNativeSymbolInfo(entry_points[2]), getNativeSymbolInfo("inner_perc_table")) [1] TRUE Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On > Behalf > Of Charles Determan Jr > Sent: Thursday, May 24, 2012 10:02 AM > To: Duncan Murdoch > Cc: r-help@r-project.org > Subject: Re: [R] inner_perc_table? > > Thank you Duncan, > > I used the R function getNativeSymbolInfo(inner_perc_table) and the object > is not found: > > Error in is.vector(X) : object 'inner_perc_table' not found > Done. > > I also tried the defining the directory to the nlme.dll file and it had the > same error. > Is there something I missed while doing this? > > Thanks, > Charles > > > On Thu, May 24, 2012 at 11:51 AM, Duncan Murdoch > wrote: > > > On 24/05/2012 11:35 AM, Charles Determan Jr wrote: > > > >> Hello, > >> > >> Does anyone on this list know what inner_perc_table is or where it is > >> typically found? I am trying to modify some source code and it is used > >> with the .C() function. When I try and run it, it states that > >> 'inner_perc_table is not found'. It is only called in such a way and > >> isn't > >> defined at any point in the previous code in which the function works. > >> Should this be in some other directory of the source code files? I have > >> been scouring through them and have continue to come up empty. > >> > > > > Look for the C function R_registerRoutines and/or the R function > > getNativeSymbolInfo. > > > > Duncan Murdoch > > > > [[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] R does not recognise columns and rows as they are supposed to be
Sorry, forgot the last line in the op code. write.table(extract, ...) is called every time through the loop, not just one time after it. Put it after closing '}'. Rui Barradas Rui Barradas wrote > > Hello, > > There are several things with your code, most of which are not errors. > > 1. X<-c(364:369) ; Y<-c(82:92 and later c(1:365) > Expressions of the form m:n are integer sequences, the c() is not needed. > > 2. extract<-vector() > First you create the vector, then keep extending it throughout the loop. > This is much slower. > If you know the final length (and type), create it like this: > extract <- numeric(365) # or double(365) > > 3. 'dir1' is not used. > > 4. Now the readBin instruction: > 4.1. 'file' is the name of a function, use, say, 'file1' instead. > 4.2. Option 'signed' defaults to TRUE but it's only valid for integer > types. > > 5. And, maybe this is the problem, in R, doubles are 8 bytes, not 4. Get > rid of option 'size', like the help page says, > > "If size is specified and not the natural size of the object, each element > of the vector is coerced to an appropriate type before being written or as > it is read." > > You are NOT reading doubles. > > As for the rest, it seems ok to me. (Keep option 'n' in readBin.) > > Hope this helps, > > Rui Barradas > > Jonsson wrote >> >> Dear All, >> The code given bellow is to extract values of one region and write that >> to a text file(there are 365 binary files in the directory). >> The problem which I am facing is that all my files are binary with size >> of 360 rows and 720 columns. >> I specified that in this line: >> file2<-matrix(data=file,ncol=720,nrow=360) but I got an error : Error in >> mean(file2[X, Y], na.rm = TRUE) : subscript out of bounds. >> and then I rewrote the above line as >> :file2<-matrix(data=file,ncol=360,nrow=720.I put ncol=360 and nrows =720 >> which is not right.But that worked and I didn't get any error.however,the >> results were not correct. >> Any help please >> X<-c(364:369) ; Y<-c(82:92) # for sellected region >> extract<-vector() >> dir1<- list.files("C:\\Users\\aalyaari\\Desktop\\New folder >> (10)\\Climate_Rad_f_GAMMA_%d.img", full.names = TRUE) >> listfile<-dir() >> for (i in c(1:365)) { >> conne <- file(listfile[i], "rb") >> file<- readBin(conne, double(), size=4, n=720*360, signed=T) >> file2<-matrix(data=file,ncol=720,nrow=360) >> extract[i]<-mean(file2[X,Y],na.rm=TRUE) >> close(conne) >> write.table(extract,"C:\\Users\\aalyaari\\Desktop\\New folder >> (10)\\sam.txt")} >> > -- View this message in context: http://r.789695.n4.nabble.com/R-does-not-recognise-columns-and-rows-as-they-are-supposed-to-be-tp4631217p4631229.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] R does not recognise columns and rows as they are supposed to be
Hello, There are several things with your code, most of which are not errors. 1. X<-c(364:369) ; Y<-c(82:92 and later c(1:365) Expressions of the form m:n are integer sequences, the c() is not needed. 2. extract<-vector() First you create the vector, then keep extending it throughout the loop. This is much slower. If you know the final length (and type), create it like this: extract <- numeric(365) # or double(365) 3. 'dir1' is not used. 4. Now the readBin instruction: 4.1. 'file' is the name of a function, use, say, 'file1' instead. 4.2. Option 'signed' defaults to TRUE but it's only valid for integer types. 5. And, maybe this is the problem, in R, doubles are 8 bytes, not 4. Get rid of option 'size', like the help page says, "If size is specified and not the natural size of the object, each element of the vector is coerced to an appropriate type before being written or as it is read." You are NOT reading doubles. As for the rest, it seems ok to me. (Keep option 'n' in readBin.) Hope this helps, Rui Barradas Jonsson wrote > > Dear All, > The code given bellow is to extract values of one region and write that > to a text file(there are 365 binary files in the directory). > The problem which I am facing is that all my files are binary with size of > 360 rows and 720 columns. > I specified that in this line: > file2<-matrix(data=file,ncol=720,nrow=360) but I got an error : Error in > mean(file2[X, Y], na.rm = TRUE) : subscript out of bounds. > and then I rewrote the above line as > :file2<-matrix(data=file,ncol=360,nrow=720.I put ncol=360 and nrows =720 > which is not right.But that worked and I didn't get any error.however,the > results were not correct. > Any help please > X<-c(364:369) ; Y<-c(82:92) # for sellected region > extract<-vector() > dir1<- list.files("C:\\Users\\aalyaari\\Desktop\\New folder > (10)\\Climate_Rad_f_GAMMA_%d.img", full.names = TRUE) > listfile<-dir() > for (i in c(1:365)) { > conne <- file(listfile[i], "rb") > file<- readBin(conne, double(), size=4, n=720*360, signed=T) > file2<-matrix(data=file,ncol=720,nrow=360) > extract[i]<-mean(file2[X,Y],na.rm=TRUE) > close(conne) > write.table(extract,"C:\\Users\\aalyaari\\Desktop\\New folder > (10)\\sam.txt")} > -- View this message in context: http://r.789695.n4.nabble.com/R-does-not-recognise-columns-and-rows-as-they-are-supposed-to-be-tp4631217p4631227.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] applying cbind (or any function) across all components in a list
I'm confused why I haven't made clear what I am asking for help with. I have two different lists with two (or many) components, [[1]] and [[2]]. One of the list has components with dim=c(2,3) and the other has dim=c(2,2). I want to create a new list with components dim=c(2,4) by binding together the averages of the columns using > cbind((l1[[1]][,1]+l2[[1]][,1])/2, (l1[[1]][,2]+l2[[1]][,1])/2, > (l1[[1]][,2]+l2[[1]][,2])/2, (l1[[1]][,3]+l2[[1]][,2])/2) which should put out: [,1] [,2] [,3] [,4] [1,]1234 [2,]2345 my problem is finding out how I can apply this function to all the components within the list in the same function. -- View this message in context: http://r.789695.n4.nabble.com/applying-cbind-or-any-function-across-all-components-in-a-list-tp4631128p4631228.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] Fwd: help needed
--- segue il messaggio inoltrato (the forwarded message follows) --- --- Begin Message --- Dear R users, i am very new to R. i am working on the following data. 01.01.1967 0.87 02.01.1967 0.87 03.01.1968 0.87 04.01.1968 0.87 05.01.1969 0.87 06.01.1969 0.87 07.01.1970 0.87 08.01.1970 0.87 09.01.1971 0.87 10.01.1971 0.87 11.01.1972 0.87 12.01.1972 0.87 13.01.1973 0.69 14.01.1973 0.70 15.01.1974 0.71 16.01.1974 0.72 I want to reshape it in the following FORMAT 19671968196919701971197219731974 1 0.870.870.870.870.71 2 0.870.870.870.870.72 OBVIOUSLY, I HAVE A LARGE AMOUNT OF DATA TO WORK WITH.i would also like to take into account the effect of leap year. For example if 1969 in a leap year then the column under it, has to have 1 extra reading. Thanks in advance --- End Message --- __ 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] inner_perc_table?
My apologies, feel a little foolish. I forgot the "" for the object. This worked, however, once I have this information, I am not sure what I should be using to fix the issue. I can determine the location in the nlme.dll file but how can I use this information to have the other function work? Regards, Charles On Thu, May 24, 2012 at 11:51 AM, Duncan Murdoch wrote: > On 24/05/2012 11:35 AM, Charles Determan Jr wrote: > >> Hello, >> >> Does anyone on this list know what inner_perc_table is or where it is >> typically found? I am trying to modify some source code and it is used >> with the .C() function. When I try and run it, it states that >> 'inner_perc_table is not found'. It is only called in such a way and >> isn't >> defined at any point in the previous code in which the function works. >> Should this be in some other directory of the source code files? I have >> been scouring through them and have continue to come up empty. >> > > Look for the C function R_registerRoutines and/or the R function > getNativeSymbolInfo. > > Duncan Murdoch > [[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] set tkscale by tkentry
Greg Snow <538280 gmail.com> writes: > > I believe that what is happening is that when you try to edit the > entry widget any intermediate values get sent to the slider widget > which then checks to see if they are in the allowable range and if it > is not then it sets the value to either the minimum and maximum and > sends that back to the entry widget while you are still trying to edit > it. Even if you highlight a single digit and try to replace it with a > different digit it first deletes the highlighted digit resulting in a > number smaller than the minimum of the slider which then updates the > entry widget to the minimum before the new digit can go in, then > adding the new digit makes it larger than the slider maximum. > Greg is right. You might try validating on focusout, rather than the key, but this is easy enough to do in R code, rather than let tcl do that work: a <- 306870; b <- 3026741 tt<-tktoplevel() varalpha <- tclVar(a) charalpha <- tclVar(as.character(a)) scale <- tkscale(tt, from=a, to=b, resolution=1, label="alpha", variable=varalpha, showvalue=TRUE, orient="horiz") ed <- tkentry(tt, textvariable=charalpha) tkpack(ed) tkpack(scale) ## connect tkconfigure(scale, command=function(...) { tclvalue(charalpha) <- as.character(tclvalue(varalpha)) }) valid_input <- function(...) { val <- as.numeric(tclvalue(charalpha)) if(a <= val & val <= b) { message("set to ", val) tclvalue(varalpha) <- val } else { message("not valid...") tkfocus(ed) } } tkbind(ed, "", valid_input) tkbind(ed, "", valid_input) __ 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] Exclude when sd=0
Assuming you mean that the column is constant so that the sd is 0 test <- matrix(rnorm(50), nrow=10, ncol=5) test[,2] <- 1 test[,4] <- 2 # test[,which(apply(test, 2, sd) > 0)] could fail on rounding errors Test2 <- test[, which(apply(test, 2, sd) > 1e-10)] -- David L Carlson Associate Professor of Anthropology Texas A&M University College Station, TX 77843-4352 > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- > project.org] On Behalf Of Chris Burns > Sent: Wednesday, May 23, 2012 7:45 PM > To: r-help@r-project.org > Subject: [R] Exclude when sd=0 > > How do I trim a matrix to exclude columns that have no standard > deviation? > > Thanks, > > Chris > > [[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] inner_perc_table?
