[R] Mapping help

Dear all, I have to make a last minute map...my thesis is due in a few days and our GIS lady fell ill, so my supervisor asked me to try in R., but I am a beginner. I've searched online, but have not found something at the global scale. I have a global dataset of dependent values (eg. rate of comm

Re: [R] Storing Big Matrix in R memory

Hello, If you can, you probably should upgrade for R version 3.0.1. Regards, Pascal 2013/8/27 Peter Maclean > I would like to store a big spatial weight matrix in R memory to do more > calculation. I know there are memory issue for 32 bit computer and I have > tried increasing the memory to

Re: [R] return the name of source.

See findSourceTraceback() of R.utils. /Henrik On Mon, Aug 26, 2013 at 10:58 PM, peter dalgaard wrote: > > On Aug 27, 2013, at 07:12 , nevil amos wrote: > >> Is there a fuction that will allow me to retrun the filename for a script >> from within that script. > > Not a standard one, but you shou

Re: [R] return the name of source.

On Aug 27, 2013, at 07:12 , nevil amos wrote: > Is there a fuction that will allow me to retrun the filename for a script > from within that script. Not a standard one, but you should be able to get at it via a bit of breaking and entering: sys.status() and friends (sys.calls, sys.frames) give

[R] return the name of source.

Is there a fuction that will allow me to retrun the filename for a script from within that script. fir instance If I have a script "myscript.r": FileName<-unknown.fucntion() print(FileName) and run it source("myscript.r") will return "myscript.r" Thanks Nevil Amos [[alternative

Re: [R] Storing Big Matrix in R memory

I would like to store a big spatial weight matrix in R memory to do more calculation. I know there are memory issue for 32 bit computer and I have tried increasing the memory to maximum without success. I am using R version 2.15.2 and window vista. The data has about 15,000 observations and I wo

[R] predict.glm with constant non-zero response intercept

hi all -- i'm running into a strange problem that i can't seem to easily get around, but i'm probably just missing something obvious. i have a model to which some data is fit using glm with no intercept term (using my data variables "x" and "y" and a specific link function "mylink"):

Re: [R] 'format' behaviour in a 'apply' call depending on 'options(digits = K)'

Thanks to Aleksey Vorona and Duncan Murdoch, this bug is now fixed in R-devel! Mathieu. Le 08/01/2013 01:47 PM, William Dunlap a écrit : You could report it as a bug at https://bugs.r-project.org/bugzilla3/ Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message

Re: [R] Sorting data

Hi, Hope this is what you meant.. with(dat1,mean(V21[(V2==1|V2==0) & V24<25],na.rm=TRUE))  #[1] 2.8125  sapply(0:1,function(i) with(dat1,mean(V21[V2==i & V24<25],na.rm=TRUE))) #[1] 2.75 3.00 ". who are in condition 1 or 0 (V2) and then vice versa ..."  dat1[,21] # [1]   NA 3.40 3.00 3

Re: [R] labcurve - use of putKey function

Hello, That is why I changed 2 lines in the code. Because points are misplaced if you keep points with the lines. Regards, Pascal 2013/8/27 Igor Ribeiro > Because I'm already controlling points using points function. So I don't > want labcurve to change anything on the lines... Just draw the

Re: [R] sbusetting data by rows (every 69 rows)

Hi R.L. No problem. Try this: #slightly modified the example: set.seed(24)  dat1<- as.data.frame(matrix(sample(c(1:10,Inf,-Inf),2000*3,replace=TRUE),ncol=3))   lst1<-split(dat1,((seq_len(nrow(dat1))-1)%/%69)+1)  lst2<-lapply(lst1,function(x) {colnames(x)<-letters[1:3];x}) lst3<-lapply(lst2,funct

Re: [R] If else loop problem: the condition has length > 1 and only the first element will be used

Hi, Your attachment didn't came through.. Using the same example, that I used before: set.seed(549)  dat1<- data.frame(t.tr=sample(c(paste("Pst", c(24, 48, 72)), paste("Pto", c(24, 48, 72)), paste ("Pm", c(24, 48, 72))), 50, replace=TRUE), signal2 = sample(600:700, 50, replace=TRUE)) dat2<- data

Re: [R] labcurve - use of putKey function

Because I'm already controlling points using points function. So I don't want labcurve to change anything on the lines... Just draw the legend the way I need. On Aug 26, 2013 8:35 PM, "Pascal Oettli" wrote: > Hello, > > Please keep the r-list included when you reply. > > Why do you want to add po

