similar to: lm.boot function gives error

I understand that the hat matrix is a function of the predictor variable alone. So, in the following example why do the values on the diagonal of the hat matrix change when I go from an unweighted fit to a weighted fit? Is the function hatvalues giving me something other than what I think it is? library(ISwR) data(thuesen) attach(thuesen) fit <- lm(short.velocity ~ blood.glucose)

Dear all, I have the following output generated by linear regression. Since there is only one regression intercept and one slope for one set of data, what is the meaning of std. error for intercept and that of slope? Thanks in advance. Sincerely, Minghua > data(thuesen) > attach(thuesen) > lm(short.velocity~blood.glucose) Call: lm(formula = short.velocity ~ blood.glucose)

I have a work around for this, but can someone explain why the first example does not work properly? I believed it worked in the previous version of R, by selecting just the rows=200525 and omitting the na's. I just upgraded to 2.13. I am also concern with the row numbers being different in the selections, should I be worried? FYI, I just selected the first few rows for demonstration, please

Hi, I am new to R so forgive me if the following query is somewhat simple.  I have a small tab-separated file with n records of glucose values, one record per day, seven measurements per day. It looks like this: date    sober    no    vm    nm    va    na    vs 20091229    NA    6.8    NA    2.7    11.7    2.7    6.2 I'd like to make a graph on which the glucose day curves are plotted

Greetings R users, I am trying to renumber my groups within the file shown below. The groups are currently set as 8,9,10,etc. I would like to renumber this as 1,2,3,etc. I have searched the help files and only come across using the rownames to renumber the values but I need to match values. Any assistance is always appreciated, Regards, Charles structure(list(Group = structure(c(1L, 1L, 1L,

Hi all, I would like to test the homoegeneity of variances between several linear model for some analysis of covariance. It seems that the Bartlett's test is a good test to use but I am having problem using with linear model and I cannot find any examples on the internet. There are some examples for comparisons of variances but not linear models. If I take the hellung data set, which is

Hi everybody! I want to find a closer neighbourins observation. This is my code: ########################## library(klaR) library(ipred) library(mlbench) data(PimaIndiansDiabetes2) dane=na.omit(PimaIndiansDiabetes2)[,c(2,5,9)] dane[,2]=log(dane[,2]) dane[,1:2]=scale(dane[,1:2]) zbior.uczacy=sample(1:nrow(dane),nrow(dane)/2,F)

Hello, If you want to bootstrap a statistic, I suggest you use base package boot. You would need the data in a data.frame, see how you could do it. library(boot) bootMedianSE <- function(data, indices){ d <- data[indices, ] fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d) ypred <- predict(fit) y <- d$crp median(y - ypred)^2 } dat <-

I forgot, you should also set.seed() before calling boot() to make the results reproducible. Rui Barradas On 5/22/2018 10:00 AM, Rui Barradas wrote: > Hello, > > If you want to bootstrap a statistic, I suggest you use base package boot. > You would need the data in a data.frame, see how you could do it. > > > library(boot) > > bootMedianSE <- function(data,

Dear R-experts, I am trying to bootstrap (and average) the median squared error evaluation metric for a robust regression. I can't get it. What is going wrong ? Here is the reproducible example. ############################# install.packages( "quantreg" ) library(quantreg) crp <-c(12,14,13,24,25,34,45,56,25,34,47,44,35,24,53,44,55,46,36,67) bmi

Hi! I'm a bit lost while performing multiple comparisons on a lme model of that type: lmeglu=lme(glucose~Ath*tim,random=~1|Vol,na.action=na.omit,data=data) multc = glht(lmeglu, linfct = mcp(Ath = "Tukey", tim = "Tukey")) This works fine for identifying the effect of each factor. However, when I look for their interactions, l only obtain error messages. For example this

Greetings, I am using rpart for classification with "class" method. The test data is the Indian diabetes data from package mlbench. I fitted a classification tree firstly using the original data, and then exchanged the order of Body mass and Plasma glucose which are the strongest/important variables in the growing phase. The second tree is a little different from the first one. The

Hi, I'm trying to figure out how anova works in R by translating the examples in Sokal And Rohlf's (1995 3rd edition) Biometry. I've hit a snag with planned comparisons, their box 9.4 and section 9.6. It's a basic anova design: treatment <- factor(rep(c("control", "glucose", "fructose", "gluc+fruct",

Hi I want to create a nice picture about my result of k -the nearest neighbours algorithm. Here is my easy code: ################################# library(klaR) library(ipred) library(mlbench) data(PimaIndiansDiabetes2) dane=na.omit(PimaIndiansDiabetes2)[,c(2,5,9)] dane[,2]=log(dane[,2]) dane[,1:2]=scale(dane[,1:2]) zbior.uczacy=sample(1:nrow(dane),nrow(dane)/2,F)

Hello, Right! I copied from the OP's question without thinking about it. Corrected would be bootMedianSE <- function(data, indices){ d <- data[indices, ] fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d) ypred <- predict(fit) y <- d$crp median((y - ypred)^2) } Sorry, rui Barradas On 5/22/2018 11:32 AM, Daniel Nordlund wrote: > On 5/22/2018

Hi All, I have about one hundred patients and all the patients had their glucose measured on three different days. The days are all the same for all he patients. So I have three measurement for each patient . I want to know whether the day when the glucose was measured has an effect on the measurements. I was thinking to use a single factor analysis of variance but I am not sure how to do it in

Hi there, >library(faraway) >pima pregnant glucose diastolic triceps insulin bmi diabetes age test 1 6 148 72 35 0 33.6 0.627 50 1 2 1 85 66 29 0 26.6 0.351 31 0 >data(pima) >pima pregnant glucose diastolic triceps insulin bmi diabetes age test 1 6 148 72 35 0 33.6

Hello, I'm trying to plot a graph of blood glucose versus date. I also record conditions, such as missing the previous night's medications, and missing exercise on the previous day. My data looks like: > b2[68:74,] # A tibble: 7 ? 5 Date Time bg missed_meds no_exercise <date> <time> <dbl> <lgl> <lgl> 1 2023-10-17 08:50

I do not know if I have a problem or not. The R script at the end of this email seems to run properly and a I get a boxplot that looks proper but I get the long string of messages during execution of the script looking like: ... The following object(s) are masked from dat ( position 8 ) : sugar The following object(s) are masked from dat ( position 9 ) :

Greetings, I checked the Indian diabetes data again and get one tree for the data with reordered columns and another tree for the original data. I compared these two trees, the split points for these two trees are exactly the same but the fitted classes are not the same for some cases. And the misclassification errors are different too. I know how CART deal with ties --- even we are using the