similar to: How does R compute sums of squares?

Hi. I know that R computes sums of squares based on the diagonal of t(Q) %*% y %*% t(y) %*% Q, where Q comes from the QR-decomposition of the model matrix. Does anyone know where I can find a proof for this result? All Best and Happy New Year, Ethan [[alternative HTML version deleted]]

Hi. I want to create a plot with Pantone654 as the background. The RGB for this color is (0,61,121), which corresponds to a hex of #003D79. How do I specify the bg parameter for this? All Best, Ethan [[alternative HTML version deleted]]

Hi. I want to use the text() command to produce graphics text in the (Windows) font Tahoma, but I don't know how to do this. I appreciate any suggestions/ideas you may have. All Best, Ethan [[alternative HTML version deleted]]

The lm() function in R seems to handle the inversion of singular X'X matrices (where there is collinearity between regression inputs) in a way where one of the inputs is dropped and this also seems to be the default behavior in SAS (please let me know if i'm wrong about this). In some other packages (i.e. octave ols() function) the pseudo inverse is computed where singular values less

Hi, a colleague ran a stepwise discriminant analysis twice in a row and got different results, suggesting some "sochasticity" in the algorithms involved. I looked at her data and found that there was a lot of collinearity, so that I reckoned that maybe "stepclass" (klaR) cannot find a clear winner when trying to include a new variable and makes a random choice. Is that true?

I would like to create a tree model with at most one split point per variable using tree, rpart or other routine. Its OK if a variable enters at more than one node but if it does then all splits for that variable should be at the same point. The idea is that I want to be able to summarize the data as binary factors with the chosen split points. I don't want to have three level or more

I am working with a RCB 2x2x3 ANCOVA, and I have noticed a difference in the calculation of sum of squares in a Type III calculation. Anova output is a follows: > Anova(aov(MSOIL~Forest+Burn*Thin*Moisture+ROCK,data=env3l),type=3) Anova Table (Type III tests) Response: MSOIL Sum Sq Df F value Pr(>F) (Intercept) 22.3682 1 53.2141 3.499e-07 *** Forest

Hi everyone. Based on a dependent variable (y), I'm trying to generate some independent variables with a specified correlation. For this there's no problems. However, I would like that have all my "regressors" to be orthogonal (i.e. no correlation among them. For example, y = x1 + x2 + x3 where the correlation between y x1 = 0.7, x2 = 0.4 and x3 = 0.8. However, x1, x2 and x3

Dear R community, as stated in Breiman et.al. (1984) and De'Ath & Fabricius (2000) using sums of absolute deviations about the median as an impurity measure gives robust trees. I would like to use this method in rpart. Has somebody already tried this method in rpart? Is there maybe already a script available somewhere? I am aware of the possibility to define usersplits myself with

Ok guys How do i check MD5 SUM in the ISO Directory Mike -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.centos.org/pipermail/centos/attachments/20090822/eac4e80f/attachment-0003.html>

Does anyone know how to get a vector of column sum from a data frame? You can use colSums(), but this gives you a object of type "numeric" with the column labels in the first row, and the sums in the second row. I just want a vector of the sums, and I can't figure out a way to index the "numeric" object. Thanks!

Hello, I just downloaded the very great CentOS LiveCD and have some issue with the sum. when I do md5sum -c md5sum.livecd they simply don't match. But when check the size file with the file size posted by johnny in the announce http://lists.centos.org/pipermail/centos-announce/2006-May/012905.html They both match (in 717729792 bytes) Probably I am doing something wrong, just to be sure

Is it possible to calculate partial sums in R? I have the equation: Px = (k / x) * SUM_{from j >= x to infinity} Pj x=1,2,3,... where k is a parameter, Pj is the Poisson probability, and x is a rank. (This is for rank-frequency data.) In other words, the frequency of an entity at rank x depends on the sum of frequencies of all entities at ranks j > x. Many thanks, Andrew

anova.lm() gives the sequential tests: > set.seed(1) > dat <- data.frame(y=rnorm(10), x1=runif(10), x2=runif(10)) > anova(lm(y ~ x1 + x2, dat)) Analysis of Variance Table Response: y Df Sum Sq Mean Sq F value Pr(>F) x1 1 1.1483 1.1483 2.0943 0.1911 x2 1 0.4972 0.4972 0.9068 0.3727 Residuals 7 3.8383 0.5483 > anova(lm(y ~ x2 + x1,

I've recently run into the problem of using aov with nested factors, and wanting to get the type II and III sums of squares. Normally Anova from the car package would do fine, but it doesn't like having an Error included, so my.aov <-aov(Response ~ Treatment + Error(Treatment:Replicate)) Anova(my.aov, type="II") yields Error in Anova(nested.anova) : no applicable method

I'm trying to make more compact the definition of a function as for example: f <- function(x) 2/x+3/x by simply defining the array of coefficients arr = c(2,3) and setting: g <- function(x) sum((arr/x)) Everything seems to work fine because the values returned by f and g result coincident for different values of their argument, but when I try to plot the function g using: x =

Hi, first I want to thank all of you for the quick aid which is provided here on the list during all times. Thanks a lot for that! Then, I have a problem using summaryBy which most probably is a problem of wrong use by me or the like: I use this command: summaryBy(total+total.inf~gr, aE, FUN=sum) where aE is a > str(aE) 'data.frame': 127880 obs. of 16 variables: $ gr

If I want to test for a salesman effect on average sale, I model it as "Sale ~ Salesman". But what if I want to test for a salesman effect on total sales? How do I model that? Given a significant salesman effect, how do I follow up with the equivalent of Tukey's HSD to determine who the (indistinguishly) top salesmen are? If I add another factor, say "region" to the

Hello, I'm trying to perform an integration on a function that contains a sum. Similarly I'm hoping whatever acts as a fix will be good enough to aid in a 2nd case as well but I've already put in a work-around for that situation. My code example is here: watermelon=c(0,1,2,3) w<-function(x){ sum(sin(x-watermelon))} integrate(function(x){w(x)},0,2)

Dear listserv, Please consider the following dataset: x <- matrix(nrow = 8, ncol = 2) colnames(matrix) <- c("classification", "soluble_fiber") x[1:4,1] <- "bagel" x[5:8,1] <- "donut" How would I simulate a dataset for a one-way fixed-effect ANOVA (where "classification" is the treatment variable and "soluble_fiber" is the