[R] Programmatically give file name to a matrix
I have a situation when I need to save matrix with file names that are programmatically created. for (i in levels(mergeTrn$Continent)) { matrix here # I want to save this matrix with a file name that carries i from for loop. The following does not work. paste(plotroc_GBM_Trn_, i, sep=) - matrix } Thanks, Kumar -- Section of Integrative Biology University of Texas at Austin Austin, Texas 78712, USA [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Add axes to a 3D scene (afer makeTriangles)
Hi all, I am drawing some 3D surfaces using the Triangle tools (package misc3) and drawScene.rgl. Do you know if it is possible to add axes and graduation on the scene? Christophe -- View this message in context: http://r.789695.n4.nabble.com/Add-axes-to-a-3D-scene-afer-makeTriangles-tp4657025.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Programmatically give file name to a matrix
Hello, ?assign assign(paste(plotroc_GBM_Trn_, i, sep=), matrix(rnorm(100),10,10)) HTH, Pascal Le 30/01/2013 17:04, Kumar Mainali a écrit : I have a situation when I need to save matrix with file names that are programmatically created. for (i in levels(mergeTrn$Continent)) { matrix here # I want to save this matrix with a file name that carries i from for loop. The following does not work. paste(plotroc_GBM_Trn_, i, sep=) - matrix } Thanks, Kumar __ 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] Creating dummy variables in r
On Jan 30, 2013, at 04:58 , Bert Gunter wrote: You almost never need dummy variables in R. R creates them automatically from factors given model and possibly contrasts specification. ?contrasts ## for some technical details. If you have not read An Introduction to R do so now. Pay particular attention to the chapter on modeling and categorical variables. You can also google around to find appropriate tutorials. Here is one: http://www.ats.ucla.edu/stat/r/modules/dummy_vars.htm I repeat: DO not create dummy variablesby hand in R unless you have understood the above and have good reason to do so. In this case it's a cutpoint-type situation, and the user might be excused for not wanting to deal with the mysteries of cut() (yet). More importantly, the main issue here seems to be a lack of understanding of where new variables are located. I.e., if the data set is called dd, you need dd$prev1 - (etc) and if you use attach(), do it _after_ modifying the data (or detach() and reattach). Otherwise, new variables end up in the global environment. (This is logical enough once you realize that the result of a computation does not necessarily fit into the dataset.) By the way: You don't need ifelse(): as.numeric(ret1 = .5) or even just (ret1 = .5) works. -- Bert On Tue, Jan 29, 2013 at 7:21 PM, Joseph Norman Thomson thoms...@email.arizona.edu wrote: Hello, Semi-new r user here and still learning the ropes. I am creating dummy variables for a dataset on stock prices in r. One dummy variable is called prev1 and is: prev1 - ifelse(ret1 = .5, 1, 0) where ret1 is the previous day's return. The variable prev1 is created fine and works in my regression model and for running conditional statistics. However, when I call the names() function on the dataset the freshly created variable (prev1) doesn't show up; also, when I export the dataset the prev1 variable doesn't show up in the exported file. Is there a way to make the variable show up on both the call function but more importantly on the exported file? Or am I forced to create dummy variables elsewhere(much tougher)? Thanks, Joe __ 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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ 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. -- 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] problem wih plotrix:cluster.overplot
On 01/29/2013 06:53 PM, Jean Véronis wrote: Hello, I am trying to use cluster.overplot from package plotrix and I get an error message when I add the away parameter: require(plotrix) distance- read.table(distance.txt) cmd- cmdscale(distance) cp- cluster.overplot(cmd, away=2) Error in if (sum(overplots) 1) { : missing value where TRUE/FALSE needed If I don't use away, everything works fine. I don't see anything in the documentation that explains this error. Any help will be appreciated. Hi Jean, The away argument needs to have both an x and y distance to offset the points. Try: cp-cluster.overplot(cmd,away=c(2,2)) I'll add this to the docs, thanks. 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] Relative Risk in logistic regression
Hi all, I am very grateful to all those who write to me 1) how i can obtain relative risk (risk ratio) in logistic regression in R. 2) how to obtain the predicted risk for a certain individual using fitted regression model in R. Many thanks, in advance, for your help. Amin. __ 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] Arima model estimated by Maximum likelihood conditional on a zero initial residual
Hello, I would like to estimate Arima model by Maximum likelihood conditional on a zero initial residual. I am using: model - arima(y[1:N], order=c(1, 0, 1), method=ML, include.mean=TRUE) model What I have to use to estimate arima model by Maximum likelihood conditional on a zero initial residual? Important is conditional on a zero initial residual. Thank you. [[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] substring from behind
Hello together, i have a question for substring. I know i can filter a number like this one: bill$No-substring(bill$Customer,2,4) in this case i get the 2nd, 3rd and 4th number of my Customer ID. But how can i do this, if i want the 2nd, 3rd and 4th number of a column. Like this one. I have: Mercedes_02352 Audi_03555 and now i want to filter this data.frame to 235 355 can you help me? Thanks. Mat -- View this message in context: http://r.789695.n4.nabble.com/substring-from-behind-tp4657029.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] problem wih plotrix:cluster.overplot
great! many thanks! Le 30 janv. 2013 à 09:52, Jim Lemon j...@bitwrit.com.au a écrit : On 01/29/2013 06:53 PM, Jean Véronis wrote: Hello, I am trying to use cluster.overplot from package plotrix and I get an error message when I add the away parameter: require(plotrix) distance- read.table(distance.txt) cmd- cmdscale(distance) cp- cluster.overplot(cmd, away=2) Error in if (sum(overplots) 1) { : missing value where TRUE/FALSE needed If I don't use away, everything works fine. I don't see anything in the documentation that explains this error. Any help will be appreciated. Hi Jean, The away argument needs to have both an x and y distance to offset the points. Try: cp-cluster.overplot(cmd,away=c(2,2)) I'll add this to the docs, thanks. 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.
Re: [R] substring from behind
Hi Mat, The following should get you started: s - Mercedes_02352 substr(s, nchar(s) - 3, nchar(s) - 1) [1] 235 # defining a function foo - function(x, a = 3, b = 1) substr(s, nchar(x) - a, nchar(x) - b) foo(s) [1] 235 HTH, Jorge.- On Wed, Jan 30, 2013 at 8:05 PM, Mat wrote: Hello together, i have a question for substring. I know i can filter a number like this one: bill$No-substring(bill$Customer,2,4) in this case i get the 2nd, 3rd and 4th number of my Customer ID. But how can i do this, if i want the 2nd, 3rd and 4th number of a column. Like this one. I have: Mercedes_02352 Audi_03555 and now i want to filter this data.frame to 235 355 can you help me? Thanks. Mat -- View this message in context: http://r.789695.n4.nabble.com/substring-from-behind-tp4657029.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] recoding variables again :(
Hello everybody! I have again a rather simple question concerning recoding of variables: I have a variable/data-frame column BIRTHPLACE containing abbreviations of the 26 swiss counties (AG, AI, AR, BE, ZH, ... ) as well as international country codes (USA, GER, ESP, etc.) and another variable RES_STA indicating the residence status (A, B, C, X, Y) My goal is now to create a new variable VARNEW under the following conditions: - should be the RESIDENCE_STATUS - except: - if RESIDENCE_STATUS is X and at the same time BIRTHPLACE is one of the 26 swiss counties then it should be swiss - otherweise it should be unknown I have already tried the following code: mydata$VARNEW-mydata$RESIDENCE_STATUS # setting VARNEW as RESIDENCE_STATUS idx-(mydata$RESIDENCE_STATUS==X !(# TRUE: unknown; FALSE: swiss mydata$BIRTHPLACE==AG | mydata$BIRTHPLACE==BE | mydata$BIRTHPLACE==AR ... ) ) and then? Thank you for any help! David [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] recoding variables again :(
Hi David, Check ?%in% for a simpler approach. Regards, Jorge.- On Wed, Jan 30, 2013 at 8:42 PM, David Studer wrote: Hello everybody! I have again a rather simple question concerning recoding of variables: I have a variable/data-frame column BIRTHPLACE containing abbreviations of the 26 swiss counties (AG, AI, AR, BE, ZH, ... ) as well as international country codes (USA, GER, ESP, etc.) and another variable RES_STA indicating the residence status (A, B, C, X, Y) My goal is now to create a new variable VARNEW under the following conditions: - should be the RESIDENCE_STATUS - except: - if RESIDENCE_STATUS is X and at the same time BIRTHPLACE is one of the 26 swiss counties then it should be swiss - otherweise it should be unknown I have already tried the following code: mydata$VARNEW-mydata$RESIDENCE_STATUS # setting VARNEW as RESIDENCE_STATUS idx-(mydata$RESIDENCE_STATUS==X !(# TRUE: unknown; FALSE: swiss mydata$BIRTHPLACE==AG | mydata$BIRTHPLACE==BE | mydata$BIRTHPLACE==AR ... ) ) and then? Thank you for any help! David [[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.
[R] Fwd:
http://www.consultadifesapdl.it/6cmpqy.php __ 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] Relative Risk in logistic regression
Hi all, I am very grateful to all those who write to me 1) how i can obtain relative risk (risk ratio) in logistic regression in R. 2) how to obtain the predicted risk for a certain individual using fitted regression model in R. Many thanks, in advance, for your help. Amin. __ 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] Netcdf and Raster Package Questions, Need .asc File for GIS
Douglas M. Hultstrand dmhultst at metstat.com writes: Hello R-Group, I am new working with netcdf files and the raster package in R.I am trying to read in a netcdf file using the package ncdf.I am able to get the lat, lon and parameter I need and can plot using fill.contour. Please post on R-sig-geo, where the raster developers often handle questions like this. Roger __ 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] recoding variables again :(
Hi, set.seed(125) dat1-data.frame(BIRTHPLACE=sample(c(AG,AI,AR,BE,ZH,USA,GER,ESP),20,replace=TRUE),RES_STA=sample(LETTERS[c(1:3,24:25)],20,replace=TRUE)) dat1$VARNEW-ifelse(dat1$RES_STA==X dat1$BIRTHPLACE%in%c(AG,AI,AR,BE,ZH),swiss,unknown) A.K. - Original Message - From: David Studer stude...@gmail.com To: r-help@r-project.org Cc: Sent: Wednesday, January 30, 2013 4:42 AM Subject: [R] recoding variables again :( Hello everybody! I have again a rather simple question concerning recoding of variables: I have a variable/data-frame column BIRTHPLACE containing abbreviations of the 26 swiss counties (AG, AI, AR, BE, ZH, ... ) as well as international country codes (USA, GER, ESP, etc.) and another variable RES_STA indicating the residence status (A, B, C, X, Y) My goal is now to create a new variable VARNEW under the following conditions: - should be the RESIDENCE_STATUS - except: - if RESIDENCE_STATUS is X and at the same time BIRTHPLACE is one of the 26 swiss counties then it should be swiss - otherweise it should be unknown I have already tried the following code: mydata$VARNEW-mydata$RESIDENCE_STATUS # setting VARNEW as RESIDENCE_STATUS idx-(mydata$RESIDENCE_STATUS==X !( # TRUE: unknown; FALSE: swiss mydata$BIRTHPLACE==AG | mydata$BIRTHPLACE==BE | mydata$BIRTHPLACE==AR ... ) ) and then? Thank you for any help! David [[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] I think you misunderstood my explantation.
