search for: 6,3

Dear useRs, Trying to replace the diagonal of a matrix is not working for me. I want a matrix with .6 on the diag and .4 elsewhere. The following code looks like it should work--when I lookk at mps and idx they look how I want them too--but it only replaces the first element, not each element on the diagonal. mps <- matrix(rep(.4, 3*3), nrow=n, byrow=TRUE) idx <- diag(3) mps idx mps[idx] &l...

Dear R users; Is it possible to get the row and column number of a particular entry in a dist object? Let's say that I want to find the position of the value 1.1837 (the last entry on the dist object below), that is [6,3]. Can I get those values without transforming the object to a matrix?, i.e. working with the dist object only. 1 2 3 2 0.23935864 3 0.56655914 0.71923104 4 0.15272561 0.37926989 0.43931332 5 0.17728654 0.13355685 0.73025495 6 0.61783536 0.52055379 1....

...ng lattice's xyplot using code like that below. How do I specify a symbol and color for the variable 'prob' and different one's for 'll.prob'? Thanks, Matt xyplot( prob + ll.prob ~ time.eff |stat.id + time.out ,data = OUT, allow.multiple = TRUE, layout = c(6,3), as.table = TRUE , panel = function(x,y){panel.abline(h = 0) panel.xyplot(x,y)}, strip = TRUE, ) Matt Pocernich NCAR - Research Applications Program 303-497-8312

...Further, as the two graphs are related, I want to put a small or even no margin between the two graphs. I manipulate the mar and oma variables to this end. Sadly, I am not getting the results that I want. > par(mfcol=c(2,1)) # 2 plot window > par(mar=c(0,4,0,2), oma=c(0,2,0.5,2), pin=c(6,0.5)) # Set all variables at once for the upper plot > plot(1:9) # sample upper plot > par(pin=c(6,3)) # set window size for 2nd plot > plot(1:9) # sample lower plot This method produces the two box plots of varying heights, but does not change the margins....

Hola a todos, no acierto a convertir una cadena "6,3" en el double 6.3 sin perder la parte decimal. Me sirve también que desde el csv lo lea como double, pero lo convierte siempre a factor y desde ahi lo paso a caracter porque creo que será más facil. El caso es que con as.double el resultado sale redondeado y no veo cómo recuperar los decimal...

...I need your help to create a code for this. Thank you very much, Cecília Carmo (Universidade de Aveiro) #My reproducible example: firm<-sort(rep(1:3,5),decreasing=F) year<-rep(2000:2004,3) X<-rnorm(15) data1<-data.frame(firm,year,X) data1 firm<-sort(rep(4:6,3),decreasing=F) year<-rep(2001:2003,3) X<-rnorm(9) data2<-data.frame(firm,year,X) data2 finaldata<-rbind(data1,data2) finaldata [[alternative HTML version deleted]]

Hi, In the exactRankTest package, I've become aware that you can get anomalous p-values (i.e., above 1) from the wilcox.exact method, as in: > wilcox.exact(c(-0.6,0.8,-0.5)) Exact Wilcoxon signed rank test data: c(-0.6, 0.8, -0.5) V = 3, p-value = 1.25 alternative hypothesis: true mu is not equal to 0 This is disturbing. Has anyone encountered this before, and if so is there an obvious reason why this should happen? Thanks, Roger

Hi folks, I have two dataframes like these: DF <- data.frame(ID=c("AA1234","AB3233","AC4353","AD2345","AE7453"), CK32344=c(1,3,2,4,1), CK32664=c(2,1,1,2,3), CK33422=c(2,2,1,3,2)) VAL <- data.frame(num=rep(6,3), type=c("CK32344", "CK32664", "CK33422"), number=c("32","452","234")) I want to replace the values in DF$CK32344, DF$CK32664 DF$CK33422 with the corresponding number...

Hello All, I am drawing a graph having 18 small graphs inside using par(mfrow = c(6,3)) command. My problem is how to specify the margins of the whole 18 graphs. I used par(mar=c(6.5, 6.5, 1.5, 1.5)) for each graph separately already but it does not left any margins for the 'mtext()' for the margins of the whole 18 graphs. Any comments please. Thanks, MJK [[alternative...

This fixes a long standing issue where it was not possible to do "make usr/klibc/arch/x86_64/longjmp.o" in the kernel. The principle is that all .o files to be part of klibc are listed with klib-y. For each directory a klib.list file is made that specify all .o file and the final AR then adds all .o files to create libc.a. This patch introduce the infrastructure and converts x86_...

Hi all, I have a data frame with column over which I would like to run repeated functions for data analysis. Currently I am only running recursively over two columns where I column 1 has two states over which I split and column two has 3 states. The function therefore runs 2 x 3 = 6 times as shown when running the following code: mydata <- data.frame(userid = c(5, 6, 5, 6, 5, 6), taskid = c(1, 1, 2, 2, 3, 3), stuff = 11:16) mydata mydata <- mydata[with(mydata, order(userid, taskid)), ] mydata lapply(split(mydata, mydata[,1]), function(x){ lapply(split(x, x[,2]),...

