similar to: How to get the minimum ?

Dear R-ians, I am looking for the simplification of a formula to improve the calculation speed of my program. Therefore I want to simplify the following formula: H = Si (Sj ( sqrt [ (Ai - Aj)?? + (Bi - Bj)?? ] ) ) where: A, B = two vectors (with numerical data) of length n sqrt = square root Si = summation over i (= 0 to n) Sj = summation over j (= 0 to n) Ai = element of A with index i Aj

Hi I just started using RGui.exe under widnows. I have a text file containing date arranged in columns and rows, each column has the same format, each row with different formats. 3 of the columns are something like this 1/12/2006 3:59:45 PM I need to calculate the different in seconds between 2 selected periods using their row’s index My solution: Read the file in a data frame and

Hello everbody, i'm currently using the anova()-test for a small data.frame of 40 rows and 2 columns. It works well, but is there any preconditions for a valid variance analysis, that i should consider? Thank you for your answer, Daniel

hi list, i have to ask you again, having tried and searched for several days... i want to do a TukeyHSD after an Anova, and want to get the adjusted p-values after the Tukey Correction. i found the p.adjust function, but it can only correct for "holm", "hochberg", bonferroni", but not "Tukey". Is it not possbile to get adjusted p-values after

Dear R-core There is a different output for the apply function due to the change of unlist as mentioned in the R news. Newly, applying as.factor() (or factor()) in str(dat <- data.frame(x = 1:10, f1 = gl(2,5,labels = c("A", "B")))) (d1 <- apply(dat,2,as.factor)) newly returns a character matrix while in R-2.3.1 the same command resulted in an integer matrix that was

hi all, I wish to draw on the same graphic device 3 functions. But i don't want them to be on different graph, i want to compare them on the same I don't need mfrow or mfcol, I need something else... 1 graph on 1 device inside this graph 3 ploted function. I saw something unsing data.frame, but i think it's overkill, and something less complicated must exist, if not why? why not

Hi In mixed-model with lme() How can I obtain Type II SS or Type III SS for fixed effect? Thanks Julien

Hi, I'm studying math, and i have to make an analysys using Randomised Block Design. I have two factors, i know how to do this in Statistica, but how to do this in R, i read some manuals but the only thing that i have found was 2 factor ANOVA. Please could someone help me, or give some usefull links ?? Krzytsztof Suwada

Hi All. I need a bit of help extracting the residual error variance from the VarCorr structure from lmer. #Here's a 2-way random effects model lmer.1 <- lmer(rating ~ (1|person)+(1|rater), data = dat) #Get the structure vc.fit <- VarCorr(lmer.1) #results in..... $person 1 x 1 Matrix of class "dpoMatrix" (Intercept) (Intercept) 0.7755392 $rater 1 x 1 Matrix

useRs, A no doubt simple question, but I am baffled. Indeed, I think I once knew the answer, but can't recover it. The default contrasts for aov (and lm, and...) are contr.treatment and contr.poly for unordered and ordered factors, respectively. But, how does one invoke the latter? That is, in a data.frame, how does one indicate that a factor is an *ordered* factor such that

Hi all, I just want to understand how R computes p-value in a simple linear regression model? The reason is that in Matlab in the function which "evaluate standard errors for multivariate normal regression", it just provide estimates and standard errors, without giving out p-value, It computes t-statistics as follows: abs(beta_hat/std_beta_hat) how to go further to get p-value?

Hi, I use anova() to find interaction effect and then I need to find data with main effect from those data with interaction effect. How to do that? I used : anova(lm(data~factor1*factor2)), then select data with interaction effect. Then I need to select those data also with main effect of factor1 or factor2, from previous selected data. How to do that? Many thanks for your time on my

Hey, I'm generating a .jpeg file for a web application passing parameters to R via cgi. R gets the parameters by using commandArgs, which are passed to a variable named j and z j<-commandArgs()[3] z<-commandArgs()[4] Later I want to use the characters strings of the argument which the variables are holding for my filename, e.g.: jpeg(file="d:/data/images/jz.jpeg",...)

Hi, The description of all.equal states "is.all.equal should be used for programming, typically in if expressions. It is a simple wrapper using identical as shown in the documentation there.", but is.all.equal is not explicitly defined there (although there is a hint in the comments that is.all.equal <- function(x,y) isTRUE(all.equal(x,y))). Could the documentation be corrected? (or

Dear R-list I have a question concerning the argument "adjustSigma" in the function "lme" of the package "nlme". The help page says: "the residual standard error is multiplied by sqrt(nobs/(nobs - npar)), converting it to a REML-like estimate." Having a look into the code I found: stdFixed <- sqrt(diag(as.matrix(object$varFix))) if (object$method

Dear R-users, I computed a simple mixed models which was: mod<-lmer(nb ~ site + (1|patelle),tr) The output was: Linear mixed-effects model fit by REML Formula: nb ~ site + (1 | patelle) Data: tr AIC BIC logLik MLdeviance REMLdeviance 1157.437 1168.686 -574.7184 1164.523 1149.437 Random effects: Groups Name Variance Std.Dev. patelle

I have some doubts about the validity of my procedure to estimeate linear contrasts ina a factorial design. For sake of semplicity, let's imagine a one way ANOVA with three levels. I am interested to test the significance of the difference between the first and third level (called here contrast C1) and between the first and the seconda level (called here contrast C2). I used the following

All, Happy World Cup and Wimbledon. This morning finds me with the first of my many daily questions. I am running a ks.test on residuals obtained from a regression model. I use this code: > ks.test(Year5.lm$residuals,pnorm) and obtain this output One-sample Kolmogorov-Smirnov test data: Year5.lm$residuals D = 0.7196, p-value < 2.2e-16 alternative hypothesis: two.sided Warning

Good morning, I used in R contr.sum for the contrast in a lme model: > options(contrasts=c("contr.sum","contr.poly")) > Septo5.lme<-lme(Septo~Variete+DateSemi,Data4.Iso,random=~1|LieuDit) > intervals(Septo5.lme)$fixed lower est. upper (Intercept) 17.0644033 23.106110 29.147816 Variete1 9.5819873 17.335324 25.088661 Variete2 -3.3794907 6.816101 17.011692 Variete3

I would like to annotate my plot with a little box containing the slope, intercept and R^2 of a lm on the data. I would like it to look like... +----------------------------+ | Slope : 3.45 +- 0.34 | | Intercept : -10.43 +- 1.42 | | R^2 : 0.78 | +----------------------------+ However I can't make anything this neat, and I can't find out how to combine this