search for: oberv

Hello ! I am attempting to switch from being a long time SAS user to R, and would really appreciate a bit of help ! The first thing I do in getting a large dataset (thousands of obervations and hundreds of variables) is to run a SAS command PROC CONTENTS VARNUM command - this provides me a table with the name of each variable, its type and length; then I run a PROC MEANS - for numerical variables it gives me a table with the number of non-missing values, min, max, mean and std....

...ose I have a long format for a longitudinal data id time x 1 1 10 1 2 11 1 3 23 1 4 23 2 2 12 2 3 13 2 4 14 3 1 11 3 3 15 3 4 18 3 5 21 4 2 22 4 3 27 4 6 29 I want to select the x values for each ID when time is equal to 3. When that observation is not observed, then I want to replace it with the obervation at time equal to 4. otherwise just use NA. How can I implement this with a quick command? [[alternative HTML version deleted]]

hello! my question conserns with use of "panel" package (written by R.C.Gentlman) (unfortunately the manual and help sites are very short) 1. is it possible to do analysis just without a(ny) covariate? i suggest do it by introducing a covariate with level=0 in all obervations, this because of Q(z)=Q_o exp(beta*z), but it seemingly doesn't work 2. in the option gamma in the call of panel function: do you mean an initial value for parameter vector gamma? say if i have 3 theta-parameters, so i have to initialize gamma=c(xxx,xxx,xxx), correct? 3. are the (fi...

I have 1000 oberved binary reponses Y, and I also have 1000 predictited probabilities predicted_prob (between 0 and 1) for those observed reponses. How can I plot a gain chart in R? Thanks. -- View this message in context: http://www.nabble.com/how-to-plot-a-gains-chart-in-R--tp25083628p25083628.html Sent from th...

....0000 2.62500 0.0000 0.18750 0.0000 0.00000 0.0000 0.40625 [,19] [,20] [1,] 48.90625 47.5 [2,] 0.00000 0.0 The problem is: I want to group the data. I want to have ten groups. The first two bars should be [1,1] and [2,1] together in one bar and in the second bar of the first obervation should be [1,2] and [2,2] (stacked with beside =TRUE). Therefore the first observation is [1,1],[1,2],[2,1] and [2,2]. For a better understanding: I want to split the darkblue bar into a green and a blue bar (please see picture) per observation. Has anyone a suggestion how I can do this?...

Hi, > Say we have a large-sized vector of observation A and a time vector with > the same size. The latter vector basically is the time flag when ith > obervation of A occurs, hence A[i] occurs at T[i]. > > I have no trouble with doing: T[A <= 10 | A >=20] for example > > Now, what I am confused: I tried to pull out when the largest value of A > with: > > T[A == max(A, na.rm=TRUE)] > > this statement comes up with a...

...,7,7,16,4,5,12,18,4,5,12,18,5,12,18,12,18,18,15,16,15,16) id<-data.frame(cbind(id1,id2)) where a pair means that the records belong to the same individual (e.g., record 4 and record 8; 17 and 18...). My problem now is to get a list with all records that belong to the same person (in the example, obervations 1,3,4,5,6,7,8,12, 17 and 18 are all from the same person). The problem is to find the link between 1 and 8 (only through 1 and 4 and 4 and 8) and the link between 1 and 17 (through 18). I can do it in my head, but I am missing the code that would work its way through too many records. Any...

...,580,795,855, 1366,1577,2060,2412,2486,2796,2802,2934,2988), Dth=c(rep(1,43), 0,0, rep(1,39), rep(0,6))) CoxG0 <- coxph(Surv(time,Dth) ~ Gp, Gastric) srvGastA <- survfit(Surv(Gastric$time,Gastric$Dth)~1) ## 88 distinct times #Gastric$time is all the time points either death or largest obervation time. #srvGastA$time is all the unique times newGas <- data.frame(start=0, stop=1, Dth=1, Ploidy=1, tim=0) #newGas <- r(0,1,1,1,0) for (i in 2:90) { timind <- match(Gastric$time[i],srvGastA$time) tmpmat <- array(0, dim=c(timind,5))#build an array with dim('index',5)...