Re: [R] labcurve - use of putKey function

Hello, Please keep the r-list included when you reply. Why do you want to add points to lines only in the legend? If so, the legend would be incorrect. Regards, Pascal 2013/8/26 Igor Ribeiro > Hi Pascal, > Thank you very much - your solution works partially - it will include the > pch in the

Re: [R] Naming columns

Your question prompts more questions than answers. Not sure what "imported from notepad" means (clipboard? delimited how?). Don't know what you obtained in R (what does the str function tell you? What about dput(head(yourdata)))? You really need to tell us what R code you executed and data you

[R] Naming columns

Hi , I just imported a large data set from notepad. I want to label the columns in R. I used 'import Dataset' to bring in my data set Now, I would like to label V1,V2,V3 etc?? Thanks -- View this message in context: http://r.789695.n4.nabble.com/Naming-columns-tp4674595.html Sent from the R

Re: [R] Loop for converting character columns to Numeric

Thanks a lot. That works great. I have another question, I will send you another email.   Best,Farnoosh Sheikhi Cc: R help Sent: Monday, August 26, 2013 12:06 PM Subject: Re: Loop for converting character columns to Numeric Hi, Suppose you created a datafr

[R] factor() without abstract types

How do you extend factor() without abstract data types? The idea is to have factor (x, TYPE) to transform all or selected components of x to factor TYPE. Value: a data structure of same type as x, with factor-like components transformed to be like TYPE. In the Pascal family, you would

[R] SimpleR

Hi, I am new to R, and am trying to use the "SimpleR" manual. The manual is excellent. However, when I try to install the SimpleR package with R version 3.x.x, I get errors. The manual says that the package is good with R versions 1.5.0 and above. So: 1) Is there an updated package for Sim

Re: [R] how to apply summation sign formula to R

Hi David Thanks for your help. I tried it for a simplified example with vectors instead of matrices. Once again the formula:  C = (Σ(from i=0 to i)  A^i ) x B x (Σ(from i=0 to i) A^i )’ I applied it at follows – but couldn’t figure out what’s missing: library(expm) i <- c(0,1,2,3,4,5,6,7,8,9,10)

[R] Pasting excell spreedsheet into notepad for R

Hi all, Does anyone know how to export an excell spreedsheet into notepad in R format as a linear vector as opposed to columns? I.e. 2,3,4,5,6,7 etc. Thanks, Mary -- View this message in context: http://r.789695.n4.nabble.com/Pasting-excell-spreedsheet-into-notepad-for-R-tp4674568.html Sent

Re: [R] If else loop problem: the condition has length > 1 and only the first element will be used

Match worker perfectly. Thanks for the help! Michael Fethe On Aug 26, 2013, at 11:10 AM, "arun" wrote: > Hi, > > You could try: > set.seed(549) > dat1<- data.frame(t.tr=sample(c(paste("Pst", c(24, 48, 72)), paste("Pto", > c(24, 48, 72)), paste ("Pm", c(24, 48, 72))), 50, replace=TRUE), sign

Re: [R] detect multivariate outliers with aq.plot {mvoutliers} high dimensions

OK, thanks:) Maybe for my purpose a visible evaluation is also fine. But in general I imagine it to happen quite often that in ecological data, people have e.g. more species than sites. -- View this message in context: http://r.789695.n4.nabble.com/detect-multivariate-outliers-with-aq-plot-mvo

Re: [R] If else loop problem: the condition has length > 1 and only the first element will be used

Thank you all for your suggestions. However, Dennis since your method seemed the easiest. I gave it a shot with no success while it produces NAs. The R inferno is not helpful for this part of my data normalization. I need to apply this to my whole data.frame before I perform statistical tests (

Re: [R] Daily Time series setting

Thank you let me try my luck, Sudheer On Monday, August 26, 2013, Jim Lemon wrote: > On 08/26/2013 12:35 PM, Sudheer Joseph wrote: > >> Hi, >> I am trying to load a daily time series as a r time series >> object in the script at below link using the data at second link. I tried >> s

[R] Questions before installation

Hi, As we are in thinking of installing and using R as a data analysis software on our machines, we have few questions before going ahead with the installation. Please find below the questions: 1) Does R allow data to be sent out or processed out of the organization network (e.g. ability to

Re: [R] hide unused labels and resize panels in lattice barchart

Hi Noah I have modified to give you an idea of panel widths. Also for factors you must have the same factors for all panels although not all will have the full complement. I usually do this by hand as latticeExtra does not do exactly what i want. library(lattice) library(latticeExtra) tab <- s