Hi, Your dataset had already some missing values. So, I need to subset only those rows that are not missing. !is.na(temp$ACTIVE_KWH) # [1] TRUE TRUE TRUE FALSE TRUE FALSE FALSE TRUE TRUE TRUE TRUE FALSE #[13] TRUE TRUE TRUE temp$ACTIVE_KWH[!is.na(temp$ACTIVE_KWH)] #[1] 1201.9 1202.2 1202.8 1203.9 12.0 1206.0 1206.3 1206.5 1207.3 1207.9 #[11] 1208.4 ?diff() will get the differences between successive values diff(temp$ACTIVE_KWH[!is.na(temp$ACTIVE_KWH)]) #[1] 0.3 0.6 1.1 -1191.9 1194.0 0.3 0.2 0.8 0.6 #[10] 0.5 #Here, the length is 1 less than the previous case as the first value is removed. diff(temp$ACTIVE_KWH[!is.na(temp$ACTIVE_KWH)])0 # [1] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE #Added `FALSE` at the beginning to make the length equal to subset data indx- c(FALSE,diff(temp$ACTIVE_KWH[!is.na(temp$ACTIVE_KWH)])0) indx #[1] FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE #Using this index, further subset the already subset data for differences of values 0 temp$ACTIVE_KWH[!is.na(temp$ACTIVE_KWH)][indx] #[1] 12 temp$ACTIVE_KWH[!is.na(temp$ACTIVE_KWH)][indx]- NA #changed to NA #Similarly for REACTIVE_KWH Hope this helps. A.K. From: 남윤주 jamansymp...@naver.com To: arun smartpink...@yahoo.com Sent: Wednesday, January 30, 2013 12:51 AM Subject: Re: I think you misunderstood my explantation. Oh, I forgot to ask about those code. Can u expain what dose that mean? Using the first dataset temp: temp$ACTIVE_KWH[!is.na(temp$ACTIVE_KWH)][c(FALSE,diff(temp$ACTIVE_KWH[!is.na(temp$ACTIVE_KWH)]) 0)]-NA temp$REACTIVE_KWH[!is.na(temp$REACTIVE_KWH)][c(FALSE,diff(temp$REACTIVE_KWH[!is.na(temp$REACTIVE_KWH)]) 0)]-NA -Original Message- From: arunsmartpink...@yahoo.com To: 남윤주jamansymp...@naver.com; Cc: R helpr-help@r-project.org; Sent: 2013-01-30 (수) 10:37:18 Subject: Re: I think you misunderstood my explantation. Hi, Sorry, I didn't check your codes previously. I hope this works for you (especially the 0). Using the first dataset temp: temp$ACTIVE_KWH[!is.na(temp$ACTIVE_KWH)][c(FALSE,diff(temp$ACTIVE_KWH[!is.na(temp$ACTIVE_KWH)]) 0)]-NA temp$REACTIVE_KWH[!is.na(temp$REACTIVE_KWH)][c(FALSE,diff(temp$REACTIVE_KWH[!is.na(temp$REACTIVE_KWH)]) 0)]-NA temp # ID CTIME ACTIVE_KWH REACTIVE_KWH #1 HM001 201212121301 1201.9 1115.5 #2 HM001 201212121302 1202.2 1115.8 #3 HM001 201212121303 1202.8 1115.8 #4 HM001 201212121304 NA 1116.1 #5 HM001 201212121305 1203.9 1116.7 #6 HM001 201212121306 NA 1116.7 #7 HM001 201212121307 NA 1116.7 #8 HM001 201212121308 NA NA #9 HM001 201212121309 1206.0 1118.2 #10 HM001 201212121310 1206.3 1118.6 #11 HM001 201212121311 1206.5 1118.8 #12 HM001 201212121312 NA NA #13 HM001 201212121313 1207.3 NA #14 HM001 201212121314 1207.9 1121.1 #15 HM001 201212121315 1208.4 1121.3 temp1$ACTIVE_KWH[!is.na(temp1$ACTIVE_KWH)][c(FALSE,diff(temp1$ACTIVE_KWH[!is.na(temp1$ACTIVE_KWH)]) 0)]-NA #Similarly with the second dataset: temp1$ACTIVE_KWH[!is.na(temp1$ACTIVE_KWH)][c(FALSE,diff(temp1$ACTIVE_KWH[!is.na(temp1$ACTIVE_KWH)]) 0)]-NA temp1$REACTIVE_KWH[!is.na(temp1$REACTIVE_KWH)][c(FALSE,diff(temp1$REACTIVE_KWH[!is.na(temp1$REACTIVE_KWH)]) 0)]-NA A.K. From: 남윤주 jamansymptom@naver.com To: arun smartpink111@yahoo.com Sent: Tuesday, January 29, 2013 7:42 PM Subject: I think you misunderstood my explantation. Hi, Assume that first CTIME value is '20120101'. It means ACTIVE_KWH measured from '20120101' to present. show example below row. 1 HM001 201212121301 1201.9 1115.5 1 row's ACTIVE_KWH accumulated value that measured from '20120101' to '201212121301'. when CTIME is '201212121301', ACTIVE_KWH is '1201.9'. And, when CTIME is '201212121302', ACTIVE_KWH is '1202.2'. It means that 0.3 is measured during 1 minute. And ACTIVE_KWH is a accumulated value. Thus, ACTIVE_KWH must increase, as CTIME increases. You got it? So, I have to define strange value as subtraction value like ( temp$ACTIVE_KWH[i] - temp$ACTIVE_KWH[i-1]) 50). '50' can be chagned. - for(i in 2:m){ temp$ACTIVE_KWH[i]- ifelse(temp$ACTIVE_KWH[i]- temp$ACTIVE_KWH[i-1]0,NA, temp$ACTIVE_KWH[i]) } -- But, in this case, critical error occured. If temp$ACTIVE_KWH[3] is NA, posterior data (temp$ACTIVE_KWH[4], [5], [6]...) is imputed as NA. Last mail contains Detailed source code and result. Can you recommend better idea to avoid imputed dataset as a successive NA. -Original Message- From: arunsmartpink111@yahoo.com To: 남윤주jamansymptom@naver.com; Cc: R helpr-help@r-project.org; Sent:
Re: [R] Fastest way to compare a single value with all values in one column of a data frame
Hi Dimitri, Does this help? k1-data.frame(item=sample(rep(letters),10,replace=T),a=c(1:10),b=11:20) k2-data.frame(item=f,a=3,b=10) merge-function(y,x) { if(y$amin(x$a)) { x-rbind(x,y) x-x[-which.min(x$a),] } return(x) } merge(k2,k1) or much faster way would be to refer library(sqldf). --- On Wed, 30/1/13, Dimitri Liakhovitski dimitri.liakhovit...@gmail.com wrote: From: Dimitri Liakhovitski dimitri.liakhovit...@gmail.com Subject: [R] Fastest way to compare a single value with all values in one column of a data frame To: r-help r-help@r-project.org Date: Wednesday, 30 January, 2013, 2:41 AM Hello! I have a large data frame x: x-data.frame(item=letters[1:5],a=1:5,b=11:15) # in actuality, x has 1000 rows x$item-as.character(x$item) I also have a small data frame y with just 1 row: y-data.frame(item=f,a=3,b=10) y$item-as.character(y$item) I have to decide if y$a is larger than the smallest of all the values in x$a. If it is, I want y to replace the whole row in x that has the lowest value in column a. This is how I'd do it. if(y$amin(x$a)){ whichmin-which(x$a==min(x$a)) x[whichmin,]-y[1,] } I am wondering if there is a faster way of doing it. What would be the fastest possible way? I'd have to do it, unfortunately, many-many times. Thank you very much! -- Dimitri Liakhovitski gfk.com http://marketfusionanalytics.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. [[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] Integration of mixed normal distribution
Hello, You could do something like the following. fun - function(x, mean, sd1, sd2, p) dnorm(x, mean, sd1)*p + dnorm(x, mean, sd2)*(1 - p) fun2 - function(x1, x2, mean, sd1, sd2, p){ p1 - pnorm(x2, mean, sd1) - pnorm(x1, mean, sd1) p2 - pnorm(x2, mean, sd2) - pnorm(x1, mean, sd2) p1*p + p2*(1 - p) } integrate(fun, 0, 1, mean = 0, sd1 = 1, sd2 = 2, p = 0.5) fun2(0, 1, mean = 0, sd1 = 1, sd2 = 2, p = 0.5) Hope this helps, Rui Barradas Em 30-01-2013 09:19, Johannes Radinger escreveu: Hi, I already found a conversation on the integration of a normal distribution and two suggested solutions (https://stat.ethz.ch/pipermail/r-help/2007-January/124008.html): 1) integrate(dnorm, 0,1, mean = 0, sd = 1.2) and 2) pnorm(1, mean = 0, sd = 1.2) - pnorm(0, mean = 0, sd = 1.2) where the pnorm-approach is supposed to be faster and with higher precision. I want to integrate a mixed normal distribution like: normaldistr_1 * p + normaldistr_2 * (1-p) where p is between 0 and 1 and the means for both distributions are 0 but the standard deviations differ. In addition, I want to get the integrals from x to infinity or from - infinity to x for the mixed distribution. Can that be done with high precision in R and if yes how? best regards, Johannes __ 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] Fastest way to compare a single value with all values in one column of a data frame
If you wanted this for all values in x that are smaller, i'd use x[x$a y$a,] - y for just the smallest: x[intersect(which(x$a y$a),which.min(x$a)),] - y On 29.01.2013, at 22:11, Dimitri Liakhovitski wrote: Hello! I have a large data frame x: x-data.frame(item=letters[1:5],a=1:5,b=11:15) # in actuality, x has 1000 rows x$item-as.character(x$item) I also have a small data frame y with just 1 row: y-data.frame(item=f,a=3,b=10) y$item-as.character(y$item) I have to decide if y$a is larger than the smallest of all the values in x$a. If it is, I want y to replace the whole row in x that has the lowest value in column a. This is how I'd do it. if(y$amin(x$a)){ whichmin-which(x$a==min(x$a)) x[whichmin,]-y[1,] } I am wondering if there is a faster way of doing it. What would be the fastest possible way? I'd have to do it, unfortunately, many-many times. Thank you very much! -- Dimitri Liakhovitski gfk.com http://marketfusionanalytics.com/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sweave files generating miktex errors
On 13-01-29 6:04 PM, Troy S wrote: Ok, yes I realize it. So let me try to fix it: I removed the sty files, and set TEXINPUTS to 'C:\R\R-2.15.2\share\texmf\tex\latex I am back where I started: the tex file will not process. Please let me know what I should do to fix the issue. Follow my advice in the original message? Duncan Murdoch Troy On Tue, Jan 29, 2013 at 2:20 PM, Duncan Murdoch murdoch.dun...@gmail.com mailto:murdoch.dun...@gmail.com wrote: On 13-01-29 4:55 PM, Troy S wrote: Duncan, Copying all 3 sty files to my working directory solved the problem. I'm still curious why this was necessary in my new install. Duncan, many thanks! You do realize that you have now done a Very Bad Thing, I hope. Duncan Murdoch Troy On Tue, Jan 29, 2013 at 10:58 AM, Duncan Murdoch murdoch.dun...@gmail.com mailto:murdoch.dun...@gmail.com mailto:murdoch.duncan@gmail.__com mailto:murdoch.dun...@gmail.com wrote: On 29/01/2013 1:37 PM, Troy S wrote: Dear useRs-- I have been using Sweave with miktex for years, but on a new install on Windows XP, miktex seems to be hung up on single quotes. See example below. Digging through stackexchange, I found using \usepackage[noae]{Sweave} in the tex file solved the problem. My questions are: --Why would this happen? I have the ae package installed. --If inserting [noae] is the solution, how do I have this done by Sweave? I'll start with the error messages followed by the tex file outputted by Sweave. Any help, please?\ I don't know if this is the cause of your problem, but you are doing one Bad Thing: you have a copy of Sweave.sty in the same directory as your document. This is Bad because it is very easy to end up with Sweave.sty from R version X while using Sweave in R version Y, where X != Y. You should leave Sweave.sty where it is in the R installation, and use R CMD Sweave --pdf testsweave.Rnw (or one of the many equivalents) to process it. Duncan Murdoch Troy This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) entering extended mode (C:\Documents and Settings\v\My Documents\testsweave.tex LaTeX2e 2011/06/27 Babel v3.8m and hyphenation patterns for english, afrikaans, ancientgreek, ar abic, armenian, assamese, basque, bengali, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, ga lician, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, iceland ic, indonesian, interlingua, irish, italian, kannada, kurmanji, latin, latvian, lithuanian, malayalam, marathi, mongolian, mongolianlmc, monogreek, ngerman, n german-x-2012-05-30, nynorsk, oriya, panjabi, pinyin, polish, portuguese, roman ian, russian, sanskrit, serbian, slovak, slovenian, spanish, swedish, swissgerm an, tamil, telugu, turkish, turkmen, ukenglish, ukrainian, uppersorbian, usengl ishmax, welsh, loaded. (c:\miktex2.9.4757\tex\latex\ base\article.cls Document Class: article 2007/10/19 v1.4h Standard LaTeX document class (c:\miktex2.9.4757\tex\latex\base\size10.clo)) (C:\Documents and Settings\v\My Documents\Sweave.sty (c:\miktex2.9.4757\tex\latex\base\ifthen.sty) (c:\miktex2.9.4757\tex\latex\graphics\graphicx.sty (c:\miktex2.9.4757\tex\latex\graphics\keyval.sty) (c:\miktex2.9.4757\tex\latex\graphics\graphics.sty (c:\miktex2.9.4757\tex\latex\graphics\trig.sty) (c:\miktex2.9.4757\tex\latex\00miktex\graphics.cfg) (c:\miktex2.9.4757\tex\latex\pdftex-def\pdftex.def (c:\miktex2.9.4757\tex\generic\oberdiek\infwarerr.sty) (c:\miktex2.9.4757\tex\generic\oberdiek\ltxcmds.sty))))
Re: [R] rpart
Carol, Actually, you have only five nodes, numbered 1, 2, 3, 6, and 7. And all five nodes are included in your plot. Nodes 1 and 3 are branching nodes; nodes 2, 6, and 7 are terminal nodes. Try typing just the name of the rpart object for a very brief text version of the tree. rpart.res Jean On Sun, Jan 27, 2013 at 1:31 PM, carol white wht_...@yahoo.com wrote: Hi, When I look at the summary of an rpart object run on my data, I get 7 nodes but when I plot the rpart object, I get only 3 nodes. Should the number of nodes not match in the results of the 2 functions (summary and plot) or it is not always the same? Look forward to your reply, Carol summary(rpart.res) Call: rpart(formula = mydata$class ~ ., data = as.data.frame(t(mydata))) n= 62 CP nsplit rel errorxerror xstd 1 0.6363636 0 1.000 1.000 0.1712469 2 0.1363636 1 0.3636364 0.6818182 0.1532767 3 0.010 2 0.2272727 0.7727273 0.1596659 Variable importance Hsa.627 Hsa.692 Hsa.692.2 Hsa.3306 Hsa.601 Hsa.831 Hsa.1832 Hsa.2456 1913111010 8 6 6 Hsa.8147 Hsa.1131 Hsa.692.1 6 5 5 Node number 1: 62 observations,complexity param=0.6363636 predicted class=t expected loss=0.3548387 P(node) =1 class counts:2240 probabilities: 0.355 0.645 left son=2 (14 obs) right son=3 (48 obs) Primary splits: Hsa.62759.83to the left, improve=15.05376, (0 missing) Hsa.8147 1696.23 to the right, improve=14.46790, (0 missing) Hsa.37937 379.39 to the right, improve=13.75358, (0 missing) Hsa.692.2 842.305 to the right, improve=12.38710, (0 missing) Hsa.1832 735.805 to the right, improve=11.90495, (0 missing) Surrogate splits: Hsa.692.2 1086.655 to the right, agree=0.903, adj=0.571, (0 split) Hsa.3306 170.515 to the left, agree=0.887, adj=0.500, (0 split) Hsa.60188.065 to the left, agree=0.887, adj=0.500, (0 split) Hsa.6921251.99 to the right, agree=0.871, adj=0.429, (0 split) Hsa.831281.54 to the left, agree=0.871, adj=0.429, (0 split) Node number 2: 14 observations predicted class=n expected loss=0 P(node) =0.2258065 class counts:14 0 probabilities: 1.000 0.000 Node number 3: 48 observations,complexity param=0.1363636 predicted class=t expected loss=0.167 P(node) =0.7741935 class counts: 840 probabilities: 0.167 0.833 left son=6 (7 obs) right son=7 (41 obs) Primary splits: Hsa.8147 1722.605 to the right, improve=4.915215, (0 missing) Hsa.1832 681.145 to the right, improve=4.915215, (0 missing) Hsa.1410 49.985 to the left, improve=4.915215, (0 missing) Hsa.2456 186.195 to the right, improve=4.915215, (0 missing) Hsa.11616 969.085 to the right, improve=4.915215, (0 missing) Surrogate splits: Hsa.1832 681.145 to the right, agree=1.000, adj=1.000, (0 split) Hsa.2456 186.195 to the right, agree=1.000, adj=1.000, (0 split) Hsa.6921048.375 to the right, agree=0.979, adj=0.857, (0 split) Hsa.692.1 1136.75 to the right, agree=0.979, adj=0.857, (0 split) Hsa.1131 1679.54 to the right, agree=0.979, adj=0.857, (0 split) Node number 6: 7 observations predicted class=n expected loss=0.2857143 P(node) =0.1129032 class counts: 5 2 probabilities: 0.714 0.286 Node number 7: 41 observations predicted class=t expected loss=0.07317073 P(node) =0.6612903 class counts: 338 probabilities: 0.073 0.927 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] remove label from specific axis
Readers, For a graph plot instruction: plot(seq(10:50),type='h',yaxt='n',yaxs='i',lab=c(20,2,2),xlab='x axis label',bty='l',main='graph title') how to remove y-axis label and keep the x-axis label? _ r2151 __ 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] remove label from specific axis
Hello, Just use ylab = . Hope this helps, Rui Barradas Em 30-01-2013 13:33, e-letter escreveu: Readers, For a graph plot instruction: plot(seq(10:50),type='h',yaxt='n',yaxs='i',lab=c(20,2,2),xlab='x axis label',bty='l',main='graph title') how to remove y-axis label and keep the x-axis label? _ r2151 __ 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] recoding variables again :(
