search for: hets

For the following code is there a way to make the jitter all line up horizontally, instead of them being just randomly spread around a value. So for ex if there are multiple values at 63 for genotype wt then all the values should be plotted on the same y value of 63 but spaced apart by a certain factor or noise.. library(lattice); dataFrame <- as.data.frame(t(structure(c( 64,'wt',

Thanks Bert for all the help. I got the legend figured out Friday but left early becoz of long weekend so didn't get a chance to reply.. I modified the plot margins a little bit and Here's what I finally had... par(mar=c(c(10, 6, 6, 10) + 0.1)); par(xpd=FALSE); with (dataFrame, stripchart(marbles_buried ~ genotype, method="jitter", vertical=TRUE, col = c('blue',

-----Original Message----- From: Ghosh, Sandeep Sent: Thursday, June 30, 2005 5:43 PM To: 'Berton Gunter' Subject: plot legend outside the grid Thanks for the pointers... I managed to get everything to look and feel the way I want except for the legend to plot outside the grid... Thanks for the note on the par, but I'm not able to it to plot outside the plot grid.. dataFrame <-

Ik ben afwezig vanaf 29/07/2006 en ik ben niet eerder terug dan 23/08/2006. Ik ben met verlof van 31 juli tot en met 22 augustus. Voor dringende zaken kan je mailen naar helpdesk@vlafo.be. --------------------------------------------------------------------------------- DISCLAIMER : De personeelsleden van het agentschap doen hun best om in e-mails betrouwbare informatie te geven. Toch kan

- Deze mail is doorgezonden omdat iemand dacht dat u wellicht interesse zou kunnen hebben. Mocht u na het lezen van deze mail geen interesse hebben dan kunt u de boodschap verwijderen, u ontvangt geen verdere berichten meer tenzij u daar uitdrukkelijk om verzoekt. Computervrienden De club computervrienden gaat zich lanceren in mei 2002. Voordat deze datum wordt bereikt willen wij het volgende

Hi, I'm analyzing a large number of simulations using lm(), a sample of the resulting data is pasted below. In some simulations, the response variable doesn't vary, ie: > tmp[[2]]$richness [1] 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 When I analyze this using R version 2.8.0 (2008-10-20) on a linux cluster, I get an appropriate result: ## begin R

I am trying to bootstrap the parameters for a model that is estimated through the optim() function and find that when I make the call to boot, it runs but returns the exact same estimate for all of the bootstrap estimates. I managed to replicate the same problem using a glm() model but was able to fix it when I made a call to the variables as data frame by their exact names. But no matter how I

[image001.gif] Diploma's en titels zonder studie op basis van eerder verworven competenties Geachte heer of mevrouw, Graag willen wij u attenderen op onze unieke dienstverlening. Wellicht dat er voor u mogelijkheden zijn om zonder studie op basis van eerder verworven competenties via een External Programme in het bezit te komen van een academische Bachelor, Master

How to do the attachment in mail like the yahoo and gmail does? --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Ruby on Rails: Talk" group. To post to this group, send email to rubyonrails-talk-/JYPxA39Uh5TLH3MbocFFw@public.gmane.org To unsubscribe from this group, send email to

Hello, On my Chakra linux machine the json gem would not get compiled, I already made a issue for that. But for continuing my project can I have ruby on rails without the json gem ? And if so, how can I do this ? Roelof -- You received this message because you are subscribed to the Google Groups "Ruby on Rails: Talk" group. To post to this group, send email to

Beste, ik heb getracht om Matlab (versie 6.0.0.88 Release 12 op Win98 op aparte partitie op mijn hd) mbv wine te draain op mijn RedHat 7.0. Het betreft codeweavers-wine, versie 20010112. Matlab start heel fijn op, wine spuwt een proces of 15 en vreet zowat al mijn geheugen op maar voor de rest lijkt hij wel te werken. (Andere applicaties zoals notepad, wordpad, freecell, etc werkten perfect..)

Geachte gebruiker, U probeert een attachment te versturen met een virus. Het betreft het attachment movie0045.pif dat u heeft gestuurd naar . Het attachment is besmet met het virus: VIRUS= Win32.Sobig.F (CA(Vet),NAI,Kaspersky). Het bericht "Re: That movie" was verzonden vanaf samba@samba.org. De status van de attachment is: Purged. Deze mail is uitsluitend ter informatie, u hoeft

Hello. I have a vector and within that vector is one expression. When I display this vector it comes up as expression(NA_character_, NA_character_, "Null Effect", "Pooled effect", NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, paste("Het Contours ", I^2, sep = ""), 0.4, 0.41, 0.42, 0.45, NA_character_) Where the

R-users, I'm seeking any suggestions on optimizing some code for speed. Here's the setup: the code below is part of a larger chunk that is calculating Fst values across loci and alleles. This chunk is designed to calculate the proportion ('p.a') of an allele ('a') at a locus in each population ('p') and the proportion of individuals heterozygous for that

Hi there, The results have just been in and the Ruby-on-Rails team from Finalist IT Group reached an incredible _second_ place! We''re actually so surprised, that I personally expect the start of RoR projects here soon. We''ve already announced some RoR courses. And that while we''re mainly a Java-only shop! You can read a report from one of the participating

I am trying to replace values of a vector (consisting of 15 values) by a value that is related to a matching value in a dataset (consisting of 17 rows). Here's an example The vector: v <- c(f,a,e,d,m,o,e,f,i,n,e,i,b,a,o) The dataset's columns consist of the following values d[,1] <- c(a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q) d[,2] <- 1:17 So I want to end up with a vector that

Dear R users, Could you help my with the following problem? I want to repeat a glm analysis with 2 independent variables for all 900 variables (snps) in my data set. So, I want to check whether snp1 has a different effect on my outcome variable in patients and controls(phenotype). And repeat that for snp2 to snp900. Is there an easy way to get a summary of the data, e.g. a list of P values of all

Hey gang, I find myself struggling with multiple step definitions in cucumber all the time. I regularly test the contents of my flash[:notice] in steps, since I think that''s part of the behaviour of the app. So, while implementing a new feature, I ran into the following error: /opt/local/lib/ruby/gems/1.8/gems/cucumber-0.1.15/bin/../lib/cucumber/ step_mother.rb:81:in

Dear Massimo, It seems straightforward to use weighted.mean() in a dplyr context library(dplyr) mydf %>% group_by(date_time, type) %>% summarise(vel = weighted.mean(speed, n_vehicles)) Best regards, ir. Thierry Onkelinx Statisticus / Statistician Vlaamse Overheid / Government of Flanders INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR NATURE AND FOREST Team

Hi All, Probably simple question: I want to make a share for sharing an lmhosts file which gets included by the clients. This share has to be very public and accept nullsessions (whatever that me be (no username and password?)). I want this because I need to generate aliases on the fly for for our fileserver for some clients (terminalservers specifically). So I thought that having an server