Re: [R] If else loop problem: the condition has length > 1 and only the first element will be used

Hi better approach is to use match. Let have some lookup table called test > test onetwo 1 Pst 24 -17.29 2 Pst 48 -43.93 3 Pto 24 -29.39 and original data called big > big<-data.frame(one=sample(test$one, 20, replace=TRUE), two=100) > big one two 1 Pto 24 100 2 Pst 48 100 3 Ps

Re: [R] how to apply summation sign formula to R

On Aug 26, 2013, at 11:59 AM, Sebastian Hersberger wrote: Hi David Thanks for your help. I tried it for a simplified example with vectors instead of matrices. Once again the formula: C = (Σ(from i=0 to i) A^i ) x B x (Σ(from i=0 to i) A^i )’ I applied it at follows – but couldn’t figure out

[R] matrix into vector with vertex names - shortestpath reshape2

> > Two years ago, as shown in the script at the end, I created shortestpaths > using Igraph, then using Melt in Reshape2 I converted the the resulting > matrix into > three column vectors - "vertex1", "vertex2", "shortestpath". It worked > then. > However, in the meantime I have installed new v

Re: [R] lags of a variable, with a factor

On 08/26/2013 11:44 PM, Michael Friendly wrote: ... Thanks for trying again, but that doesn't work either with a by= variable. Note that your function is recursive, and also k=k should be passed in the else{ ... lags() }. Hi Michael, You are correct about the k=, and I had used a separate objec

Re: [R] fitted

I don't fully understand what you are looking for, but you may want to check out ?predict, ?predict.lm On Sat, Aug 24, 2013 at 7:43 AM, alR wrote: > I have fitted a multiple regression model to the row of a matrix using lm: > > ft<-lm(datos[i, ]-r1 + r2+ r3 + r4,keep.data = TRUE,model=TRUE) > >

Re: [R] Citing Package Contributing Authors

On Mon, 26 Aug 2013, Charles Determan Jr wrote: Greetings, I am familiar with the function cite('packageName') which provides the output generated from the DESCRIPTION file. ...unless the package provides a CITATION file. Then, citation('pkg') shows the content of the CITATION. citation('pk

Re: [R] Citing Package Contributing Authors

Thank you for your reply Stephan, I like to be very thorough and make sure all names are attributed so in the case that I check the url of a package and it lists contributing authors that aren't provided with citation() would it be appropriate to cite it like this: Smith, J. [pr] and Johnson, J.

Re: [R] Citing Package Contributing Authors

Hi, it usually is a good idea to look at the output of citation() (which, however, also often is auto-generated) or at the authors listed in package vignettes. And thanks for citing R package authors. When I review papers, I often have to remind authors of this... Best Stephan On 26.08.20

[R] Citing Package Contributing Authors

Greetings, I am familiar with the function cite('packageName') which provides the output generated from the DESCRIPTION file. In most cases this is sufficient but I was wondering if there are contributing authors (in addition to the primary) also listed on the CRAN page. Is there a proper way to

Re: [R] Loop for converting character columns to Numeric

Hi, Suppose you created a dataframe like this: set.seed(28)   dat1<-as.data.frame(simplify2array(list(letters[1:5],sample(1:20,5,replace=TRUE),6:10)),stringsAsFactors=FALSE)  str(dat1) #'data.frame':    5 obs. of  3 variables: # $ V1: chr  "a" "b" "c" "d" ... # $ V2: chr  "1" "2" "10" "18" ... #

[R] Bivariate skew normal cdf; very slow

Dear all, I am calculating the bivariate skew normal cdf in "sn" package using "pmsn" function. Although it is quite convenient ( thanks to prof. Azzalini) but it seems to be slow. For example, it takes about 1 minute in calculation of 100k of such cdf values. I am thinking to write a c++ code fo

Re: [R] Pasting excell spreedsheet into notepad for R

Hi, May be this helps: dat1<- as.data.frame(matrix(1:30,6,5)) lapply(seq_len(ncol(dat1)),function(i) dat1[,i]) [[1]] [1] 1 2 3 4 5 6 [[2]] [1]  7  8  9 10 11 12 [[3]] [1] 13 14 15 16 17 18 [[4]] [1] 19 20 21 22 23 24 [[5]] [1] 25 26 27 28 29 30 #from excel dataset  (the same data) library(X

Re: [R] plot categorical variable with percentage infomation

Here is a way of getting the percents (and total if you want it): a=as.factor(sample(1:3,10,replace=T)) # use 'barplot' since it return x-values for putting labels on bars aTable <- table(a) # count the factors xPos <- barplot(aTable) # put percent at top bar text(xPos, aTable, sprintf("%.0f%%",