Thank you Duncan, I used the R function getNativeSymbolInfo(inner_perc_table) and the object is not found: Error in is.vector(X) : object 'inner_perc_table' not found Done. I also tried the defining the directory to the nlme.dll file and it had the same error. Is there something I missed while doing this? Thanks, Charles On Thu, May 24, 2012 at 11:51 AM, Duncan Murdoch wrote: > On 24/05/2012 11:35 AM, Charles Determan Jr wrote: > >> Hello, >> >> Does anyone on this list know what inner_perc_table is or where it is >> typically found? I am trying to modify some source code and it is used >> with the .C() function. When I try and run it, it states that >> 'inner_perc_table is not found'. It is only called in such a way and >> isn't >> defined at any point in the previous code in which the function works. >> Should this be in some other directory of the source code files? I have >> been scouring through them and have continue to come up empty. >> > > Look for the C function R_registerRoutines and/or the R function > getNativeSymbolInfo. > > Duncan Murdoch > [[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] inner_perc_table?
On 24/05/2012 11:35 AM, Charles Determan Jr wrote: Hello, Does anyone on this list know what inner_perc_table is or where it is typically found? I am trying to modify some source code and it is used with the .C() function. When I try and run it, it states that 'inner_perc_table is not found'. It is only called in such a way and isn't defined at any point in the previous code in which the function works. Should this be in some other directory of the source code files? I have been scouring through them and have continue to come up empty. Look for the C function R_registerRoutines and/or the R function getNativeSymbolInfo. 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.
[R] transform 1 col to 2 col
Hi, Just an addendum. Suppose if you want to add more columns and do the split, you can try, inscrutable.df2<-data.frame(inscrutable.df1,V2=(c(rnorm(1),"GTDF","SQ:1234","DFFD","DFDSE")[rep(c(1,2,3,4,5),times=4)])) inscrutable3.df3<-data.frame(matrix(inscrutable.df1$V1,ncol=2,byrow=TRUE),matrix(inscrutable.df2$V2,ncol=2,byrow=TRUE)) A.K. - Original Message - From: "MacQueen, Don" To: Soheila Khodakarim Cc: "r-help@r-project.org" Sent: Thursday, May 24, 2012 10:19 AM Subject: Re: [R] transform 1 col to 2 col See insertion below -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 5/24/12 1:43 AM, "Jim Lemon" wrote: >On 05/24/2012 05:10 PM, Soheila Khodakarim wrote: >> Dear All >> >> How can I transform 1 column to 2 columns in R? >> >> 211217_s_at >> >> GO:0005249 >> >> 211217_s_at >> >> GO:0005251 >> >> 211217_s_at >> >> GO:0005515 >> >> 211217_s_at >> >> GO:0015271 >> >> 211217_s_at >> >> GO:0030955 >> >> >> >> >> >> 211217_s_at >> >> GO:0005249 >> >> 211217_s_at >> >> GO:0005251 >> >> 211217_s_at >> >> GO:0005515 >> >> 211217_s_at >> >> GO:0015271 >> >> 211217_s_at >> >> GO:0030955 >> >Hi Soheila, >Let us begin with the assumption that your "col" is roughly what you >posted, and that text resides in a file named "inscrutable.dat". Let us >first read that text into a data frame in R: > >inscrutatble.df<-read.table("inscrutable.dat") > inscrutable.df > V1 > >1 211217_s_at >2 GO:0005249 >3 211217_s_at >4 GO:0005251 >5 211217_s_at >6 GO:0005515 >7 211217_s_at >8 GO:0015271 >9 211217_s_at >10 GO:0030955 >11 211217_s_at >12 GO:0005249 >13 211217_s_at >14 GO:0005251 >15 211217_s_at >16 GO:0005515 >17 211217_s_at >18 GO:0015271 >19 211217_s_at >20 GO:0030955 > >Now we will further assume that you want to place every odd element in >one "col" and every even element in the other "col". Or perhaps: matrix(inscrutable.df$V1, ncol=2, byrow=TRUE) or data.frame(matrix(inscrutable.df$V1, ncol=2, byrow=TRUE)) to get a data frame. -Don > >inscrutable2.df<-data.frame( > col1=inscrutable.df$V1[seq(1,length(inscrutable.df$V1),by=2)], > col2=inscrutable.df$V1[seq(2,length(inscrutable.df$V1),by=2)]) >inscrutable2.df > col1 col2 >1 211217_s_at GO:0005249 >2 211217_s_at GO:0005251 >3 211217_s_at GO:0005515 >4 211217_s_at GO:0015271 >5 211217_s_at GO:0030955 >6 211217_s_at GO:0005249 >7 211217_s_at GO:0005251 >8 211217_s_at GO:0005515 >9 211217_s_at GO:0015271 >10 211217_s_at GO:0030955 > >There we are - easy as pie. > >Jim > >__ >R-help@r-project.org mailing list >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide >http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Upper an lower bound in linear programing
On Thu, May 24, 2012 at 07:16:20AM -0700, Haio wrote: > Hello everyone! > I´m trying to solve a linear optimization problem with r using the LPsolve > function and i want to set upper and lower bounds for the obejctive > function. Any idea´s on how to do this?? Hello: The formulation of an LP problem allows to use the objective function also as a row in the constraint matrix. So, it is possible to set a constraint on it. For example, maximizing x + y under the constraints x/4 <= y <= 2 + x/3 x + y <= 9 (constraint on the objective function) may be done as follows library(lpSolve) crit <- c(1, 1) mat <- rbind( c(-1/4, 1), c(-1/3, 1), c( 1, 1)) dir <- c(">=", "<=", "<=") rhs <- c(0, 2, 9) out <- lp("max", objective.in=crit, const.mat=mat, const.dir=dir, const.rhs=rhs) out Success: the objective function is 9 out$solution [1] 7.2 1.8 Hope this helps. Petr Savicky. __ 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] svychisq using two frames
Hello, I’m hoping you have a few minutes to help out someone very new to R. I’ve done some searching, but cannot find this particular issue. I have survey data from two different time periods (years). Both years are stratified samples and have the same variables (and variable names), but are different people in the community answering in the different years. Everything loads into the survey design fine and I can calculate the means and CIs I need. I want to calculate whether there was a change in a single variable between years. The variable of interest has binary yes/no answers. I was planning to use the ‘svychisq’ function, but cannot figure out how to do this across what is currently two data frames. I haven’t actually tried to run a svychisq with the two frames yet as I am waiting on the data for the second frame. I’m trying to plan out the logic ahead of time. So, let’s say I have survey design frame Year1 with variables Wt1 and MyVar, and survey design frame Year2 with variables Wt2 and MyVar – remember the variable name is the same in both frames. My first inclination is that I would use a statement like: svychisq(~MyVar+MyVar, Year1+Year2, statistic=”Chisq”) However, it seems from reading the help files that I can only use the ‘svychisq’ function if I declare a single data frame in the design statement, rather than referencing a frame for Year1 and a different frame for Year2. So, I could use ‘rbind’ to stack the two years of data together with their appropriate weights into a single data frame. svychisq(~MyVar+MyVar, BothYears, statistic=”Chisq”) But now I have a problem that the variable name is the same across years, so how do I differentiate the different time periods in the syntax for the formula? Do I need to also create two new variables per: MyVar.1 [year = 1] <- MyVar MyVar.2 [year = 2] <- MyVar svychisq(~MyVar.1+MyVar.2, BothYears, statistic=”Chisq”, na.rm= TRUE) (I feel like I may be overthinking this and the answer is much simpler) -- View this message in context: http://r.789695.n4.nabble.com/svychisq-using-two-frames-tp4631220.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] How to evaluate R things from Visual Studio?
