search for: sire

...earner. 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 <- read.table(textConnection(" sire dam adg 1 1 2.24 1 1 1.85 1 2 2.05 1 2 2.41 2 1 1.99 2 1 1.93 2 2 2.72 2 2 2.32 3 1 2.33 3 1 2.68 3 2 2.69 3 2 2.71 4 1 2.42 4 1 2.01 4 2 1.86 4 2 1.79 5 1 2.82 5 1 2.64 5 2 2.58 5 2 2.56"), header = TRUE) # my R practice codes require (lme4) lmer(adg...

...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 nested within Sire (random) Host * Sire (random) Host * Dam within Sire (random) So without the interactions I have: hogmodel = lme(gain ~ host, random = ~1|sire/dam) If I understand correctly, that "sire/dam" term gives me both Sire and Dam within Sire as random factors....

...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 to nmle. I have data from a classical quantitative genetics experiment, with 33 sires mated each to 2 dams, with 15 progeny from each dam raised on 5 host plants (3 larvae per host). So the model I would like to run has the following factors: Host (fixed) Sire (random) Dam [nested within sire] (random) Host * Sire (random interaction) Host * Dam [nested within sire] (random inte...

I am trying to run a simple nested manova with two levels of nesting, Sires, and Dams within Sires. The goal is to extract the among sires covariance matrix and secondarily, the among Dams within Sires covariance matrix. Both sires and dams are random effects. At present there are four dependent variables, but that may change. This is part of a larger bootstrapping pr...

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 i=1 to 744; do j=745 to 1540; m[j]=0; if sire[j]=anml[i] then do; if m1[j]=m1[i] and m2[j]=m2[i] then m[j]=0; else if m1[j]=m1[i] then m[j] =...

...lves are pedigree information). I created a new dataset (same number of rows as the pedigree dataset, 2 colums) and I use a looping functions to assign haplotypes according to a standrd biological reprodictive process (i.e. meiosis, sexual reproduction). My code is someting like: off = function(sire, dam){ # simulation of reproduction, two inds sch.toll = round(runif(1, min = 1, max = 2)) dch.toll = round(runif(1, min = 1, max = 2)) s.gam = sire[,sch.toll] d.gam = dam[,dch.toll] offspring = cbind(s.gam,d.gam) # offspring } for (i in 1:dim(new)[1]){ if(ped[i,3] != 0 & ped[i,5] !=...

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 individua...

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...

...I realize my code should be written better to get rid of the loops, so if anyone has suggestions there I would appreciate this as well. Thanks in advance. Code to calculate generations forward and backward: #queryIds holds the unique Ids for parents of the index animals queryIds = unique(c(ped$Sire, ped$Dam)); for(i in 1:gens){ if (length(queryIds) == 0){break}; #allPed is the dataframe with Id,Dam,Sire and Sex for animals in our colony newRows <- subset(allPed, Id %in% queryIds); queryIds = c(newRows$Sire, newRows$Dam); ped <- unique(rbind(newRows,ped)); } #build...

Estimados Estoy probando una alternativa, la cual no tiene ranef Pero puedo saber cuáles son desde la siguiente matriz, 4 , 5, 6, 7, 8 son hijos de sire1, sire3 y sire4. Obtengo los resultados, pero ¿como puedo tomar los nombres de filas y columnas, sire1, sire2, sire3, 4, 5, 6, 7, y 8, de forma tal de hacer legible el resultado para cada animal? Lógicamente, un listado de números sin decir que animal es, convengamos que no se entiende mucho. &gt...

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 1361 1003_1 510711 635246 627321 168 F 754 1004 715272 62356 61380 67 F > dfb...

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 consistent. > &gt...

...ed 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 crossed in a diallel design with reciprocals but without self-crosses. Two individuals were measured for each combination of sire and dam. The goal is to partition the phenotypic variance based on comparisons among full and half-siblings, and draw inferences about the additive and dominant components of variation. The data look like this (complete dataset is at the end of this message) >summary(df) val sire dam block...

...check if the value of drow and srow are >0 for each line... in practical terms, I am attributing haplotypes to a pedigree, so I have to give the haplotypes to the parents before I give them to the offspring. The vectors *zippo* and *zappo* are the chances of getting one or the other hap from the sire and dam respectively. *gp* is the vectors of non-ancestral animals. *new* is a two col matrix where the haps are stored. Cheers, Federico -- Federico C. F. Calboli Department of Epidemiology and Public Health Imperial College, St Mary's Campus Norfolk Place, London W2 1PG Tel +44 (0)20 75...

...s 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 ~ temp*food, random =~1|POP/SIRE/DAM, na.action=na.exclude)) summary(reg1) which shows me the random effects as standard deviations. I then want the SIRE and POP variance components (as Vsire and Vpop respectively) to put into the calculation of Qst as Qst <- Vpop/(Vpop + 8*Vsire) And obviously writing down the component...

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 use all the data in column under Score 1. What should I do to calculate acf...

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 connection to a file of the genotypes for each SNP. It is set up as a continues string of genotype (0, 1, 2), the first 169 for the first SNP, the second 169 for the second SNP and so on. I need a way of dete...

...servations, Factors A and B are fixed with ma and mb levels respectively, Covariates 1 to p are fixed regressions and p is a general number but less than 20. Heritability is a variable and should be requested by the program. You are given a list of the animal IDs of animals with records, and their sires and dams. Sire and dam IDs are smaller than the IDs of their progeny, and all IDs go from 1 to n where n can be up to 10 million animals. Tasks: 1) Estimate the EBVs of all animals using BLUP and mixed model equations and list the top 10 largest EBVs with associated animal IDs, first ten l...

...ll 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 it's something simple! Thanks, -- Toby Seaman FWD 63143 (eventually!)