Re: [R] plot categorical variable with percentage infomation

Hi Jean, I would like to give a histogram for a categorical variable, with x-axis be different levels, and a number of percentage showed on top of each bar. Thanks. Best wishes, Jie On Mon, Aug 26, 2013 at 12:43 PM, Adams, Jean wrote: > Jie, > > I'm not exactly sure what you're after. Perhaps

Re: [R] plot categorical variable with percentage infomation

Jie, I'm not exactly sure what you're after. Perhaps this will help you get started. count <- table(a) prop <- count/length(a) b <- plot(a) text(b, count, prop, pos=1) Jean On Mon, Aug 26, 2013 at 11:27 AM, Jie wrote: > Dear All, > > Suppose I have a categorical variable > a=as.factor(sampl

Re: [R] Get Index of variables from stepAIC

Here's one way to do it ... # create example data frame y <- rnorm(30) gene_subset <- data.frame(y, x1=rnorm(30), x2=rnorm(30), x3=100*y+rnorm(30)) # fit a full linear model fit <- lm(y ~ ., df) # reduce the model reduced_model <- stepAIC(fit, trace=FALSE) # NON-omitted variables (excluding the

[R] plot categorical variable with percentage infomation

Dear All, Suppose I have a categorical variable a=as.factor(sample(1:3,10,replace=T)) plot(a) and hist(as.numeric(a),freq=F) would give the histogram of it. But I do not know how to add the counts or percentage information for plot.factor(). hist() can do it but as a numeric variable, the x-axis

Re: [R] POSIXct bug for conversion of specific combinations of date and time

I deal with non-daylight-savings time data all the time using Windows with its system time set to daylight time. Sys.setenv(TZ="Etc/GMT+4") sets the zone for the R process only and does not affect the system time settings. Using this method lets me handle data from all around the world, with or

Re: [R] If else loop problem: the condition has length > 1 and only the first element will be used

Hi, You could try:  set.seed(549)  dat1<- data.frame(t.tr=sample(c(paste("Pst", c(24, 48, 72)), paste("Pto", c(24, 48, 72)), paste ("Pm", c(24, 48, 72))), 50, replace=TRUE), signal2 = sample(600:700, 50, replace=TRUE)) dat2<- data.frame(t.tr=c(paste("Pst", c(24, 48, 72)), paste("Pto", c(24, 48,

Re: [R] sbusetting data by rows (every 69 rows)

Hi R.L., No problem. You may try: set.seed(24)  dat1<- as.data.frame(matrix(sample(1:10,2000*3,replace=TRUE),ncol=3))   lst1<-split(dat1,((seq_len(nrow(dat1))-1)%/%69)+1)  lst2<-lapply(lst1,function(x) {colnames(x)<-letters[1:3];x}) res<-lapply(lst2,function(x) {x$z<-with(x,(a-b)/c);x}) head(res

Re: [R] Randomization

Silvano, I am a little confused as to what you are looking for. Do you want each group to have approximately the same mean and variance? Randomly assigning groups should be sufficient for the means and variances to be somewhat similar. I'm not sure what your goal would be to randomly split your

Re: [R] If else loop problem: the condition has length > 1 and only the first element will be used

Suggestion: Don't do the ifelse stuff below. See ?switch instead. -- Bert On Sun, Aug 25, 2013 at 11:32 PM, arun wrote: > HI, > > It may be better to provide an example dataset using ?dput(). > dput(head(dataset),20) > > Try: > signal3<- ifelse(t.tr=="Pst 24", signal2-17.29, ifelse(t.tr=="Pst

Re: [R] transform variables

Hi, Also, you could also try: vec1<-rep(yrmon[,1],each=(ncol(yrmon)-1)) vec2<-as.vector(t(yrmon[,-1]))  vec3<- rep(1:12,nrow(yrmon)) res2<-data.frame(year=vec1,Month=vec3,Values=vec2) row.names(res1)<- row.names(res2) attr(res1,"row.names")<- attr(res2,"row.names")  identical(res1,res2) #[1] TRUE

Re: [R] lags of a variable, with a factor

On 8/25/2013 8:12 PM, Jim Lemon wrote: lags <- function(x, k=1, prefix='lag', by) { if(missing(by)) { n <- length(x) res <- data.frame(lag0=x) for (i in 1:k) { res <- cbind(res, c(rep(NA, i), x[1:(n-i)])) } colnames(res) <- paste0(prefix, 0:k) } else { for(levl in lev