I'll answer line by line so: "You ask for help from people who have used R.NET, give them time to answer." I have.. I'm not sure why you seem so certain I didn't. But it is beside the point anyway "Also, what about folks who have not used it? "Does not work" is not enough information. What did you try? Give enough detail so that someone might follow your footsteps, including VS versions, R location and version, assumptions made, and links to resources used on the way." Folks who haven't used it won't be able to help me anyway. People who did get it working will be able to help me, and those are the people I'm counting on. The sole reason I did not disclose any R.NET related info is to avoid making an already large message, larger. When I find someone, if I find someone who does have R.NET and can help me, I will gladly give any and all needed information "FWIW, I got the R.NET example to work on Windows 7 (64-bit OS) with Visual Studio 10 Express (32-bit only) with their example program here: http://rdotnet.codeplex.com/ I downloaded this file, and set the VS reference to it: http://rdotnet.codeplex.com/downloads/get/211446 I used R-2.12.0 as per their example REngine.SetDllDirectory(@"C:\Program Files\R\R-2.12.0\bin\i386");" It's worth much, believe me. I really needed another person who got R.NET working. It does work on R 2.12.0, I agree. but i need it to work any latest R. I can't make my users use old and obsolete versions of R. And people have made it work on 2.15.0 (currently latest version) as well. "I set R_HOME and R_LIBS to paths for the R-2.12.0 installation." Can you please tell me exactly what string you put inside R_HOME and R_LIBS environment variables? (i.e. "C:\Progs\R\R-2.15.0\bin"? "C:\Progs\R\R-2.15.0"? what exactly?) Also, it was YOU who created those Environmental Variables, right? Because they do not exist in my computer and i didn't delete them. Safe to assume that we're talking about System -> Advanced -> Environment Variables -> System Variables. correct? "I read this page: http://rdotnet.codeplex.com/documentation?referringTitle=Home I had a quick stab at using a recent R build, but dropped it to try the stated version in their example. There are error messages printed from the Windows application when it fails and they are worth reading (and reporting when asking for help)." I've already read every page on rdotnet.codeplex.com Of course there are. I need not repeat myself. I explained above why no direct R.NET info were given. Now that I have found one guy who has R.NET working, I shall :) My R version is: 2.15.0 and is installed on "C:\Progs\R\R-2.15.0" instead of "C:\Program Files\R\R-2.15.0" (but it doesnt work on that either, tried it) The code I try to run on R.NET version 1.4.1 (changeset 6d2c3f161801) is the example they have on the official page: Private Sub Form1_Load(sender As System.Object, e As System.EventArgs) Handles MyBase.Load Dim Rdo As REngine = REngine.CreateInstance("RDotNet", "-q") ' quiet mode ' .NET Framework array to R vector. Dim group1 As NumericVector = Rdo.CreateNumericVector(New Double() {30.02, 29.99, 30.11, 29.97, 30.01, 29.99}) Rdo.SetSymbol("group1", group1) ' Direct parsing from R script. Dim group2 As NumericVector = Rdo.Evaluate("group2 <- c(29.89, 29.93, 29.72, 29.98, 30.02, 29.98)").AsNumeric() ' Test difference of mean and get the P-value. Dim testResult As GenericVector = Rdo.Evaluate("t.test(group1, group2)").AsList() Dim p As Double = testResult("p.value").AsNumeric().First() End Sub The Error i get is this: System.DllNotFoundException: Dll was not found. at RDotNet.NativeLibrary.UnmanagedDll..ctor(String dllName) in C:\Users\N1h1l1sT\Downloads\rdotnet_6d2c3f161801\RDotNet.NativeLibrary\UnmanagedDll.cs:line 42 at RDotNet.REngine..ctor(String id, String dll) at RDotNet.REngine.CreateInstance(String id, String dll) at RAttemptFour.Form1.Form1_Load(Object sender, EventArgs e) in C:\Users\N1h1l1sT\Dropbox\Visual Basic 2010\Test Projects\RAttemptFour\RAttemptFour\Form1.vb:line 35 No matter what i put inside the HOME or R_Home environment variables on my windows, it just won't work... the line 35 refers to this line: "Dim Rdo As REngine = REngine.CreateInstance("RDotNet", "-q") ' quiet mode" --- Giannis Mamalikidis e-mail: giannis_mamaliki...@msn.com Member of the STAINS Research Group (STAtistics & INformation Systems Group) of the Aristotle University of Thessaloniki Site: http://stains.csd.auth.gr B.Sc Student in Electrical and Mechanical Engineering __ 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 comment
[R] Manually modifying an hclust dendrogram to remove singletons
Dear R-Help, I have a clustering problem with hclust that I hope someone can help me with. Consider the classic hclust example: hc <- hclust(dist(USArrests), "ave") plot(hc) I would like to cut the tree up in such a way so as to avoid small clusters, so that we get a minimum number of items in each cluster, and therefore avoid singletons. e.g. in this example, you can see that Hawaii is split off onto its own at quite a high level. I would like to avoid having a single item clustered on its own like this. How can I achieve this? I have tried manually modifying the tree using dendrapply but have not been able to produce a valid solution thus far.. Suggestions are welcome. Best wishes, Mark __ 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] use list as function arguments
?do.call On Thu, May 24, 2012 at 9:32 AM, Alexander Shenkin wrote: > Hello Folks, > > Is there any way to pass a list into a function such that the function > will use the list as its arguments? I haven't been able to figure that out. > > The background: I'm trying to build a function that will apply another > function multiple times, each time with a different set of specified > arguments. I'm trying to figure out how to pass in a list of argument > lists, then loop through the top-level list, passing in the lower-level > list as function arguments. > > pseudocode: > > b = list( list( arg1 = 1, arg2 = 2 ), > list( arg1 = 3, arg2 = 4 ) > ) > > a <- apply_function(arglist) { > for (i in length(arglist)) { > b(arglist[i]) > } > } > > > Specifically, the actual use I'm trying to implement is a function to > format columns of data frames and matrices independently. What I have > so far is below, but it's not working. Perhaps I'm going about this the > wrong way? > > > format_cols <- function(x, format_list = list()) { > # usage: length(format_list) must equal ncol(x) > # format list should be a list of lists of key=value pairs > corresponding to format settings for each column > > if (is.data.frame(x)) { > newout = data.frame() > } else if (is.matrix(x)) { > newout = matrix() > } > > for (i in 1:ncol(x)){ > newout = cbind(newout, format(x,format_list[[i]])) > x[,i] = format(x,format_list[[i]]) > } > return(newout) > } > > Thanks, > Allie > > __ > 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. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.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] R does not recognise columns and rows as they are supposed to be
On May 24, 2012, at 11:51 AM, Jonsson wrote: Dear All, The code given bellow is to extract values of one region and write that to a text file(there are 365 binary files in the directory). The problem which I am facing is that all my files are binary with size of 360 rows and 720 columns. I specified that in this line:file2<- matrix(data=file,ncol=720,nrow=360) but I got an error : Error in mean(file2[X, Y], na.rm = TRUE) : subscript out of bounds. and then I rewrote the above line as :file2<-matrix(data=file,ncol=360,nrow=720.I put ncol=360 and nrows =720 which is not right.But that worked and I didn't get any error.however,the results were not correct. Any help please dir1<- list.files("C:\\Users\\aalyaari\\Desktop\\New folder (10)\\Climate_Rad_f_GAMMA_%d.img", full.names = TRUE) listfile<-dir() for (i in c(1:365)) { conne <- file(listfile[i], "rb") file<- readBin(conne, double(), size=4, n=720*360, signed=T) file2<-matrix(data=file,ncol=720,nrow=360) extract[i]<-mean(file2[X,Y],na.rm=TRUE) close(conne) write.table(extract,"C:\\Users\\aalyaari\\Desktop\\New folder (10)\\sam.txt")} Please do not cross post to SO and Rhelp. (See Posting Guide.) (But if you do, rhelp generally expects reproducible examples which this is not.) -- David Winsemius, MD 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.
[R] R does not recognise columns and rows as they are supposed to be
Dear All, The code given bellow is to extract values of one region and write that to a text file(there are 365 binary files in the directory). The problem which I am facing is that all my files are binary with size of 360 rows and 720 columns. I specified that in this line:file2<-matrix(data=file,ncol=720,nrow=360) but I got an error : Error in mean(file2[X, Y], na.rm = TRUE) : subscript out of bounds. and then I rewrote the above line as :file2<-matrix(data=file,ncol=360,nrow=720.I put ncol=360 and nrows =720 which is not right.But that worked and I didn't get any error.however,the results were not correct. Any help please dir1<- list.files("C:\\Users\\aalyaari\\Desktop\\New folder (10)\\Climate_Rad_f_GAMMA_%d.img", full.names = TRUE) listfile<-dir() for (i in c(1:365)) { conne <- file(listfile[i], "rb") file<- readBin(conne, double(), size=4, n=720*360, signed=T) file2<-matrix(data=file,ncol=720,nrow=360) extract[i]<-mean(file2[X,Y],na.rm=TRUE) close(conne) write.table(extract,"C:\\Users\\aalyaari\\Desktop\\New folder (10)\\sam.txt")} -- View this message in context: http://r.789695.n4.nabble.com/R-does-not-recognise-columns-and-rows-as-they-are-supposed-to-be-tp4631217.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] inner_perc_table?
Hello, Does anyone on this list know what inner_perc_table is or where it is typically found? I am trying to modify some source code and it is used with the .C() function. When I try and run it, it states that 'inner_perc_table is not found'. It is only called in such a way and isn't defined at any point in the previous code in which the function works. Should this be in some other directory of the source code files? I have been scouring through them and have continue to come up empty. Many thanks for any assistance, Charles [[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] use list as function arguments
Hello Folks, Is there any way to pass a list into a function such that the function will use the list as its arguments? I haven't been able to figure that out. The background: I'm trying to build a function that will apply another function multiple times, each time with a different set of specified arguments. I'm trying to figure out how to pass in a list of argument lists, then loop through the top-level list, passing in the lower-level list as function arguments. pseudocode: b = list( list( arg1 = 1, arg2 = 2 ), list( arg1 = 3, arg2 = 4 ) ) a <- apply_function(arglist) { for (i in length(arglist)) { b(arglist[i]) } } Specifically, the actual use I'm trying to implement is a function to format columns of data frames and matrices independently. What I have so far is below, but it's not working. Perhaps I'm going about this the wrong way? format_cols <- function(x, format_list = list()) { # usage: length(format_list) must equal ncol(x) #format list should be a list of lists of key=value pairs corresponding to format settings for each column if (is.data.frame(x)) { newout = data.frame() } else if (is.matrix(x)) { newout = matrix() } for (i in 1:ncol(x)){ newout = cbind(newout, format(x,format_list[[i]])) x[,i] = format(x,format_list[[i]]) } return(newout) } Thanks, Allie __ 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] Memory allocation error
Dear All, I am running R in a system with the following configuration *Processor: Intel(R) Xeon(R) CPU X5650 @ 2.67GHz OS: Ubuntu X86_64 10.10 RAM: 24 GB* The R session info is * R version 2.14.1 (2011-12-22) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=en_US.UTF-8LC_MESSAGES=en_US.UTF-8 [7] LC_PAPER=C LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C * I have a matrix of dimensions 12 rows X 29318 columns. The matrix contains numeric as well as NA values. I am using the* rcorr *function from the * Hmisc* package to get correlation information from the matrix (* rcorr(matrix)*). During the calculation I get the error "*cannot allocate vector of size 6.7 GB*". When I check the memory allocation of my R session I get the following information *gc() used (Mb) gc trigger (Mb) limit (Mb) max used (Mb) Ncells 249638 13.4 467875 25.0 NA 407500 21.8 Vcells 1499217 11.52335949 17.9 7000 1970005 15.1 *Can someone please help me in finding a workaround to the problem. -Regards -- Swaraj Basu PhD Student (Bioinformatics - Functional Genomics) Animal Physiology and Evolution Stazione Zoologica Anton Dohrn Naples [[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] Using NA as a break point for indicator variable?
