search for: samp2

...the stats package 'dist' function, using euclidean distance. I took a subset of this dataset (3 samples x 3 loci) to test how euclidean distance is calculated: 3x3 subset used Locus1 Locus2 Locus3 Samp1 GG <NA> GG Samp2 AG CA GA Samp3 AG CA GG The euclidean distance function is defined as: sqrt(sum((x_i - y_i)^2)) My assumption was that the difference between x_i and y_i would be the number of allelic differences at each base pair site...

.... . . . . . . I want to select rows belonging to 7 random plots for 100 times. (There are 50 plots in total) So I created a list of 100 vectors, each vector has 7 elements. samp <- lapply(1:100, function(i) sample(LETTERS)) samp2 <- lapply(samp2, "[", 1:7) How can I select the 26 plots from 'data' using 'samp'? samp3 <- sample(LETTERS, 7) samp4 <- subset(data, plot %in% samp3) # this works samp5 <- subset(data, plot %in% samp2[[1]]) # this works as well, but I used a for loop to get i...

...e t test vs various #### non-parametric alternatives sim.size <- 200 sample.size <- 10 set.seed(231) mu1 <- 0 delta <- seq(-2,2, length=50) for (j in 1:length(delta)) { mu2 <- mu1 + delta[j] for (i in 1:sim.size) { # Generate ith sample samp1 <- rnorm(mean=mu1,sample.size) samp2 <- rnorm(mean=mu2,sample.size) # Perform ith set of tests test1 <- t.test(samp1, samp2,alternative = c("two.sided")) pt.test[i] <- (test1$p.value < 0.05) test2 <- wilcox.test(samp1, samp2,alternative = c("two.sided"), exact = TRUE)...

...-levels also converted to  dummy variables). So I have worked with them in two ways: i created a variable X1 = dgen*dtoe  and I get an error "Error in dgen * dtoe : non-conformable arrays"then i run a glm, binomial using that interaction variable and i get :                  logit_x = glm(samp2$STATUS ~ dgen*dtoe, data=samp2,family = binomial("logit")) > summary(logit_x) Call: glm(formula = samp2$STATUS ~ dgen * dtoe, family = binomial("logit"),     data = samp2) Deviance Residuals:     Min       1Q   Median       3Q      Max  -2.6594   0.2431   0.2563   0.2563...

...sults of each iteration for 86 variables. I run 10,000 iterations. So, the matrix samp is 10,000 x 86. I want to use the gelman-rubin test to check for convergence. To do that, I need at least two chains. If I run second chain with different starting values and seed, I could save to the matrix 'samp2'. So, I will have two matrices 10,000x86. I want to use the function gelman.diag(x, confidence = 0.95, transform=FALSE), where x: An 'mcmc.list' object with more than one chain, and with starting values that are overdispersed with respect to the posterior distribution. How do I create m...

Dear All, I have a dissimilarity matrix which I happily convert to a distance object by running: X <- as.dist(Y) and I can happily now run either hclust(X) or agnes(X). So that the various bits of output are labelled correctly I would dearly like to be able to give names to the columns and rows of X, as would happen if I ran: mydata<-read.table("clipboard",header=T)

> 3x3 subset used > Locus1 Locus2 Locus3 > Samp1 GG <NA> GG > Samp2 AG CA GA > Samp3 AG CA GG > > The euclidean distance function is defined as: sqrt(sum((x_i - y_i)^2)) My > assumption was that the difference between x_i and y_i would be the number > of allelic difference...

Hi, Is there a command to get the indices of intersecting elements of two vectors as intersect() will give the elements and not its indices. Thanks in advance. Praveen Surendran School of Medicine and Medical Sciences University College Dublin Belfield, Dublin 4 Ireland. [[alternative HTML version deleted]]

...7, 1.846162, -4.60517, 2.121427, 1.973118, -4.60517, 2.251568, -4.60517, 2.270724, 0.70338, 0.963816, -4.60517, 0.023703, -4.60517, 2.043382, 1.070586, 2.768289, 1.085169, 0.959334, -0.02428, -4.60517, 1.371895, 1.533227) "zzzanova" <- structure(list(Intensity = c(t(Samp1), t(Samp2), t(Samp3), t(Samp4)), Group = structure(c(1,1,1,1,1,1,1,1,1, 2,2,2,2,2,2,2,2, 3,3,3,3,3,3,3,3,3, 4,4,4,4,4,4,4,4,4,4, 5,5,5,5,5,5,5,5,5, 6,6,6,6,6,6,6,6,6), .Label = c("Group1", "Group2", "Group3", "Group4", &quo...

...e to estimate the relationship between the probability of being a student and number of siblings (alive). However, I need to include a number of relevant covariates. My code is below: fm3 <- mob(Student ~ age + alive + sex2 + cwa + cha + cym | Religion+Servant + Literacy, control = ctrl, data = samp2, model = glinearModel, family =binomial()) plot(fm3, tp_args = list(which = "alive"),tnex = 2, type = "simple" ) Which works fine. However, the plots generated do not show the conditional relationship between the dependent and independent variable at each node. Basically, there...

...e this method (aov) assumes equal variances. How can I adjust this to do an ANOVA with unequal variances #####SCRIPT STARTS #Creates a structured list suitable for ANOVA analysis # Intensity Group (1,2,3,4) Sample(1:62) "zzzanova" <- structure(list(Intensity = c(t(Samp1), t(Samp2), t(Samp3), t(Samp4)), Group = structure(c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1, 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2, 3,3,3,3,3,3,3,3,3,3,3,3,3,3, 4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4), .Label = c("Group1", "Group2", "Group3", "Group4"), class =...

...h iteration for 86 variables. I run 10,000 iterations. So, the >matrix samp is 10,000 x 86. I want to use the gelman-rubin test to check for >convergence. To do that, I need at least two chains. If I run second chain >with different starting values and seed, I could save to the matrix 'samp2'. >So, I will have two matrices 10,000x86. I want to use the function >gelman.diag(x, confidence = 0.95, transform=FALSE), where x: An 'mcmc.list' >object with more than one chain, and with starting values that are >overdispersed with respect to the posterior distribution. H...

Hi I am very new to R and statistical programming in general. I am trying to reorder data from a .csv file. I have managed to import the data and create a zero matrix. I am now trying to fill the matrix. There seems to be some problem with this section of my code. I have highlighted the dodgy code in red. Please help if possible. ###################################################### ###########