similar to: R-alpha: User friendly functions

Hi, I encountered a booting problem with memdisk 2.83, USB and IBM Thinkpad T61, apparently the same issue as described here: http://syslinux.zytor.com/archives/2008-April/009850.html The boot process always stops after "Loading boot sector... booting...". With debug tracers enabled, the last few output lines are: Loading boot sector... FR<p>Dbooting...

Full_Name: Glenn Stone Version: 1.5.1 and 1.6.0 OS: win2000 Submission from: (NULL) (168.140.227.9) The following code produces incorrect fitted values in version 1.5.1 and an error in 1.6.0 Error in "contrasts<-"(*tmp*, value = "contr.treatment") : contrasts apply only to factors In addition: Warning message: variable ihalf is not a factor in:

dear professor: thank you for your help,witn your help i develop the nomogram successfully. after that i want to do the internal validation to the model.i ues the bootpred to do it,and then i encounter problem again,just like that.(´íÎóÓÚerror to :complete.cases(x, y, wt) : ²»ÊÇËùÓеIJÎÊý¶¼Ò»Ñù³¤(the length of the augment was different)) i hope you tell me where is the mistake,and maybe i have

Hi R-users, I have a simple question for R heavy users. If I have a data frame like this dfr <- data.frame(id=1:16, categ=rep(LETTERS[1:4], 4), var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1)) dfr <- dfr[order(dfr$categ),] and I want to score values or points in variable named "var3" following this kind of logic: 1. the highest value of var3 within category (variable named

Hello, I think this is a simple problem but I am not coming up with a simple solution. I think it just an indexing problem. I can easily replace values in a matrix from a dataframe when the dataframe has row and column numbers. In the example below I use row and column names and I can not get it to work #make a matrix where rows and columns are the lat and long for a bounding box of Australia

The following should work: > dfr.samp <- dfr[tapply(1:nrow(dfr), dfr$x, sample, 1),] > dfr.samp x y z 10 a 10 J 2 b 2 B 9 c 9 I Andy From: Kelly Hildner > > I don't use R much, and I have been unable to figure out how > to get the > subset of my data frame that I would like. > > For example, if this were my data frame: > > > dfr <-

Hi, guys I am Wenjin from Graduate School of Chinese Academy of Science, pursing a master degree and my current research interests including using Data mining and Information retrieve technology to analysis software engineering (SE) data and support SE. I have great interested in "Weight Schemes" project. and in the last few days I have learnt some detail about DFR model family by

Would someone be so kind as to explain in English what the ANOVA code (anova.lm) is doing? I am having a hard time reconciling what the text books have as a brute force regression and the formula algorithm in 'R'. Specifically I see: p <- object$rank if (p > 0L) { p1 <- 1L:p comp <- object$effects[p1] asgn <-

Hi R-users, How do I calculate a number of NA's in a row of every second column in my data frame? As a starting point: dfr <- data.frame(sapply(x, function(x) sample(0:x, 6, replace = TRUE))) dfr[dfr==0] <- NA So, I would like to count the number of NA in row one, two, three etc. of columns X1, X3, X5 etc. Thanks in advance Lauri [[alternative HTML version deleted]]

Hello, I have been trying to make a graph that have error bars and text at specific position. I used the following code from the help file of xYplot(Hmisc) as an example except I add a myPanel function, which is just supposed to add letters from the alphabet at the position aligned at y = 3. It constantly gives me error: "Error using packet 1 argument "subscripts" is

Hi all: Can someone advice me on how to hold the residuals Mean sq value on a string so it can be used in other calculations. I was trying something like this: Msquare<-dfr$Mean sq but fails..Thanks dfr <- read.table(textConnection("percentQ Efficiency 1.565 0.0125 1.94 0.0213 0.876 0.003736 1.027 0.006 1.536 0.0148 1.536 0.0162 2.607 0.02 1.456 0.0157 2.16 0.0103

OTOH, > sapply(1:9, function(i){ + sum(dfr$time <= quantile(dfr$time, 1./3., type = i)) + }) [1] 8 8 6 6 6 6 8 6 6 Only the default (type = 7) and the first two types give the result lines() gives now. I think there is plenty of reasons to give why any of the other 6 types might be better suited in Tukey's method. So to my mind, chaning the definition of line() to give sensible

Dear useRs, I have a data frame and I want to plot all rows. Each row is represented as a line that links the values in each column. The plot looks like this: dfr <- data.frame(A=sample(1:50,10),B=sample(1:50,10), C=sample(1:50,10),D=sample(1:50,10)) xa <- 10*1:4 plot(c(10,40),c(0,50)) for (i in 1:nrow(dfr)) { lines(xa,dfr[i,],pch=20,type="o") } Things get more complicated

Here is a piece of code fitting a model to a (part) of a dataset, just for illustration. I can extract the random interaction and the residual variance in group meth==1 using VarCorr, but how do I get the other residual variance? Is there any way to get the other variances in numerical form directly - it seems a litte contraintuitive to use "as.numeric" when extracting estimates,

Hi list, I have a data set - something like this dfr <- data.frame(A=factor(letters[1:25]),B=runif(25), C=sample(LETTERS[1:4],25,replace=TRUE)) and I want to create a dotplot: library(lattice)dotplot(A ~ B | C, data=dfr, scales=list(y=list(relation="free"))) but this puts uneven spaces along the y-axis in each panel. drop.unused.levels=TRUE will drop conditioning

Matthew Wiener <mcw@ln.nimh.nih.gov> writes: > t1 <- data.frame(matrix(rnorm(16), nc=4)) > t1$X1 <- 1 > t1$X2 <- 2 > print(t1) > Error: dim<- length of dims do not match the length of object Well, it is prototype-compatible. Splus 5.3 does likewise. A way out is t1<-data.frame(unclass(t1)) However, we do seem to have a bug in the area: > t1 <-

Matthew Wiener <mcw@ln.nimh.nih.gov> writes: > t1 <- data.frame(matrix(rnorm(16), nc=4)) > t1$X1 <- 1 > t1$X2 <- 2 > print(t1) > Error: dim<- length of dims do not match the length of object Well, it is prototype-compatible. Splus 5.3 does likewise. A way out is t1<-data.frame(unclass(t1)) However, we do seem to have a bug in the area: > t1 <-

Can anyone explain to me why the AIC values are so different when using glm.nb and glm with a negative.binomial family, from the MASS library? I'm using R 1.8.1 with Mac 0S 10.3.4. >library(MASS) > dfr <- data.frame(c=rnbinom(100,size=2,mu=rep(c(10,20,100,1000),rep(25,4))), + f=factor(rep(seq(1,4),rep(25,4)))) > AIC(nb1 <- glm.nb(c~f, data=dfr)) [1] 1047 >

I don't use R much, and I have been unable to figure out how to get the subset of my data frame that I would like. For example, if this were my data frame: > dfr <- data.frame(x=rep(letters[1:3], 4), y=(1:12), z=(LETTERS[1:12])) > dfr x y z 1 a 1 A 2 b 2 B 3 c 3 C 4 a 4 D 5 b 5 E 6 c 6 F 7 a 7 G 8 b 8 H 9 c 9 I 10 a 10 J 11 b 11 K 12 c 12 L I would like to

Seriously, if a method gives a wrong result, it's wrong. line() does NOT implement the algorithm of Tukey, even not after the patch. We're not discussing Excel here, are we? The method of Tukey is rather clear, and it is NOT using the default quantile definition from the quantile function. Actually, it doesn't even use quantiles to define the groups. It just says that the groups