This method worked perfectly! The rle() function was key and I was completely unfamiliar with it. Thanks so much, Max On Thu, May 24, 2012 at 8:52 AM, Rui Barradas wrote: > Hello, > > Assuming that 'd' is your original data.frame and that you've set entire > rows to NA, try this > > > d$leak_num <- NA > ix <- !is.na(d[, 1]) # any column will do, entire row is NA > ## alternative, if other rows may have NAs, due to something else > #ix <- apply(d, 1, function(x) all(!is.na(x))) > r <- rle(ix) > v <- cumsum(r$values) > d$leak_num[ix] <- rep(v[r$values], r$lengths[r$values]) > d > > > Hope this helps, > > Rui Barradas > > Em 24-05-2012 11:00, Max Brondfield escreveu: > >> Date: Wed, 23 May 2012 16:42:02 -0400 >> From: Max Brondfield >> > >> To:r-help@r-project.org >> Subject: [R] Using NA as a break point for indicator variable? >> Message-ID: >>> gmail.com >> > >> Content-Type: text/plain >> >> >> Hi all, >> I am working with a spatial data set for which I am only interested in >> high >> concentration values ("leaks"). The low values (< 90th percentile) have >> already been turned into NA's, leaving me with a matrix like this: >> >> < CH4_leak >> >> lonlatCH4 >> 1 -71.11954 42.35068 2.595834 >> 2 -71.11954 42.35068 2.595688 >> 3 NA NA NA >> 4 NA NA NA >> 5 NA NA NA >> 6 -71.11948 42.35068 2.435762 >> 7 -71.11948 42.35068 2.491003 >> 8 NANA NA >> 9 -71.11930 42.35068 2.464475 >> 10 -71.11932 42.35068 2.470865 >> >> Every time an NA comes up, it means the "leak" is gone, and the next valid >> value would represent a different leak (at a different location). My goal >> is to tag all of the remaining values with an indicator variable to >> spatially distinguish the leaks. I am envisioning a simple numeric >> indicator such as: >> >> lonlatCH4leak_num >> 1 -71.11954 42.35068 2.595834 1 >> 2 -71.11954 42.35068 2.595688 1 >> 3 NA NA NA NA >> 4 NA NA NA NA >> 5 NA NA NA NA >> 6 -71.11948 42.35068 2.435762 2 >> 7 -71.11948 42.35068 2.491003 2 >> 8 NANA NA NA >> 9 -71.11930 42.35068 2.064475 3 >> 10 -71.11932 42.35068 2.070865 3 >> >> Does anyone have any thoughts on how to code this, perhaps using the NA >> values as a "break point"? The data set is far too large to do this >> manually, and I must admit I'm completely at a loss. Any help would be >> much >> appreciated! Best, >> >> Max >> >>[[alternative HTML version deleted]] >> >> >> -- Max Brondfield Research Assistant Department of Geography & Environment Boston University [[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] applying cbind (or any function) across all components in a list
The "do.call" function may be what you want, but it is not completely clear. If that does not solve your problem then try giving us an example of what your list looks like (the dput function can help) and what you want your end result to look like. On Thu, May 24, 2012 at 5:38 AM, Hans Thompson wrote: > The combination of column means in cbind is what I am trying to do. I just > don't know how to do it for every component (2 in this case) of the list. > I've been able to work with R to apply a function across all components when > there is just one list but I am having trouble working with multiple lists > at the same time. > > -- > View this message in context: > http://r.789695.n4.nabble.com/applying-cbind-or-any-function-across-all-components-in-a-list-tp4631128p4631187.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. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.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] package metafor: specify weights?
At the moment, there is no possibility of specifying the weights with the rma() function. While the main model fitting part could be easily adapted to incorporate user-specified weights, the problem comes in with all the additional statistics that can be computed based on a fitted model. How should the predict() function now work? What would be the definition of I^2 now? How would one generalize the influence and outlier statistics to that case? Just to give some examples. Of course, I could leave out such things when the user has specified the weights, but then things also get confusing for the user. For example, it is already less than ideal that you can only use the trim and fill method with models that do not incorporate moderators. Nobody has (as of yet) generalized the trim and fill method to that case. But when there are too many "special cases", the package becomes unusable. I did consider user-specified weights at one point, but it opened up so many cans of worms that I preferred to quickly put the lid pack on those cans. That item is written down in my to-do list, but to be honest, it is somewhere at the very end of that list. If you are hesitant to combine the results from those two types of studies, what about simply using a moderator to distinguish the two groups? Best, Wolfgang -- Wolfgang Viechtbauer, Ph.D., Statistician Department of Psychiatry and Psychology School for Mental Health and Neuroscience Faculty of Health, Medicine, and Life Sciences Maastricht University, P.O. Box 616 (VIJV1) 6200 MD Maastricht, The Netherlands +31 (43) 388-4170 | http://www.wvbauer.com > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] > On Behalf Of Anke Stein > Sent: Thursday, May 24, 2012 15:59 > To: r-help@r-project.org > Subject: [R] package metafor: specify weights? > > Dear R-experts, > Dear Wolfgang, > > Weighted model fitting in metafor uses the inverse of the study specific > variances as weights. > I am wondering if it is possible to specify different weights. > > In my meta-analysis, there are two types of studies with (intrinsic) > differences in their range of sample sizes (which are used to calculate > the variances of Fisher's z). > I would like to try normalizing the sample sizes within each set of the > two study types and use these normalized sample sizes as weights. > Would that be possible with rma()? So far, I only found the option > "weighted = TRUE/FALSE", but no possibility to specify which weights > should be used. > > Many thanks in advance, > Anke > > > > > > > -- > __ > Anke Stein (Dipl.-Biol.) > > Biodiversity, Macroecology & Conservation Biogeography Head Prof. Dr. > Holger Kreft Georg-August University of Göttingen Büsgenweg 2 | 37077 > Göttingen | Germany > > phone +49(0)551-39-13761 > fax +49(0)551-39-3618 > ast...@uni-goettingen.de > http://www.uni-goettingen.de/biodiversity > > __ > 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] set tkscale by tkentry
I believe that what is happening is that when you try to edit the entry widget any intermediate values get sent to the slider widget which then checks to see if they are in the allowable range and if it is not then it sets the value to either the minimum and maximum and sends that back to the entry widget while you are still trying to edit it. Even if you highlight a single digit and try to replace it with a different digit it first deletes the highlighted digit resulting in a number smaller than the minimum of the slider which then updates the entry widget to the minimum before the new digit can go in, then adding the new digit makes it larger than the slider maximum. There may be a way to slow down the updates or validations so that you can edit the entry before the slider tries to validate and change the value. Or you can just include 0 as the minimum of the slider so that the intermediate values in the entry widget are in the allowable range. On Thu, May 24, 2012 at 2:20 AM, Alexander wrote: > Hi, I am working under Windows and I am using R2.11 > I want to use tkscale in my GUI. As the interval is quite big, I can't set > the scale to a certain specific value. Therefore I want to add tkentry to > allow the user to set tkscale to a certain value. > > Here is the code > > library(tcltk) > > tt<-tktoplevel() > tkpack(m1<-tkscale(tt,from=306870.00, to=3026741, label="alpha", > variable="varalpha",showvalue=TRUE, resolution=1, > orient="horiz"),side="bottom") > tkpack(tkentry(tt,width="10",textvariable="varalpha",validate="key",validatecommand="string > is double %P"),side="bottom") > > As you can see, varalpha is a global variable (I think in tcl, not in R), > and if you change the parameter by the scale, the value in tkentry is > updated. If I want to set now the value of varalpha by tkentry, it bugs. > Does anyone have an idea, why? I already tried the interval from 0 to any > number, which works fine... > > library(tcltk) > > tt<-tktoplevel() > tkpack(m1<-tkscale(tt,from=0.00, to=1000, label="alpha", > variable="varalpha",showvalue=TRUE, resolution=1, > orient="horiz"),side="bottom") > tkpack(tkentry(tt,width="10",textvariable="varalpha",validate="key",validatecommand="string > is double %P"),side="bottom") > > Thanks in advance, > Alexander > > -- > View this message in context: > http://r.789695.n4.nabble.com/set-tkscale-by-tkentry-tp4631174.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. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.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] Using NA as a break point for indicator variable?
> Does anyone have any thoughts on how to code this, perhaps using the NA > values as a "break point"? You can count the cumulative number of NA breakpoints in a vector with cumsum(is.na(vector)), as in > cbind(d, LeakNo=with(d, cumsum(is.na(lon)|is.na(lat)|is.na(CH4 lon lat CH4 LeakNo 1 -71.11954 42.35068 2.595834 0 2 -71.11954 42.35068 2.595688 0 3 NA NA NA 1 4 NA NA NA 2 5 NA NA NA 3 6 -71.11948 42.35068 2.435762 3 7 -71.11948 42.35068 2.491003 3 8 NA NA NA 4 9 -71.11930 42.35068 2.464475 4 10 -71.11932 42.35068 2.470865 4 Add 1 if you want to start with 1. If you only want to increase the count after each sequence of NA's then you could use rle() or > na <- with(d, is.na(lon)|is.na(lat)|is.na(CH4)) > cbind(d, LeakNo=cumsum(c(TRUE, na[-1] < na[-length(na)]))) lon lat CH4 LeakNo 1 -71.11954 42.35068 2.595834 1 2 -71.11954 42.35068 2.595688 1 3 NA NA NA 1 4 NA NA NA 1 5 NA NA NA 1 6 -71.11948 42.35068 2.435762 2 7 -71.11948 42.35068 2.491003 2 8 NA NA NA 2 9 -71.11930 42.35068 2.464475 3 10 -71.11932 42.35068 2.470865 3 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On > Behalf > Of Max Brondfield > Sent: Wednesday, May 23, 2012 1:42 PM > To: r-help@r-project.org > Subject: [R] Using NA as a break point for indicator variable? > > Hi all, > I am working with a spatial data set for which I am only interested in high > concentration values ("leaks"). The low values (< 90th percentile) have > already been turned into NA's, leaving me with a matrix like this: > > < CH4_leak > > lonlatCH4 > 1 -71.11954 42.35068 2.595834 > 2 -71.11954 42.35068 2.595688 > 3 NA NA NA > 4 NA NA NA > 5 NA NA NA > 6 -71.11948 42.35068 2.435762 > 7 -71.11948 42.35068 2.491003 > 8 NANA NA > 9 -71.11930 42.35068 2.464475 > 10 -71.11932 42.35068 2.470865 > > Every time an NA comes up, it means the "leak" is gone, and the next valid > value would represent a different leak (at a different location). My goal > is to tag all of the remaining values with an indicator variable to > spatially distinguish the leaks. I am envisioning a simple numeric > indicator such as: > > lonlatCH4leak_num > 1 -71.11954 42.35068 2.595834 1 > 2 -71.11954 42.35068 2.595688 1 > 3 NA NA NA NA > 4 NA NA NA NA > 5 NA NA NA NA > 6 -71.11948 42.35068 2.435762 2 > 7 -71.11948 42.35068 2.491003 2 > 8 NANA NA NA > 9 -71.11930 42.35068 2.064475 3 > 10 -71.11932 42.35068 2.070865 3 > > Does anyone have any thoughts on how to code this, perhaps using the NA > values as a "break point"? The data set is far too large to do this > manually, and I must admit I'm completely at a loss. Any help would be much > appreciated! Best, > > Max > > [[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] transform 1 col to 2 col
See insertion below -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 5/24/12 1:43 AM, "Jim Lemon" wrote: >On 05/24/2012 05:10 PM, Soheila Khodakarim wrote: >> Dear All >> >> How can I transform 1 column to 2 columns in R? >> >> 211217_s_at >> >> GO:0005249 >> >> 211217_s_at >> >> GO:0005251 >> >> 211217_s_at >> >> GO:0005515 >> >> 211217_s_at >> >> GO:0015271 >> >>211217_s_at >> >>GO:0030955 >> >> >> >> >> >> 211217_s_at >> >> GO:0005249 >> >> 211217_s_at >> >> GO:0005251 >> >>211217_s_at >> >> GO:0005515 >> >> 211217_s_at >> >> GO:0015271 >> >>211217_s_at >> >> GO:0030955 >> >Hi Soheila, >Let us begin with the assumption that your "col" is roughly what you >posted, and that text resides in a file named "inscrutable.dat". Let us >first read that text into a data frame in R: > >inscrutatble.df<-read.table("inscrutable.dat") > inscrutable.df > V1 > >1 211217_s_at >2 GO:0005249 >3 211217_s_at >4 GO:0005251 >5 211217_s_at >6 GO:0005515 >7 211217_s_at >8 GO:0015271 >9 211217_s_at >10 GO:0030955 >11 211217_s_at >12 GO:0005249 >13 211217_s_at >14 GO:0005251 >15 211217_s_at >16 GO:0005515 >17 211217_s_at >18 GO:0015271 >19 211217_s_at >20 GO:0030955 > >Now we will further assume that you want to place every odd element in >one "col" and every even element in the other "col". Or perhaps: matrix(inscrutable.df$V1, ncol=2, byrow=TRUE) or data.frame(matrix(inscrutable.df$V1, ncol=2, byrow=TRUE)) to get a data frame. -Don > >inscrutable2.df<-data.frame( > col1=inscrutable.df$V1[seq(1,length(inscrutable.df$V1),by=2)], > col2=inscrutable.df$V1[seq(2,length(inscrutable.df$V1),by=2)]) >inscrutable2.df > col1 col2 >1 211217_s_at GO:0005249 >2 211217_s_at GO:0005251 >3 211217_s_at GO:0005515 >4 211217_s_at GO:0015271 >5 211217_s_at GO:0030955 >6 211217_s_at GO:0005249 >7 211217_s_at GO:0005251 >8 211217_s_at GO:0005515 >9 211217_s_at GO:0015271 >10 211217_s_at GO:0030955 > >There we are - easy as pie. > >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-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] Upper an lower bound in linear programing
Hello everyone! I´m trying to solve a linear optimization problem with r using the LPsolve function and i want to set upper and lower bounds for the obejctive function. Any idea´s on how to do this?? -- View this message in context: http://r.789695.n4.nabble.com/Upper-an-lower-bound-in-linear-programing-tp4631202.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] package metafor: specify weights?
