similar to: Get distribution of positive/negative examples for each cluster

Hi All I am trying to translate a proc mixed into a lme() syntax. It seems that I was able to do it for part of the model, but a few things are still different. It is a 2-level bivariate model (some call it a pseudo-3-level model). PROC MIXED DATA=psdata.bivar COVTEST METHOD = ml; CLASS cluster_ID individual_id variable_id ; MODEL y = Dp Dq / SOLUTION NOINT; RANDOM Dp Dq / SUBJECT = cluster_ID

Hello all, I what to print the banner plot that is output from the mona method in the cluster package but the problem is I dont want to print all that red ink. Here is an example: data(animals) ma <- mona(animals) ma ## Plot similar to Figure 10 in Struyf et al (1996) plot(ma) I can change the bar color by using the argument col=c(0,0) - plot(ma,col=c(0,0)) - but then the variable labels also

Hello there, I'm trying to do lots of plots in one for-loop. But somehow ggplot does not evaluate arguments as expected. Here is an example: library(lattice) library(ggplot2) pl <- list() pl2 <- list() cDat <- as.data.frame(cbind(x1=0:100,x2=0:10,x3=1:20)) for(obs in c("x1", "x2")) { pl[[obs]] <- xyplot(cDat[,obs] ~ cDat[,"x3"], main=obs)

I'm puzzled by the following problem, which appears when attempting to run an analysis on part of a dataset: If I try: csubset <- dat$Diagnosis==0 cont <- lme(fixed=cform, random = ~1|StudyName, data=dat,subset=csubset,na.action=na.omit) Then I get: Error in eval(expr, envir, enclos) : Object "csubset" not found But if I do

Hi everybody, I have tested CTDB+GFS2+CMAN under Debian. It works good but I do not understand some points. It is possible to run the CTDB defining it under services section in cluster.conf but running it on the second node shuts down the process at the first one. My CTDB configuration implies 2 active-active nodes. Does CTDB care if the node starts with clean_start="0" or

hi, all I do a hotplug detach a network in a thread, because virDomainDetachDeviceFlags maybe asynchronous, so I do like follow: cb_para->cluster_id = info->cluster_id; cb_para->group_id = info->group_id; cb_para->vsys_id = info->vsysid; cb_para->vnf_id = info->vnf_id; cb_para->conn = conn; cb_para->time_out = 20*1000;//20s

Hi all, Sorry for the re-post, I sent my previous e-mail before it was complete. I am trying to model seroprevalence using the differential equation: dP/dt = beta*seronegative*.001*(seropositive)-0.35*(0.999)*(seropositive)-r*seropositive. I would like to estimate my two parameters, beta and r, using maximum likelihood methods. I have included my code below:

Hi, I have a table for an 1 week exam result for many classes in school, like this: Day Class_ID Test Result 1 Monday 1 Paper Passed 2 Tuesday 1 Oral Passed 3 Friday 1 Paper Passed 4 Monday 3 Paper Passed 5 Sunday 3 Oral Passed 6 Monday 3

==================================================================== "Harmony makes small things grow, lack of it makes great things decay." Sallust ==================================================================== Release Announcements --------------------- This is is the latest stable release of Samba 4.0. Major enhancements in Samba 4.0.3 include: o

==================================================================== "Harmony makes small things grow, lack of it makes great things decay." Sallust ==================================================================== Release Announcements --------------------- This is is the latest stable release of Samba 4.0. Major enhancements in Samba 4.0.3 include: o

Dear R experts: Here is my problem: #Data 1 Y <- c(0.5, 0.1, 0.5, 1.3, 1.4, 1.6, 1.65, 2.4, 2.6, 3.4, 3.6, 4.3, 4.42, 4.8, 4.7, 3.4, 3.3, 2.8, 2.8, 1.2, 1.1, 0.5, 0.2, 0.1, -0.2, -1.5, -2.5, -1.3, -0.5, -0.1) X <- seq(1:30) X1 <- c(rep("T1", 24), rep("T2", 6)) dat1 <- data.frame(Y, X, X1) require(lattice) mcol <- c("green", "red")

Hi, I want to insert a key into each panel of a trellis plot, which I can do with a custom panel function that calles draw.key. The problem arises because I want the top right hand corner of the key to start in the top right hand corner of the panel. If you run my code below, you can see that the key appears in the center of each panel. This is because the default viewport in draw.key is the

so it could be zero in both cases when given column has 0 set in it -- View this message in context: http://r.789695.n4.nabble.com/how-to-find-smallest-non-negative-and-biggest-non-positive-number-in-a-matrix-column-tp3077870p3077870.html Sent from the R help mailing list archive at Nabble.com.

Hi, I'm doing some work with linear models, and I've scaled my data using the scale(dataset) function. This was great at removing the skew, but I now can't perform the Box Cox transformation on the data set (using the boxcox(dataset) function), as the scaling has returned negative values. So my question is: how can I get the scale function to return a positive set of data (so I can

Hi R, I have a vector with all positive values and have fit the GEV distribution (shape parameter=0) to the data and I get the negative value of the location parameter. Is this possible? library(evd) r=read.table("clipboard",header=F)[,1] > length(r) [1] 2087 > sum(r>=0) #All data points are positive [1] 2087 > sum(r>=1000000) #All data points are positive

Hello all, I know this is a pretty easy question but I can't find it in S poetry or R help. How can I make a negative number positive. Such as -5 to be +5 I tried +(-5), but that didn't work. So no, I don't mean taking a -5^2 just to get a positive number. This is in a function so it's not just -5 it's x. :) Thanks, Paul -- Research Technician Mass Spectrometry o The

This is a very simple question but I couldn't form a site search quesry that would return a reasonable result set. Say I have a vector: x <- c(0,2,3,4,5,-1,-2) I want to replace all of the values in 'x' with the log of x. Naturally this runs into problems since some of the values are negative or zero. So how can I replace all of the positive elements of x with the log(x) and the

Hi everybody, I really appeciate the help you have given me. I am getting around R comfortably now and am able to import research project data, do mathematical analysis of the data and produce graphs. My PhD project is going really well in its early days, already collecting data and proposal completed soon! There is one question I have though. I can see that worksheets and history can be saved,

Hi Group, I have a matrix, and I would like to replace numbers less than 0 by the smallest minimum number. Below is an small matrix, and the loop I used. I would like to get suggestions on the "R way" to do this. Thanks, Juliet # example data set mymat <- structure(c(-0.503183609420937, 0.179063475173256, 0.130473004669938, -1.80825226960127, -0.794910626384209, 1.03857280868547,

Hello users, I've noticed a problem when creating a stacked column plot when a column contains a negative and a positive value. e.g. series1<-c(-1,-2, 3, 4, 5) series2<-c( 5, -4,-3,-2, 1) data<-rbind(series1,series2) barplot(as.matrix(data), beside=FALSE) In these cases (i.e. first, third and fifth columns) the plotting is not handled correctly. Compare this output with that