similar to: Need R code

Dear r-users, Suppose I have three datasets: Dataset-1: Date x y Jan-1,2005 120 230 Jan-2,2005 123 -125 Jan-3,2005 -110 300 Jan-4,2005 114 -21 Jan-7,2005 11 299 Mar-5,2005 200 311 Dataset-2: Date x y Jan-2,2005 123 -125 Jan-3,2005 -110 300 Jan-4,2005 114 -21

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:

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...

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

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

A loose idea for *post*-0.50 development I've been giving a some (but not all that many) thoughts to whether some of the conceptual difficulties facing newcomers could be avoided by having simplified functions for common operations. We already have parts of this, e.g. in Kurts ctest routines. Specifically, I was thinking about data frames: How about

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

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]]

I would be incredibly grateful to anyone who'll help me translate some SAS code into R code. Say for example that I have a dataset named "dat1" that includes five variables: wshed, site, species, bda, and sla. I can calculate with the following SAS code the mean, CV, se, and number of observations of "bda" and "sla" for each combination of

dear professor: I have a problem about the nomogram.I have got the result through analysing the dataset "exp2.sav" through multinominal logistic regression by SPSS 17.0. and I want to deveop the nomogram through R-Projject,just like this : > n<-100 > set.seed(10) > T.Grade<-factor(0:3,labels=c("G0", "G1", "G2","G3")) >

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

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 <-

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

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

Dear all, Sorry for the simplicity of the question, but how does one go about removing duplicated rows in a data.frame? I'm looking for a quick and simple solution, as my data.frames are relatively large (50000 by 50). I've racked my brain and searched the help files and found nothing useful or quick, only duplicated() and unique() which work only work on lists. Thanks Gary.

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 <-

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

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

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

I have had the same problem and I wrote this function rmulti <- function(n, size, p) { NrDim <- length(p) if(NrDim<2) stop("The simulated variabel has to be at least 2-dimensional") res <- matrix(data=NA, nrow=n, ncol=NrDim) p <- p/sum(p) TempSize <- size for(i in 1:NrDim) { TempP <- p[i]/sum(p[i:NrDim]) TempBin <- rbinom(n=n, size=TempSize,