similar to: Matrix multiplication precision

Anyone knows hat might be the cause of this error? Thanks for any help! >library(MASS) > dif.mns = function(x2,tr1=.2,tr2=.3){ + #generates four different 'means' using + #difference scores from x2, an n x 2 matrix + #for use w/ bootstrap comparisons + diffs = apply(x2,1,diff) + mn1=mean(diffs) + mn2=mean(diffs,tr=.2) + mn3=mean(diffs,tr=.3) +

Hi list, In the object returned by summary.nlsList, what's the difference between "coefficients" and "parameters"? The have the same "Estimate", different se (therefore t value), but same p values. R.2.8.0 on winxp with nlme_3.1-89 Thanks, ...Tao +++++++++++++++++++++++++++++++++++++++++++++++++ > library(nlme) > fm1 <- nlsList(uptake ~

Hi, I'm very new to R and absolutely love it. Does anyone know how to use something in R that functions like a BY command in SAS? For example, let's say you have a variable x, and you want to see the mean. Easy... > mean(x) But what if you want to see the mean of x conditional on another discrete variable? My best attempts so far are something like... > mean(x, y_cat=1)

Hi, I just started a new course this semester on R, I never used it in my life and i'm stuck on these questions from 3 days, it would be really nice if someone could explain me the answers with the relative commands. thanks a lot in advance The following 7 questions are based on the CO2 dataset of R. 1) How many of the plants in CO2 are Mc2 for Plant? 2) How many are either Mc2 or Mn2?

Hi all, I am looking for a function for following calculation. start.month = "July" end.month = "January" months = f(start.month, end.month, by=1) * f is the function that I am looking for. Actually I want to get months = c("July", "August",.............."January") If start.month = 6 and end.month = 1 then I could use (not properly) seq()

Dear R-Help, My problem/bug came to light,when fitting a linear model using stepwise selection. I'd started with the straightfoward command step(lm(y~., dataset)) This worked fine, but because this starts with all the possible explanatory variables, it results in a model with too many explanatory variables. Hence I wanted to start with just a constant and do forward selection, to get a

Dear list, I have the following matrix. How can I make the following plot? 1. The x-axis has index 1:7, and the first column is plotted against index 1, second against 2, and so on. 2. I want the points from the left upper conner including the antidiagonal to be plotted with col=2, and the rest with col=3 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [1,] 0.589 0.857 0.923 0.944 0.954 0.963

We are having a problem remounting a Windows network share. Our server is running RHEL4 ES that mounts a Windows network shares as a drive. The Windows network shares are on Windows 2003 servers. The network share can be mounted but when the Windows 2003 servers are rebooted, the mounted drives goe down. I try to reissue the mount commands but it doesn't work, it sits there and eventually

Hola amigos, esta es mi primera duda, espero que no sea demasiado fácil. Tengo unos datos de dos variables y quiero mostrar recta de regresión y valor de correlación serie0 <- c(0.651, 0.712, 0.614, 0.645, 0.559, 0.647, 0.642, 0.534, 0.616, 0.621, 0.623) serie1 <- c(0.572, 0.641, 0.565, 0.596, 0.518, 0.604, 0.602, 0.501, 0.58, 0.589, 0.596) data <- cbind(serie0, serie1) colnames(data)

I am trying to do some verification across a large dataset, cuData, that has 23 columns. Column 23 (similarity) is the outcome 0 or 1 and the other columns are the features. I do this: verificationglm.model <- glm(formula = similarity ~ ., family=binomial, data=cuData[1:1000,]) and produce the model: > summary(verificationglm.model) Call: glm(formula = similarity ~ ., family =

I am trying to use t.test on the following data: date type INTERVAL nCASES MTF SDF MTO SDO nFST MF nOBS MO MB BIASCV BIASEV ME MAE RMSE CRCF 2001-06-15 avn GE1.00 4385 0.246 0.300 1.502 0.556 1367 1.373 4385 1.502 1.471 0.285 0.164 -1.256 1.266 1.399 0.056 2001-06-15 avn

Hi, When I look at the summary of an rpart object run on my data, I get 7 nodes but when I plot the rpart object, I get only 3 nodes. Should the number of nodes not match in the results of the 2 functions (summary and plot) or it is not always the same? Look forward to your reply, Carol -------------------------------------------- ?summary(rpart.res) Call: rpart(formula = mydata$class ~ ., data

I've created a short program to print a table of learning curve factors. However, I cannot figure out how to format the table to: 1) Get rid of the [1]s in the first column and replace it with the values of N. 2) Line up the first row with the factors (decimal fractions). Thanks for any help. The complete program and output is as follows: > Lc<-seq(0.70,0.95,0.05) #Specify learning

Hello, It would be nice if we could get a full list of packages which have a wrong sha1 checksum. I don't have the bandwidth for a full mirror of all of Dag's rpms and i also don't have shell access to such a mirror. So a small request for a mirror admin: The following checks the files repodata/*.xml.gz against the sha1 sums in repomd.xml and checks the RPMS/*.rpm files against

Hi List, I'm running a heavily loaded squid server that uses ntlm_auth to provide NTLM authentication. As load has increased over time, I've found the need to increase the number of ntlm_auth processes available to squid as well as the "winbind max clients" value in the smb.conf file. This has worked well up until now but seems I've hit some sort of limit. If I keep the

Hi all, I can't find the error in the binomial GLM I have done. I want to use that because there are more than one explanatory variables (all categorical) and a binary response variable. This is how my data set looks like: > str(data) 'data.frame': 1004 obs. of 5 variables: $ site : int 0 0 0 0 0 0 0 0 0 0 ... $ sex : Factor w/ 2 levels "0","1": NA NA NA

Dear R-help, I am having trouble with your scatterplot3d program. For help with this problem I was directed to your address by Martin Maechler at " r-core-bounces at r-project.org." I'm also sending a CC to " r-core-owner at r-project.org" as I'm not yet certain of the proper address to use for this. I have R version 2.8.1 and have downloaded 'scatterplot3d.'

Dear R People: Suppose I have the following data frame: x1 x2 x3 1 -0.1582116 0.06635783 1.765448 2 -1.1407422 0.47235664 0.615931 3 0.8702362 2.32301341 2.653805 > str(xx) 'data.frame': 3 obs. of 3 variables: $ x1: num -0.158 -1.141 0.87 $ x2: num 0.0664 0.4724 2.323 $ x3: num 1.765 0.616 2.654 I can exclude the second column nicely via: >

Hi there, Everytime when I paste My R output in WORD, the columns couldn't line up nicely like they appear in R console. Is there a way to fix this problem? Thanks for any help! time n.risk n.event survival std.err lower 95% CI upper 95% CI 6 21 3 0.857 0.0764 0.707 1.000 7 17 1 0.807 0.0869 0.636 0.977 10 15 1

Dear helpers I constructed a data frame with this structure > str(dados1) `data.frame': 485 obs. of 16 variables: $ Emissor : int 1 1 1 1 1 1 1 1 1 1 ... $ Marisca.Rio : int 1 1 1 1 1 1 1 1 1 1 ... $ Per?odo : int 1 1 1 1 1 1 1 1 1 1 ... $ Reproducao : int 3 3 3 3 3 3 3 3 3 3 ... $ Estacao : int 2 2 2 2 2 2 2 2 2 2 ... $ X30cm : int