search for: vienn

...VER box and the Linux box from which I smbmount is 2.2.13-7mdk. Everything is fine : [root@lima linux]# df Filesystem 1k-blocks Used Available Use% Mounted on /dev/hda6 1981000 1373871 504718 73% / /dev/hda7 2873619 1746320 978700 64% /persist vienne:/usr3 2044000 1691220 352780 83% /usr3 vienne:/usr6 2044000 976348 1067652 48% /usr6 /dev/hda1 3068383 1883199 1185183 61% /mnt/nt //MYSERVER/Archives 6835584 5318400 1517184 78% /mnt/Archives during a couple of hours...

Dear List, I have the following data: >> test <- data.frame(date = as.Date(c('2007-01-01','2008-03-24','2003-03-02','2008-05-03','2002-05-23','2001-06-30','2005-12-04')), nr = c(2000,2000,2000,2001,2002,2003,2003)) test date nr 1 2007-01-01 2000 2 2008-03-24 2000 3 2003-03-02 2000 4 2008-05-03 2001 5 2002-05-23 2002 6

...x. Is anyone has experience with that ? Some headers seems rather clear and interesting, but binding may not be so easy.... Actually, my question is: will I have to manipulate complex R objects in C to access to raw statistic functions ? Thanks for your experience. Regards, ++ Jul. -- Julien VIENNE CLIMPACT Institut Pierre-Simon Laplace Universit? Pierre et Marie Curie 4, place Jussieu Tour 45-46, 5?me ?t. Bureau 507 75252 Paris Cedex 5 Tel : 01 44 27 34 31 Email: jv@climpact.com Web: http://www.climpact.com/

My uncle recently bought a new computer and he wants me to help him install Guitar Pro. The thing is, he doesn't have Windows, and I'm thinking of installing Ubuntu with wine on the comp. Quiz time! Can you use wine, with GP, and if you can, deos it perform poorly in any ways?

hi ,everyone, I have a script shown as below: > bj2=bjerrdata$tyerr[bjyearnum[2]+1:bjyearnum[3]] > length(bj2) [1] 448 while > b=bjyearnum[3] > a=bjyearnum[2]+1 > bj1=bjerrdata$tyerr[a:b] > length(bj1) [1] 169 it is different with bj2 and bj1 . and the array bjyearnum is [1] 0 279 448 633 1021 1365 1813 2237 2839 3314 3798 4157 12 why bj1 is 169 while bj2 is 448

this blog entry http://www.viggie.com/blog/software/opensuse-ubuntu-usage-experience , if credible , would seem to suggest that there is no good reason to choose Suse. I really don't have time for such nonsense, maybe I'll just reinstall as Ubuntu. Also, noticed that GCC was not installed when Suse installed. That's just weird, GCC is part of Linux. And if you try to install GCC,

The following code isn't working and we can't figure out why.. letters = c("A","B","C","D","E","F","G","H","I","J") numbers = 1:3 for(i in 1:6){ #6 letters for (j in 1:3) { #3 numbers for (k in -1:1) {

Hi, I recently bought a MGE protection center 500. I am running a gentoo, and I am currently trying to enable the ups handling by the os, with no success. First, I compiled and installe nut 2.0.2 (with usb support) and psp 3.0. I made ebuild for them that I will submit soon to bugs.gentoo.org. The ups is connected via usb, and lsusb confirms that the device is detected (cf end of the mail for

I am trying to reshape a data set. Could someone please help me with the reshape, cast, and melt functions? I am new to R and I have tried reading up on how to use the reshape package, but I am very confused. Here is an example of what I am trying to do: subject coder score time [1,] 1 1 20 5 [2,] 1 2 30 4 [3,] 2 3 10 10 [4,] 2 2

Hi all, I use "write.csv" and "write.table" to write a data frame in a file like following: write.csv2(allRandomTestCase_XDroped, "allRandomTestCase.csv") But in the created file "allRandomTestCase.csv" an additional column with consecutive numbers is automatically added to the column of the data frame "allRandomTestCase_XDroped". That is why my

I have a bit of a quandy. I'm working with a data set for which I have sampled sites at a variety of dates. I want to use this data, and get a running average of the sampled values for the current and previous date. I originally thought something like ddply would be ideal for this, however, I cannot break up my data by date, and then apply a function that requires information

I have a simple 1 way anova coded like summary(ANOVA1way <- aov(Value ~ WellID, data = welldata)) How can I use the BY function to do this ANOVA for each group using another variable in the dataset?? I tried coding it like this, but it doesn't seem to work. summary(ANOVA1way <- by(welldata, Analyte, function(x) aov(Value ~ WellID, data = welldata))) In SAS I would code it like this:

Dear R Users, I'm trying to use the 'cast' function in the 'reshape' package to convert column-format data to gridded-format data. A sample of my dataset is as follows: head(finalframe) Latitude Longitude Temperature OrigLat p-value Blaney 1 -90 -38.75 NA -87.75 17.10167 NA 2 -90 135.75 NA -87.75 17.10167 NA 3 -90 80.25

I have a large dataset grouped by a factor and I want to perform a regression on each data subset based on this factor. There are many ways to do this, posted here and elsewhere. I have tried several. However I found one method posted on the R wiki which works exactly as I want, and I like the elegance and simplicity of the solution, but I don't understand how it works. Its all in the formula

Dear R-helpers, I want to make a series of boxplots on several numeric univariates with two group variables (species and population, population nested in species, and with population as the X-axis). In order to get a proper order of the individual populations in X-axis, I need to assign a wanted order to the factor (population). I used the levels() function to do this assignment, but it seemed