similar to: Wording of "R Installation and Administration"

Hi, Suppose I have: > summary(manova(plank.man)) Df Pillai approx F num Df den Df Pr(>F) plankton.new[, 1] 1 0.5267 9.8316 6 53 2.849e-07 *** Residuals 58 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 My understanding is the MANOVA summary returns a list.

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 <-

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

Hello- I am trying to download packages from CRAN to my Window-based system. I downloaded the most recent version of Perl, as instructed in the Install Manual and the "car" package, just as a trial. Where should they "reside" within my computer: the unzipped Perl I have placed at ...R/rw1070; the "car" package I unzipped and placed at ...R/rw1070/library. Is this

Hi all: Thanks in advance for your assistance. I just started learning R. I'm trying to use the Help and the downloadable manuals. I am stuck on trying to multiply matrices. Can anyone please supply a couple of lines of code that I can plug into a fresh console to see how a double precision (1x3) matrix is multiplied by a double precision (3x3) matrix? I keep getting an error

I suspect my problem is similar to having an order and wanting to have order items as part of the order, so I''ll explain the problem in those more familiar terms and then explain what if anything makes my actual situation different at the end. In an Order lets say you have: shipping_address and a shipped bool. OrderItem has quantity and name So Order has_many :OrderItems, and OrderItem

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, 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? --

Hi, my SIP provider sends me all the numbers that are dialed, i.e. when a number is appended to the phone number proper, it gets appended to the incoming number, like (my number is 0123456789, and I append a "5"): To: <sip:491234567895@sipgate.net>;tag=as1174b008 Question is, how can I use the trailing number in extensions.conf? This is ideal for a direct dial through to an

Hi, What does this mean when I have something like: > qda.boot <- boot(train, qda.bootstrap, R = 500) Error in qda.default(structure(data.matrix(x), class = "matrix"), ...) : Rank deficiency in group M with my qda.bootstrap() looks something like: > qda.bootstrap <- function(data, index) { + boot.qda <- qda(x = data[index, 2:9], group = data[index, 1]) + qda.pred

Hi, I'm doing a presentation on Neural Networks and Tree-Based Models in two weeks, at the moment I'm looking for a data set to use in the presentation. What I would like to use is a good old data, like the Iris data, that is already known by every statisticians. MASS4 uses the cpus data in Chapter 8.10 and the Cushing's syndrome in Chapter 12.4. These two data sets plus the

e.g. class Asset < ActiveRecord::Base validates_presence_of :asset_number, :make, :model, :location, :name, :serial_number validates_numericality_of :asset_number validates_uniqueness_of :asset_number belongs_to :user belongs_to :location belongs_to :asset_type, :foreign_key => ''type_id'' end class Location < ActiveRecord::Base validates_presence_of :name

I have a list of TimeSheet objects from an ealier AR query, which I can sort by the full name of the user like so: @time_sheet_entries.sort! { |a,b| a.user.full_name <=> b.user.full_name } I can sort by the started time like so: @time_sheet_entries.sort! { |a,b| a.start_time <=> b.start_time } My question is how do I do a sort so that the list is sorted primarily by user.full_name

Hi, If I have: foo <- survfit(y ~ x) where y is a survival object and x is a n-level factor. The documentation says when I plot(foo), the confidence intervals will not be plotted (which I guess is understandable as otherwise the plot will get really messy). I tried to plot with confidence intervals by using: plot(foo, conf.int = TRUE) and indeed the resulting plot is messy. However

Hi, Suppose I have: # Fit a base model d1.ph <- coxph(Surv(start, stop, event)~ ejec + diavol + score + smoking + beta + surg.done, data = data.frame(foo)) summary(update(d1.ph, . ~ . + td1)) summary(update(d1.ph, . ~ . + td2)) As I have many columns in my data frame, foo, called td's. e.g. td1, td2, td3, .... And I'd like to

Hi, This may actually be a theoretical question. When I tried to do the following: ########################################################## > colnames(rating.adclms) [1] "usage" "mileage" "sex" "excess" "ncd" [6] "primage" "minage" "drivers" "district" "cargroup" [11]

Perhaps I am making this too hard, but how does one regress y(t) on a constant, x(t-1) and y(t-1)? I've tried the manuals and until I get Dalgaard's book (just ordered on Amazon), I am stuck! Thanks to all in advance for your patience and consideration.

Hi, I have a column in a dataframe in the form of: > as.vector(SLDATX[1:20]) [1] "1/6/1986" "1/17/1986" "2/2/1986" "2/4/1986" "2/4/1986" [6] "2/21/1986" "3/6/1986" "3/25/1986" "4/6/1986" "4/10/1986" [11] "4/23/1986" "4/30/1986" "5/8/1986"

Hi, Suppose I use the following codes to read in a data set. ############################################### > rating <- scan("../Data/Rating.csv", + what = list( + usage = "", + mileage = 0, + sex = "", + excess = "", + ncd = "", +