...1 2 10/02/2000 3 c 1 2 13/02/2000 4 n 2 1 11/02/2000 5 n 2 1 15/02/2000 6 n 2 1 23/02/2000 I want to select the observations whose difference of "var4" is less than 7 for the cases with the same values of var1,var2 andvar3. The obervations have the same var1, var2 and var3 are, part1 (obs2 and obs3) and part2 (obs4,obs5, and obs6). For obs2 and obs3, their date difference is less than 7, so we donot need to delete any of them. For obs4,obs5, and obs6,we can see that obs6 should be deleted becuase its date is over 7 dyas lon...

...? ? ? ? ? ? ? ? ? 7.8 >3 ? ? ? ? ? ? ? ? 0 ? ? ? ? ? ? ? ? ? ? ? ?1.6 >3 ? ? ? ? ? ? ? ? 1 ? ? ? ? ? ? ? ? ? ? ? ? 1.2 >3 ? ? ? ? ? ? ? ?2 ? ? ? ? ? ? ? ? ? ? ? ? 1.2 >3 ? ? ? ? ? ? ? ?4 ? ? ? ? ? ? ? ? ? ? ? ?1.5 > >How can i get the result like this? In reality, i have 4000 obervations, so it's difficult to add the lines manuelly. > >Thank you so much.

Dear List members, I would like to test whether an observed occupancy of lakes in a landscape has occurred randomly (by chance) or not. How can I do that? The problem is that it concerns only a single species and I would like to use binary data only. At first I thought of generating null models and test the observed occupancy against the randomly generated one. However, this needs more than one

Dear R-users I have a dataset of 243 lines with replicate information for 20 different individuals (ID). I would like to randomly sample this dataset 100 times with a selection of unique IDs in each sample. First to create a random sample I have; cc<-read.table(blah.blah.blah) names(cc) [1] "CALL" "CONTEXT" "ORDER" "ID"

I was reading on the wiki that Asterisk runs very solid on NetBSD. Can anyone comment? What is the definition of solid? Who is running Asterisk on NetBSD and which version of Asterisk are you running? Also, I know there is limited support for Digium cards on NetBSD, but is there any support at all? Would a TE410P work in NetBSD? I want to build a very simple VoIP to TDM gateway. My idea

I am experiencing a change in behaviour of my Queues in 1.4.9 vs 1.4.8. I do not pass the 'n' option to any call to Queue() in my dialplan. Yet since I upgraded to 1.4.9, I have occasionally seen this on my console: -- Nobody picked up in 20000 ms -- Exiting on time-out cycle That log message "Exiting on time-out cycle" is exclusive to the logic in app_queue meant to

Just wondering why Event.stopObserving() doesn''t remove the itself from the Event.obervers array? Is there a reason for this? --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Ruby on Rails: Spinoffs" group. To post to this group, send email to rubyonrails-spinoffs-/JYPxA39Uh5TLH3MbocFFw@publi...

Dear collegues, I´ ve worked with near infrared (NIR) spectroscopy to assess chemical, physical, mechanical and anatomical properties of wood. I use "The Unscrambler" software to correlate the matrix of dependent variables (Y) with the matrix of spectral data (X) and I would like to migrate to R. The matrix of spectral variables is very large (2345 columns and n lines, where n =

Dear R-users, Following is part of my data, where slide has 36 levels and block 48 levels. I have done boxplot for each slide on the same graph. There are outliers for each slide and I tried to use indentify functtion to identify outliers in such a way that when I click on an outlier or point, the points will be labelled by either their block or ID or by both but without success. How can I make

Hi, I''m a newbie in both Ruby and Ruby on Rails. I finished the Ruby on Rail Tutorial on Todo list and it was great. But, now I''m trying to understand the logic behind it, that is something that is not obvious on how each one of them ties together. Is there a flowchart that shows how the inner working of Ruby on Rails? In another words, let''s suppose we consider the

...000? > 0.40625 > > ? ? ? ? [,19] [,20] > [1,] 48.90625? 47.5 > [2,]? 0.00000???0.0 > > The problem is: I want to group the data. I want to have > ten groups. The > first two bars should be [1,1] and [2,1] together in one > bar and in the > second bar of the first obervation should be [1,2] and > [2,2] (stacked with > beside =TRUE). Therefore the first observation is > [1,1],[1,2],[2,1] and > [2,2].? ? > > For a better understanding: I want to split the darkblue > bar into a green > and a blue bar (please see picture) per observation. &gt...