Dear R-experts, Dear Wolfgang, Weighted model fitting in metafor uses the inverse of the study specific variances as weights. I am wondering if it is possible to specify different weights. In my meta-analysis, there are two types of studies with (intrinsic) differences in their range of sample sizes (which are used to calculate the variances of Fisher's z). I would like to try normalizing the sample sizes within each set of the two study types and use these normalized sample sizes as weights. Would that be possible with rma()? So far, I only found the option "weighted = TRUE/FALSE", but no possibility to specify which weights should be used. Many thanks in advance, Anke -- __ Anke Stein (Dipl.-Biol.) Biodiversity, Macroecology & Conservation Biogeography Head Prof. Dr. Holger Kreft Georg-August University of Göttingen Büsgenweg 2 | 37077 Göttingen | Germany phone +49(0)551-39-13761 fax +49(0)551-39-3618 ast...@uni-goettingen.de http://www.uni-goettingen.de/biodiversity __ 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] Question on if i am allowed to do something
I see. It is decided then. As it might be possible that the license might be violated, I certainly will not include R in my programme. Thank you all for your help, you've been very helpful. --- Giannis Mamalikidis e-mail: giannis_mamaliki...@msn.com Member of the STAINS Research Group (STAtistics & INformation Systems Group) of the Aristotle University of Thessaloniki Site: http://stains.csd.auth.gr B.Sc Student in Electrical and Mechanical Engineering __ 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] building transformation macro
On May 24, 2012, at 9:06 AM, David Winsemius wrote: On May 24, 2012, at 4:43 AM, davv.vikas wrote: Hi All, I am writing a macro which perform following functionality 1) take a file say excel or csv 2)read all the labels 3) dynamically genrate square,square-root,log of each column with appropiate column name and write it to a new file, i am stuck at dynamically genrateing the column names like Revenue_square etc. Exmaple " i have a excel with 3 columns Revenue, cost & profit Now my Macro should be able to read the values and for each coulmn and perform square,square root and log of Revenue, cost & profit and write to excel with column names as Revenue_square,Revenue_squareroot, Revenue_log cost_square etc... Below is my code test$Rev_square = test[c(1)]^2 I would use test[[1]]^2. Generally one wants to pull the vector out of the list "container". data=read.delim2("//ARLMSAN01/CTRX_Data/vikasK.sharma/Desktop/ balancesheet_example.csv",header=T,sep=",") headings = names(data) show (headings) HEADINGS = toupper(headings) # Perhaps replacing your for-loop body with this for ( i in 1:length(HEADINGS)) { assign(paste(HEADINGS[i], "Rev_Sqr", sep=""), data[[i]]^2) } The generalization to additional transformations seems obvious. The solution offered in StackOverflow by Greg Snow is better. AND please stop cross-posting. It is not specifically deprecated on SO, but it is on Rhelp. -- David Winsemius, MD 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] How to avoid numbering while printing list element?
Will this do it for you: > x <- list(1,2,3) > x[[1]] [1] 1 > cat(x[[1]], '\n') 1 > On Thu, May 24, 2012 at 5:50 AM, Manish Gupta wrote: > HI, > > I am printing list element. > > print(desc[[2]]) > [1] XXXxx > > I want to remove [1] > Output should be > XXXxx > > How can i implement it? > > Thanks > > > -- > View this message in context: > http://r.789695.n4.nabble.com/How-to-avoid-numbering-while-printing-list-element-tp4631180.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. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ 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 retrieve value form dataframe by Key?
Hello, Try ?subset subset(dtfr, Col1 == "xxx") ?"[" dtfr[Col1 == "xxy", ] Hope this helps, Rui Barradas Em 24-05-2012 11:00, Manish Gupta escreveu: Date: Thu, 24 May 2012 01:09:58 -0700 (PDT) From: Manish Gupta To:r-help@r-project.org Subject: [R] How to retrieve value form dataframe by Key? Message-ID:<1337846998835-4631173.p...@n4.nabble.com> Content-Type: text/plain; charset=us-ascii Hi, I have one datagrame Col1 Col2Col3 Col3 xxx erThis is third record of 1st line. This is 4th record of first line. xxy erThis is third record of 1st line. This is 4th record of second line xyx erThis is third record of 1st line. This is 4th record of third line I want to get records corresponding to key (here Col1). How can i implement it? e.g. if i key in xxx i should get xxx erThis is third record of 1st line. This is 4th record of first line. if i key in xxy i should get xxy erThis is third record of 1st line. This is 4th record of second line Regards -- View this message in context:http://r.789695.n4.nabble.com/How-to-retrieve-value-form-dataframe-by-Key-tp4631173.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] Question on if i am allowed to do something
Your answers were very helpful, thanks :) But, you see, the program I'm writing will be a team effort - and while I have not problem on making this program an open source - we don't really see eye to eye with the team on this. So no source code will be available - just freeware. R will not be at all altered! the only thing that R will do is be used to show a couple of plots, and make some t.test and ks.test. Does that qualify as a derivative work of R? Or would derivative work be if I altered R's Source code? I've read the license but English is not my Native language, hence some terminology is unclear to me. And I'm just an undergraduate with no income in the middle of an economy crisis. I don't really see me paying a lawyer to ask the question. --- Giannis Mamalikidis e-mail: giannis_mamaliki...@msn.com Member of the STAINS Research Group (STAtistics & INformation Systems Group) of the Aristotle University of Thessaloniki Site: http://stains.csd.auth.gr B.Sc Student in Electrical and Mechanical Engineering --- -Original Message- From: Marc Schwartz Sent: Thursday, May 24, 2012 3:58 AM To: Duncan Murdoch Cc: Giannis Mamalikidis ; r-help@r-project.org Subject: Re: [R] Question on if i am allowed to do something On May 23, 2012, at 6:51 PM, Duncan Murdoch wrote: On 12-05-23 1:31 PM, Giannis Mamalikidis wrote: Hello all. I would like to know: provided that I absolutely state that R is not mine and I also include the R’s License which will be shown so people know R and R’s license, (provided the above) am I allowed to include R’s folder (the folder that has its binaries) on my freeware program or not? R is licensed under the GPL v 2 or 3. The recommended packages that come with it are separately licensed, but are also (mainly) GPL. So just follow the terms of that license, which is included with R. Duncan Murdoch Just to add on to Duncan's reply, you might want to review the GPL FAQ (and possibly consult a lawyer): http://www.gnu.org/licenses/gpl-faq.html The use and distribution of R within your 'freeware' can place certain obligations on you, including possibly requiring that you license your application with a GPL compatible license and make the source code of your application available in situations where your application is 'linked' to R. If you do not intend to make your own source code available, you should be very clear on what the GPL requires of you before proceeding. Regards, Marc Schwartz __ 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] edgeR: design matrix setup
On 05/23/2012 07:24 AM, rmje wrote: I have a data frame like this: T0h T0.25h T0.5h T1h T2h T3h T6h T12h T24h T48h C0h C0.25h C0.5h C1h C2h C3h C6h C12h C24h C48h NM_001001130 68 9556 43 66 62 68 90 63 89 65 8558 49 81 49 76 73 48 77 NM_001001152 7912952 50 24 45 130 154 112 147 56 8567 33 52 31 93 77 65 109 ... ... ... It has about 29000 rows and 20 columns (incl col.name). How would you set this up in edgeR if you want to compare the 10 first with the 10 last? Please ask questions about Bioconductor packages on the Bioconductor mailing list http://bioconductor.org/help/mailing-list/ There have been recent time-course questions on the mailing list, so it's good to browse recents posts. Remember to include sessionInfo(). Have you looked at the extensive vignette for this package http://bioconductor.org/packages/2.10/bioc/html/edgeR.html and limma, and taken a stab at answering your own question? It helps to see where you're coming from...; for this particular question and if you're statistically naive it might help to have a conversation with someone local who is a little versed in statistics. Martin -- View this message in context: http://r.789695.n4.nabble.com/edgeR-design-matrix-setup-tp4631094.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. -- Computational Biology Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: M1-B861 Telephone: 206 667-2793 __ 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] building transformation macro
On May 24, 2012, at 4:43 AM, davv.vikas wrote: Hi All, I am writing a macro which perform following functionality 1) take a file say excel or csv 2)read all the labels 3) dynamically genrate square,square-root,log of each column with appropiate column name and write it to a new file, i am stuck at dynamically genrateing the column names like Revenue_square etc. Exmaple " i have a excel with 3 columns Revenue, cost & profit Now my Macro should be able to read the values and for each coulmn and perform square,square root and log of Revenue, cost & profit and write to excel with column names as Revenue_square,Revenue_squareroot, Revenue_log cost_square etc... Below is my code test$Rev_square = test[c(1)]^2 I would use test[[1]]^2. Generally one wants to pull the vector out of the list "container". data=read.delim2("//ARLMSAN01/CTRX_Data/vikasK.sharma/Desktop/ balancesheet_example.csv",header=T,sep=",") headings = names(data) show (headings) HEADINGS = toupper(headings) # Perhaps replacing your for-loop body with this for ( i in 1:length(HEADINGS)) { assign(paste(HEADINGS[i], "Rev_Sqr", sep=""), data[[i]]^2) } The generalization to additional transformations seems obvious. -- David Winsemius, MD 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] Question on if i am allowed to do something
