search for: 0.31

I?m confused by the difference in the fit of a gam model (in package mgcv) when I specify an interaction in different ways. I would appreciate it if someone could explain the cause of these differences. For example: x <- c(105, 124, 124, 124, 144, 144, 150, 176, 178, 178, 206, 206, 212, 215, 215, 227, 229, 229, 229, 234, 234, 254, 254, 290, 290, 303, 334, 334, 334, 344,

CentOS Errata and Bugfix Advisory 2017:0070 Upstream details at : https://rhn.redhat.com/errata/RHBA-2017-0070.html The following updated files have been uploaded and are currently syncing to the mirrors: ( sha256sum Filename ) x86_64: 00f8b320bf53b12e6666d28aff3838a69a42d0605e8b9f3050faad3dfa095908 spice-glib-0.31-6.el7_3.2.i686.rpm

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,

OK, I'm tired now but this is a new release an it owns ;). It should be quite stable but I didn't test it. So what has changed: 0.31 One more Release, I'm glad about this one - Some smaller an greater Fixes - After Compiling, the Script creates a helperfile to help mode 4 - Sections in Logfile - Trap included to handle ctrl+c - Implemented a

Hi all, I have 2 questions: 1)How do I calculate the mean on an imported txt file? I've imported the file below and that's what it looks like imported. How do I then calcuate the mean, median, or mode on the column LeafArea using the desktop R package? Any help would be greatly appreciated!! Thanks, Nat LeafType Leaflets LeafArea ShapeRatio LeafWeight LeafThickness 1 1

Hi 2 all ! I have problems with concurrent filesystem actions on a ocfs2 filesystem which is mounted by 2 nodes. OS=RH5ES and OCFS2=1.2.6 F.e.: If I have a LV called testlv which is mounted on /mnt on both servers and I do a "dd if=/dev/zero of=/mnt/test.a bs=1024 count=1000000" on server 1 and do at the same time a du -hs /mnt/test.a it takes about 5 seconds for du -hs to execute: 270M

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")

This version contains some minor timeout-related fixes. Sorry, I still haven't gotten a chance to incorporate the command-line-driven tftp clients... bad me :( Changes in 0.31: Put in a check to make sure xinetd (in particular) doesn't pass us an IPv6 socket. Fix some problems related to timeout negotiation. Allow the user to set the default timeout

Dear R users, I'm trying figuring out with forest plot and rmeta. Here's my code... library(rmeta) tabletext <- structure(c("", NA, "Sex", "[F]", "[M]", NA, "Age class", "(0,60]", "(60,80]", NA, "Karnofsky class", "[70;90]",

Hi listers, I have a simple doubt... I need to manipulate the data base as: var1 var2 7 0.1 7 0.1 8 0.12 10 0.15 12 0.18 20 0.31 I would like to aggregate the equal cases as var1 making the sum of var2... I would get... var1 var2 7 0.2 8 0.12 10 0.15 12 0.18 20 0.31 Thanks in advance, Marcio -- View this message in context:

I'm having trouble reshaping a data.frame from long to wide. (I think that's the right terminology; feel free to educate me.) I've looked at the reshape function and package and plyr package, but I can't quite figure out how to do this after a dozen variations. I have a data.frame with more levels than this, but similar to: > tst K1 K2 K3 V1 V2 V3 1 10 D a 0.08 99

Dear all I was thinking about efficient reading data into R and tried several ways to test if load(file.Rdata) or readRDS(file.rds) is faster. The files file.Rdata and file.rds contain the same data, the first created with save(d, ' file.Rdata', compress=F) and the second with saveRDS(d, ' file.rds', compress=F). First I used the function microbenchmark() and was a astonished

$ R --version R 1.6.1 (2002-11-01). So I would like to perform principal components analysis on a 16X16 correlation matrix, [princomp(cov.mat=x) where x is correlation matrix], the problem is princomp complains that it is not non-negative definite. I called eigen() on the correlation matrix and found that one of the eigenvectors is close to zero & negative (-0.001832311). Is there any way

#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,

Hi, I?m trying to get smooth curves connecting points in a plot using "spline" but I don?t get what I whant. Eg.: x<-1:5 y <- c(0.31, 0.45, 0.84, 0.43, 0.25) plot(x,y) lines(spline(x,y)) Creates a valley between the first and second points, then peaks at 3rd, and another valley between 4th and 5th. I?m trying to get a consistently growing curve up to the 3rth point and then a

Hi, consider this one as an FYI, or a seed for further discussion. I am aware that many traps on sample() have been reported over the years. I know that these are also documents in help("sample"). Still I got bitten by this while writing sample(units, size=length(units)); where 'units' is an index (positive integer) vector. It works in all cases as expected (=I expect)

Hi I have two columns in data frame. First column is based on "ID" assigned to each group of my data (similar ID depicts one group). From second column, I want to identify highest value among each group and want to assign the same ID to that highest value. Right now the data looks like: ID Value 1 0.69 1 0.31 2 0.01 2 0.99 3 1.00 4 NA 4

Hello all, I've become confused by the output produced by a call to anova(model1,model2). First a brief background. My model used to predict final tree height is summarised here: Df Sum Sq Mean Sq F value Pr(>F) Treatment 2 748.35 374.17 21.3096 7.123e-06 *** HeightInitial 1 0.31 0.31 0.0178 0.89519

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)

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