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Hi, I hope somebody can help me with this issue: I am doing population pyramids using the barplot command, so in the left side I have male age structure and in the right side the female age structure. To plot the male age structure I put the data in negative numbers. Now, I want to change the sign in the bar plot in such way that I have no-sign numbers, both in left and right side of the graph. I

I have a matrix containing means and CIs (lower and upper in two columns, so three columns for every data point) for several points. I have to build a graph of these means accompained by the CIs (as wiskers). No problems with making the graph of means, but I don't know how to introduce CIs. Can anybody advise? -- View this message in context:

...mented in R yet. This discussion is form March 2006 and there seemed to be no package or function implemented at the time. Has this changed? Is there a package that provides nonparametric manova as in McArdle and Anderson (2001) and Anderson (2001) now? Thanks, Daniel ------------------------- cuncta stricte discussurus

...I do not know how to tell "sub" between which of them to extract. The string looks like this 12/01/03/08 The extracted variables should look like: x1=12 x2=01 x3=03 x4=08 If anybody could help or point me to useful help, I would be greatful. Cheers, Daniel ------------------------- cuncta stricte discussurus

Hello, I'm trying to do cor(x1,x2) and I get the following error: Error in cor.default(x1, x2) : missing observations in cov/cor A few things: 1. I've used cor() many times and have never encountered this error. 2. length(x1) = length(x2) 3. is.numeric(x1) = is.numeric(x2) = TRUE 4. which(is.na(x1)) = which(is.na(x2)) = integer(0) {the same goes for is.nan()} 5. I also try

...reported in summary(reg). The difference of 1% seems nonmarginal. Should I have multiplied var(e)*solve(t(z)%*%z) by n and divided by n-1? But even if I do this, I still observe a difference. Can anybody help me out what the source of this difference is? Cheers, Daniel ------------------------- cuncta stricte discussurus

...within the text. #This is done using: n=100 #choose n large enough length(which(is.na(gregexpr("Saint",x,ignore.case=TRUE)[[1]][1:n])==FALSE)) But again, if the text is large, I cannot assign it to x. I'd be grateful for any suggestions. Cheers, Daniel ------------------------- cuncta stricte discussurus

...x b2=x[x>a] ## assign all x greater than a b1 and b2 should be equal as I see it. Anybody who can explain why they are not makes my day, especially the one who can tell me how I can get b1 equal to b2 (without sacrificing the ifelse condition). Thanks much, Daniel ------------------------- cuncta stricte discussurus

The myoglobin sequence, with reference number NM_005368 in Gen bank, has the following frequencies of DNA nucleotides: A C G T 237 278 309 242 Do these data provide sufficient evidence, at the 1% level of significance, that the DNA nucleotides have an unequal distribution, that is the DNA nucleotides are not evenly utilised? Clearly state your hypothesis, test statistic and conclusion.

Hi So my particular problem is this: I have a row vector of length 5200 elements - specifically created by x<-rbinom(5200,1,0.5) y<-matrix(x,nrow=1,ncol=5200) y now, each element is either a 0 or a 1 - e.g. it could be (0,1,1,1,1,0,0,0,1,1,1) e.t.c. when the element is a 1, i need to multiply a number (say 1000) by 1.005, and if it is 1 again, multiply it _again_ by 1.005. so for

I am writing a rd doc, and need to use "#" in a url adress. This would make: \url{http://www.xxxx.org/myfolder/#myanchor} Of course, I suppose this will not work because # is a special character starting a comment line in the rd dialect. I did not found a similar example in "Writing R exentions". I am not sure bout using \dQuote{a quotation}), and use \sQuote and \dQuote

I have time-series data on approval ratings of British Prime Ministers. The prime ministers dating from MacMillan onward till today are coded as dummy variables and the approval ratings are entered for each month. I want to know the mean value of the approval rating of each Prime Minister in the dataset and the approval rating during his/her first month and last month as PM. What R code should

...d meeting deadlines. Interested people send their all contact details and interest at: &nbsp; bhangisolutions at gmail.com If it is possible to chat at Gtalk will be more great. MSN, Yahoo, SkyPe is possible and also telephone. Best regards, Mahesh ------------------------- cuncta stricte discussurus ------------------------- -----Urspr?ngliche Nachricht----- Von: David Winsemius [mailto:dwinsemius at comcast.net] Gesendet: Saturday, October 03, 2009 11:56 PM An: Daniel Malter Betreff: offlist Re: AW: [R] Urgently needed Exercise solutions related to PracticalData Analysis...

I am trying to get the function "Models" to work each time there is an instance of k. This code will stop after the first model is complete. I need it to come back and pass the next value of c into the "Initial.State" function. any ideas? col<-c(23:28) #Setup for(k in col){ Initial.State(Response=zample[,c(k,29)], Explanatory=zample[,variable_columns],

I have a data set like the following: subject visit x1 1 1 0.5 1 2 1.2 1 3 0.7 2 1 0.4 2 2 0.6 2 3 1.0 ..... where x1 is the interval between the two visits. Now I want to calculate the cumulative intervals since the beinging, for example subject visit x1 cum 1 1 0.5 0.5 1 2 1.2 0.5+1.2 1 3 0.7 0.5+1.2+0.7 2 1 0.4 0.4 2 2 0.6 0.4+0.6 2 3 1.0 0.4+0.6+1.0 ..... is there an easy to generate the

I'm trying to fit a Cox proportional hazards model to some hospital admission data. About 25% of the patients have had at least one admission, and of these, 40% have had two admissions within the 12 month period of the study. Each patients has had one of 4 treatments, and one of the treatment groups has had no admissions for the period. I used:

Hi R people, I have a very basic question to ask - I'm sorry if it's been asked before, but I searched the archives and could not find an answer. All the examples I found were much more complicated/nuanced versions of the problem - my question is much more simple. I have data with multiple, nested fixed effects (as I understand it, fixed effects are specified by the experimental design

Dear R users, I was wondering if anyone knows how to obtain(subset) the first and/or the last record of a subject in a longitudinal setup. Normally in SAS one uses first.variable1 and last.variable1. So my question is that is there an R way of doing this. Regards, -- Luwis Diya, Phd student (Biostatistics), Biostatistical Center, School Of Public Health, Catholic University of Leuven, U.Z. St

...te, if lme fits with "restricted maximum likelihood," I think I remember that you cannot compare them. You have to fit them with "maximum likelihood." I am pointing this out because lmer with restricted maximum likelihood by standard, so lme might too. ------------------------- cuncta stricte discussurus ------------------------- -----Urspr?ngliche Nachricht----- Von: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] Im Auftrag von John Sorkin Gesendet: Monday, September 07, 2009 4:00 PM An: r-help at r-project.org Betreff: [R] Omnibus test for main effec...

i have a series of regressions i need to run where everything is the same except for the dependent variable, e.g.: lm(y1 ~ x1+x2+x3+x4+x5, data=data) lm(y2 ~ x1+x2+x3+x4+x5, data=data) lm(y3 ~ x1+x2+x3+x4+x5, data=data) is it possible to run all these regs with a single command? given that the bulk of the work for linear regressions is inverting a matrix that depends only on the independent