search for: 0.852

Dear R fans, I would like to create a scatterplot showing the relationship between 4 continuous variables. I thought of using the package "scatterplot 3d" to have a 3-dimensional plot and then using the size of the symbols to represent the 4th variable. Does anybody know how to do this? I already tried to create this graph using the colour of the symbols, but I was unable to generate

Hello, my test.R file contains two huge arrays (>3000 entries), from which R needs to calculate the Pearson Correlation, if I look at the file the numbers look correct. if I run R R < test.R --no-save I see things like this: 0.723, 0.838, 1.002, 0.364, 0.357, 0.227, 0.982+ , 0.963, 0.535, 1.214, 1.270, 0.832, 1.033, 0.632, 2.482, 1.239, 0.743, 1.077, 0.962, 1.052, 1.075, 1.427, 1.395,

I have the following xts objetct "temp" > str(temp) An ?xts? object from 2010-12-26 to 2011-03-05 containing: Data: num [1:70, 1] 2.95 0.852 -0.139 1.347 2.485 ... - attr(*, "dimnames")=List of 2 ..$ : NULL ..$ : chr "t_n" Indexed by objects of class: [POSIXct,POSIXt] TZ: GMT xts Attributes: NULL > temp t_n 2010-12-26

Dear Thomas, The head of my dataset > head(wsuv) parcel sp time censo treatment species 1 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 2 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 3 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 4 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1

Hi all, I can't find the error in the binomial GLM I have done. I want to use that because there are more than one explanatory variables (all categorical) and a binary response variable. This is how my data set looks like: > str(data) 'data.frame': 1004 obs. of 5 variables: $ site : int 0 0 0 0 0 0 0 0 0 0 ... $ sex : Factor w/ 2 levels "0","1": NA NA NA

Dear useRs, I reran an analysis with bam (mgcv, version 1.7-17) originally conducted using an older version of bam (mgcv, version 1.7-11) and this resulted in the same estimates, but much lower standard errors (in some cases 20 times as low) and lower p-values. This obviously results in a larger set of significant predictors. Is this result expected given the improvements in the new version? Or

Hi everyone, I'm sorry if this turns out to be more a statistical question than one specifically about R - but would greatly appreciate your advice anyway. I've been using a logistic regression model to look at the relationship between a binary outcome (say, the odds of picking n white balls from a bag containing m balls in total) and a variety of other binary parameters:

Hi, I have recently been attempting to find the LD50 from two predicted fits (For male and females) in a Generalised linear model which models the effect of both sex + logdose (and sex*logdose interaction) on proportion survival (formula = y ~ ldose * sex, family = "binomial", data = dat (y is the survival data)). I can obtain the LD50 for females using the dose.p() command in the MASS

Hi, I am new to R. My question is: how to get the 'hazard ratio' using the 'coxph' function in 'survival' package? thanks, karena -- View this message in context: http://r.789695.n4.nabble.com/About-hazard-ratio-urgent-tp3595527p3595527.html Sent from the R help mailing list archive at Nabble.com.

R gurus: I'm trying to get another round of rconifers out and I need some advice/help crushing differences in the examples test. I'm trying to make sure the max sdi values are being respected. I've added a tests/rconifers-Ex.Rout.save (from windows i386-pc-mingw32) and when I ran R CMD check (both R-2.13.0), I got the following results: * using log directory

Hi everyone! situation: Asterisk-server A: 192.168.11.6 Asterisk-server B: 192.168.2.44 server B contains a register => username:password@192.168.11.6 But... when I boot it, I get: Registered to '192.168.11.6', who sees us as 10.138.3.2:4569 Why doesn't server A see server B as 192.168.2.44?? All other traffic going over these lines has no problems with this. The

Hi, I am generating large Kac matrices (also known as Clement matrix). This a tridiagonal matrix. I was wondering whether there is a vectorized solution that avoids the `for' loops to the following code: n <- 1000 Kacmat <- matrix(0, n+1, n+1) for (i in 1:n) Kacmat[i, i+1] <- n - i + 1 for (i in 2:(n+1)) Kacmat[i, i-1] <- i-1 The above code is fast, but I am curious about

Hello, I did a cross-validation using cvlm from DAAG package but wasn't sure how to assess the result. Does this result means my model is a good model? I understand that the overall ms is the mean of sum of squares. But is 0.0987 a good number? The response (i.e. gailRel5yr) has min,1st Quantile, median, mean and 3rd Quantile, and max as follows: (0.462, 0.628, 0.806, 0.896, 1.000, 2.400) ?

Hiya, I''m trying to track down some throughput latency that our customer seems to be attributing to our product, I can''t see what he''s talking about, but I want to try and get some deeper granularity than I might get with something like smokeping, and maybe even see if its down to something tunable on our end. I''ve been looking for some examples on how

Hi, dear all: I am a beginner. I appreciate any help or hint from you. I am trying to do calculation with matrices. I have 3 matrices. One is matrixA, 2nd is matrixB, and last is matrixC. Here is matrixA: 1.8511.40.0831.001 0.8771.30.1161.33 1.9021.21.1020.302 0.8640.1261.110.252 1.8230.2161.0020.307 Next is matrixB: 0.8761.770.1930.328 0.8911.0090.2381.004

Full_Name: Adrian Custer Version: 0.90.0 OS: Linux on Thinkpad (pentium) and desktop (K6) Submission from: (NULL) (128.32.251.234) When I create a boxplot, and then try to overlay a lowess fit or just the points, the points do not appear in the highest level and the lowess curve does not reach the highest level. However, if I add one to each of the models, the problem is solved. I tried this

Dear Researchers, Sorry for this email but I am not a statistician, and for this I have this problem to understand. Thanks in Advance for help and suggestions. Gianni I have 21 classes (00, 01, 02, 04, ....,020) with different length. I did a kruskal wall test in R with the following code kruskal.test(m.class.l, m.class.length.lf) Kruskal-Wallis rank sum test data: m.class.l and

> > If one compares the random effect estimates, in fact, one sees that > > they are in the correct proportion, with the expected signs. They are > > just approximately eight orders of magnitude too small. Is this a bug? > > BLUPs are essentially shrinkage estimates, where shrinkage is > determined with magnitude of variance. Lower variance more > shrinkage towards

When I just run a for loop it works. But if I am going to run a for loop every time for large vectors I might as well use C or any other language. The reason R is powerful is becasue it can handle large vectors without each element being manipulated? Please let me know where I am wrong. for(i in 1:length(news1o)){ + if(news1o[i]>s2o[i]) + s[i]<-1 + else + s[i]<--1 + } --

Hi all R users I did a logistic regression with my binary variable Y (0/1) and 2 explanatory variables. Now I try to draw my nomogram with predictive value. I visited the help of R but I have problem to understand well the example. When I use glm fonction, I have a problem, thus I use lrm. My code is: modele<-lrm(Y~L+P,data=donnee) fun<- function(x) plogis(x-modele$coef[1]+modele$coef[2])