similar to: Discriminant plot

I'm confused on how is the right way to plot a discriminant analysis made by lda function (MASS package). (I had attached my data fro reproduction). When I plot a lda object : X <- read.table("data", header=T) lda_analysis <- lda(formula(X), data=X) plot(lda_analysis) #the above plot is completely different to: plot(predict(lda_analysis)$x,

Dear R users, When I use aggregate with table as FUN, I get what I would call a strange behaviour if it involves numerical vectors and one "level" of it is not present for every "levels" of the "by" variable: --------------------------- > df <- data.frame(A=c(1,1,1,1,0,0,0,0),B=c(1,0,1,0,0,0,1,0),C=c(1,0,1,0,0,1,1,1)) >

Dear fellow users Are there any Vietnamese language resources for beginners of R? If so, I would be interested in hearing from people who have had experience with them and which are better (if there is more than one). I am involved with an aid project in Vietnam, and would like to move the scientists involved from using Excel for 'analysis' to R. Thanks .... Peter Alspach The

Hi, I created a principal component plot using the first two principal components. I used the function princomp() to calculate the scores. now, I would like to superimpose an ellipse representing the center and the 95% confidence interval of a series of points in my plot (as to illustrate the grouping of my samples). I looked at the ellipse() function in the ellipse package but can't get it

Dear R-users, I am using R(a package igraph) to calculate certain topological features of networks. When I try to draw a plot between these features I get an error. Following is the code I am using : *> plot(met_eco_deg,met_eco_bet) > lmout<-lm(met_eco_bet ~ met_eco_deg) > abline(lmout) Error in int_abline(a = a, b = b, h = h, v = v, untf = untf, ...) : plot.new

I'm trying to assign NAs to values that satisfy certain conditions (more complex than shown below) and it gives the right result, but breaks the loop having done the first one viz: new<-c(rep(5,4),6) for (i in 1:6) {new[new[i]>5.5][i]<-NA} gives the correct result, though an error message appears which causes a break if it's in a loop. If I can get rid of the error message and

Hello, I am using the recode() function in Rcmdr and the result is not what I expect so I am almost sure I did something wrong but what... > test <- data.frame(x=1:10) > library(car) > recode(test$x,'1:5=0 ; else=1', as.factor.result=TRUE) [1] 0 0 0 0 0 1 1 1 1 1 Levels: 0 1 BUT > library(Rcmdr) # recode from the car package is now masked Now I recode test$x

Dear R helpers I have one data as given below. date                     value1          value2             value3 30-Nov-2010           100                 40                 61 25-Nov-2010           108                 31                 88 14-Sep-2010            11                 180               56 I want the following output date                 name       amount 30-Nov-2010      value1

Dear Forum, I have a class variable 1 (populations A-E), and two other class variables, variable 2 and variable 3. What I want is to see if the combination of var 2 and var 3, will give me a pattern that allows to distinguish populations. I found several packages like ade4, with pcaiv function and factoMineR. but there are not working. Using the ade4 package, when I try to build the pca: pca1

Don't use aggregate's simplify=TRUE when FUN() produces return values of various dimensions. In your case, the shape of table(subset)'s return value depends on the number of levels in the factor 'subset'. If you make B a factor before splitting it by C, each split will have the same number of levels (2). If you split it and then let table convert each split to a factor, one

Dear all, i'm trying to label specific individuals (supplementary ones) after a PCA with the FactoMiner package. There is not much details (possibilities?) in the R-help of the plot.pca function. There is indeed a "label" parameter but i could only manage to label the supplementary individuals with there "row.names" (i.e. label="indiv.sup") and not with the

Dear list, I have a list of three matrices : i = list(matrix(1:4,2,2), matrix(3:6,2,2), matrix(9:12,2,2)) I would like to sum the matrices, as follows : [,1] [,2] [1,] 13 19 [2,] 16 22 I used this code : k <- i[[1]] for (j in (2:length(i))) { k <- k + i[[j]]} But, is it possible to sum without a loop ? Thanks in advance, Carlos [[alternative HTML version deleted]]

Hi everyone, I 'm building a function, in the middle it controls the sign of a variable x. If x < 0 the function write a warning (Error: negative value!). At this point I want the function stops without execute the remaining code. How can I do to terminate the function before your ending? Thanks in advance. Paolo

Dear all, let suppose I have following vector:   > dat1 <- c(rep("asd", 5), rep("xyz", 12), rep("erd", 17)) > dat1 <- dat1[sample(1:length(dat1), length(dat1), replace=F)] > dat1  [1] "erd" "xyz" "erd" "asd" "asd" "erd" "xyz" "asd" "erd" "erd"

Hi R-List, I have a question regarding R-language formats, I think. I am producing a series of graphs (using plot, barplot, barchart, and bwplot, using either text or mtext to place values on the graphs) and tables for a Francophone country. In fact, I have already done so. However, while they are pleased with the results they've requested I convert all of my decimal points into the French

Hi all, I hope it's not too trivial for the list - I'm trying to concatenate two factor arrays, and obtain the following: > f1<-factor(c("a","a","b")) > f1 [1] a a b Levels: a b > f2<-factor(c("b","b","a")) > f2 [1] b b a Levels: a b > c(f1,f2) [1] 1 1 2 2 2 1 Instead of getting: [1] a a b b b a Levels: a

Hello everyone. When I execute the following in R > (18-46)/(45-93) [1] 0.5833333 I get small precision for what I am trying to deal with . Is it possible to increase the precision for this and for other operations? For example openoffice calc for this operation returns 0.58333333333333300000 I I would like to thank you for your help [[alternative HTML version

This is my first post to the mailing list and I guess it's a pretty stupid question but I can't figure it out. I hope this is the right forum for these kind of questions. Before I started using R I was using STATA to run a Wilcoxon signed-rank test on two variables. See data below:

Dear R Users, If we issue simple plot command in R we don't get legend of the plot automatically. For example, following lines plots two curves, but to write a legend of these two curves there is no simple command. I checked with ?legend but it seems bit complicated for me. Does anyone know how to get a legend in a simple way for following R plot. Thanks, Yogesh >plot (x,y,

I know the following may sound too basic but I thought the mailing list is for the benefit of all levels of people. I ran a simple if statement on two numeric vectors (news1o and s2o) which are of equal length. I have done an str on both of them for your kind perusal below. I am trying to compare the numbers in both and initiate a new vector s as 1 or 0 depending on if the elements in the arrays