Giannis, While your economic situation is appreciated, that you plan to give away your application and not generate any income from it, is irrelevant to the question. The nuances as to how you are interfacing with R and that you plan to include R (importantly, possibly only a part of it) as a component of your application are. Not altering R's source code is similarly only one part of the issue, though if you were, you would be clearly crossing the line to a derivative work. However, even in that scenario there are other considerations as to whether or not those changes impact your application versus just the modified R code. You are asking a legal question on a list that is focused on technical considerations and expecting the list participants to provide free legal advice, based upon which you may be making life altering decisions. While you may not be able to afford to pay a lawyer now, you and you partners may find yourselves needing to pay a lawyer in the future to defend yourselves against a lawsuit for copyright and licensing infringement. A situation, by the way, that won't be at all helpful to your future career. You may wish to see if there are any lawyers locally or at your institution with specific experience in this domain who may be will to provide you with pro bono advice. Lacking expert legal advice, the only thing we can tell you is "don't do it" because you may be violating the GPL license and therefore may expose yourself and your partners to future legal risk. Regards, Marc On May 24, 2012, at 7:23 AM, Giannis Mamalikidis wrote: > Your answers were very helpful, thanks :) > > But, you see, the program I'm writing will be a team effort - and while I > have not problem on making this program an open source - we don't really see > eye to eye with the team on this. > So no source code will be available - just freeware. > > R will not be at all altered! the only thing that R will do is be used to > show a couple of plots, and make some t.test and ks.test. > > Does that qualify as a derivative work of R? Or would derivative work be if I > altered R's Source code? > I've read the license but English is not my Native language, hence some > terminology is unclear to me. > > And I'm just an undergraduate with no income in the middle of an economy > crisis. I don't really see me paying a lawyer to ask the question. > > --- > Giannis Mamalikidis > e-mail: giannis_mamaliki...@msn.com > Member of the STAINS Research Group > (STAtistics & INformation Systems Group) of the > Aristotle University of Thessaloniki > Site: http://stains.csd.auth.gr > B.Sc Student in Electrical and Mechanical Engineering > --- > -Original Message- From: Marc Schwartz > Sent: Thursday, May 24, 2012 3:58 AM > To: Duncan Murdoch > Cc: Giannis Mamalikidis ; r-help@r-project.org > Subject: Re: [R] Question on if i am allowed to do something > > > On May 23, 2012, at 6:51 PM, Duncan Murdoch wrote: > >> On 12-05-23 1:31 PM, Giannis Mamalikidis wrote: >>> Hello all. >>> >>> I would like to know: provided that I absolutely state that R is not mine >>> and I also include the R’s License which will be shown so people know R and >>> R’s license, >>> (provided the above) am I allowed to include R’s folder (the folder that has >>> its binaries) on my freeware program or not? >> >> R is licensed under the GPL v 2 or 3. The recommended packages that come >> with it are separately licensed, but are also (mainly) GPL. So just follow >> the terms of that license, which is included with R. >> >> Duncan Murdoch > > > Just to add on to Duncan's reply, you might want to review the GPL FAQ (and > possibly consult a lawyer): > > http://www.gnu.org/licenses/gpl-faq.html > > The use and distribution of R within your 'freeware' can place certain > obligations on you, including possibly requiring that you license your > application with a GPL compatible license and make the source code of your > application available in situations where your application is 'linked' to R. > If you do not intend to make your own source code available, you should be > very clear on what the GPL requires of you before proceeding. > > Regards, > > Marc Schwartz > __ 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] Using NA as a break point for indicator variable?
Hello, Assuming that 'd' is your original data.frame and that you've set entire rows to NA, try this d$leak_num <- NA ix <- !is.na(d[, 1]) # any column will do, entire row is NA ## alternative, if other rows may have NAs, due to something else #ix <- apply(d, 1, function(x) all(!is.na(x))) r <- rle(ix) v <- cumsum(r$values) d$leak_num[ix] <- rep(v[r$values], r$lengths[r$values]) d Hope this helps, Rui Barradas Em 24-05-2012 11:00, Max Brondfield escreveu: Date: Wed, 23 May 2012 16:42:02 -0400 From: Max Brondfield To:r-help@r-project.org Subject: [R] Using NA as a break point for indicator variable? Message-ID: Content-Type: text/plain Hi all, I am working with a spatial data set for which I am only interested in high concentration values ("leaks"). The low values (< 90th percentile) have already been turned into NA's, leaving me with a matrix like this: < CH4_leak lonlatCH4 1 -71.11954 42.35068 2.595834 2 -71.11954 42.35068 2.595688 3 NA NA NA 4 NA NA NA 5 NA NA NA 6 -71.11948 42.35068 2.435762 7 -71.11948 42.35068 2.491003 8 NANA NA 9 -71.11930 42.35068 2.464475 10 -71.11932 42.35068 2.470865 Every time an NA comes up, it means the "leak" is gone, and the next valid value would represent a different leak (at a different location). My goal is to tag all of the remaining values with an indicator variable to spatially distinguish the leaks. I am envisioning a simple numeric indicator such as: lonlatCH4leak_num 1 -71.11954 42.35068 2.595834 1 2 -71.11954 42.35068 2.595688 1 3 NA NA NA NA 4 NA NA NA NA 5 NA NA NA NA 6 -71.11948 42.35068 2.435762 2 7 -71.11948 42.35068 2.491003 2 8 NANA NA NA 9 -71.11930 42.35068 2.064475 3 10 -71.11932 42.35068 2.070865 3 Does anyone have any thoughts on how to code this, perhaps using the NA values as a "break point"? The data set is far too large to do this manually, and I must admit I'm completely at a loss. Any help would be much appreciated! Best, Max [[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] Question on if i am allowed to do something
On 12-05-24 8:23 AM, Giannis Mamalikidis wrote: Your answers were very helpful, thanks :) But, you see, the program I'm writing will be a team effort - and while I have not problem on making this program an open source - we don't really see eye to eye with the team on this. So no source code will be available - just freeware. R will not be at all altered! the only thing that R will do is be used to show a couple of plots, and make some t.test and ks.test. Does that qualify as a derivative work of R? Or would derivative work be if I altered R's Source code? I would say that makes your program a derivative work, so you should not distribute R as part of it. But I am not a lawyer; in fact, I'm an interested party: a copyright holder on parts of R. So I don't know if the law is on my side, but you can be assured that you are violating the wishes of at least one R copyright holder if you do distribute it in a closed source package. Duncan Murdoch I've read the license but English is not my Native language, hence some terminology is unclear to me. And I'm just an undergraduate with no income in the middle of an economy crisis. I don't really see me paying a lawyer to ask the question. --- Giannis Mamalikidis e-mail: giannis_mamaliki...@msn.com Member of the STAINS Research Group (STAtistics& INformation Systems Group) of the Aristotle University of Thessaloniki Site: http://stains.csd.auth.gr B.Sc Student in Electrical and Mechanical Engineering --- -Original Message- From: Marc Schwartz Sent: Thursday, May 24, 2012 3:58 AM To: Duncan Murdoch Cc: Giannis Mamalikidis ; r-help@r-project.org Subject: Re: [R] Question on if i am allowed to do something On May 23, 2012, at 6:51 PM, Duncan Murdoch wrote: On 12-05-23 1:31 PM, Giannis Mamalikidis wrote: Hello all. I would like to know: provided that I absolutely state that R is not mine and I also include the R’s License which will be shown so people know R and R’s license, (provided the above) am I allowed to include R’s folder (the folder that has its binaries) on my freeware program or not? R is licensed under the GPL v 2 or 3. The recommended packages that come with it are separately licensed, but are also (mainly) GPL. So just follow the terms of that license, which is included with R. Duncan Murdoch Just to add on to Duncan's reply, you might want to review the GPL FAQ (and possibly consult a lawyer): http://www.gnu.org/licenses/gpl-faq.html The use and distribution of R within your 'freeware' can place certain obligations on you, including possibly requiring that you license your application with a GPL compatible license and make the source code of your application available in situations where your application is 'linked' to R. If you do not intend to make your own source code available, you should be very clear on what the GPL requires of you before proceeding. Regards, Marc Schwartz __ 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 cbind (or any function) across all components in a list
The combination of column means in cbind is what I am trying to do. I just don't know how to do it for every component (2 in this case) of the list. I've been able to work with R to apply a function across all components when there is just one list but I am having trouble working with multiple lists at the same time. -- View this message in context: http://r.789695.n4.nabble.com/applying-cbind-or-any-function-across-all-components-in-a-list-tp4631128p4631187.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] quotes in Rscript -e through system
I figured out how to use quotes and parentheses when using Rscript -e (on a bash shell): Rscript -e write\(1,\"a.txt\"\) --> Question 1: why do the parentheses need to be escaped in the shell? (More a shell than an R question) Then I figured out how to use quotes and parentheses when calling Rscript through system: system('Rscript -e write\\(1,\\\'a.txt\\\'\\)') --> Question 2: why exactly is it necessary to use double and triple escapes? I kinda understand it, but not completely. Thanks for your insights! Adi -- View this message in context: http://r.789695.n4.nabble.com/quotes-in-Rscript-e-through-system-tp4631186.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] How to avoid numbering while printing list element?