...ovy University of Virginia Department of Psychology USPS: P.O.Box 400400 Charlottesville, VA 22904-4400 Parcels: Room 102 Gilmer Hall McCormick Road Charlottesville, VA 22903 Office: B011 +1-434-982-4729 Lab: B019 +1-434-982-4751 Fax: +1-434-982-4766 WWW: http://www.people.virginia.edu/~mk9y/

Hello, Is this a bug in the lmer routine? > library(lme4) > ### test case based on rats data from Crawley > a<-rnorm(36);b<-rep(1:3,each=12);c<-rep(1:2,each=6,3);d<-rep (1:3,each=2,6) > > ### mixed model works when c & d are numeric, lmer assumes they are factors > m <- lmer(a ~ b + (1|c/d)) > > ### but bails out when they are actually specified as factors > c<-factor(c)...

Dear List, i know there are some solutions for this in the archive, but they're not very good for numeric matrices, since they usually convert rows/columns to character strings. Is there an easy way to do $subject for numeric matrices properly, or i need to do it by hand? Thanks, Gabor

...... but I try to do it in R ;-) : I would like to generate a matrix Q satisfying (for a given Z, X and W) the two following conditions: t(Q)%*%Q=Z (1) XQ=W (2) where: Q is m rows and r columns X is p rows and m columns D is p rows and r columns C is r rows and r columns with m>p,r e.g: m=6, p=2 r=3 Z=matrix(c(1,.2,.5,.2,1,.45,.5,.45,1),3,3) X=matrix(c(.1,.3,.5,.6,.2,.1,.8,1,.4,.2,.2,.9),2,6) W=matrix(c(0,.8,.4,.6,.2,0),2,3) #Create a matrix satisfying (1) is easy: A=matrix(runif(18),6,3) Q1=svd(A)$u%*%chol(Z) #For the second condition (2), a solution is given by Q2=A%*%ginv(X%*...

...K in the input data files and data file 14th is not the biggest input file. Here is the code: for(Br in 1:3) { if(Br == 1) B <- "U" if(Br == 2) B <- "A" if(Br == 3) B <- "O" ofn <- sprintf("%s.png", B) bitmap(ofn, type = "png256", width = 21, height = 30, pointsize = 30, bg = "white", res = 300) <<<<when I decrease res to 50(res=50), everything is OK with no error message par(cex.main = 1.2, cex.lab = 1.8, cex.axis = 1.4) par(mfrow = c(6,3)) for(C in 1:18) { if( ( C + 2 ) %% 3 ==...

...le(main=hazanth[3,1],cex.main=0.9) # plot for Hiwi with(hazanth[4,3:7],silsurv(a1=a1,b1=b1,a2=a2,a3=a3,b3=b3));title(main=hazanth[4,1],cex.main=0.9) # plot for !Kung with(hazanth[5,3:7],silsurv(a1=a1,b1=b1,a2=a2,a3=a3,b3=b3));title(main=hazanth[5,1],cex.main=0.9) # plot for Yanomamo with(hazanth[6,3:7],silsurv(a1=a1,b1=b1,a2=a2,a3=a3,b3=b3));title(main=hazanth[6,1],cex.main=0.9) # plot for Tsimane ###############################

...early always results in p-values above 95%. I wrote an example programme which demonstrates this effect. I uploaded a PDF showing the results Here is the code which produces the PDF file: ------------------------------------------------------------------------------------- s <- matrix(runif(1600,0,1000), nrow=20) a.res1 <- pvclust(t(s), method.hclust="complete", method.dist="euclidian", nboot=500) a.res2 <- pvclust(t(s), method.hclust="complete", method.dist="maximum", nboot=500) a.res3 <- pvclust(t(s), method.hclust="complete",...

...symbols to draw individual points, CI and the lines connect group means from different time points; 2. Add jitters to x axis to allow CIs not be overlapped Could anyone modify the attached code to achieve this? Thanks library(lattice) set.seed(123) src <- data.frame(time = rep(c(1, 6, 24), rep(6,3)), trt = rep(rep(c('C','T'), each=3), 3)) src <- transform(src, x=sqrt(time)+2*(trt=='C')+rnorm(18), trt = ordered(trt, levels=c('C','T'))) src <- src[-5,] src$m <- ave(src$x, src[c('time','trt')], FUN = mean) src$sd &lt...

.... The next stage will be to make the animals move in reference to each other using a moving average variable. Therefore I think I need all animals simulated within one subroutine (loop). I probably haven't explained thie very well but here is the code I have been working on.... #creation of a 6*6 matrix with the transition probabilities transitions<-matrix(0,nrow=6,ncol=6); transitions[1,1]<- (0/6); transitions[1,2]<- (0/6); transitions[1,3]<-(0/6); transitions[1,4]<-(0/6);transitions[1,5]<-(0/6); transitions[1,6]<-(6/6);transitions[2,1]<- (1/6);transitions[2,2]&l...