similar to: factor documentation issue

Is there a way in R (12.x) to avoid the implicit coercion of factors to integers in the context of subscripts? If this is not possible, is there a way to get at least a warning, if any coercion of this type happens, given that the action of this coercion is almost never what is wanted? Of course, in the rare case that as.integer() is applied explicitly onto a factor, the warning is not needed,

Hi using this code example: library(nlme) fm1 <- lme(Orthodont, random = ~1) plot(augPred(fm1)) is there any way to have the plots in each cell labelled and ordered according to Orthodont$Sex? I.e. in addition to the bar with the label for Orthodont$Subject there is another bar labelling the Sex of the subject? thanks a lot christoph --

I would like to use lattice graphics to plot multiple functions (or groups or subpopulations) on the same plot region, using different line types "lty" or colors "col" to distinguish the functions (or groups). In traditional graphics, this seems straightforward: First plot all the data using 'type="n"', and subsequently execute a series of "points"

I'm having difficulties with plot.locfit.3d, at least I think that is the problem. I have a large dataframe (about 4 MM cases) and was hoping to see a non-parametric estimate of the hazard plotted against two variables: > fit <- locfit(~surv.yr+ ur_protein + ur_creatinine, data=TRdta, cens = 1-death, family = "hazard", xlim=c(0,10)) # it took somewhere between 1 and 2

I'm trying a recommendation on the help page for 'factor': > x <- c(1, 2, 1, 2) > x <- factor(x, labels = c("one", "two")) > x [1] one two one two Levels: one two > as.numeric(levels(x))[x] [1] NA NA NA NA Warning message: NAs introduced by coercion Also, > as.numeric(as.character(x)) [1] NA NA NA NA Warning message: NAs introduced by

Hola! The R FAQ says: 7.12 How do I convert factors to numeric? It may happen that when reading numeric data into R (usually, when reading in a file), they come in as factors. If f is such a factor object, you can use as.numeric(as.character(f)) to get the numbers back. More efficient, but harder to remember, is as.numeric(levels(f))[as.integer(f)] In any case, do not call as.numeric()

Hi, I have a dataframe that I imported from a .txt file by: skogTemp <- read.delim2("Skogaryd_shoot_data.txt", header=TRUE, fill=TRUE) and the data are factors, how can avoid factors from the beginning? Although the file contains both characters and numbers. I tried to convert some of the columns from factor to numeric and as I understood it you can not use only as.numeric but

Hi all: My data is in the attachment. I want to analysis the mean difference of y between 2 sex. My code: result_lm<-lm(y~factor(sex) + x1 + x2) summary(result_lm) The result of "factor(sex)m" 136.83, is the mean difference of y between 2 sex,and the corresponding p value is 0.07618. My question is: how to get the mean y of sex(m) and sex(f) respectively via lm function? Many

How do I fix this error ? I tried coercion to a vector but that didn't work. msci <-read.csv("..MSCIexUS.csv", header=TRUE) head(msci) Date index 1 Dec 31, 1969 100 2 Jan 30, 1970 97.655 3 Feb 27, 1970 96.154 4 Mar 31, 1970 95.857 5 Apr 30, 1970 85.564 6 May 29, 1970 79.005 > str(msci) 'data.frame': 510 obs. of 2 variables: $ Date : Factor w/ 510

Dear people, I am including an example of a dataframe: mydataframe<-data.frame(X=c(1:4),total_bill=c(16.99,10.34,21.01,23.68),tip=c(1.01,1.66,3.50,3.31),sex=c("Male","Male","Male","Female")) When I use the sapply function getting the information about the factors works: sapply(mydataframe,function(x)is.factor(x)) X total_bill tip

Hi! Yesterday I accidentally discovered this: > a <- LETTERS[1:5] > a [1] "A" "B" "C" "D" "E" > > a[1] <- factor(a[1]) > a [1] "1" "B" "C" "D" "E" BUT: > b <- factor(LETTERS[1:5]) > b [1] A B C D E Levels: A B C D E > b[1] <- factor(b[1]) > b [1] A B C D E

Hi, I could not use 'exlcude=' option in factor() to exclude a level from a existing factor. x is a factor: > x [1] a b c Levels: a b c > factor(x,exclude="c") [1] a b c Levels: a b c Warning message: NAs introduced by coercion However, "c" is not coded as NA. The following does not work either: >

I am basically fine with the change. How about using just the following? if(!is.character(exclude)) exclude <- as.vector(exclude, typeof(x)) # may result in NA x <- as.character(x) It looks simpler and is, more or less, equivalent. In factor.Rd, in description of argument 'exclude', "(when \code{x} is a \code{factor} already)" can be removed. A larger

Dear all, I have a data frame with one factor and four numeric variables and wish to obtain the var-cor matrix separately by factor. I tried by() and sapply() but getting nowhere. I understand this can be done by subsetting the dataframe, but there should have some sleek ways of doing it. Here is a simulated dataframe; s <- rep(c("A","B","C"), c(25,22,18)) d

Having spent the last few weeks trying to decipher R, I feel I may finally be getting somewhere, but i'M still in need of some advice and all my tutors seem to be on holiday! Basically a bit of background, I have data collected on a population of Lizards which includes age,sex, and body condition. I collected data myself this year and I have data previously collected from 1999, 2002 and

I simply can''t figure this out. I have been reading and re-reading Agile book and wiki.rubyonrails.org - all sorts of validation methods and still, it doesn''t work. Controller code def create @client = Client.new(params[:client]) if @client.save! flash[:notice] = ''Client was successfully created.'' redirect_to :action =>

Dear group, I know this issue has been already covered, and before you reply I must say I have read the R-FAQ and search the mailing list archive. I still can't manage to change my factor to numeric as I couldn't find any clear answer. Here is my df : Pose1 <- structure(list(DESCRIPTION = structure(c(1L, 2L, 3L, 4L, 5L, 8L), .Label = c(" SUGAR NO.11 May/10 ", "COTTON

Hi, I don't know how to make prediction with a factor in the linear model. Say yd=c(1,2,3,4,5) > sl=c(2,3,4,5,6) > sex=c("male","male","female","female","male") > sex=factor(sex) > m=lm(sl~yd+sex+sex:yd) How to make a prediction with new data like yd=c(4,5,6,7,8) for male and female separately ? Thank you very

I am executing a Repeated Measures Analysis of Variance with 1 DV (LOCOMOTOR RESPONSE), 2 Within-Subjects Factors (AGE, ACOUSTIC CONDITION), and 1 Between-Subjects Factor (SEX). Does anyone know whether the between-subjects factor (SEX) belongs in the Error Term of the aov or not? And if it does belong, where in the Error Term does it go? The 3 possible scenarios are listed below: e.g., 1.

Hello, I''m trying to fit a linear mixed effects model of the form: lme(y ~ x * Sex * Year, random=x|subject) where Sex and Year are factors with two and three levels respectively. I want to compare the fixed effects for each level to the overall mean, but the default in R is to compare to the first level. This can be changed by adding the term -1 to the righthand side of the model