search for: dfr

...ntly 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... FR<p>DFR<p>DFR<p>DFR<p>DFR<p>DFR<p><DFR<p>DFR<p>DFR<p>DFR<p>DFR<p>DFR<p>DFR<p>DFR <p>DFR<p>DFR<p>DFR<p>DFR<p>DFR<p>DFR<p>DFR<p>DFR<p>DFR<p>DFR<p>DFR<p>DFGDFIDFD...

...ly only to factors In addition: Warning message: variable ihalf is not a factor in: model.frame.default(object, data, xlev = xlev) The problem in 1.5.1 seems to be model.matrix return inconsistently ordered interaction terms. There are abviously several work arounds. ### Start of example dfr <- expand.grid(i=0:154,j=0:154) dfr <- dfr[with(dfr,i+j<=154),] dfr$logj <- with(dfr,log(j+0.5)) dfr$jmin6 <- with(dfr, pmin(j,6)) dfr$ihalf <- with(dfr, factor(floor(i/6))) dfr$i100 <- with(dfr, ifelse(i<100,1,0)) dfr$j0 <- with(dfr, ifelse(j==0,1,0)) Ai <- runif(le...

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 "...

...ving 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 dfr.sel<-select.frame(dfr,age,sex,pain1:pain5) with the : specifying a range, instead of dfr.sel<-dfr[,c(1,2,5,6,7,8,9)] or attach(dfr) dfr.sel<-data.frame(age,sex,pain1,pain2,pain3,pain4,pain5) detach(dfr) (which isn't even safe if there are variables of the same name in the glob...

...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 chosen the wrong function. thank you turly yours ...... load package 'rms' > ddist <- datadist(dfr) > options(datadist='ddist') > n<-100 > set.seed(10) > T.Grade<-factor(0:3,labels=c("G0", "G1", "G2","G3")) > Sex<-factor(sample(0:1, 100, replace=TRUE),labels=c("F","M")) > Smoking<-factor(sample(0:1...

...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 and all elements have the value of -9990 bb<-matrix(c(rep(-9999,691*886)),nrow=691 ,ncol=886,dimnames=list(seq(-10,-44.50,by=-0.05),seq(112,156.25,by=0.05))) #dfr with row names and col names and values to be replaced in the matrix dfr <- data.frame(cbind(x=seq(120,125,by=0.05), y=-25, var.1=1)) #insert the values from the dfr into the matrix bb[dfr$x,dfr$y]<-d$var.1 Thanks for your help, Daisy

...ote that indexing could be involved). So assignments to non-existing variables just create them as local variables within the function. Making a virtue out of necessity, that might actually be considered a feature... ---------------------------------------- "select.frame" <- function (dfr, ...) { subst.call <- function(e) { if (length(e) > 1) for (i in 2:length(e)) e[[i]] <- subst.expr(e[[i]]) e } subst.expr <- function(e) { if (is.call(e)) subst.cal...

Hey guys,hi.:) I've finished implementing the PL2 scheme . The bounds I have implemented for it are as good as I could, given the nature of the scheme and my mathematical skills.However,tight bounds for other named DFR schemes will be easier to implement because their forumlas are quite simpler compared to PL2 . Will send in a pull request in a couple of days once I'm done with the tests and the documentation. I'll now start working on the DPH scheme as described in Section 3 here:- http://trec.nist.go...

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 exampl...

...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 chosen the wrong function. thank you turly yours ...... load package 'rms' > ddist <- datadist(dfr) > options(datadist='ddist') > n<-100 > set.seed(10) > T.Grade<-factor(0:3,labels=c("G0", "G1", "G2","G3")) > Sex<-factor(sample(0:1, 100, replace=TRUE),labels=c("F","M")) > Smoking<-factor(sample(0:1...

...te.cases(x, y, wt) : ???????????(the length of the augment was different)) i hope you tell me where is the mistake,and maybe i have chosen the wrong function. thank you turly yours ...... load package 'rms' > ddist <- datadist(dfr) > options(datadist='ddist') > n<-100 > set.seed(10) > T.Grade<-factor(0:3,labels=c("G0", "G1", "G2","G3")) > Sex<-factor(sample(0:1, 100, replace=TRUE),labels=c("F","M")) > Smoking<-factor(sample(0:1...

...rsing 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 reading some papers and web page. I find that Terrier Project (http://terrier.org/) has implement most of DFR scheme in Java language, and briefly read related source of Terrier's package( org.terrier.matching.models), I think "weight scheme" can imitate that package...

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]]

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 1.698 0.0196 1.64 0.0098684 1.814 0.0183 2.394 0.0107 2.469 0.0221 3.611 0.019...

...sign[object$qr$pivot][p1] nmeffects <- c("(Intercept)", attr(object$terms, "term.labels")) tlabels <- nmeffects[1 + unique(asgn)] ss <- c(unlist(lapply(split(comp^2, asgn), sum)), ssr) df <- c(unlist(lapply(split(asgn, asgn), length)), dfr) } else { ss <- ssr df <- dfr tlabels <- character(0L) } ms <- ss/df f <- ms/(ssr/dfr) P <- pf(f, df, dfr, lower.tail = FALSE) I think I understand the check for 'p' being non-zero. 'p' is essentially the number of...

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 because I want the columns to be rescaled so as to fit nicely on a...

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")) >

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, ch...

...sition aligned at y = 3. It constantly gives me error: "Error using packet 1 argument "subscripts" is missing, with no default". The same code can work with xyplot from the lattice package (does not have error bar though). Can somebody kindly offer some help? Thanks, Wen dfr <- expand.grid(month=1:12, continent=c('Europe','USA'), sex=c('female','male')) set.seed(1) dfr <- upData(dfr, y=month/10 + 1*(sex=='female') + 2*(continent=='Europe') + runif(48,-.15,.15),...