similar to: convert list to data frame

Dear R users: > append(1:5, 0:1, after=2) [1] 1 2 0 1 3 4 5 If I want to repeat the appended value every 2 like the following: [1] 1 2 0 1 3 4 0 1 5 How should I modify? Thank you for any help.

I'm trying to write a function to find the means over factors of the responses in a mlm (something I would do easily in SAS with PROC SUMMARY). The not-working stub of a function to do what I want is below, and my problem is that I don't know how to call aggregate (or some other function) in the context of terms in a linear model extracted from a lm/mlm object. means.mlm <-

Dear R Users,i am very new to R. I want your help on an issue regarding distance matrix and cluster analysis i had discharge data of 4 rivers(a,b,c,d) in 4 vectors each having 364 values > dput(qmu)structure(list(a = c(0.26, 0.25, 0.25, 0.25, 0.24, 0.23, 0.22, 0.21, 0.21, 0.21, 0.2, 0.19, 0.19, 0.19, 0.19, 0.18, 0.18, 0.18, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17,

Hi, I have a 3d array as below, I want to make this array to a matrix of p=50(rows) and n=20(columns) with the coverage values . The code before the array is: library(binom) Loading required package: lattice pi.seq<-seq(from = 0.01, to = 0.5, by = 0.01) no.seq<-seq(from = 5, to = 100, by = 5) cp.all = binom.coverage( p = pi.seq, n = no.seq , conf.level = 0.95, method = "exact")

Folks: I consider my reply below rather clumsy: One has to keep track of index numbers other than that which is inserted and must separately change column names. Is there as "essentially better" way to do this, either via base R or via an R package. I leave it to you to define "essentially better." Thanks. Cheers, Bert On Mon, Aug 1, 2011 at 10:17 AM, Bert Gunter

I am trying to build Miredo 1.1.5 (http://www.remlab.net/miredo/devel.shtml.en) I have followed the rpmbuild instructions from: http://www.owlriver.com/tips/non-root/, and have the miredo source in ~/build/miredo-1.1.5. I run ./configure (as the INSTALL text file tells me to do) and get the error: checking for Judy.h usablity... no checking for Judy.h presence... no checking for Judy.h...

Hi, all: a<-data.frame(x=c(1:10),y=c(10:1)) edit(a) how to quit the edit in Linux, and then do other computing? As In Windows, just close the edit window, but how to do in Linux? Thanks in advanced !!

Since upgrading my server from CentOS 4.5 to 4.6 I've been getting the following error from amanda backups: mutilate /home lev 1 FAILED [compress got signal 11, /bin/tar got signal 13] I was away from the house for most of the end of December and had a couple of other issues that came up that could have been related but apparently weren't (why is it that several things all go wrong

I have a couple of hundred American Community Survey Summary Files files containing rectangular arrays of data, mainly though not exclusively numeric. Each file is referred to as a sequence (henceforth "seq"). From these files I am trying to extract particular subsets (tables) consisting of a sets of columns. These tables are defined by three numbers (now in columns in a data frame):

Hi there, I'm dealing with a pretty big dataset (~22,000 entries) with numerous entries for every day over a period of several years. I have a column "judy" (for Julian Day) with 0 beginning on Jan. 1st of every new year (I want to compare tendencies between years). However, in order to control for a leap year (2004), I simply need to subtract 1 from every judy value for the year

I want to create a user "samba" that has admin privileges over the directory where Win95 apps will be installed, but does not have root privileges. Are their any problems with doing so? Thanks, Carey =====================================================================e <> Carey F. Cox, PhD | PHONE: (409) 880-8770 <> <> Assistant

#this is a daily series of precipitation data. I would like to condense it into weekly means. How can I do this #as a side note I would like to do this same thing to two years worth of fifteen minute interval data and make it into #a series of daily averages (there are 96 readings per day) #is aggregate the right function? or... y <- c(1.23, 0, 0, 0, 0, 0, 0, 0, 0, 0.27, 0, 0.29, 0, 0, 0,

Hello. I know that it is recommended to run smbd as a standalone daemon and to avoid inetd. Can you please tell me why inetd is discouraged and what problems it imposes? Also, I have one user who is having problems accessing her personal files on a MacOSX 10.3.2 via smb. Any ideas what may be causing it? Judy Lin NACS-DCS

Dear R list, I have the following data: set.seed(1) n <- 5 # number of subjects tt <- 3 # number of repeated observation per subject numco <- 2 # number of covariates x <- matrix(round(rnorm(n*numco),2), ncol=numco) # the actual covariates x > x [,1] [,2] [1,] -0.63 -0.82 [2,] 0.18 0.49 [3,] -0.84 0.74 [4,] 1.60 0.58 [5,] 0.33 -0.31 I need to form a matrix

I have a problem with R. I try to compute the confidence interval for my df. When I want to create the plot I have this problem: Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differ. I try this code: library(dplR) df.rwi <- detrend(rwl = df, method = "Spline",nyrs=NULL) write.table(df.rwi,file="rwi.txt",quote=FALSE,row.names=TRUE)

On Mon, Dec 16, 2013 at 04:16:29PM -0800, Michael Dalton wrote: > Commit 2613af0ed18a ("virtio_net: migrate mergeable rx buffers to page frag > allocators") changed the mergeable receive buffer size from PAGE_SIZE to > MTU-size, introducing a single-stream regression for benchmarks with large > average packet size. There is no single optimal buffer size for all >

On Mon, Dec 16, 2013 at 04:16:29PM -0800, Michael Dalton wrote: > Commit 2613af0ed18a ("virtio_net: migrate mergeable rx buffers to page frag > allocators") changed the mergeable receive buffer size from PAGE_SIZE to > MTU-size, introducing a single-stream regression for benchmarks with large > average packet size. There is no single optimal buffer size for all >

Is there a mechanism to interate through a vector of strings? Say I have a data frame of 50 variables (VAR1 to VAR50), each with 100 measurements along with some coding factors (EXP and DOSE). I want to calculate the mean of a subset of each of VAR1 to VAR 50 (selecting by EXP and DOSE). Rather than just copy/pasting the same code 50 times, I thought of using a for loop to go through a

Hi All, I have a function that is used with data frames having multiple id's per row and it aggregates the data down to 1 id per row. It also randomly selects one of the within-id values of a variable (mod), which often differ within-id. Assume this data frame (below) is much larger and I want to repeat this function, say 100 times, and then derive the mean values of r over those 100

Forgive my novice question, but I am a new student of Linux working on presenting the ext3 journaling filesystem to my class. I seek any advice on how to visibly demonstrate (including a purposeful crash of a Linux box) the benefits of ext3 over ext2. I am not worthy to lick the bootstraps of this group, but I beg for any help! The problem I am having extends to even locating the .journal file