dat1$VARNEW-rep(unknown,nrow(dat1)) dat1$VARNEW[dat1$RES_STA==X dat1$BIRTHPLACE %in% c(AG,AI,AR,BE,ZH)]-swiss --- On Wed, 30/1/13, arun smartpink...@yahoo.com wrote: From: arun smartpink...@yahoo.com Subject: Re: [R] recoding variables again :( To: stude...@gmail.com stude...@gmail.com Cc: R help r-help@r-project.org Date: Wednesday, 30 January, 2013, 5:06 PM Hi, set.seed(125) dat1-data.frame(BIRTHPLACE=sample(c(AG,AI,AR,BE,ZH,USA,GER,ESP),20,replace=TRUE),RES_STA=sample(LETTERS[c(1:3,24:25)],20,replace=TRUE)) dat1$VARNEW-ifelse(dat1$RES_STA==X dat1$BIRTHPLACE%in%c(AG,AI,AR,BE,ZH),swiss,unknown) A.K. - Original Message - From: David Studer stude...@gmail.com To: r-help@r-project.org Cc: Sent: Wednesday, January 30, 2013 4:42 AM Subject: [R] recoding variables again :( Hello everybody! I have again a rather simple question concerning recoding of variables: I have a variable/data-frame column BIRTHPLACE containing abbreviations of the 26 swiss counties (AG, AI, AR, BE, ZH, ... ) as well as international country codes (USA, GER, ESP, etc.) and another variable RES_STA indicating the residence status (A, B, C, X, Y) My goal is now to create a new variable VARNEW under the following conditions: - should be the RESIDENCE_STATUS - except: - if RESIDENCE_STATUS is X and at the same time BIRTHPLACE is one of the 26 swiss counties then it should be swiss - otherweise it should be unknown I have already tried the following code: mydata$VARNEW-mydata$RESIDENCE_STATUS # setting VARNEW as RESIDENCE_STATUS idx-(mydata$RESIDENCE_STATUS==X !( # TRUE: unknown; FALSE: swiss mydata$BIRTHPLACE==AG | mydata$BIRTHPLACE==BE | mydata$BIRTHPLACE==AR ... ) ) and then? Thank you for any help! David [[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. [[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] Sweave files generating miktex errors
On 1/29/2013 6:04 PM, Troy S wrote: Ok, yes I realize it. So let me try to fix it: I removed the sty files, and set TEXINPUTS to 'C:\R\R-2.15.2\share\texmf\tex\latex I am back where I started: the tex file will not process. Please let me know what I should do to fix the issue. Troy Don't do that either. Go to Programs -Miktex - Maintenance (Admin) - Settings and on the Roots tab, add a folder, e.g., c:\localtexmf having the same structure as the default texmf tree. E.g., I have there \tex\latex\..., \bibtex\{bib,bst} ... where I put all things tex that aren't installed directly by Miktex -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. Chair, Quantitative Methods York University Voice: 416 736-2100 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ 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] Relative Risk in logistic regression
Relative risk = exp(coef(model)) --- On Wed, 30/1/13, aminreza Aamini amin.r@gmail.com wrote: From: aminreza Aamini amin.r@gmail.com Subject: [R] Relative Risk in logistic regression To: R-help R-help@r-project.org Date: Wednesday, 30 January, 2013, 4:19 PM Hi all, I am very grateful to all those who write to me 1) how i can obtain relative risk (risk ratio) in logistic regression in R. 2) how to obtain the predicted risk for a certain individual using fitted regression model in R. Many thanks, in advance, for your help. Amin. __ 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] Sweave files generating miktex errors
On 13-01-30 8:52 AM, Michael Friendly wrote: On 1/29/2013 6:04 PM, Troy S wrote: Ok, yes I realize it. So let me try to fix it: I removed the sty files, and set TEXINPUTS to 'C:\R\R-2.15.2\share\texmf\tex\latex I am back where I started: the tex file will not process. Please let me know what I should do to fix the issue. Troy Don't do that either. Go to Programs -Miktex - Maintenance (Admin) - Settings and on the Roots tab, add a folder, e.g., c:\localtexmf having the same structure as the default texmf tree. E.g., I have there \tex\latex\..., \bibtex\{bib,bst} ... where I put all things tex that aren't installed directly by Miktex I think this is better than making copies of Sweave.sty in every project in that it's only one place that needs fixing when you upgrade R, but why not make use of the facilities in R to handle this? Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fastest way to compare a single value with all values in one column of a data frame
Hi, I guess you could also use: x[match(min(x$a),x$a[x$ay$a]),]- y x # item a b #1 f 3 10 #2 b 2 12 #3 c 3 13 #4 d 4 14 #5 e 5 15 A.K. - Original Message - From: Dimitri Liakhovitski dimitri.liakhovit...@gmail.com To: r-help r-help@r-project.org Cc: Sent: Tuesday, January 29, 2013 4:11 PM Subject: [R] Fastest way to compare a single value with all values in one column of a data frame Hello! I have a large data frame x: x-data.frame(item=letters[1:5],a=1:5,b=11:15) # in actuality, x has 1000 rows x$item-as.character(x$item) I also have a small data frame y with just 1 row: y-data.frame(item=f,a=3,b=10) y$item-as.character(y$item) I have to decide if y$a is larger than the smallest of all the values in x$a. If it is, I want y to replace the whole row in x that has the lowest value in column a. This is how I'd do it. if(y$amin(x$a)){ whichmin-which(x$a==min(x$a)) x[whichmin,]-y[1,] } I am wondering if there is a faster way of doing it. What would be the fastest possible way? I'd have to do it, unfortunately, many-many times. Thank you very much! -- Dimitri Liakhovitski gfk.com http://marketfusionanalytics.com/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Relative Risk in logistic regression
Example from linear regression help (?lm) ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) group - gl(2,10,20, labels=c(Ctl,Trt)) weight - c(ctl, trt) lm.D9 - lm(weight ~ group) lm.D90 - lm(weight ~ group - 1) prediction_error= lm.D90$residuals --- On Wed, 30/1/13, aminreza Aamini amin.r@gmail.com wrote: From: aminreza Aamini amin.r@gmail.com Subject: [R] Relative Risk in logistic regression To: R-help R-help@r-project.org Date: Wednesday, 30 January, 2013, 4:19 PM Hi all, I am very grateful to all those who write to me 1) how i can obtain relative risk (risk ratio) in logistic regression in R. 2) how to obtain the predicted risk for a certain individual using fitted regression model in R. Many thanks, in advance, for your help. Amin. __ 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] substring from behind
Hi, You could use: dat1-data.frame(col1=c(Mercedes_02352,Audi_03555)) dat1[,1]-as.numeric(gsub(.*_\\d{1}(.*)\\d{1}$,\\1,dat1[,1])) #if the number of digits are the same dat1 # col1 #1 235 #2 355 A.K. - Original Message - From: Mat matthias.we...@fnt.de To: r-help@r-project.org Cc: Sent: Wednesday, January 30, 2013 4:05 AM Subject: [R] substring from behind Hello together, i have a question for substring. I know i can filter a number like this one: bill$No-substring(bill$Customer,2,4) in this case i get the 2nd, 3rd and 4th number of my Customer ID. But how can i do this, if i want the 2nd, 3rd and 4th number of a column. Like this one. I have: Mercedes_02352 Audi_03555 and now i want to filter this data.frame to 235 355 can you help me? Thanks. Mat -- View this message in context: http://r.789695.n4.nabble.com/substring-from-behind-tp4657029.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] temporal correlogram
Hi all, Ive just started to learn working with R. I wonder if it is possible to make a temporal correlogram in R and how to do it. Actually what I have are movement tracks of birds, existing as XY-positions every 5 minutes for a duration of at least 60 minutes (but up to 8 hours). From these tracks, I calculated step lengths (ie distances travelled within each 5 min-step) and turning angles (also per 5 min-step) for each tracking session (I tracked each individual bird several times). I manually calculated the correlation between step-legths at time t and t+5min and they are strongluy correlated. I want to know for how many time steps this correlation holds. Therefore,I want to calculate autocorrelations (between step lengths, between turning angles - and eventually also between step lengths and turning angles, but thats more complex) for subsequent steps (so I need a moving window frame for delta time/step). On the Y-axis, I would then have rho (autocorrelation coefficient for step length or turning angle), and on the X-axis, delta time (as time steps 0, 5, 10,15, ... minutes). Ideal would be to have the average rho + CI (or SD) displayed over all individuals (or tracking sessions). Is there a package to do this i R? I searched the web but found only correlograms for spatial data (sncf-package), not temporal. Thank you very much for your kind assistance! Valérie -- Valérie Lehouck Assistant Professor Research Unit Terrestrial Ecology Ghent University KL Ledeganckstraat 35 9000 Ghent Belgium Tel.: +32(0)9/264 52 58 website: http://www.ecology.ugent.be/terec/personal.php?pers=vl [[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] (no subject)
Dear Team, I am getting the following error message when try to run vb application The program was running fine in 32 windows 7 machine. When i moved the same program to 64 bit windows 8 machine i am getting the following error Error in inDL(x, as.logical(local), as.logical(now), ...) : unable to load shared object 'C:/Program Files/R/R-2.15.1/library/stats/libs/i386/stats.dll': LoadLibrary failure: The specified module could not be found. During startup - Warning message:package 'stats' in options(defaultPackages) was not found [[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] Percentages in bar plot
Dear R-users, Though it's a silly thing to ask, but I'm not getting a way out. I wish to find the percentage distribution for a data vector 'stop'. The coomand below is giving the frequency distribution. May I know the option to see the percentages instead of frequencies. Similarly, what option I should put in a histogram plot to see the percentages instead of frequencies? - stop-c(8,6,6,6,8,6,6,8,8,6,6,6,8,8,8) barplot(table(stop)) Many thanks for your time. Regards, Jamil. [[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] substring from behind
It's possible the direct way in R: customer - c(Mercedes_02352, Audi_03555) substr(customer, nchar(customer) - 3, nchar(customer) - 1) Isidro -Mensaje original- De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] En nombre de Jorge I Velez Enviado el: miércoles, 30 de enero de 2013 10:50 Para: Mat CC: r-help@r-project.org Asunto: Re: [R] substring from behind Hi Mat, The following should get you started: s - Mercedes_02352 substr(s, nchar(s) - 3, nchar(s) - 1) [1] 235 # defining a function foo - function(x, a = 3, b = 1) substr(s, nchar(x) - a, nchar(x) - b) foo(s) [1] 235 HTH, Jorge.- On Wed, Jan 30, 2013 at 8:05 PM, Mat wrote: Hello together, i have a question for substring. I know i can filter a number like this one: bill$No-substring(bill$Customer,2,4) in this case i get the 2nd, 3rd and 4th number of my Customer ID. But how can i do this, if i want the 2nd, 3rd and 4th number of a column. Like this one. I have: Mercedes_02352 Audi_03555 and now i want to filter this data.frame to 235 355 can you help me? Thanks. Mat -- View this message in context: http://r.789695.n4.nabble.com/substring-from-behind-tp4657029.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-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] Percentages in bar plot
Hello, Try barplot(table(stop)/sum(table(stop))) Hope this helps, Rui Barradas Em 30-01-2013 11:34, Naser Jamil escreveu: Dear R-users, Though it's a silly thing to ask, but I'm not getting a way out. I wish to find the percentage distribution for a data vector 'stop'. The coomand below is giving the frequency distribution. May I know the option to see the percentages instead of frequencies. Similarly, what option I should put in a histogram plot to see the percentages instead of frequencies? - stop-c(8,6,6,6,8,6,6,8,8,6,6,6,8,8,8) barplot(table(stop)) Many thanks for your time. Regards, Jamil. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] texture mapping images on the faces of a cuboid with rgl
Hi, I am interested in mapping 6 different images on the faces of a cube. I found the full code for doing this here : http://rwiki.sciviews.org/doku.php?id=graph_gallery:cube But I am unable to adapt the code for my purpose. I would appreciate it if I could get some help on the following points : 1. My images are jpeg images with dimensions of 1936*2584*3. Must I seperately process the grey and rgb attributes as in the code in the above page. Would appreciate it if I could get some explicit help on processing the image matrix. I don't follow in detail how a square portion of the graphic is extracted also. 2. Instead of a cube, I am interested in mapping images on a cuboid (l!=b!=h). Could I get tips on how this can be done. How should the dimensions of the image matrix be tailored then? 3. The following line in the code appears to identify and fix the corner points : coords - rbind( c(0 , 2, 3), c(1 , 3, 2), c(3, 0 , 2), c(2, 1 , 3), c(2, 3, 0 ), c(3, 2, 1 ) ) How should I read and understand the orientation of the cube corners from this? A simple tip can get me started. I would like to add some example code (to complement Romain Francois' code on the rwiki page referred to above). But, as can be seen from my questions, I don't understand this sufficiently to do this. But I hope that I have given sufficient information to explain my problem. Thanks, Ravi [[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] Change rows and columns
Please reply with context to the list. Most R-help readers do not use nabble. I don't quite understand your question but do you mean something like ?names? John Kane Kingston ON Canada -Original Message- From: matthias.we...@fnt.de Sent: Tue, 29 Jan 2013 07:41:21 -0800 (PST) To: r-help@r-project.org Subject: Re: [R] Change rows and columns thx, that worked perfekt, but is there any way to convert my Number 1 row # 123 up as column description? -- View this message in context: http://r.789695.n4.nabble.com/Change-rows-and-columns-tp4656965p4656969.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. FREE ONLINE PHOTOSHARING - Share your photos online with your friends and family! Visit http://www.inbox.com/photosharing to find out more! __ 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] pROC in R
I don't see what is happening from your code but you have a typo in the emai if not in your code. It should be library (pROC) John Kane Kingston ON Canada -Original Message- From: feth...@yahoo.fr Sent: Mon, 28 Jan 2013 14:44:50 + (GMT) To: r-help@r-project.org Subject: [R] pROC in R B Dear, I would like to use pROC software for my study, butB I could not uploaded it in R. Could you please help me to overcome this problem? This is the message when I write Library (pROC) : Le chargement a nC)cessitC) le package : plyr Type 'citation(pROC)' for a citation. Attachement du package : b??pROCb?? The following object(s) are masked from b??package:statsb??: B B B cov, smooth, var Thank you in advance Best Regards Fethi Student at Geneva 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. FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks orcas on your desktop! __ 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] Sweave files generating miktex errors
Michael, I added C:\R\R-2.15.2\share\texmf to the list of roots and hit apply and that solved the problem. Thanks! Duncan, I did try the R CMD Sweave you suggested and that did not solve the issue. The problem was in running LaTeX so how was your suggestion help me? It would have to copy the .sty files somewhere to help the situation. Thanks guys. I appreciate the help! Troy On Wed, Jan 30, 2013 at 5:52 AM, Michael Friendly frien...@yorku.ca wrote: On 1/29/2013 6:04 PM, Troy S wrote: Ok, yes I realize it. So let me try to fix it: I removed the sty files, and set TEXINPUTS to 'C:\R\R-2.15.2\share\texmf\**tex\latex I am back where I started: the tex file will not process. Please let me know what I should do to fix the issue. Troy Don't do that either. Go to Programs -Miktex - Maintenance (Admin) - Settings and on the Roots tab, add a folder, e.g., c:\localtexmf having the same structure as the default texmf tree. E.g., I have there \tex\latex\..., \bibtex\{bib,bst} ... where I put all things tex that aren't installed directly by Miktex -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. Chair, Quantitative Methods York University Voice: 416 736-2100 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA [[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] Fastest way to compare a single value with all values in one column of a data frame
