search for: 0.44

Dear R Users, Could you please help me on the following issue? I have a real large yearly data set. For each year I have 365 flow values. Some of the flow values are not known and that’s why you will see NA written in those slots. I wanted to know, is there a way that I can estimate those values? I tried approx command but it seems least helpful for the kind of issue I am up against.

I assume that I have missed something fundamental and that it is there in front of me in "An Introduction to R", but I need someone to point me in the right direction. > x1 <- "1159 1129 1124 -5 -0.44 -1.52" > x2 <- c("1159","1129","1124","-5","-0.44","-1.52") > x3 <- unlist(strsplit(x1,"

Dear arun and all R users, I will first of all try to simply define my issue.. I have data in the following format Year Discharge dd/mm/yyyy x .. … … … There are some NA values in the discharge which I would like to predict by using “predict command”. I cant figure out the way to write the coding for that. Could you please help me on that??? I have also ,written

I want to compute B=A^{1/2} such that B*B=A. For example a=matrix(c(1,.2,.2,.2,1,.2,.2,.2,1),ncol=3) so > a [,1] [,2] [,3] [1,] 1.0 0.2 0.2 [2,] 0.2 1.0 0.2 [3,] 0.2 0.2 1.0 > a%*%a [,1] [,2] [,3] [1,] 1.08 0.44 0.44 [2,] 0.44 1.08 0.44 [3,] 0.44 0.44 1.08 > b=a%*%a i have tried to use singular value decomposion > c=svd(b) > c$u%*%diag(sqrt(c$d))

Hello everyone,           I have a basic question regarding logical operators. > x<-seq(-1,1,by=0.02) > x   [1] -1.00 -0.98 -0.96 -0.94 -0.92 -0.90 -0.88 -0.86 -0.84 -0.82 -0.80 -0.78  [13] -0.76 -0.74 -0.72 -0.70 -0.68 -0.66 -0.64 -0.62 -0.60 -0.58 -0.56 -0.54  [25] -0.52 -0.50 -0.48 -0.46 -0.44 -0.42 -0.40 -0.38 -0.36 -0.34 -0.32 -0.30  [37] -0.28 -0.26 -0.24 -0.22 -0.20 -0.18 -0.16

Hi all, I have pushed out tftp-hpa-0.44 with the following changes: Changes in 0.44: Allow the client to specify a range of local port numbers, just like the server can. Fix sending SIGHUP to update the regular expression table. -hpa

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)

Inkscape Announces 0.44 Release :: http://www.inkscape.org The Inkscape community today officially released the latest version of its vector graphic drawing software. Inkscape 0.44 adds many new features, is faster and more usable, better supports SVG, is translated into more languages (17 on last count), and adds a lot of polish and refinement. Major new features include Layers dialog, support

t<--4:4 y<-c(5,7,10,13,15,16,14,12,11) plot(t,y,type="l") how can i add a curve y=0.83*t-0.44*t^2 in the graph? [[alternative HTML version deleted]]

Hi, First, when I try the example Prostate with bound 0.44 (as in the manual), I got a different result: > l1c.P <- l1ce(lpsa ~ ., Prostate, bound=0.44) > l1c.P .... Coefficients: (Intercept) lcavol lweight age lbph svi 1.0435803 0.4740831 0.1953156 0.0000000 0.0000000 0.3758199 lcp gleason pgg45 0.0000000 0.0000000

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")

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, I'm a database admin for a database which manage chromatographic results of products during stability studies. I use R for the reporting of the results in MS Word through R2wd. But now I think I need your help: suppose we have the following data frame: ID rrt Mnd Result 1 0.45 0 0.10 1 0.48 0 0.30 1 1.24 0 0.50 2 0.45 3 0.20 2 0.48 3 0.60 2 1.22 3 0.40 3

In the package lasso2, there is a Prostate Data. To find coefficients in the prostate cancer example we could impose L1 constraint on the parameters. code is: data(Prostate) p.mean <- apply(Prostate, 5,mean) pros <- sweep(Prostate, 5, p.mean, "-") p.std <- apply(pros, 5, var) pros <- sweep(pros, 5, sqrt(p.std),"/") pros[, "lpsa"] <-

There are updates available, for the below multimedia packages (CentOS 5.5, 32 bit), but I'm getting missing dependency errors from yum update for these. ---> Package directfb.i386 0:1.2.10-1.el5.rf set to be updated ---> Package ffmpeg.i386 0:0.6.1-1.el5.rf set to be updated ---> Package ffmpeg-libpostproc.i386 0:0.6.1-1.el5.rf set to be updated ---> Package

Dear all,I am a new user of WinBUGS and need your help. After running the following code, I got parameters of beta0 through beta4 (stats, density), but I don't know how to get the prediction of the last value of h, the variable I set to NA and want to model it using the following code.Does anyone can given me a hint? Any advice would be greatly appreciated.Best

Hi All, I'm wondering whether there is a quick way to know the position of a specific charcater in a long character: for example frg="((D:41.04,I:41.04):45.05,(((E:2.32,((G:0.67,J:0.67):0.44,H:1.11):1.21)" and I would like to know that the 1st, 2nd, 26th, 27th, 28th ... character is "(", is there a quick way to do that? Thanks! Hua

$ 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

Re-submitted for 0.44. b.t.w. I notice that the new 0.44 'R' option is not documented in the tftp man page. ----------------------------------------- I have encountered a situation where I would like to use tftp-hpa to retrieve a file that resides within an absolute path containing a ':' character. Ala, "tftp foobar -c get C:2/tftpdir/myfile". Since the tftp client