similar to: data frame component replacement: feature or bug? (PR#266)

Displaying 20 results from an estimated 3000 matches similar to: "data frame component replacement: feature or bug? (PR#266)"

1999 Sep 05
1
data frame component replacement: feature or bug?
Hi, all. The following does not behave as I think it should, and as it seems to me it has in the past (although I can't check this easily). I know it happens in both R-0.64.2 and R-0.65.0 on an old Power Computing running Linux-PPC 1999, and in R-0.64.2 on an SGI running Irix 6.5. Try the following: t1 <- data.frame(matrix(rnorm(16), nc=4)) > t1 X1 X2 X3 X4 1 -0.7206945
1999 Sep 18
1
lambda error update (PR#282)
Matthew Wiener <mcw@ln.nimh.nih.gov> writes: > > parameters (xxdefun in gram.y calls lang4). The arg count got changed > > from 0.64.2 to 0.65.0 (from 3 to 4) to accommodate the stored source > > attribute. I suppose that a function stored in a saved workspace might > > cause some kind of mess when restored. > > > > If I understand this correctly, it
2000 Dec 13
0
comparing ancova models: summary
Thanks to John Fox, Brian Ripley, and Peter Dalgaard for responding. The short answer (as in Peter Dalgaard's reply, already posted to the list) is that the models I'm concerned with can in fact be compared using ancova. The key fact is that while the parameters may not be nested, the subspaces I'm examining are. An additional note from Prof. Ripley on AIC and BIC (which I quote in
2000 Dec 13
1
comparing ancova models
Hello, all. I've got what is probably a simple question about comparison of models using anova, specifically about the situations in which it's valid. I understand, I think, what's going on when the models are strictly nested (as most are in the demo(lm) examples). My question involves what happens when the models aren't strictly nested. In my particular case, I'm doing
2009 Jul 12
1
Booting problem with memdisk + Thinkpad + USB
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...
2002 Oct 24
3
model.matrix (via predict) (PR#2206)
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:
1997 Apr 08
1
R-alpha: User friendly functions
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
1999 Dec 07
1
Bug list summary (automatic post)
================================================= This is an automated summary of the status of the R-bugs repository. Note that this may be neither complete nor perfectly correct at any given instance: Not all bugs are reported, and some reported bugs may have been fixed, but the repository not yet updated. Some bug fixes are difficult to verify because they pertain to specific hardware or
2010 Oct 04
2
i have aproblem --thank you
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
2007 May 18
4
Simple programming question
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$categ),] and I want to score values or points in variable named "var3" following this kind of logic: 1. the highest value of var3 within category (variable named
2011 Jun 29
2
Indexing to Insert values from a dataframe into a matrix
Hello, I think this is a simple problem but I am not coming up with a simple solution. I think it just an indexing problem. I can easily replace values in a matrix from a dataframe when the dataframe has row and column numbers. In the example below I use row and column names and I can not get it to work #make a matrix where rows and columns are the lat and long for a bounding box of Australia
2006 Apr 20
1
Randomly selecting one row for each factor level [Broadca st]
The following should work: > dfr.samp <- dfr[tapply(1:nrow(dfr), dfr$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 <-
2011 Mar 28
2
GSoC 2011 Weighting Schemes
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
2010 Feb 27
1
Newbie help with ANOVA and lm.
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 <-
2007 May 20
2
Number of NA's in every second column
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]]
2008 Mar 06
2
How to hold a value(Mean sq) with a string
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$Mean sq but fails..Thanks dfr <- read.table(textConnection("percentQ Efficiency 1.565 0.0125 1.94 0.0213 0.876 0.003736 1.027 0.006 1.536 0.0148 1.536 0.0162 2.607 0.02 1.456 0.0157 2.16 0.0103
2008 May 30
1
Question about adding text to xYplot(Hmisc)
Hello, I have been trying to make a graph that have error bars and text at specific position. I used the following code from the help file of xYplot(Hmisc) as an example except I add a myPanel function, which is just supposed to add letters from the alphabet at the position aligned at y = 3. It constantly gives me error: "Error using packet 1 argument "subscripts" is
2017 May 31
2
stats::line() does not produce correct Tukey line when n mod 6 is 2 or 3
OTOH, > sapply(1:9, function(i){ + sum(dfr$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
2005 Apr 06
3
looking for a plot function
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
2007 Jan 02
1
How to extract the variance componets from lme
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,