HI, Sorry, my previous solution doesn't work. This should work for your dataset: set.seed(1851) x- data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F) y- data.frame(item=f,a=3,b=10,stringsAsFactors=F) x[x$a%in%which.min(x[x$ay$a,]$a),]- y #if there are multiple minimum values set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e4,replace=TRUE),a=sample(1:30,1e4,replace=TRUE),b=sample(1:100,1e4,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 330 system.time({x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1}) # user system elapsed # 0.000 0.000 0.001 length(x1$a[x1$a==1]) #[1] 0 #For some reason, it is not working when the multiple number of minimum values some value set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e5,replace=TRUE),a=sample(1:30,1e5,replace=TRUE),b=sample(1:100,1e5,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 3404 x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1 length(x1$a[x1$a==1]) #[1] 3404 #not getting replaced #However, if I try: set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e6,replace=TRUE),a=sample(1:5000,1e6,replace=TRUE),b=sample(1:100,1e6,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 208 system.time(x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1) #user system elapsed # 0.124 0.016 0.138 length(x1$a[x1$a==1]) #[1] 0 #Tried Jessica's solution: set.seed(1851) x- data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F) y- data.frame(item=f,a=3,b=10,stringsAsFactors=F) x[intersect(which(x$a y$a),which.min(x$a)),] - y x # item a b #1 a 8 25 #2 a 10 26 #3 f 3 10 #replaced #4 e 15 26 #5 b 13 20 #6 a 5 23 #7 d 4 29 #8 e 2 24 #9 c 7 30 #10 e 14 24 #11 d 2 20 #12 e 10 21 #13 c 13 27 #14 d 12 23 #15 b 11 26 #16 e 5 22 #17 c 1 26 #it is not replaced #18 a 8 21 #19 e 10 26 #20 c 2 22 A.K. - Original Message - From: Dimitri Liakhovitski dimitri.liakhovit...@gmail.com To: r-help r-help@r-project.org Cc: Sent: Tuesday, January 29, 2013 4:11 PM Subject: [R] Fastest way to compare a single value with all values in one column of a data frame Hello! I have a large data frame x: x-data.frame(item=letters[1:5],a=1:5,b=11:15) # in actuality, x has 1000 rows x$item-as.character(x$item) I also have a small data frame y with just 1 row: y-data.frame(item=f,a=3,b=10) y$item-as.character(y$item) I have to decide if y$a is larger than the smallest of all the values in x$a. If it is, I want y to replace the whole row in x that has the lowest value in column a. This is how I'd do it. if(y$amin(x$a)){ whichmin-which(x$a==min(x$a)) x[whichmin,]-y[1,] } I am wondering if there is a faster way of doing it. What would be the fastest possible way? I'd have to do it, unfortunately, many-many times. Thank you very much! -- Dimitri Liakhovitski gfk.com http://marketfusionanalytics.com/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Relative Risk in logistic regression
On 01/30/2013 09:02 AM, nalluri pratap wrote: Relative risk = exp(coef(model)) Only if you fit using the log link. Using the logit link, this gives odds ratios. --- On Wed, 30/1/13, aminreza Aamini amin.r@gmail.com wrote: From: aminreza Aamini amin.r@gmail.com Subject: [R] Relative Risk in logistic regression To: R-help R-help@r-project.org Date: Wednesday, 30 January, 2013, 4:19 PM Hi all, I am very grateful to all those who write to me 1) how i can obtain relative risk (risk ratio) in logistic regression in R. 2) how to obtain the predicted risk for a certain individual using fitted regression model in R. Many thanks, in advance, for your help. Amin. __ 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. -- Kevin E. Thorpe Head of Biostatistics, Applied Health Research Centre (AHRC) Li Ka Shing Knowledge Institute of St. Michael's Assistant Professor, Dalla Lana School of Public Health University of Toronto email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016 __ 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 use ...
There is now a blog post that attempts to answer the question in the subject line: http://www.burns-stat.com/the-three-dots-construct-in-r/ Pat On 17/01/2013 14:36, Ivan Calandra wrote: Dear users, I'm trying to learn how to use the I have written a function (simplified here) that uses doBy::summaryBy(): # 'dat' is a data.frame from which the aggregation is computed # 'vec_cat' is a integer vector defining which columns of the data.frame should be use on the right side of the formula # 'stat_fun' is the function that will be run to aggregate stat.group - function(dat, vec_cat, stat_fun){ require(doBy) df - summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))), data=dat, FUN=stat_fun) return(df) } Example, works fine: my_data - structure(list(cat = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L), .Label = c(A, B), class = factor), varnum = c(-0.754816565434373, -1.94101630973709, -0.102461836059522, -0.519952759645808, -1.73772800855664, -1.13939178585609, 0.522356715260142, -0.701428514907824, 1.45197576541159, 0.0844567413828095)), .Names = c(cat, varnum), row.names = c(NA, -10L), class = data.frame) stat.group(dat=my_data, vec_cat=1, stat_fun=mean) Now summaryBy() has an ... argument and I would like to use it. For example, I would like to be able to add the trim argument to my call like this: stat.group(dat=my_data, vec_cat=1, stat_fun=mean, trim=0.2) I know I can do it using this ... but I have no idea how to do it. I've tried to search for it, but a search with ... doesn't yield interesting results! Thank you in advance for your help! Ivan -- Patrick Burns pbu...@pburns.seanet.com twitter: @burnsstat @portfolioprobe http://www.portfolioprobe.com/blog http://www.burns-stat.com (home of: 'Impatient R' 'The R Inferno' 'Tao Te Programming') __ 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] Relative Risk in logistic regression
I am not sure why one would want a relative risk from a logistic regression. The measure of association from a logistic regression is the odds ratio, not the relative risk. John John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Kevin E. Thorpe kevin.tho...@utoronto.ca 1/30/2013 11:14 AM On 01/30/2013 09:02 AM, nalluri pratap wrote: Relative risk = exp(coef(model)) Only if you fit using the log link. Using the logit link, this gives odds ratios. --- On Wed, 30/1/13, aminreza Aamini amin.r@gmail.com wrote: From: aminreza Aamini amin.r@gmail.com Subject: [R] Relative Risk in logistic regression To: R-help R-help@r-project.org Date: Wednesday, 30 January, 2013, 4:19 PM Hi all, I am very grateful to all those who write to me 1) how i can obtain relative risk (risk ratio) in logistic regression in R. 2) how to obtain the predicted risk for a certain individual using fitted regression model in R. Many thanks, in advance, for your help. Amin. __ 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. -- Kevin E. Thorpe Head of Biostatistics, Applied Health Research Centre (AHRC) Li Ka Shing Knowledge Institute of St. Michael's Assistant Professor, Dalla Lana School of Public Health University of Toronto email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016 __ 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. Confidentiality Statement: This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential and privileged information. Any unauthorized use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and destroy all copies of the original 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] Relative Risk in logistic regression
On 01/30/2013 11:17 AM, John Sorkin wrote: I am not sure why one would want a relative risk from a logistic regression. The measure of association from a logistic regression is the odds ratio, not the relative risk. John Yes, the natural measure, when using the logit link, is the OR. I always use that myself, but if you use the log link, you get RR. As for why you would, I would suggest it's because physicians think they understand a RR better than and OR. Kevin E. Thorpe kevin.tho...@utoronto.ca 1/30/2013 11:14 AM On 01/30/2013 09:02 AM, nalluri pratap wrote: Relative risk = exp(coef(model)) Only if you fit using the log link. Using the logit link, this gives odds ratios. --- On Wed, 30/1/13, aminreza Aamini amin.r@gmail.com wrote: From: aminreza Aamini amin.r@gmail.com Subject: [R] Relative Risk in logistic regression To: R-help R-help@r-project.org Date: Wednesday, 30 January, 2013, 4:19 PM Hi all, I am very grateful to all those who write to me 1) how i can obtain relative risk (risk ratio) in logistic regression in R. 2) how to obtain the predicted risk for a certain individual using fitted regression model in R. Many thanks, in advance, for your help. Amin. __ 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. -- Kevin E. Thorpe Head of Biostatistics, Applied Health Research Centre (AHRC) Li Ka Shing Knowledge Institute of St. Michael's Assistant Professor, Dalla Lana School of Public Health University of Toronto email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016 __ 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] Relative Risk in logistic regression
If you use a log link, you are not, I believe, performing a logistic regression! John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Kevin E. Thorpe kevin.tho...@utoronto.ca 1/30/2013 11:22 AM On 01/30/2013 11:17 AM, John Sorkin wrote: I am not sure why one would want a relative risk from a logistic regression. The measure of association from a logistic regression is the odds ratio, not the relative risk. John Yes, the natural measure, when using the logit link, is the OR. I always use that myself, but if you use the log link, you get RR. As for why you would, I would suggest it's because physicians think they understand a RR better than and OR. Kevin E. Thorpe kevin.tho...@utoronto.ca 1/30/2013 11:14 AM On 01/30/2013 09:02 AM, nalluri pratap wrote: Relative risk = exp(coef(model)) Only if you fit using the log link. Using the logit link, this gives odds ratios. --- On Wed, 30/1/13, aminreza Aamini amin.r@gmail.com wrote: From: aminreza Aamini amin.r@gmail.com Subject: [R] Relative Risk in logistic regression To: R-help R-help@r-project.org Date: Wednesday, 30 January, 2013, 4:19 PM Hi all, I am very grateful to all those who write to me 1) how i can obtain relative risk (risk ratio) in logistic regression in R. 2) how to obtain the predicted risk for a certain individual using fitted regression model in R. Many thanks, in advance, for your help. Amin. __ 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. -- Kevin E. Thorpe Head of Biostatistics, Applied Health Research Centre (AHRC) Li Ka Shing Knowledge Institute of St. Michael's Assistant Professor, Dalla Lana School of Public Health University of Toronto email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016 Confidentiality Statement: This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential and privileged information. Any unauthorized use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and destroy all copies of the original 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] Programmatically give file name to a matrix
This is FAQ 7.21. The most important part of that answer is at the end where it says that it is better to use a list. Your code could be something like: plotroc - list() for (i in levels(mergeTrn$Continent) { # matrix defined here plotroc[[ paste(plotroc_GBM_TRN_,i, sep=) ]] - matrix } now all of your created matrices are in the plotroc list (or whatever you want to call it) and can be operated on using functions like sapply and lapply. On Wed, Jan 30, 2013 at 1:04 AM, Kumar Mainali kpmain...@gmail.com wrote: I have a situation when I need to save matrix with file names that are programmatically created. for (i in levels(mergeTrn$Continent)) { matrix here # I want to save this matrix with a file name that carries i from for loop. The following does not work. paste(plotroc_GBM_Trn_, i, sep=) - matrix } Thanks, Kumar -- Section of Integrative Biology University of Texas at Austin Austin, Texas 78712, USA [[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. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.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] Relative Risk in logistic regression
On 01/30/2013 11:26 AM, John Sorkin wrote: If you use a log link, you are not, I believe, performing a logistic regression! I guess strictly speaking, that is true. I was being a little sloppy in terminology. Kevin E. Thorpe kevin.tho...@utoronto.ca 1/30/2013 11:22 AM On 01/30/2013 11:17 AM, John Sorkin wrote: I am not sure why one would want a relative risk from a logistic regression. The measure of association from a logistic regression is the odds ratio, not the relative risk. John Yes, the natural measure, when using the logit link, is the OR. I always use that myself, but if you use the log link, you get RR. As for why you would, I would suggest it's because physicians think they understand a RR better than and OR. Kevin E. Thorpe kevin.tho...@utoronto.ca 1/30/2013 11:14 AM On 01/30/2013 09:02 AM, nalluri pratap wrote: Relative risk = exp(coef(model)) Only if you fit using the log link. Using the logit link, this gives odds ratios. --- On Wed, 30/1/13, aminreza Aamini amin.r@gmail.com wrote: From: aminreza Aamini amin.r@gmail.com Subject: [R] Relative Risk in logistic regression To: R-help R-help@r-project.org Date: Wednesday, 30 January, 2013, 4:19 PM Hi all, I am very grateful to all those who write to me 1) how i can obtain relative risk (risk ratio) in logistic regression in R. 2) how to obtain the predicted risk for a certain individual using fitted regression model in R. Many thanks, in advance, for your help. Amin. __ 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]] -- Kevin E. Thorpe Head of Biostatistics, Applied Health Research Centre (AHRC) Li Ka Shing Knowledge Institute of St. Michael's Assistant Professor, Dalla Lana School of Public Health University of Toronto email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016 __ 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] RandomForest and Missing Values
If you wish to remove missing values, you can use the option na.action=na.omit.If you wish to Impute you can use rfImpute. --- On Mon, 28/1/13, Lorenzo Isella lorenzo.ise...@gmail.com wrote: From: Lorenzo Isella lorenzo.ise...@gmail.com Subject: [R] RandomForest and Missing Values To: r-h...@stat.math.ethz.ch Date: Monday, 28 January, 2013, 10:07 PM Dear All, I would like to use a randomForest algorithm on a dataset. The set is not particularly large/difficult to handle, but it has some missing values (both factors and numerical values). According to what I found https://stat.ethz.ch/pipermail/r-help/2005-September/078880.html https://stat.ethz.ch/pipermail/r-help/2007-January/123117.html the randomForest package has a problem with missing data (essentially you have to resort to some trick to introduce them into your dataset --a median value, the most common factor, a linear interpolation etc...). Seen that I could not find a clear workaround for this (but I cannot be the only one who has in mind to do a randomForest on a less than perfect data set), can anyone help me out? I am concerned about the consequences of introducing the missing values into the data set. The cforest function in the Party package does not seem to have this limitation, but on the other hand the randomForest package has passed the test of timeso should I drop it in this case? Any suggestion is appreciated. Cheers Lorenzo __ 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] how to use ...
