similar to: Extracting variance components from a Manova

Hi R community, I am new bird to R and moved recently from SAS. I am no means expert on either but very curious learner. So your help crucial for me to learn R. I have already got positive expression. I was trying to fit a mixed model in animal experiment but stuck at simple point. The following similar example is from SAS mixed model pp 212. # data genetic_evaluation <-

Hi Everyone, I posted a similar question about a week ago, but haven't gotten any replies -- I'm afraid that's because my previous question was too vague. Let me try again with a more specific question, and I hope someone can help. NOTE, I know I should be using the newer lme4 package, I just haven't had a chance to update my version of R yet, so the question below relates

-------- Original Message -------- Subject: [AGDG-LIST:405] R Computing Contest Date: Sat, 21 Oct 2006 12:08:13 -0400 From: Larry Schaeffer <lrs at uoguelph.ca> Reply-To: lrs at uoguelph.ca To: Animal Geneticist's Discussion <agdg-list at colostate.edu> For those that are interested only: R Computer Programming Challenge Given: y = Factor A + Factor B + b1(Covariate1) +

Hello, I am trying to make the switch from SAS, and I have a fairly elemental problem with syntax using the nlme package for analyzing mixed models. There was a previous question on this topic posted to this list, so I apologize for redundancy, but I didn't understand the advice given to that inquiry. The model I want to run has the following factors: Host (fixed) Sire (random) Dam

Ron Crump wrote: > Hi, > > I have a dataframe that contains pedigree information; > that is individual, sire and dam identities as separate > columns. It also has date of birth. > > These identifiers are not numeric, or not sequential. > > Obviously, an identifier can appear in one or two columns, > depending on whether it was a parent or not. These should > be

Hi, I have a dataframe that contains pedigree information; that is individual, sire and dam identities as separate columns. It also has date of birth. These identifiers are not numeric, or not sequential. Obviously, an identifier can appear in one or two columns, depending on whether it was a parent or not. These should be consistent. Not all identifiers appear in the individual column - it is

I have a data frame containing the Id, Mother, Father and Sex from about 10,000 animals in our colony. I am interested in graphing simple family trees for a given subject or small number of subjects. The basic idea is: start with data frame from entire colony and list of index animals. I need to identify all immediate relatives of these index animals and plot the pedigree for them. We're

Hi All, I'd like to ask for a few clarifications. I am doing some calculations over some biggish datasets. One has ~ 23000 rows, and 6 columns, the other has ~620000 rows and 6 columns. I am using these datasets to perform a simulation of of haplotype coalescence over a pedigree (the datestes themselves are pedigree information). I created a new dataset (same number of rows as the pedigree

I want to use a diallel analysis in R, for some of my own data. I've been through the primary literature and textbooks, and remain stumped as to how to implment this in R. I can illustrate the problem using a published example dataset: [Cockerham and Weir (1977) Quadratic Analyses of Reciprocal Crosses. Biometrics, Vol. 33, No. 1 pp. 187-203] In this study, 8 different individuals were

Hi, I'm running into a problem subsetting a data frame that I have never encountered before: > dim(chkPd) [1] 3213 6 > df = head(chkPd) > df PN WB Sire Dam MG SEX 601 1001 715349 61710 61702 67 F 969 1001_1 511092 616253 615037 168 F 986 1002_1 511082 616253 623905 168 F 667 1003 715617 61817 61441 67 F

I assume I'm missing something obvious here... The short form of my main question is: how do I extract variance components from an lme object? The longer form (plus optional supplementary question!): I'm looking at some quantitative genetics, and want to estimate two variance components so that I can then calculate a statistic called Qst from them. So I have this: reg1 <- lme(y ~

Dear lists, I want to construct a loop in R, but don't know how to do it. I can do it in SAS, but I prefer in R (which I am hoping I will off SAS for good soon). Could anyone help me to convert the SAS codes to equivalent R codes. Basically, the following codes were written to establish the sire gametes or phases for daughter design for one markers two alleles. Here are the SAS code: do

Hi All, is there any chance of vectorising the two ifelse() statements in the following code: for(i in gp){ new[i,1] = ifelse(srow[i]>0, new[srow[i],zippo[i]], sample(1:100, 1, prob =Y1, rep = T)) new[i,2] = ifelse(drow[i]>0, new[drow[i]>0,zappo[i]], sample(1:100, 1, prob =Y1, rep = T)) } Where I am forced to check if the value of drow and srow are >0 for each line... in

Dear R helpers I have a table and i need to make new table table1: sire snp1 snp2 snp3 snp4 snp5 snp6 snp7 snp8 snp9 snp10 snp11 snp12 snp13 snp14 snp15 8877 -1 -1 -1 -1 0 0 -1 -1 -1 0 1 1 1 -1 -1 7765 1 1 1 0 0 0 -1 1 1 1 0 0 0 1 0 8766 1 1 -1 0 -1 -1 0 -1 0 -1 -1 -1 0 1 0 6756 0 1 0 -1 1 -1 -1 0 0 0 0 -1 0 1 1 5644 -1 0 1 -1 0 0 0 0 -1 -1 0 0 0 0 1 I have table2 sire

Hi everyone, since my last gem update, I get the following error when using webgen: /home/dams/opt/ruby/gems/webgen-0.5.14/lib/webgen/cli/utils.rb:10: Use RbConfig instead of obsolete and deprecated Config. /usr/lib/ruby/1.9.1/psych.rb:154:in `parse'': (<unknown>): couldn''t parse YAML at line 21 column 9 (Psych::SyntaxError) from /usr/libb/ruby/1.9.1/psych.rb:154:in

Dear ladies and gentlemen, I would like to calculate autocovarinace and cross-covariance scores 1, 2 and 3 of four classes A, B, C and D. I am using acf and ccf from time sires library. My problem is that I can not separate my data among the classes A, B, C and D. When I calculated acf for Score 1, I got a wrong result. The reason being that instead of using ony 60, 40 and 20, the program

Hi, I'm using R for a QTL analysis of SNP data. I was wondering if anyone had any advice on fitting a dominance effect into the following function; > myfun4 function (x) { x <- scan(con, nmax=169) y <- unique(x[which(!is.na(x))]) if(length(y)>1) { summary(lme(Ad ~ x, random= ~1|sire, na.action="na.omit")) } else {print("no.infomation")} } Con is the

Just the sort of newbie question we all hate ;-) I'm a bit stuck with MOH. I think all is done right and I've read everyhing I can find, but whenever * tries to do MOH, all that happens is '-z: No such file or directory' Yes, I am on redHat. Yes I have installed real mpg123. Yes, it does seem to work from the command line. Any suggestions would be greta, I'm sire

Hi R help, I wanted to set a fixed heritability for the object "Trait" left of ~ operator. Is this possible using lme, or would I need to use something like gls? This is the model line contained within a function; lme.out <- summary(lme(Trait ~ genotype, random = ~1|sire, weights=~number, na.action="na.omit")) Many thanks for any help.