Re: [R] transform variables

HI, You could also try: res<-reshape(yrmon,varying=!grepl("year",colnames(yrmon)),v.names="Values",timevar="Month",direction="long")[,-4] res1<- res[order(res$year,res$Month),]  head(res1) #    year Month Values #1.1 1901 1 -0.2557446 #1.2 1901 2 -0.2318646 #1.3 1901 3 -0.1961822 #1

Re: [R] Analogues to my data and prediction problem

Hi Ben, This question apparently has nothing to do with R and is therefore off-topic for this list. You should post this question on a statistics forum, or seek local help. Best, Ista On Mon, Aug 26, 2013 at 1:50 AM, Ben Harrison wrote: > Hello, I am quite a novice when it comes to predictive

Re: [R] transform variables

Catalin, first, keep the communication on the list, so it gets documented also for others. Second, my code does work for an example I made up myself, but if you don't "provide commented, minimal, self-contained, reproducible code" (as the posting guide asks you to; see last line of this e-ma

Re: [R] transform variables

Have a look at the packages reshape and reshape2 They were written with this type of problems in mind. On Aug 26, 2013, at 1:04 PM, catalin roibu wrote: > Dear all! > > I have a data frame composed by 13 columns (year, and 12 months). I want to > transform this data base in another like this >

[R] Partial correlation test

Dear All, Sorry to bother you again. As my previous mail was messy to understand, please find it again to give me a solution. I'd like to do a partial correaltion test ['pcor.test ()' or 'parcor()'] between Irid.area and Casa.PC1 variables controlling the influence of SL (co-variate) according to

Re: [R] transform variables

On 08/26/2013 09:04 PM, catalin roibu wrote: Dear all! I have a data frame composed by 13 columns (year, and 12 months). I want to transform this data base in another like this year month values 1901 1 1901 2 1901 3 . 1901 12 1902 1 1902 2 1902 12 Is there a possibility to succeed t

Re: [R] transform variables

Hello, Catalin, assume your data frame is as simple as A <- data.frame( year = , month01 = , , + month12 = ) then, e.g., reshape( A, varying = c( "month01", , "month12"), + v.names = "Values", timevar = "Month", direction = "long") should do what you want

Re: [R] Daily Time series setting

On 08/26/2013 12:35 PM, Sudheer Joseph wrote: Hi, I am trying to load a daily time series as a r time series object in the script at below link using the data at second link. I tried setting the frequency as 365, but the data is not getting loaded with correct number of samples whic

[R] transform variables

Dear all! I have a data frame composed by 13 columns (year, and 12 months). I want to transform this data base in another like this year month values 1901 1 1901 2 1901 3 . 1901 12 1902 1 1902 2 1902 12 Is there a possibility to succeed that in R? Thank you! best regards! CR -- --

Re: [R] writing spdiags function for R

Dear R-list, l have been working on a translation of a matlab library into R, it took me a while but I am almost there to submit it to CRAN...however, for this library to be computationally competitive I need to solve an issue with the home-made R version of the spdiags.m function (an old issue co

Re: [R] arbitrary subset of dataframe

You've misplaced the comma. mysample <- df[, sample(ncol(df), 50, replace=FALSE)] ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium

[R] arbitrary subset of dataframe

Dear all! I want to arbitrary subset a data frame by variables. I try this code, but the subset work only for rows, but I want to subset by variables. # sample without replacement mysample <- df[sample(1:ncol(df), 50, replace=FALSE),] Please help me to solve this probl

Re: [R] Time axis formatting.

Thank you. with best regards, Sudheer On Mon, Aug 26, 2013 at 4:36 AM, Jim Lemon wrote: > On 08/26/2013 01:09 AM, Sudheer Joseph wrote: > >> Attached is a plot with a time series. If I have a time series object in >> R. >> How do I get the plot in the attached format of time axis?. When I issue

Re: [R] Time axis formatting.

Thank you Jim, I was trying few options but was not able to get it done. Thanks a lot. with best regards, Sudheer On Mon, Aug 26, 2013 at 3:40 AM, jim holtman wrote: > Forgot the year on the plot: > > > n <- 100 > x <- data.frame(time = seq(from = as.Date('2010-1-1')

[R] Daily Time series setting

Hi, I am trying to load a daily time series as a r time series object in the script at below link using the data at second link. I tried setting the frequency as 365, but the data is not getting loaded with correct number of samples which is 2192. Can any one help me on this how to set