Because R can be interactive, I find that a little exploring through the use of strategically placed browser() calls (?browser if you are unfamiliar with this handy debugging tool) is often the fastest way to solve little R puzzles like this. For example, try this (in an R GUI): f2 - function(y,...){ browser(); list(...) } f1 - function(x,...){ browser(); f2(x,...) } Now at the command line: f1( x=1, z= 2:4) ## In the f1 browser do ls() to see what's in the f1 environment ## and then do list(...) to see what's in the dots argument ## then issue the browser command c to continue ## and do the same in the f2 browser . ## then try: f1( x=1, y= 2:4) ## and f1( 2:4) f1( y=2:4) ## this will produce an error in f2 I think this help give you a better idea of what's going on (or totally confuse?!). -- Bert On Wed, Jan 30, 2013 at 8:15 AM, Patrick Burns pbu...@pburns.seanet.com wrote: There is now a blog post that attempts to answer the question in the subject line: http://www.burns-stat.com/the-three-dots-construct-in-r/ Pat On 17/01/2013 14:36, Ivan Calandra wrote: Dear users, I'm trying to learn how to use the I have written a function (simplified here) that uses doBy::summaryBy(): # 'dat' is a data.frame from which the aggregation is computed # 'vec_cat' is a integer vector defining which columns of the data.frame should be use on the right side of the formula # 'stat_fun' is the function that will be run to aggregate stat.group - function(dat, vec_cat, stat_fun){ require(doBy) df - summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))), data=dat, FUN=stat_fun) return(df) } Example, works fine: my_data - structure(list(cat = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L), .Label = c(A, B), class = factor), varnum = c(-0.754816565434373, -1.94101630973709, -0.102461836059522, -0.519952759645808, -1.73772800855664, -1.13939178585609, 0.522356715260142, -0.701428514907824, 1.45197576541159, 0.0844567413828095)), .Names = c(cat, varnum), row.names = c(NA, -10L), class = data.frame) stat.group(dat=my_data, vec_cat=1, stat_fun=mean) Now summaryBy() has an ... argument and I would like to use it. For example, I would like to be able to add the trim argument to my call like this: stat.group(dat=my_data, vec_cat=1, stat_fun=mean, trim=0.2) I know I can do it using this ... but I have no idea how to do it. I've tried to search for it, but a search with ... doesn't yield interesting results! Thank you in advance for your help! Ivan -- Patrick Burns pbu...@pburns.seanet.com twitter: @burnsstat @portfolioprobe http://www.portfolioprobe.com/blog http://www.burns-stat.com (home of: 'Impatient R' 'The R Inferno' 'Tao Te Programming') __ 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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ 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] Sweave files generating miktex errors
On 13-01-30 11:00 AM, Troy S wrote: Michael, I added C:\R\R-2.15.2\share\texmf to the list of roots and hit apply and that solved the problem. Thanks! Duncan, I did try the R CMD Sweave you suggested and that did not solve the issue. The problem was in running LaTeX so how was your suggestion help me? It would have to copy the .sty files somewhere to help the situation. No, if you use the --pdf option (as I suggested!!) it would run LaTeX for you, after figuring out what distribution of LaTeX you're using, and putting together the right incantation to tell it to find the style files. Please do not post your solution on a blog somewhere. There is way too much bad advice about running Sweave on the net already. Duncan Murdoch Thanks guys. I appreciate the help! Troy On Wed, Jan 30, 2013 at 5:52 AM, Michael Friendly frien...@yorku.ca mailto:frien...@yorku.ca wrote: On 1/29/2013 6:04 PM, Troy S wrote: Ok, yes I realize it. So let me try to fix it: I removed the sty files, and set TEXINPUTS to 'C:\R\R-2.15.2\share\texmf\__tex\latex I am back where I started: the tex file will not process. Please let me know what I should do to fix the issue. Troy Don't do that either. Go to Programs -Miktex - Maintenance (Admin) - Settings and on the Roots tab, add a folder, e.g., c:\localtexmf having the same structure as the default texmf tree. E.g., I have there \tex\latex\..., \bibtex\{bib,bst} ... where I put all things tex that aren't installed directly by Miktex -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. Chair, Quantitative Methods York University Voice: 416 736-2100 x66249 tel:416%20736-2100%20x66249 Fax: 416 736-5814 tel:416%20736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ 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] Sweave files generating miktex errors
Duncan, Alas, I had left out the --pdf flag when I tried the solution. I am glad I have an answer now, and a better understanding of the problem. Troy On Wed, Jan 30, 2013 at 9:50 AM, Duncan Murdoch murdoch.dun...@gmail.comwrote: On 13-01-30 11:00 AM, Troy S wrote: Michael, I added C:\R\R-2.15.2\share\texmf to the list of roots and hit apply and that solved the problem. Thanks! Duncan, I did try the R CMD Sweave you suggested and that did not solve the issue. The problem was in running LaTeX so how was your suggestion help me? It would have to copy the .sty files somewhere to help the situation. No, if you use the --pdf option (as I suggested!!) it would run LaTeX for you, after figuring out what distribution of LaTeX you're using, and putting together the right incantation to tell it to find the style files. Please do not post your solution on a blog somewhere. There is way too much bad advice about running Sweave on the net already. Duncan Murdoch Thanks guys. I appreciate the help! Troy On Wed, Jan 30, 2013 at 5:52 AM, Michael Friendly frien...@yorku.ca mailto:frien...@yorku.ca wrote: On 1/29/2013 6:04 PM, Troy S wrote: Ok, yes I realize it. So let me try to fix it: I removed the sty files, and set TEXINPUTS to 'C:\R\R-2.15.2\share\texmf\__**tex\latex I am back where I started: the tex file will not process. Please let me know what I should do to fix the issue. Troy Don't do that either. Go to Programs -Miktex - Maintenance (Admin) - Settings and on the Roots tab, add a folder, e.g., c:\localtexmf having the same structure as the default texmf tree. E.g., I have there \tex\latex\..., \bibtex\{bib,bst} ... where I put all things tex that aren't installed directly by Miktex -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. Chair, Quantitative Methods York University Voice: 416 736-2100 x66249 tel:416%20736-2100%20x66249 Fax: 416 736-5814 tel:416%20736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA [[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] Fastest way to compare a single value with all values in one column of a data frame
Thank you, everyone! I'll try to test those different approaches. Really appreciate your help! Dimitri On Wed, Jan 30, 2013 at 11:03 AM, arun smartpink...@yahoo.com wrote: HI, Sorry, my previous solution doesn't work. This should work for your dataset: set.seed(1851) x- data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F) y- data.frame(item=f,a=3,b=10,stringsAsFactors=F) x[x$a%in%which.min(x[x$ay$a,]$a),]- y #if there are multiple minimum values set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e4,replace=TRUE),a=sample(1:30,1e4,replace=TRUE),b=sample(1:100,1e4,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 330 system.time({x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1}) # user system elapsed # 0.000 0.000 0.001 length(x1$a[x1$a==1]) #[1] 0 #For some reason, it is not working when the multiple number of minimum values some value set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e5,replace=TRUE),a=sample(1:30,1e5,replace=TRUE),b=sample(1:100,1e5,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 3404 x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1 length(x1$a[x1$a==1]) #[1] 3404 #not getting replaced #However, if I try: set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e6,replace=TRUE),a=sample(1:5000,1e6,replace=TRUE),b=sample(1:100,1e6,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 208 system.time(x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1) #user system elapsed # 0.124 0.016 0.138 length(x1$a[x1$a==1]) #[1] 0 #Tried Jessica's solution: set.seed(1851) x- data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F) y- data.frame(item=f,a=3,b=10,stringsAsFactors=F) x[intersect(which(x$a y$a),which.min(x$a)),] - y x # item a b #1 a 8 25 #2 a 10 26 #3 f 3 10 #replaced #4 e 15 26 #5 b 13 20 #6 a 5 23 #7 d 4 29 #8 e 2 24 #9 c 7 30 #10e 14 24 #11d 2 20 #12e 10 21 #13c 13 27 #14d 12 23 #15b 11 26 #16e 5 22 #17c 1 26 #it is not replaced #18a 8 21 #19e 10 26 #20c 2 22 A.K. - Original Message - From: Dimitri Liakhovitski dimitri.liakhovit...@gmail.com To: r-help r-help@r-project.org Cc: Sent: Tuesday, January 29, 2013 4:11 PM Subject: [R] Fastest way to compare a single value with all values in one column of a data frame Hello! I have a large data frame x: x-data.frame(item=letters[1:5],a=1:5,b=11:15) # in actuality, x has 1000 rows x$item-as.character(x$item) I also have a small data frame y with just 1 row: y-data.frame(item=f,a=3,b=10) y$item-as.character(y$item) I have to decide if y$a is larger than the smallest of all the values in x$a. If it is, I want y to replace the whole row in x that has the lowest value in column a. This is how I'd do it. if(y$amin(x$a)){ whichmin-which(x$a==min(x$a)) x[whichmin,]-y[1,] } I am wondering if there is a faster way of doing it. What would be the fastest possible way? I'd have to do it, unfortunately, many-many times. Thank you very much! -- Dimitri Liakhovitski gfk.com http://marketfusionanalytics.com/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dimitri Liakhovitski gfk.com http://marketfusionanalytics.com/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fastest way to compare a single value with all values in one column of a data frame