HI, I am printing list element. print(desc[[2]]) [1] XXXxx I want to remove [1] Output should be XXXxx How can i implement it? Thanks -- View this message in context: http://r.789695.n4.nabble.com/How-to-avoid-numbering-while-printing-list-element-tp4631180.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] set tkscale by tkentry
Hi, I am working under Windows and I am using R2.11 I want to use tkscale in my GUI. As the interval is quite big, I can't set the scale to a certain specific value. Therefore I want to add tkentry to allow the user to set tkscale to a certain value. Here is the code library(tcltk) tt<-tktoplevel() tkpack(m1<-tkscale(tt,from=306870.00, to=3026741, label="alpha", variable="varalpha",showvalue=TRUE, resolution=1, orient="horiz"),side="bottom") tkpack(tkentry(tt,width="10",textvariable="varalpha",validate="key",validatecommand="string is double %P"),side="bottom") As you can see, varalpha is a global variable (I think in tcl, not in R), and if you change the parameter by the scale, the value in tkentry is updated. If I want to set now the value of varalpha by tkentry, it bugs. Does anyone have an idea, why? I already tried the interval from 0 to any number, which works fine... library(tcltk) tt<-tktoplevel() tkpack(m1<-tkscale(tt,from=0.00, to=1000, label="alpha", variable="varalpha",showvalue=TRUE, resolution=1, orient="horiz"),side="bottom") tkpack(tkentry(tt,width="10",textvariable="varalpha",validate="key",validatecommand="string is double %P"),side="bottom") Thanks in advance, Alexander -- View this message in context: http://r.789695.n4.nabble.com/set-tkscale-by-tkentry-tp4631174.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] building transformation macro
Hi All, I am writing a macro which perform following functionality 1) take a file say excel or csv 2)read all the labels 3) dynamically genrate square,square-root,log of each column with appropiate column name and write it to a new file, i am stuck at dynamically genrateing the column names like Revenue_square etc. Exmaple " i have a excel with 3 columns Revenue, cost & profit Now my Macro should be able to read the values and for each coulmn and perform square,square root and log of Revenue, cost & profit and write to excel with column names as Revenue_square,Revenue_squareroot, Revenue_log cost_square etc... Below is my code test$Rev_square = test[c(1)]^2 data=read.delim2("//ARLMSAN01/CTRX_Data/vikasK.sharma/Desktop/balancesheet_example.csv",header=T,sep=",") headings = names(data) show (headings) HEADINGS = toupper(headings) for ( i in 1:length(HEADINGS)) { show(i) data$Rev_square=data[c(i)]^2 show(data$Rev_square) names(data) } -- View this message in context: http://r.789695.n4.nabble.com/building-transformation-macro-tp4631175.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] plot(summary) quantreg - Not all outputs needed
Hi Folks, I am currently trying to present some results I obtained by using the quantreg package developed by Roger Koenker. After calculating fit<-summary(rq(Y~X1+X2, tau=2:98/100) ) the function plot(fit) presents a really nice the results by showing the values for all "regressors" in the given interval tau. But in my case, I only need the output of a single variable, say X1 and I am not interested in plotting the others. Is there a way to hide the other graphics? Thank you very much for your help, Cheers Stefan -- View this message in context: http://r.789695.n4.nabble.com/plot-summary-quantreg-Not-all-outputs-needed-tp4631184.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] Mistake:number of cluster centres must lie between 1 and nrow(x)
Dear List: I write codes to get"Mean and standard deviations (Std.) for EM estimatesâ500 samples from the FM-ST model"(R package mixsmsn).It has a mistake:"number of cluster centres must lie between 1 and nrow(x) ". my codes below: mu1 <- c(2,2) Sigma1 <- matrix(c(1.5,0,0,1.5), 2,2) shape1 <-c(-5,10) nu1 <- 4 mu2 <- c(-2,-1) Sigma2 <- matrix(c(1.5,0,0,1.5), 2,2) shape2 <-c(-5,10) nu2 <- 4 pii<-c(0.6,0.4) arg1 = list(mu=mu1, Sigma=Sigma1, shape=shape1, nu=nu1) arg2 = list(mu=mu2, Sigma=Sigma2, shape=shape2, nu=nu2) y <- rmmix(n=500, p = pii, "Skew.t", list(arg1,arg2)) St.analysis <- smsn.mmix(y, nu=3, g=2, get.init = TRUE, criteria = TRUE, group = TRUE, family = "Skew.t",calc.im=TRUE) St.analysis ##skew t fit giving intial values for(i in 1:500){ mu1 <- c(0,0) Sigma1 <- matrix(c(1,0,0,1), 2,2) shape1 <-c(-1,5) nu1 <- 10 mu2 <- c(0,0) Sigma2 <- matrix(c(1,0,0,1), 2,2) shape2 <-c(-1,5) nu2 <- 10 pii<-c(0.5,0.5) mu<-list(mu1,mu2) #nu<-list(nu1,nu2) Sigma<-list(Sigma1,Sigma2) shape<-list(shape1,shape2) St.analysis[i] <- smsn.mix(y[i,],g=2,nu=10,mu=mu,shape=shape,pii = pii,get.init = TRUE, criteria = TRUE,group = TRUE, family = "Skew.t",calc.im=TRUE) print(St.analysis[i]) } -- View this message in context: http://r.789695.n4.nabble.com/Mistake-number-of-cluster-centres-must-lie-between-1-and-nrow-x-tp4631177.html Sent from the R help mailing list archive at Nabble.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.
Re: [R] select part of files from a list.files
Hi again Joshua. I tried your function. I think it's what I need. It works well in the small example of my first post. But I have difficulties to adapt it to my data. I'll try to give you another fake example with my real script and kind of data (you can just copy and paste it to try): ST1 <- data.frame(sensor1=rnorm(1:10),sensor2=c(NA,NA,NA,NA,NA,rnorm(6:10)),sensor3=c(1,NA,NA,4:10),sensor4=c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),date_time=(date())) write.table(ST1,"ST1_2012.csv",sep=";",quote=F, row.names = TRUE) ST2 <- data.frame(sensor1=c(NA,NA,NA,NA,NA,6:10),sensor2=rnorm(1:10),sensor3=c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),sensor4=c(1,NA,NA,4:10),date_time=(date())) write.table(ST2,"ST2_2012.csv",sep=";",quote=F, row.names = TRUE) ST3 <- data.frame(sensor1=c(1,NA,NA,4:10),sensor2=c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),sensor3=rnorm(1:10),sensor4=c(NA,NA,NA,NA,NA,6:10),date_time=(date())) write.table(ST3,"ST3_2012.csv",sep=";",quote=F, row.names = TRUE) ST4 <- data.frame(sensor1=c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),sensor2=c(1,NA,NA,4:10),sensor3=c(NA,NA,NA,NA,NA,6:10),sensor4=rnorm(1:10),date_time=(date())) write.table(ST4,"ST4_2012.csv",sep=";",quote=F, row.names = TRUE) filenames <- list.files(pattern="\\_2012.csv$") Sensors <- paste("sensor", 1:4,sep="") Stations <-substr(filenames,1,3) nsensors <- length(Sensors) nstations <- length(Stations) nobs <- nrow(read.table(filenames[1], header=TRUE,sep=";")) yr2008 <- array(NA,dim=c(nobs, nsensors, nstations)) for(i in seq_len(nstations)){ tmp <- read.table(filenames[i], header=TRUE, sep=";") yr2008[ , , i] <- as.matrix(tmp[, Sensors]) } dimnames(yr2008) <- list(seq.int(nobs), Sensors, Stations) cor1_5 <- lapply(Sensors, function(s) cor(yr2008[1:5, s, ],use="pairwise.complete.obs")) names(cor1_5) <- Sensors cor1_5 For the moment, it makes correlations between the same sensors of each file (only for a part of my data), whatever the number of NA or numeric data. I want it to do the same, but with your function: if (sum(!is.na(data[rows, ])) >= minpresent){ data } else {NULL} } I want it to give me the same correlation matrices for each sensors between my files, but I want it to calculate the correlation coefficient only if I have at least 3 numeric values (out of 5 in the example), and not whatever the number of these numeric values (just 1 or 2 for example). If there're less than 3 numeric values (1 or 2), give NA for correlation in the matrix. And if there're only NAs in the sensor data, do nothing with it (keep it and go to the next sensor). I tried to combinate your function with mine but it doesn't work. Hope you've understood. Thanks for your help! -- View this message in context: http://r.789695.n4.nabble.com/select-part-of-files-from-a-list-files-tp4630769p4631185.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] Optimizing problem R
On Wed, May 23, 2012 at 10:56:43PM -0700, kylmala wrote: > Hi, > > and thanks for replying this. Yes, you are right that the term > min(24/bb,26/cc) is actually min(bb/24,cc/26) - my mistake. But still I > don't get it. If the objective function is > > #min(m1,m2,m3) > f.obj <- c(1, 1, 1) > > #and we now that > # a+aa<=15 > # b+bb+bbb<=35 > # cc+ccc<=40 > > # so > > # 1a + 0b + 0c + 1aa + 0bb + 0cc + 0aaa + 0bbb + 0ccc <=15 > # 0a + 1b + 0c + 0aa + 1bb + 0cc + 0aaa + 1bbb + 0ccc <=35 > # 0a + 0b + 0c + 0aa + 0bb + 1cc + 0aaa + 0bbb + 1ccc <=40 > # and the matrix of numeric constraint coefficients is > > C = matrix(nrow=3, data=c(1,0,0,0,1,0,0,0,0,1,1,0,0,0,1,0,1,0,0,0,1)) > > #Constraits are: > > f.rhs <- c(15, 35, 40) > > #and > > f_dir = c("<=", "<=", "<=") > > Because the command lp is form lp ("max", f.obj, f.con, f.dir, f.rhs), how > can I get those other constraints in > > m1 <= a/10 > m1 <= b/11 > m2 <= aa/13 > m2 <= bb/12 > m2 <= cc/10 > m3 <= bb/24 > m3 <= cc/26 Hi. There are 12 variables, namely a, b, c, aa, bb, cc, aaa, bbb, ccc, m1, m2, m3 and 3 + 7 constraints. So, the matrix should be 10 times 12 and may be created as follows C11 <- rbind( c(1, 0, 0, 1, 0, 0, 0, 0, 0), c(0, 1, 0, 0, 1, 0, 0, 1, 0), c(0, 0, 0, 0, 0, 1, 0, 0, 1)) C12 <- matrix(0, nrow=3, ncol=3) A <- matrix(0, nrow=7, ncol=12) colnames(A) <- c("a", "b", "c", "aa", "bb", "cc", "aaa", "bbb", "ccc", "m1", "m2", "m3") A[1, c( "a", "m1")] <- c(1/10, -1) A[2, c( "b", "m1")] <- c(1/11, -1) A[3, c("aa", "m2")] <- c(1/13, -1) A[4, c("bb", "m2")] <- c(1/12, -1) A[5, c("cc", "m2")] <- c(1/10, -1) A[6, c("bb", "m3")] <- c(1/24, -1) A[7, c("cc", "m3")] <- c(1/26, -1) f.mat <- rbind(cbind(C11, C12), A) The variables "c" and "aaa" are not used at all. Is this correct? The other parameters and the call are library(lpSolve) f.obj <- c(0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1) f.rhs <- c(15, 35, 40, 0, 0, 0, 0, 0, 0, 0) f.dir = c("<=", "<=", "<=", rep(">=", times=7)) out <- lp("max", f.obj, f.mat, f.dir, f.rhs) out Success: the objective function is 2.612179 out$solution [1] 0.00 0.00 0.00 15.00 35.00 37.916667 0.00 [8] 0.00 0.00 0.00 1.153846 1.458333 The constraint on m1 requires only m1 <= min(a/10,b/11), however, since we maximize m1 + m2 + m3, the optimum will satisfy m1 = min(a/10,b/11). Similarly for m2, m3. Hope this helps. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] transform 1 col to 2 col
On 05/24/2012 05:10 PM, Soheila Khodakarim wrote: Dear All How can I transform 1 column to 2 columns in R? 211217_s_at GO:0005249 211217_s_at GO:0005251 211217_s_at GO:0005515 211217_s_at GO:0015271 211217_s_at GO:0030955 211217_s_at GO:0005249 211217_s_at GO:0005251 211217_s_at GO:0005515 211217_s_at GO:0015271 211217_s_at GO:0030955 Hi Soheila, Let us begin with the assumption that your "col" is roughly what you posted, and that text resides in a file named "inscrutable.dat". Let us first read that text into a data frame in R: inscrutatble.df<-read.table("inscrutable.dat") inscrutable.df V1 1 211217_s_at 2 GO:0005249 3 211217_s_at 4 GO:0005251 5 211217_s_at 6 GO:0005515 7 211217_s_at 8 GO:0015271 9 211217_s_at 10 GO:0030955 11 211217_s_at 12 GO:0005249 13 211217_s_at 14 GO:0005251 15 211217_s_at 16 GO:0005515 17 211217_s_at 18 GO:0015271 19 211217_s_at 20 GO:0030955 Now we will further assume that you want to place every odd element in one "col" and every even element in the other "col". inscrutable2.df<-data.frame( col1=inscrutable.df$V1[seq(1,length(inscrutable.df$V1),by=2)], col2=inscrutable.df$V1[seq(2,length(inscrutable.df$V1),by=2)]) inscrutable2.df col1 col2 1 211217_s_at GO:0005249 2 211217_s_at GO:0005251 3 211217_s_at GO:0005515 4 211217_s_at GO:0015271 5 211217_s_at GO:0030955 6 211217_s_at GO:0005249 7 211217_s_at GO:0005251 8 211217_s_at GO:0005515 9 211217_s_at GO:0015271 10 211217_s_at GO:0030955 There we are - easy as pie. 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] How to retrieve value form dataframe by Key?
