similar to: Getting the Bootstrap Error Rate of QDA

Dear list, (I've not upgraded to R1.4 yet) I have the following $class component in a predict.qda object: > unique(mod23S.qda.pred$class) [1] 12 17 8 10 4 9 5 13 14 19 20 15 6 3 7 1 23 11 18 21 16 2 22 NA Levels: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 Nevertheless, when I try to identify the individual(s) with NA, I get: >

Dear R user, I'm using qda (quadratic discriminant analysis) function (package MASS) to classify 58 explanatory variables (numeric type with different ranges) using a grouping variable (factor 2 levels "0" "1"). I'm using the qda method for class 'data.frame' (in this way I don't need to specify a formula). Using the function:

z <- qda(train, cl) save(z, file = "qda.dat") load("qda.dat") predict(qda.dat, test)$class I'm trying to save z where z <-qda(train, cl) and load z for later use in predict(z, test)$class. I think I successfully saved z by save(z, file = "qda.dat") but when I tried to load by load("qda.dat") and call predict(qda.dat, test)$class, it gives me error

You need to learn how to access code for nonexported methods. See ? "::" > methods(qda) [1] qda.data.frame* qda.default* qda.formula* qda.matrix* see '?methods' for accessing help and source code Shows you that the methods are not exported from the namespace. Hence you need to use the triple colon operator to see their code: > MASS:::qda Once you have the code, I

I guess the question that is being asked here is what is the scaling matrix that is being returned in the qda object. The help file on qda() says: ... scaling: for each group ?i?, ?scaling[,,i]? is an array which transforms observations so that within-groups covariance matrix is spherical. ... This is a bit ambiguous. I tried a few cases (spectral, QR decomposition, especially given that it is an

Hi, Does anyone know what the error message mean? > Boot2.Pillai <- function(x, ind) { + x <- as.matrix(x[,2:ncol(x)]) + boot.x <- as.factor(x[ind, 1]) + boot.man <- manova(x ~ boot.x) + summary(manova(boot.man))[[4]][[3]] + } > > man.res <- manova(as.matrix(pl.nosite) ~ + as.factor(plankton.new[,1]))$residuals > boot2.plank <-

Hello, I am Souradeep Chattopadhyay and I am a graduate student at Iowa State University Department of Statistics. Can anyone please explain the mathematical formulation behind the scaling matrix returned by the qda function in MASS package. I want to understand how this scaling matrix is derived from the inputs given to the qda function. Example Code The following example is using

Hi, the following strange error appears when I use qda: > qda1 <- qda(as.data.frame(mfilters[cvtrain,]),as.factor(traingroups)) Error: function is not a closure That's also strange: > qda1 <- qda(mfilters[cvtrain,],as.factor(traingroups)) Error in qda.default(mfilters[cvtrain, ], as.factor(traingroups)) : length of dimnames must match that of dims Some backgroud: >

Dear wxruby users I''ve been trying to get wxruby-based weft-qda 0.9.6 running on a linux from scratch 6.1 system (this list had a discussion about weft-qda last year and alex fenton''s also been trying to help me out - no luck so far). I get stuck with wxruby 0.60 (I think). I compiled wxruby against wxGTK 2.4.2 (without gtk2, without unicode - gtk+ version is 1.2.10) - after

Dear R Help Members, I am aware how to plot the LD1 vs LD2 from a lda in R, using the code: plot(baseline.systat.hat$x, col=baseline.systat.hat$class,pch=as.numeric(baseline.systat.hat$class)) However, I need to use the quadratic discriminant function, qda due to data properties. Is there a way to obtain the same sort of plot for from a qda object (and the output of predict qda). I have not

Hi, i am trying to find an appopriate R function which will estimate the bias associated with the apparent error rate of my logistic discriminant model (groups = 2, covariates = 3). I have read that bootstrapping can be used for this. Does anyone have any ideas on how I can go about doing this? I have quite a small sample size for one of my groups and I realize this may cause some problems, but

Hi, i am trying to find an appopriate R function which will estimate the bias associated with the apparent error rate of my logistic discriminant model (groups = 2, covariates = 3). I have read that bootstrapping can be used for this. Does anyone have any ideas on how I can go about doing this? I have quite a small sample size for one of my groups and I realize this may cause some problems, but

Hi all, If I run LDA on the same data (2-class classification) with default(no priors specified in the lda function) vs. "prior=c(0.5, 0.5)", the results are different. The (0.5, 0.5) priors give better 1-classify-to-1 rate, and the proportional priors(default, no priors specified in the lda function) give better 0-classify-to-0 rate, for both training and testing data sets. However,

Hi, I have been using some of the functions in r for classification purposes, chiefly lda, qda, knn and nnet. My problem is that the only one I can figure out how to represenent graphically is lda (using plot.lda). I have tried 'fooling' this function into accepting qda input for plotting but to no avail. I wonder if you have any suggestions? Thanks alot, Anne Marie Power Marine lab.

Hi, I hope I haven't got this wrong, but I don't think one needs Perl installed if they install the "precompiled" packages for Windows right? I'm just wondering if the first sentence in the "R Installation and Administration" can be re-worded a bit so newbies won't get confused.... ---------- Forwarded message ---------- Date: Mon, 2 Jun 2003 08:56:37 +1200

Hi, I'm a R newbie and I have a supervised 2-class classification problem. To find out the best representation of my data (dim = 45). I want to perform LDA und QDA on the diffrent data representations to find out, which is best to discriminate the 2 sets. For LDA there exists a method plot.lda shows (in the 2 class case) a histogramm of the data, projected onto the linear discriminants (pleas

Hi, If I do something like: biplot(x, cex.axis = .7) then it will only change the font size for axis 1 and 2, but not 3 and 4. Is there a way to change the fonts on axis 3 and 4? -- Cheers, Kevin ------------------------------------------------------------------------------ /* Time is the greatest teacher, unfortunately it kills its students */ -- Ko-Kang Kevin Wang Master of Science

Hi, I'm not sure if this can be done but.. I know that with ifelse() I can do something like: ifelse(x <= 3, 1, 2) to go through each element in my vector x, and if x_i <= 3 substitute the number with 1 else with 2. Essentially I'll get a vector with 2 levels. Can I tweak it so I can get 3-levels? For example: if(x <= 3) then 1 elseif(3 < x <= 4) then 2

Hi, I've been looking through the documentation for sample(), but can only get it to work with vectors. Is it possible to sample from a dataframe? -- Cheers, Kevin ------------------------------------------------------------------------------ /* Time is the greatest teacher, unfortunately it kills its students */ -- Ko-Kang Kevin Wang Master of Science (MSc) Student SLC Tutor and Lab

Hi, Say I have a factor with 20-levels: 1, 2, 3, ..., 20, called foo. If I do plot(foo) it will draw a barplot. But the x-axis is sorted alphanumerically, i.e. 1, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 2, 3, ..., 9, which is not what I want. I'd like to x-axis to be in the order of 1 ~ 20 numerically. So, the question is, how do I change the order on the x-axis in this case? --