similar to: Regarding pp.plot(){CircStats}

Hi, i found a bug in circ.plot of CircStats package, namely that the plot ingore the datasets with value = 0. it's simple to fix it, but i cannot seem to figure out where to submit the fix. is there a official procedure of submitting bug report/patch to the contributed packages? -- View this message in context:

Hi, I've currently to work with some circular data. Unfortunately I'm not very familiar with circular statistics and would really appreciate if I could get some help concerning the CircStats package this way. My data lies in the range 0 to 2*pi, and is transformed to radians (as expected by the CircStats methods). Calculating the mean direction (circ.mean) results for some datasets in a

circ.plot in CircStats under R 1.8.0 yields incorrect plots (easily compare hist() and rose.diag()). However, the code in circ.plot has not changed. Has anyone tracked this down yet?

Dear R community, I tried to post a query about the use of the rose.diag function from the CircStats package to the r-help forum a few days ago. It gave all appearances of having been successful. It would have been my first post; I am correctly recieving all the R-help emails, but  but the message has not appeared anywhere on the forum. Am I held up by the moderator somewhere? any help would be

Hello, I am quite new to R and although I have read some material (apparently not enough) I'm getting very frustrated. I'm trying to do some simple circular statistics in R (CircStats, primarily circ.mean) and I'm confused about the results. What is a negative radian or degree mean in R? I thought it would be a coterminal angle, but that doesn't make sense with my data

I have a vector of dimension 67, but CircStat circ.summary and circ.disp seems to conclude that the dimension is 1. I am new to R so maybe I'm just missing something. > dim(data) [1] 67 1 > circ.mean(data) [1] 0.2200071 > circ.summary(data) n mean.dir rho 1 1 0.2200071 43.12534 > circ.disp(data) n r rbar var 1 1 43.12534 43.12534 -42.12534 >

Hi, If I have two sets of directional data (in radians) and want to compare them with a multifactorial anova. Is it even legitimate to compare circular data with an anova? The books I've picked up from the library don't really say, but it looks unlikely. If it is allowable, is my having stored the data as circular (X = as.circular(A)) something the aov() function will take into account,

help . I have a file which contains a table that has 2 columns of which the first one is a time sereis. I need to extract the first column alone and do an acf plot on it. I dont know how to extract the first column and pass it to the acf function in the ts package. Can anybody pls help me. end -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list

Dear R Users, The granger.test command in the MSBVAR package estimates all possible bivariate Granger causality tests for m variables. If one passes a data frame with 3 rows, it returns 6 granger tests in two rows, one for the F-statistic and another for the p-value. For example: > a<-rnorm(1:10) > b<-c(lag(a),rnorm(1)) > c<-c(lag(b),rnorm(1)) >

Recently, I noted a post and replies on R-Help from Professor Marc Feldesman regarding a cross tabulation function that generates row and column proportions, marginal values, expected cell values and tests for independence presumably similar in a fashion to the output of the S-Plus crosstabs() function or the SAS Proc Freq. Martin Maechler had posted some code in reply for folks to update and

Folks, I'm a new user and I'm trying to build a standalone application using libvirt-cim (rather than use a full-fledged CIMOM). I need some help. While calling the CMPIInstanceMI->ft->enumerateInstances, I need to pass a reference to CMPIObjectPath. My application is crashing because I don't know what is the correct constructor for this. If I do ref = CMPIObjectPath () then

i have a data frame(dat) which has many variables.and i use the following script to get the crosstable. >danx2<-c("x1.1","x1.2","x1.3","x1.4","x1.5","x2","x4","x5","x6","x7","x8.1","x8.2","x8.3","x8.4","x11",

Hi, I've had this problem for a while and tackled it is a quite dirty way so I'm wondering is a better solution exists: If we have two vectors: v1 = c(0,1,2,3,4) v2 = c(5,3,2,1,0) How to remove one instance of the "3,1" / "1,3" double? At the moment I'm using the following solution, which is quite horrible: v1 = c(0,1,2,3,4) v2 = c(5,3,2,1,0) ft <-

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Is it an undocumented (at least I missed it if it's documented) feature of the reshape function to do numeric variables followed by character? I ask because that seems to be the case below. > str(rcw) 'data.frame': 23 obs. of 21 variables: $ ICU : int 1 18 17 9 22 19 6 16 25 26 ... $ Q6.RC.1 : chr "SM" "JF" "IW"

Dear List, I have an ascii text file with data I'd like to extract. Example: Year Built: 1873 Gross Building Area: 578 sq ft Total Rooms: 6 Living Area: 578 sq ft There is a lot of data I'd like to ignore in each record, so I'm hoping there is a way to use strings as delimiters to get the data I want (e.g. tell R to take data between "Built:" and "Gross" -

Dear expeRts, What's the easiest way to convert an ftable object to a matrix such that the row names of the ftable object are shown in the first couple of columns of the matrix? This is (typically) required, for example, when the final goal is to print the matrix via xtable. Below is a rather complicated example of how to do it... Cheers, Marius ## Goal: convert an ftable() to a

Hola, Dell es los pocos que yo sepa que ofrecen equipos con Linux instalado como SO de base: http://www.dell.com/es/empresas/p/xps-13-9350-laptop-ubuntu/pd?oc=bnx9333 Saludos, Carlos Ortega www.qualityexcellence.es El 6 de septiembre de 2016, 21:33, Ruben Tobalina Ramirez < lagrimaescrita en gmail.com> escribió: > Buenas, > > gracias a los que me ayudaron, he conseguido hacer

Sí, la primera reunión de la temporada será el 8 de este mes. Un saludo, Carlos J. Gil Bellosta http://www.datanalytics.com El 3 de septiembre de 2016, 23:52, Ruben Tobalina Ramirez < lagrimaescrita en gmail.com> escribió: > Muchas gracias a todos! > > Pues si, Eric, R a veces me vuelve loco, tiene una lógica muy particular. > Miraré el libro, que lattice no lo he usado

Y bueno, algo más bonito sí que queda con ggplot... #--------------- datIn <- read.csv("pec.csv", header = TRUE, sep = ";", dec = ",") library(ggplot2) lab_x_idx <- c(1, round(nrow(datIn)/2, 0), nrow(datIn)) lab_x <- as.vector(datIn$hora[ lab_x_idx]) fre_gg <- ggplot( datIn, aes( x = 1:nrow(datIn), y = frec)) + geom_bar(stat =