Sorry - I should have clarified: My identifiers (in column item) will always be unique. In other words, one entry in column item will never be repeated - neither in x nor in y. Dimitri On Wed, Jan 30, 2013 at 1:27 PM, Dimitri Liakhovitski dimitri.liakhovit...@gmail.com wrote: Thank you, everyone! I'll try to test those different approaches. Really appreciate your help! Dimitri On Wed, Jan 30, 2013 at 11:03 AM, arun smartpink...@yahoo.com wrote: HI, Sorry, my previous solution doesn't work. This should work for your dataset: set.seed(1851) x- data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F) y- data.frame(item=f,a=3,b=10,stringsAsFactors=F) x[x$a%in%which.min(x[x$ay$a,]$a),]- y #if there are multiple minimum values set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e4,replace=TRUE),a=sample(1:30,1e4,replace=TRUE),b=sample(1:100,1e4,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 330 system.time({x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1}) # user system elapsed # 0.000 0.000 0.001 length(x1$a[x1$a==1]) #[1] 0 #For some reason, it is not working when the multiple number of minimum values some value set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e5,replace=TRUE),a=sample(1:30,1e5,replace=TRUE),b=sample(1:100,1e5,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 3404 x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1 length(x1$a[x1$a==1]) #[1] 3404 #not getting replaced #However, if I try: set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e6,replace=TRUE),a=sample(1:5000,1e6,replace=TRUE),b=sample(1:100,1e6,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 208 system.time(x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1) #user system elapsed # 0.124 0.016 0.138 length(x1$a[x1$a==1]) #[1] 0 #Tried Jessica's solution: set.seed(1851) x- data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F) y- data.frame(item=f,a=3,b=10,stringsAsFactors=F) x[intersect(which(x$a y$a),which.min(x$a)),] - y x # item a b #1 a 8 25 #2 a 10 26 #3 f 3 10 #replaced #4 e 15 26 #5 b 13 20 #6 a 5 23 #7 d 4 29 #8 e 2 24 #9 c 7 30 #10e 14 24 #11d 2 20 #12e 10 21 #13c 13 27 #14d 12 23 #15b 11 26 #16e 5 22 #17c 1 26 #it is not replaced #18a 8 21 #19e 10 26 #20c 2 22 A.K. - Original Message - From: Dimitri Liakhovitski dimitri.liakhovit...@gmail.com To: r-help r-help@r-project.org Cc: Sent: Tuesday, January 29, 2013 4:11 PM Subject: [R] Fastest way to compare a single value with all values in one column of a data frame Hello! I have a large data frame x: x-data.frame(item=letters[1:5],a=1:5,b=11:15) # in actuality, x has 1000 rows x$item-as.character(x$item) I also have a small data frame y with just 1 row: y-data.frame(item=f,a=3,b=10) y$item-as.character(y$item) I have to decide if y$a is larger than the smallest of all the values in x$a. If it is, I want y to replace the whole row in x that has the lowest value in column a. This is how I'd do it. if(y$amin(x$a)){ whichmin-which(x$a==min(x$a)) x[whichmin,]-y[1,] } I am wondering if there is a faster way of doing it. What would be the fastest possible way? I'd have to do it, unfortunately, many-many times. Thank you very much! -- Dimitri Liakhovitski gfk.com http://marketfusionanalytics.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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dimitri Liakhovitski gfk.com http://marketfusionanalytics.com/ -- Dimitri Liakhovitski gfk.com http://marketfusionanalytics.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] Fastest way to compare a single value with all values in one column of a data frame
Hi, Any chance x$a to have the same number repeated? If `Item` and `a` are unique, I guess both the solutions should work. set.seed(1851) x- data.frame(item=sample(letters[1:20],20,replace=F),a=sample(1:45,20,replace=F),b=sample(20:50,20,replace=F),stringsAsFactors=F) y- data.frame(item=z,a=3,b=10,stringsAsFactors=F) x[intersect(which(x$a y$a),which.min(x$a)),] # item a b #17 c 1 48 x[x$a==which.min(x$a[x$ay$a]),] # item a b #17 c 1 48 #or x[x$a%in%which.min(x$a[x$ay$a]),] # item a b #17 c 1 48 x[x$a%in%which.min(x$a[x$ay$a]),]-y tail(x) # item a b #15 q 45 30 #16 g 10 23 #17 z 3 10 #18 r 15 39 #19 l 18 45 #20 t 35 33 #However, if `item` column is unique, but `a` is not, then the one I mentioned previously arise. set.seed(1851) x1- data.frame(item=sample(letters[1:20],20,replace=F),a=sample(1:10,20,replace=T),b=sample(20:50,20,replace=F),stringsAsFactors=F) y1- data.frame(item=z,a=3,b=10,stringsAsFactors=F) x1[intersect(which(x1$a y1$a),which.min(x1$a)),] # item a b #3 s 1 41 x1[x1$a==which.min(x1$a[x1$ay1$a]),] # item a b #3 s 1 41 #11 h 1 46 #17 c 1 48 x1[x1$a==which.min(x1$a[x1$ay1$a]),]- y1 A.K. From: Dimitri Liakhovitski dimitri.liakhovit...@gmail.com To: arun smartpink...@yahoo.com Cc: R help r-help@r-project.org; Jessica Streicher j.streic...@micromata.de Sent: Wednesday, January 30, 2013 1:49 PM Subject: Re: [R] Fastest way to compare a single value with all values in one column of a data frame Sorry - I should have clarified: My identifiers (in column item) will always be unique. In other words, one entry in column item will never be repeated - neither in x nor in y. Dimitri On Wed, Jan 30, 2013 at 1:27 PM, Dimitri Liakhovitski dimitri.liakhovit...@gmail.com wrote: Thank you, everyone! I'll try to test those different approaches. Really appreciate your help! Dimitri On Wed, Jan 30, 2013 at 11:03 AM, arun smartpink...@yahoo.com wrote: HI, Sorry, my previous solution doesn't work. This should work for your dataset: set.seed(1851) x- data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F) y- data.frame(item=f,a=3,b=10,stringsAsFactors=F) x[x$a%in%which.min(x[x$ay$a,]$a),]- y #if there are multiple minimum values set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e4,replace=TRUE),a=sample(1:30,1e4,replace=TRUE),b=sample(1:100,1e4,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 330 system.time({x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1}) # user system elapsed # 0.000 0.000 0.001 length(x1$a[x1$a==1]) #[1] 0 #For some reason, it is not working when the multiple number of minimum values some value set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e5,replace=TRUE),a=sample(1:30,1e5,replace=TRUE),b=sample(1:100,1e5,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 3404 x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1 length(x1$a[x1$a==1]) #[1] 3404 #not getting replaced #However, if I try: set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e6,replace=TRUE),a=sample(1:5000,1e6,replace=TRUE),b=sample(1:100,1e6,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 208 system.time(x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1) #user system elapsed # 0.124 0.016 0.138 length(x1$a[x1$a==1]) #[1] 0 #Tried Jessica's solution: set.seed(1851) x- data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F) y- data.frame(item=f,a=3,b=10,stringsAsFactors=F) x[intersect(which(x$a y$a),which.min(x$a)),] - y x # item a b #1 a 8 25 #2 a 10 26 #3 f 3 10 #replaced #4 e 15 26 #5 b 13 20 #6 a 5 23 #7 d 4 29 #8 e 2 24 #9 c 7 30 #10 e 14 24 #11 d 2 20 #12 e 10 21 #13 c 13 27 #14 d 12 23 #15 b 11 26 #16 e 5 22 #17 c 1 26 #it is not replaced #18 a 8 21 #19 e 10 26 #20 c 2 22 A.K. - Original Message - From: Dimitri Liakhovitski dimitri.liakhovit...@gmail.com To: r-help r-help@r-project.org Cc: Sent: Tuesday, January 29, 2013 4:11 PM Subject: [R] Fastest way to compare a single value with all values in one column of a data frame Hello! I have a large data frame x: x-data.frame(item=letters[1:5],a=1:5,b=11:15) # in actuality, x has 1000 rows x$item-as.character(x$item) I also have a small data frame y with just 1 row: y-data.frame(item=f,a=3,b=10) y$item-as.character(y$item) I have to decide if y$a is larger than the smallest of all the values in x$a. If it is, I want y to replace the whole row in x that has the lowest value in column a. This is how I'd do it. if(y$amin(x$a)){ whichmin-which(x$a==min(x$a))
Re: [R] starting values in glm(..., family = binomial(link =log))
Thanks for your replies! It seems, that I can fit my model now, when I can provide the right starting values; however there remain warnings, such as: 1: In log(ifelse(y == 1, 1, (1 - y)/(1 - mu))) : NaNs wurden erzeugt 2: step size truncated due to divergence 3: step size truncated: out of bounds ... That makes me feel uncomfortable and I wonder whether I can trust the fitted model. Why is this kind of regression so picky about starting values compared to logistic regression? And is there a way to explain the difference between binomial - quasibinomial to a simple mind like mine? Best, Felix Von: Ravi Varadhan [mailto:ravi.varad...@jhu.edu] Gesendet: Mittwoch, 30. Januar 2013 17:02 An: Fischer, Felix Cc: r-help@r-project.org Betreff: [R] starting values in glm(..., family = binomial(link =log)) Try this: Age_log_model = glm(Arthrose ~ Alter, data=x, start=c(-1, 0), family=quasibinomial(link = log)) Ravi Ravi Varadhan, Ph.D. Assistant Professor The Center on Aging and Health Division of Geriatric Medicine Gerontology Johns Hopkins University rvarad...@jhmi.edumailto:rvarad...@jhmi.edu 410-502-2619 [[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] Going in circles
Hi, we're trying to install rJava on SLES 11. With: R CMD INSTALL rJava_0.9-3.tar.gz It says: configure: error: One or more Java configuration variables are not set. Make sure R is configured with full Java support (including JDK). Run R CMD javareconf as root to add Java support to R. So we run: R CMD javareconf It says: Updating Java configuration in /apps/R-2.14.2/lib64/R Done. and appears to update R correctly. But if we run this again: R CMD INSTALL rJava_0.9-3.tar.gz It gives the same error message: configure: error: One or more Java configuration variables are not set. Make sure R is configured with full Java support (including JDK). Run R CMD javareconf as root to add Java support to R. So we're going in circles with this. Is javareconf not fully configuring R? Are we missing something? [[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] starting values in glm(..., family = binomial(link =log))
Try this: Age_log_model = glm(Arthrose ~ Alter, data=x, start=c(-1, 0), family=quasibinomial(link = log)) Ravi Ravi Varadhan, Ph.D. Assistant Professor The Center on Aging and Health Division of Geriatric Medicine Gerontology Johns Hopkins University rvarad...@jhmi.edumailto:rvarad...@jhmi.edu 410-502-2619 [[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] non-metric multidimensional scaling
Hello, I would like to perform an NMDS on the following: I have two independent variables, which are sites and treatments. I have 6 sites which are peatlands. I collected 5 replicates (at the same time) from each of the sites. I used each of the replicates in a treatment. There were 4 treatments. 2 were controls and the other two were a simulated 30 day drought and a simulated 60 day drought. From each of the replicates in each of the treatments I measured pH and nitrate, sulphate, dissolved organic carbon, calcium, magnesium, nickel, cobalt, aluminum, and copper concentrations. I tried performing an NMDS using the following tutorial http://strata.uga.edu/software/pdf/mdsTutorial.pdf. However, I did not take into account sites or treatments. All the data was used without making those distinctions. I would like to make an NMDS plot which distinguishes the points as sites and another plot which distinguishes the points as treatments but which still demonstrates where the pH, nitrate, sulphate, etc variables are located across the plot. Do you know how I can accomplish this? I hope I am clear in my explanation. Thank you, Myra [[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] How does predict() calculate prediction intervals?