Hi, I have one datagrame Col1 Col2Col3 Col3 xxx erThis is third record of 1st line. This is 4th record of first line. xxy erThis is third record of 1st line. This is 4th record of second line xyx erThis is third record of 1st line. This is 4th record of third line I want to get records corresponding to key (here Col1). How can i implement it? e.g. if i key in xxx i should get xxx erThis is third record of 1st line. This is 4th record of first line. if i key in xxy i should get xxy erThis is third record of 1st line. This is 4th record of second line Regards -- View this message in context: http://r.789695.n4.nabble.com/How-to-retrieve-value-form-dataframe-by-Key-tp4631173.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] Error while conecting to database.
Hi, I am getting following error message while retrieving records from mysql. *Error in if (count > 0L) { : missing value where TRUE/FALSE needed Calls: -> do.call -> -> Execution halted* con2 <- dbConnect(MySQL(), user="XXX", password="",dbname="", host="XXX.X.X.X") #if(nchar(disname) == 2) { disQuery= "select Desc from Tablename where Tableanem.field = " cat(paste("*",disname,"*","\n",sep="")) disQuery<-paste(disQuery, "'", disname ,"'", sep = "") cat(disQuery,"\n") cat(paste("\\textbf{",Col3[i],"?}\n",sep="")) cat("\\leavevmode","\n") cat("\\newline","\n") #cat(disQuery,"\n") #print(dbListTables(con2)) disres <- dbSendQuery(con2, disQuery) dout <- fetch(disres, n = 1) dataout <- ifelse(is.na(dout), "-" ,dout) cat(paste(dataout),"\n") * // getting error message here* } dbDisconnect(con2) May i know why I am getting above error message. Regards -- View this message in context: http://r.789695.n4.nabble.com/Error-while-conecting-to-database-tp4631169.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] Convolve 2 uniform distributed variables
take a look at the distr package library(distr) U <- Unif() U2 <- U + U # or U2 <- convpow(U, 2) plot(U2) # or curve(d(U2)(x), from = 0, to = 2) Best Matthias On 24.05.2012 08:29, c...@mail.tu-berlin.de wrote: Hi, I want to convolve 2 uniform distributed [0,1] variables, so that I get this graphic: http://de.wikipedia.org/wiki/Benutzer:Alfred_Heiligenbrunner/Liste_von_Verteilungsdichten_der_Summe_gleichverteilter_Zufallsvariabler (second graphic) if I do u1<-seq(0,1,length=100) fu1=dunif(u1,min=0,max=1) plot(u1,fu1,type="l",xlim=c(-2,2)) u2<-seq(0,1,length=100) fu2=dunif(u2,min=0,max=1) u1u2<-convolve(u1,rev(u1),typ="o") plot(u1u2) it does not work, could you help me please? The point is the convolve function I think, what do I have to type, to get the correct convolution of two uniform distributed [0,1] variables as to be seen in the second graphic in the given link? Thanks a lot __ 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. -- Prof. Dr. Matthias Kohl www.stamats.de __ 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] transform 1 col to 2 col
Dear All How can I transform 1 column to 2 columns in R? 211217_s_at GO:0005249 211217_s_at GO:0005251 211217_s_at GO:0005515 211217_s_at GO:0015271 211217_s_at GO:0030955 211217_s_at GO:0005249 211217_s_at GO:0005251 211217_s_at GO:0005515 211217_s_at GO:0015271 211217_s_at GO:0030955 Best Wishes, Soheila [[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] Convolve 2 uniform distributed variables
Hi, I want to convolve 2 uniform distributed [0,1] variables, so that I get this graphic: http://de.wikipedia.org/wiki/Benutzer:Alfred_Heiligenbrunner/Liste_von_Verteilungsdichten_der_Summe_gleichverteilter_Zufallsvariabler (second graphic) if I do u1<-seq(0,1,length=100) fu1=dunif(u1,min=0,max=1) plot(u1,fu1,type="l",xlim=c(-2,2)) u2<-seq(0,1,length=100) fu2=dunif(u2,min=0,max=1) u1u2<-convolve(u1,rev(u1),typ="o") plot(u1u2) it does not work, could you help me please? The point is the convolve function I think, what do I have to type, to get the correct convolution of two uniform distributed [0,1] variables as to be seen in the second graphic in the given link? Thanks a lot __ 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] Optimizing problem R
Hi, and thanks for replying this. Yes, you are right that the term min(24/bb,26/cc) is actually min(bb/24,cc/26) - my mistake. But still I don't get it. If the objective function is #min(m1,m2,m3) f.obj <- c(1, 1, 1) #and we now that # a+aa<=15 # b+bb+bbb<=35 # cc+ccc<=40 # so # 1a + 0b + 0c + 1aa + 0bb + 0cc + 0aaa + 0bbb + 0ccc <=15 # 0a + 1b + 0c + 0aa + 1bb + 0cc + 0aaa + 1bbb + 0ccc <=35 # 0a + 0b + 0c + 0aa + 0bb + 1cc + 0aaa + 0bbb + 1ccc <=40 # and the matrix of numeric constraint coefficients is C = matrix(nrow=3, data=c(1,0,0,0,1,0,0,0,0,1,1,0,0,0,1,0,1,0,0,0,1)) #Constraits are: f.rhs <- c(15, 35, 40) #and f_dir = c("<=", "<=", "<=") Because the command lp is form lp ("max", f.obj, f.con, f.dir, f.rhs), how can I get those other constraints in m1 <= a/10 m1 <= b/11 m2 <= aa/13 m2 <= bb/12 m2 <= cc/10 m3 <= bb/24 m3 <= cc/26 -- View this message in context: http://r.789695.n4.nabble.com/Optimizing-problem-R-tp4631048p4631168.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] Issues while using “lift.chart” and “adjProbScore” function from ”BCA” library
Dear List, Couple of issues while using functions from “BCA” library: 1. I am trying to use “lift.chart” function from “BCA” library, but facing issues while using model where model formula is passed as formula object in glm. When model formula is written as text, then it works fine. In my case input variables and target variables are going to change dynamically, so have to used formula as formula object as derived. Below is the sample code, taken from the package document to illustrate the issues library(BCA) data(CCS) CCS$Sample <- create.samples(CCS, est=0.4, val=0.4) CCSEst <- CCS[CCS$Sample == "Estimation",] #Fit glm model with formula written as text CCS.glm <- glm(MonthGive ~ DonPerYear + LastDonAmt + Region + YearsGive, family=binomial(logit), data=CCSEst) CCSVal <- CCS[CCS$Sample == "Validation",] lift.chart(c("CCS.glm"), data=CCSVal, targLevel="Yes", trueResp=0.01, type="incremental", sub="Validation") #Fit glm model with formula passed as formula object fm <- as.formula("MonthGive ~ DonPerYear + LastDonAmt + Region + YearsGive") CCS.glm12 <- glm(fm,family=binomial(logit), data=CCSEst) lift.chart(c("CCS.glm12"), data=CCSVal, targLevel="Yes", trueResp=0.01, type="incremental", sub="Validation") Following error occurs, Error in if (any(yvar1 != yvar1[1])) { : missing value where TRUE/FALSE needed Is there any way out to use formula object in the model and using “lift.chart” function 2. Issue using “adjProbScore” function from the “BCA” library. (adjProbScore(model="CCS.glm", data=CCSVal1, targLevel="Yes", trueResp=0.01)) Error in parse(text = paste("as.character(", ActiveDataSet(), "$", yvar, : :1:16: unexpected '$' 1: as.character( $ ^ Above error is thrown, am I doing anything wrong? Please correct. Also, as in the case-1 above, can we use model fitted with formula object in “adjProbScore” function. Thanks in advance! Ajit -- View this message in context: http://r.789695.n4.nabble.com/Issues-while-using-lift-chart-and-adjProbScore-function-from-BCA-library-tp4631158.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.