For a given linear regression, I wish to find the 2-tailed t-dist probability that Y-hat = newly observed values. I generate prediction intervals in predict() for plotting, but when I calculate my t-dist probabilities, they don't agree. I have researched the issues with variance of individual predictions and been advised to use the variance formula below (in the code). I presume my variance function differs from that used in predict(). Can someone advise me as to why I cannot reproduce the results of predict()? As a test, I calculated the prediction intervals about Y-hat by hand and compared them to predict(): X - log10(c(780,54,520,703,1356,1900,2500,741,1500,600)) Y - log10(c(0.037,0.036,0.026,0.0315,0.028,0.0099,0.00955,0.0405,0.019,0.0245)) dat - data.frame(cbind(X, Y)) mod - lm(Y ~ X, data = dat) rm(X,Y) ## model predictions, y.hat, at the new X values pred - data.frame(predict.lm(mod,newdata = obs, interval = prediction, level = 0.95)) # 3 subsequent observations obs - log10(data.frame(cbind(c(40, 200, 450), c(0.3, 0.06, .034 names(obs) - c(X, Y) ## Calculating t-dist, 2-tailed, 95% prediction intervals for new observations mse - anova(mod)[2,3] new.x - obs$X - mean(dat$X) sum.x2 - sum((dat$X - mean(dat$X))^2) y.hat - pred$fit var.y.hat - mse*(1+new.x^2/sum.x2) upr - y.hat + qt(c(0.975), df = 8) * sqrt(var.y.hat) lwr - y.hat + qt(c(0.025), df = 8) * sqrt(var.y.hat) hand - data.frame(cbind(y.hat, lwr, upr)) #The limits are not the same pred hand -- Thank you, Kurt [[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] starting values in glm(..., family = binomial(link =log))
I did not get any warnings when I ran your data/model example. From: Fischer, Felix [mailto:felix.fisc...@charite.de] Sent: Wednesday, January 30, 2013 11:19 AM To: Ravi Varadhan Cc: r-help@r-project.org Subject: AW: [R] starting values in glm(..., family = binomial(link =log)) Thanks for your replies! It seems, that I can fit my model now, when I can provide the right starting values; however there remain warnings, such as: 1: In log(ifelse(y == 1, 1, (1 - y)/(1 - mu))) : NaNs wurden erzeugt 2: step size truncated due to divergence 3: step size truncated: out of bounds ... That makes me feel uncomfortable and I wonder whether I can trust the fitted model. Why is this kind of regression so picky about starting values compared to logistic regression? And is there a way to explain the difference between binomial - quasibinomial to a simple mind like mine? Best, Felix Von: Ravi Varadhan [mailto:ravi.varad...@jhu.edu] Gesendet: Mittwoch, 30. Januar 2013 17:02 An: Fischer, Felix Cc: r-help@r-project.orgmailto:r-help@r-project.org Betreff: [R] starting values in glm(..., family = binomial(link =log)) Try this: Age_log_model = glm(Arthrose ~ Alter, data=x, start=c(-1, 0), family=quasibinomial(link = log)) Ravi Ravi Varadhan, Ph.D. Assistant Professor The Center on Aging and Health Division of Geriatric Medicine Gerontology Johns Hopkins University rvarad...@jhmi.edumailto:rvarad...@jhmi.edu 410-502-2619 [[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] Add axes to a 3D scene (afer makeTriangles)
On Jan 30, 2013, at 3:09 AM, cgenolin wrote: Hi all, I am drawing some 3D surfaces using the Triangle tools (package misc3) and drawScene.rgl. Do you know if it is possible to add axes and graduation on the scene? You offer no code or data, so a specific answer is not called for. Certainly the specific answer to the possibility of labels in rgl graph is is yes. You might learn something by looking at the work that Ben Bolker and I put into a question requesting labels on the scatter3d function in package car; http://stackoverflow.com/questions/8204972/carscatter3d-in-r-labeling-axis-better/8206320#8206320 -- David Winsemius, MD Alameda, CA, USA __ 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] Fastest way to compare a single value with all values in one column of a data frame
In realy, values in a will be not integers, but numeric. They will never be identical, but it could be that they are pretty close - I don't know after how many points after the comma matter. Dimitri On Wed, Jan 30, 2013 at 2:06 PM, arun smartpink...@yahoo.com wrote: Hi, Any chance x$a to have the same number repeated? If `Item` and `a` are unique, I guess both the solutions should work. set.seed(1851) x- data.frame(item=sample(letters[1:20],20,replace=F),a=sample(1:45,20,replace=F),b=sample(20:50,20,replace=F),stringsAsFactors=F) y- data.frame(item=z,a=3,b=10,stringsAsFactors=F) x[intersect(which(x$a y$a),which.min(x$a)),] # item a b #17c 1 48 x[x$a==which.min(x$a[x$ay$a]),] # item a b #17c 1 48 #or x[x$a%in%which.min(x$a[x$ay$a]),] # item a b #17c 1 48 x[x$a%in%which.min(x$a[x$ay$a]),]-y tail(x) # item a b #15q 45 30 #16g 10 23 #17z 3 10 #18r 15 39 #19l 18 45 #20t 35 33 #However, if `item` column is unique, but `a` is not, then the one I mentioned previously arise. set.seed(1851) x1- data.frame(item=sample(letters[1:20],20,replace=F),a=sample(1:10,20,replace=T),b=sample(20:50,20,replace=F),stringsAsFactors=F) y1- data.frame(item=z,a=3,b=10,stringsAsFactors=F) x1[intersect(which(x1$a y1$a),which.min(x1$a)),] # item a b #3s 1 41 x1[x1$a==which.min(x1$a[x1$ay1$a]),] # item a b #3 s 1 41 #11h 1 46 #17c 1 48 x1[x1$a==which.min(x1$a[x1$ay1$a]),]- y1 A.K. From: Dimitri Liakhovitski dimitri.liakhovit...@gmail.com To: arun smartpink...@yahoo.com Cc: R help r-help@r-project.org; Jessica Streicher j.streic...@micromata.de Sent: Wednesday, January 30, 2013 1:49 PM Subject: Re: [R] Fastest way to compare a single value with all values in one column of a data frame Sorry - I should have clarified: My identifiers (in column item) will always be unique. In other words, one entry in column item will never be repeated - neither in x nor in y. Dimitri On Wed, Jan 30, 2013 at 1:27 PM, Dimitri Liakhovitski dimitri.liakhovit...@gmail.com wrote: Thank you, everyone! I'll try to test those different approaches. Really appreciate your help! Dimitri On Wed, Jan 30, 2013 at 11:03 AM, arun smartpink...@yahoo.com wrote: HI, Sorry, my previous solution doesn't work. This should work for your dataset: set.seed(1851) x- data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F) y- data.frame(item=f,a=3,b=10,stringsAsFactors=F) x[x$a%in%which.min(x[x$ay$a,]$a),]- y #if there are multiple minimum values set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e4,replace=TRUE),a=sample(1:30,1e4,replace=TRUE),b=sample(1:100,1e4,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 330 system.time({x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1}) # user system elapsed # 0.000 0.000 0.001 length(x1$a[x1$a==1]) #[1] 0 #For some reason, it is not working when the multiple number of minimum values some value set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e5,replace=TRUE),a=sample(1:30,1e5,replace=TRUE),b=sample(1:100,1e5,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 3404 x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1 length(x1$a[x1$a==1]) #[1] 3404 #not getting replaced #However, if I try: set.seed(1241) x1- data.frame(item=sample(letters[1:10],1e6,replace=TRUE),a=sample(1:5000,1e6,replace=TRUE),b=sample(1:100,1e6,replace=TRUE),stringsAsFactors=F) y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F) length(x1$a[x1$a==1]) #[1] 208 system.time(x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1) #user system elapsed # 0.124 0.016 0.138 length(x1$a[x1$a==1]) #[1] 0 #Tried Jessica's solution: set.seed(1851) x- data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F) y- data.frame(item=f,a=3,b=10,stringsAsFactors=F) x[intersect(which(x$a y$a),which.min(x$a)),] - y x # item a b #1 a 8 25 #2 a 10 26 #3 f 3 10 #replaced #4 e 15 26 #5 b 13 20 #6 a 5 23 #7 d 4 29 #8 e 2 24 #9 c 7 30 #10e 14 24 #11d 2 20 #12e 10 21 #13c 13 27 #14d 12 23 #15b 11 26 #16e 5 22 #17c 1 26 #it is not replaced #18a 8 21 #19e 10 26 #20c 2 22 A.K. - Original Message - From: Dimitri Liakhovitski dimitri.liakhovit...@gmail.com To: r-help r-help@r-project.org Cc: Sent: Tuesday, January 29, 2013 4:11 PM Subject: [R] Fastest way to compare a single value with all values in one column of a data frame Hello! I have a large data frame x:
[R] arithmetic and logical operators
Why, in R, does (0.1 + 0.05) 0.15 evaluate to True? What am I missing here? How can I ensure this (ostensibly incorrect) behavior doesn't introduce bugs into my code? Thanks for your time. Dave Mitchell [[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] arithmetic and logical operators
R FAQ 7.31 (Note, this isn't R specific, rather it's a problem with the finitude of computers) MW On Wed, Jan 30, 2013 at 8:32 PM, Dave Mitchell dmmtc...@gmail.com wrote: Why, in R, does (0.1 + 0.05) 0.15 evaluate to True? What am I missing here? How can I ensure this (ostensibly incorrect) behavior doesn't introduce bugs into my code? Thanks for your time. Dave Mitchell [[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] Relative Risk in logistic regression
On Jan 30, 2013, at 5:49 AM, aminreza Aamini wrote: Hi all, I am very grateful to all those who write to me 1) how i can obtain relative risk (risk ratio) in logistic regression in R. 2) how to obtain the predicted risk for a certain individual using fitted regression model in R. You obtain the predicted probabilities with something like: predict(model, data.frame(x1=a, x2=30), type = response) See ?predict.glm This would give the odds ratios (similar but larger than the risk ratios): exp(coef(model)) -- David Winsemius, MD Alameda, CA, USA __ 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] arithmetic and logical operators
On 30-01-2013, at 21:32, Dave Mitchell dmmtc...@gmail.com wrote: Why, in R, does (0.1 + 0.05) 0.15 evaluate to True? What am I missing here? How can I ensure this (ostensibly incorrect) behavior doesn't introduce bugs into my code? Thanks for your time. R-FAQ 7.31: http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f Berend __ 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] arithmetic and logical operators
Why, in R, does (0.1 + 0.05) 0.15 evaluate to True? Because floating point arithmetic is done with a fixed number of digits. If you are working in base 10 and have 2 digits to work with you would have 1/3 - .33 2/3 - .67 so that 1/3 + 1/3 2/3 How can I ensure this (ostensibly incorrect) behavior doesn't introduce bugs into my code? You can index things with integers. E.g., instead of storing 0.1, 0.05, and 0.15, store 2, 1, and 3. Do your comparisons on the integers and multiply by 0.05 when you need to use them in that format. You can work with numbers that are exactly representable with 52 binary digits, such as 1/1024 or 10^14+1/8. You can work with smooth functions so the it doesn't matter much what the sign of 0.15 - (0.10 + 0.05) is. 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 Dave Mitchell Sent: Wednesday, January 30, 2013 12:32 PM To: r-help@r-project.org Subject: [R] arithmetic and logical operators Why, in R, does (0.1 + 0.05) 0.15 evaluate to True? What am I missing here? How can I ensure this (ostensibly incorrect) behavior doesn't introduce bugs into my code? Thanks for your time. Dave Mitchell [[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] How does predict() calculate prediction intervals?
Just look at the code of predict.lm(). It is reasonably perspicuous. In particular look at res.var. cheers, Rolf Turner On 01/31/2013 05:50 AM, Kurt Rinehart wrote: For a given linear regression, I wish to find the 2-tailed t-dist probability that Y-hat = newly observed values. I generate prediction intervals in predict() for plotting, but when I calculate my t-dist probabilities, they don't agree. I have researched the issues with variance of individual predictions and been advised to use the variance formula below (in the code). I presume my variance function differs from that used in predict(). Can someone advise me as to why I cannot reproduce the results of predict()? As a test, I calculated the prediction intervals about Y-hat by hand and compared them to predict(): X - log10(c(780,54,520,703,1356,1900,2500,741,1500,600)) Y - log10(c(0.037,0.036,0.026,0.0315,0.028,0.0099,0.00955,0.0405,0.019,0.0245)) dat - data.frame(cbind(X, Y)) mod - lm(Y ~ X, data = dat) rm(X,Y) ## model predictions, y.hat, at the new X values pred - data.frame(predict.lm(mod,newdata = obs, interval = prediction, level = 0.95)) # 3 subsequent observations obs - log10(data.frame(cbind(c(40, 200, 450), c(0.3, 0.06, .034 names(obs) - c(X, Y) ## Calculating t-dist, 2-tailed, 95% prediction intervals for new observations mse - anova(mod)[2,3] new.x - obs$X - mean(dat$X) sum.x2 - sum((dat$X - mean(dat$X))^2) y.hat - pred$fit var.y.hat - mse*(1+new.x^2/sum.x2) upr - y.hat + qt(c(0.975), df = 8) * sqrt(var.y.hat) lwr - y.hat + qt(c(0.025), df = 8) * sqrt(var.y.hat) hand - data.frame(cbind(y.hat, lwr, upr)) #The limits are not the same pred hand -- Thank you, Kurt [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] testing the multiple regression model
Hi I have 25 samples in my dataset. I have written a multiple regression model and I would like to test it. I would like to train my model on 20 samples and then test it on 5 remaining. However I would like to test the model several times, each time using different 5 samples out of 25 and check how well the model performs each time. As I am new to R, I am not sure how the script should look like. Could you please help me with this ? Thank you -- View this message in context: http://r.789695.n4.nabble.com/testing-the-multiple-regression-model-tp4657119.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] fSeries not found in R
Hello all, When I tried to install fSeries in R, I got the following error messages: install.packages(fSeries,dependencies=T) Warning message: package 'fSeries' is not available (for R version 2.15.2) Is this package changing/merging to another package? Thanks, Rebecca -- This message, and any attachments, is for the intended r...{{dropped:5}} __ 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] export figure by pdf command
Works for me, so you will have to provide more information. x11() hist(rnorm(100)) dev.copy2pdf(file='mytest.pdf') X11 2 list.files(patt='mytest') [1] mytest.pdf -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 1/29/13 1:22 PM, hp wan huaping@gmail.com wrote: Can you implement it using my provided example? I read the user guide about dev.copy2pdf but I still failed. Thanks 2013/1/30 ilai ke...@math.montana.edu ?dev.copy2pdf On Tue, Jan 29, 2013 at 1:48 PM, hp wan huaping@gmail.com wrote: Dear R mailing listers, After plotting, I wanna save it as file in pdf format using pdf(name.pdf) command. It failed, but I can do it by GUI operation (file-save as-pdf). e.g. x11() hist(x, breaks = 50, probability = FALSE) pdf(hist.pdf) [[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. __ 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] Programmatically give file name to a matrix
Thank you Pascal and Greg for the suggestion. That is exactly what I needed! - Kumar On Wed, Jan 30, 2013 at 2:14 AM, Pascal Oettli kri...@ymail.com wrote: Hello, ?assign assign(paste(plotroc_GBM_Trn_**, i, sep=), matrix(rnorm(100),10,10)) HTH, Pascal Le 30/01/2013 17:04, Kumar Mainali a écrit : I have a situation when I need to save matrix with file names that are programmatically created. for (i in levels(mergeTrn$Continent)) { matrix here # I want to save this matrix with a file name that carries i from for loop. The following does not work. paste(plotroc_GBM_Trn_, i, sep=) - matrix } Thanks, Kumar -- Section of Integrative Biology University of Texas at Austin Austin, Texas 78712, USA [[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] arithmetic and logical operators
On 30-Jan-2013 20:39:34 Berend Hasselman wrote: On 30-01-2013, at 21:32, Dave Mitchell dmmtc...@gmail.com wrote: Why, in R, does (0.1 + 0.05) 0.15 evaluate to True? What am I missing here? How can I ensure this (ostensibly incorrect) behavior doesn't introduce bugs into my code? Thanks for your time. R-FAQ 7.31: http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-num bers-are-equal_003f Berend And, to put Dave's specific query into the context of that FAQ: (0.1 + 0.05) 0.15 # [1] TRUE (0.1 + 0.05) - 0.15 # [1] 2.775558e-17 so that tiny 2.775558e-17 is the inexactitude (due to finite binary representation). As an interesting variant: (1.0 + 0.5) 1.5 # [1] FALSE (1.0 + 0.5) - 1.5 # [1] 0 and that is because 1.0 and 0.5, and also 1.5, have exact finite binary representations, e.g.: 1.0 == 1. 0.5 == 0.1000 1.5 == 1.1000 whereas 0.1, 0.5 and 0.15 are these numbers divided by 10 = 2*5; and while you can exactly do the /2 part (just shift right by one binary place), you can't exactly divide by 5 in finite binary arithmetic (any more than you can exactly divide by 3 in decimal), because 5 is not a factor of the base (2) of the binary representation (whereas, in decimal, both 2 and 5 are factors of 10; but 3 isn't). Whereas R has the function all.equal() to give the right answer for most things like (0.1 + 0.05) == 0.15 # [1] FALSE all.equal((0.1 + 0.05), 0.15) # [1] TRUE (because it tests for equality to within a very small tolerance, by default the square root of the binary precision of a double precision binary real number), R does not have a straightforward method for testing the truth of (0.1 + 0.05) 0.15 (and that is because the question is not clearly discernible from the expression, when imprecision underlies it). You could emulate all.equal() on the lines of (0.1 + 0.05) (0.15 + .Machine$double.eps^0.5) # [1] FALSE (0.1 + 0.05) (0.15 - .Machine$double.eps^0.5) # [1] FALSE (or similar). Observe that .Machine$double.eps^0.5 # [1] 1.490116e-08 .Machine$double.eps # [1] 2.220446e-16 (0.1 + 0.05) - 0.15 # [1] 2.775558e-17 Hoping this helps, Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 30-Jan-2013 Time: 23:22:53 This message was sent by XFMail __ 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] Packages with functionality related to Oil/Gas exploration
Folks, As the subject describes: I would like to know if there are packages that have functionality tailored for standard Oil/Gas exploration and monetization. Thanks, KW -- [[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] XBRL data into R?
Has anyone explored pulling XBRL formatted data into R? [[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] Integration of mixed normal distribution
On Jan 30, 2013, at 4:19 AM, Johannes Radinger wrote: Hi, I already found a conversation on the integration of a normal distribution and two suggested solutions (https://stat.ethz.ch/pipermail/r-help/2007-January/124008.html): 1) integrate(dnorm, 0,1, mean = 0, sd = 1.2) and 2) pnorm(1, mean = 0, sd = 1.2) - pnorm(0, mean = 0, sd = 1.2) where the pnorm-approach is supposed to be faster and with higher precision. I want to integrate a mixed normal distribution like: normaldistr_1 * p + normaldistr_2 * (1-p) I think if you check any calculus text you will find a theorem stating that integral( a*f(x) + b*g(x) ) = a*integral(f(x)) + b*integral(g(x)) where p is between 0 and 1 and the means for both distributions are 0 but the standard deviations differ. In addition, I want to get the integrals from x to infinity or from - infinity to x for the mixed distribution. Can that be done with high precision in R and if yes how? The application to this problem seems straightforward. The fact that you are using the range of -Inf to x should make the calculations easier. -- David Winsemius, MD Alameda, CA, USA __ 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] Packages with functionality related to Oil/Gas exploration
Keith Weintraub kw1958 at gmail.com writes: Folks, As the subject describes: I would like to know if there are packages that have functionality tailored for standard Oil/Gas exploration and monetization. Check out the sos package: it will help you answer the question yourself, probably more effectively than the people here ... __ 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 does predict() calculate prediction intervals?
Kurt: You missed including the term 1/n in your var.y.hat calculation. Do that and pred and hand are the same. Brian Brian S. Cade, PhD U. S. Geological Survey Fort Collins Science Center 2150 Centre Ave., Bldg. C Fort Collins, CO 80526-8818 email: brian_c...@usgs.gov tel: 970 226-9326 On Wed, Jan 30, 2013 at 9:50 AM, Kurt Rinehart kurt.rineh...@uvm.eduwrote: For a given linear regression, I wish to find the 2-tailed t-dist probability that Y-hat = newly observed values. I generate prediction intervals in predict() for plotting, but when I calculate my t-dist probabilities, they don't agree. I have researched the issues with variance of individual predictions and been advised to use the variance formula below (in the code). I presume my variance function differs from that used in predict(). Can someone advise me as to why I cannot reproduce the results of predict()? As a test, I calculated the prediction intervals about Y-hat by hand and compared them to predict(): X - log10(c(780,54,520,703,1356,1900,2500,741,1500,600)) Y - log10(c(0.037,0.036,0.026,0.0315,0.028,0.0099,0.00955,0.0405,0.019,0.0245)) dat - data.frame(cbind(X, Y)) mod - lm(Y ~ X, data = dat) rm(X,Y) ## model predictions, y.hat, at the new X values pred - data.frame(predict.lm(mod,newdata = obs, interval = prediction, level = 0.95)) # 3 subsequent observations obs - log10(data.frame(cbind(c(40, 200, 450), c(0.3, 0.06, .034 names(obs) - c(X, Y) ## Calculating t-dist, 2-tailed, 95% prediction intervals for new observations mse - anova(mod)[2,3] new.x - obs$X - mean(dat$X) sum.x2 - sum((dat$X - mean(dat$X))^2) y.hat - pred$fit var.y.hat - mse*(1+new.x^2/sum.x2) upr - y.hat + qt(c(0.975), df = 8) * sqrt(var.y.hat) lwr - y.hat + qt(c(0.025), df = 8) * sqrt(var.y.hat) hand - data.frame(cbind(y.hat, lwr, upr)) #The limits are not the same pred hand -- Thank you, Kurt [[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.
[R] betadisper plot
Hello, I tried to make a betadisper plot; however, it is quite messy at the moment with lines and symbols. I made two plots, one focusing on sites and the other on treatments. This is the code that I used: plot(betadisper(vegdist(y.nth,method=euclidean),site)) plot(betadisper(vegdist(y.nth,method=euclidean),treatment)) I have a few questions pertaining to how I could clean up the figures for interpretation purposes: How would I be able to place labels on the centroids, which would enable me to identify which centroid is associated with which site or treatment? How do I remove the lines emerging from each of the centroids? Is it possible to display labels of the y-variables used in making the figure. Because the centroids indicate site or treatment which are the independent variables, but would it also be possible to display the distribution of y-variables? Thank you, Myra [[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] Writing audio files (.wav) in a different directory
Hi. I'm trying to export a .wav file using the writeWave function from tuneR package in a different folder than the default getwd(). After reading through the manuals of some audio packages I couldn't figure it out. I'm picking one 3-hour .wav file and asking the function to take a sample of 1 minute (from minute 100 to minute 101 of the 3-hour file) and saving it in an object rec. Here is what I'm doing: filename-QUm0B24x_SMS05_20100831_052100.wav min.ini-100 min.fin-101 library(tuneR) rec-readWave(filename,min.ini,min.fin,minutes) writeWave(rec,cut1.wav) It is writing the file in the same directory of the filename object. What I want to do is to write the cut1.wav file in a different folder. Any suggestions? Thanks in advance! -- *Humberto Mohr* * * Mestrando do PPG-Ecologia do Instituto Nacional de Pesquisas da Amazônia (INPA) Graduate student of the Master Program in Ecology in the National Institute of Amazon Research (INPA) [[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] Writing audio files (.wav) in a different directory
Hi. I'm trying to export a .wav file using the writeWave function from tuneR package in a different folder than the default getwd(). After reading through the manuals of some audio packages I couldn't figure it out. I'm picking one 3-hour .wav file and asking the function to take a sample of 1 minute (from minute 100 to minute 101 of the 3-hour file) and saving it in an object rec. Here is what I'm doing: filename-QUm0B24x_SMS05_20100831_052100.wav min.ini-100 min.fin-101 library(tuneR) rec-readWave(filename,min.ini,min.fin,minutes) writeWave(rec,cut1.wav) It is writing the file in the same directory of the filename object. What I want to do is to write the cut1.wav file in a different folder. Any suggestions? Thanks in advance! -- *Humberto Mohr* * * Mestrando do PPG-Ecologia do Instituto Nacional de Pesquisas da Amazônia (INPA) Graduate student of the Master Program in Ecology in the National Institute of Amazon Research (INPA) [[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] Please, problem using “bcPower”
Hello, I would like to perform a Box-Cox (âbcPowerâ) transformation on my data. For this, I am determining lambda using the âpowerTransformâ function. However, with one of my variables I get the following Warning Message: In estimateTransform(x, y, NULL, ...) : Convergence failure: return code = 52 My variable is: x [1] 0.0001031130 0.0001029480 0.0001040010 0.0001037940 0.0001046280 0.927650 0.942960 [8] 0.950250 0.949780 0.988210 0.975860 0.962850 0.0001015800 0.0001009190 [15] 0.0001035350 0.0001033210 0.0001028420 0.0001023150 0.0001033000 0.0001036910 0.0001026890 [22] 0.0001032480 0.0001012160 0.0001031270 0.0001034390 0.0001028180 0.0001034820 0.0001022700 [29] 0.0001042640 0.0001046720 0.0001045860 0.0001045160 0.0001045850 0.0001038040 0.0001037020 [36] 0.0001034050 0.0001048150 0.0001017180 0.0001017920 0.0001032730 0.0001029420 0.0001025220 [43] 0.0001031490 0.0001036995 0.0001039735 0.0001014877 0.0001027960 And my code is:dlambda - powerTransform(x) #It is here where I get the warning message x_BCx - bcPower(x, lambda= dlambda$lambda, jacobian.adjusted= FALSE) Does anyone know why I may be getting this Warning Message? If you require more details please get in touch. Many thanks in advance! Bea [[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] Package name
Dear list, Can I use a character to set the name of a R package? like this (-) for example (sdp-R) Thanks, John __ 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] Writing audio files (.wav) in a different directory
Put the directory name into the filename string. The syntax for doing that is somewhat OS dependent, but for most cases you can use / as the separator. You really should Google file path and your OS and learn how to do this, because it is broadly applicable outside of R (not an R-help question). Look for info on absolute path and relative path. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Humberto Mohr humberto.m...@gmail.com wrote: Hi. I'm trying to export a .wav file using the writeWave function from tuneR package in a different folder than the default getwd(). After reading through the manuals of some audio packages I couldn't figure it out. I'm picking one 3-hour .wav file and asking the function to take a sample of 1 minute (from minute 100 to minute 101 of the 3-hour file) and saving it in an object rec. Here is what I'm doing: filename-QUm0B24x_SMS05_20100831_052100.wav min.ini-100 min.fin-101 library(tuneR) rec-readWave(filename,min.ini,min.fin,minutes) writeWave(rec,cut1.wav) It is writing the file in the same directory of the filename object. What I want to do is to write the cut1.wav file in a different folder. Any suggestions? Thanks in advance! __ 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] Add axes to a 3D scene (afer makeTriangles)
Hi David, Thanks for your answer. Here is my (simplified) code : 8 - library(misc3d) # ### Fonction that draw a point A=(x,y,z), with radius r misc3dPoint - function(A,r,color=black,alpha=1){ t1 - c(A[1]+r,A[2],A[3],A[1],A[2]+r,A[3],A[1],A[2],A[3]+r) t2 - c(A[1]+r,A[2],A[3],A[1],A[2]+r,A[3],A[1],A[2],A[3]-r) t3 - c(A[1]+r,A[2],A[3],A[1],A[2]-r,A[3],A[1],A[2],A[3]+r) t4 - c(A[1]+r,A[2],A[3],A[1],A[2]-r,A[3],A[1],A[2],A[3]-r) t5 - c(A[1]-r,A[2],A[3],A[1],A[2]+r,A[3],A[1],A[2],A[3]+r) t6 - c(A[1]-r,A[2],A[3],A[1],A[2]+r,A[3],A[1],A[2],A[3]-r) t7 - c(A[1]-r,A[2],A[3],A[1],A[2]-r,A[3],A[1],A[2],A[3]+r) t8 - c(A[1]-r,A[2],A[3],A[1],A[2]-r,A[3],A[1],A[2],A[3]-r) return(data.frame(rbind(t1,t2,t3,t4,t5,t6,t7,t8),color=color,alpha=alpha)) } # ### Fonction that draw several points MA=(A,B,C,D,...) misc3dPoints - function(MA,r,color=black,alpha=1){ dataV - data.frame() if(length(r)==1){r - rep(r,ncol(MA))}else{} if(length(color)==1){color - rep(color,ncol(MA))}else{} if(length(alpha)==1){alpha - rep(alpha,ncol(MA))}else{} for(i in 1:ncol(MA)){ dataV - rbind(dataV,misc3dPoint(MA[,i],r[i],color=color[i],alpha=alpha[i])) } return(dataV) } ### Some points (points are in colomn) MA - cbind(c(3.5,3.5,3.5),c(3,2,1),c(1.5,1,0.5),c(0.3,0.3,0.3),c(0,0,0),c(4,4,4))[3:1,] ### The radius of the points r = 0.3 ### Building the list of triangle3d from the points listOfTriangles - misc3dPoints(MA,r,color=c(1:4,1,1),alpha=c(1,1,1,1,0,0)) ### Making the scene myScene - makeTriangles( v1=as.matrix(listOfTriangles[,1:3]), v2=as.matrix(listOfTriangles[,4:6]), v3=as.matrix(listOfTriangles[,7:9]), color=listOfTriangles$color,alpha=listOfTriangles$alpha ) ### Drawing the scene drawScene.rgl(myScene,xlim=c(0,4),ylim=c(0,4),zlim=c(0,4)) ### Then, I export the triangles into an asy file (with is the final pupose of all this) library(longitudinalData) saveTrianglessASY(myScene) 8 - So I rewrite my question: is it possible to add some axes to the myScene object? Christophe On Jan 30, 2013, at 3:09 AM, cgenolin wrote: Hi all, I am drawing some 3D surfaces using the Triangle tools (package misc3) and drawScene.rgl. Do you know if it is possible to add axes and graduation on the scene? You offer no code or data, so a specific answer is not called for. Certainly the specific answer to the possibility of labels in rgl graph is is yes. You might learn something by looking at the work that Ben Bolker and I put into a question requesting labels on the scatter3d function in package car; http://stackoverflow.com/questions/8204972/carscatter3d-in-r-labeling-axis-better/8206320#8206320 -- Christophe Genolini Maître de conférences en bio-statistique Vice président Communication interne et animation du campus Université Paris Ouest Nanterre La Défense __ 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.