similar to: Regexp subexpression

I was wondering if someone can help me figure out the following: I have two patient datasets, ds1 and ds2. ds1 has fields "patid", "date", and "lab1". ds2 has "patid", "date", and "lab2". I want to find all the patids that have at least 2 dated records for each lab. I started by splitting each dataset by patid, to create ds1.list

Dear R-help, Can anyone explain the meaning of the warning, Singular precision matrix in level -1, block 1 ? Or how to track down where it comes from? More precisely, using the nlme package, I'm issued with the warning itt2 <- lme(lrna~rx.nrti+lbrna, random=~1|patid, cor=corExp(form=~days|patid,nugget=T), weights=varPower( form=~lbrna),data=rna3) Warning messages: 1: Singular

Dear list member, I have the following data example ke <- data.frame(patid=c(1,1,1,2,3,3),a=c(1,2,2,1,1,2)) I want to add another variable b, such that the max of 'a' by id is returned i.e data ke becomes ke <- data.frame(patid=c(1,1,1,2,3,3),a=c(1,2,2,1,1,2),b=c(2,2,2,1,2,2)) Any help will be appreciated. Oluwakemi [[alternative HTML version deleted]]

Dear R-users! I am using R 1.7.0, under Windows XP. Having some hospital discharge data (admission date and discharge date for each patient), I want to get the number of patients in the hospital on a given date. My data look like (simple example): > x <- data.frame(patid=c("pat1", "pat2"), adm.date = c("15.03.2002", "16.03.2002"),

I have a dataframe that contains fields such as patid, labdate, labvalue. The same patid may show up in multiple rows because of lab measurements on multiple days. Is there a simple way to obtain just the first and last record for each patient, or do I need to write some code that performs that. Thanks, Steven

Does anyone have a description of the S help format? -k =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- r-devel mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-devel-request@stat.math.ethz.ch

Hello, I'm new to R but wan't to use it to compute the statistics of my medical study. The study includes several parameters for a number of patients. Each parameter was assessed by a number of readers, once with a special condition, once without. Now I have a data.frame with colums like: PatientID, ReaderID, SpecialCond(yes/no), Parameter1, Parameter2..... the rows are not sorted, and

I've got a big data.frame from which I need to extract data based on a variable number of Id's (InfCode). Until now I've been using the following dull solution as I never needed to search for more than 5 codes. Now my needs have increased faster than my R skills did and I need to call my function with about 25 values for x. There has to be a *apply or even simpler solution which

Hello, Newbie question: how do you capture groups in a regexp in R? Let's say I have txt="blah blah start=20080101 end=20090224". I'd like to get the two dates start and end. In Perl, one would say: my ($start,$end) = ($txt =~ /start=(\d{8}).*end=(\d{8})/); I've tried: txt <- "blah blah start=20080101 end=20090224" m <-

Hi, I currently have these two lines in my dialplan to extract different parts out of a variable and I'd like to do it in one line instead. Does anyone know how to use regular expression subexpressions in the dialplan? Outputting a comma separated list that can be sent to ARRAY() would be nice too (tried that, didn't work -- only got the first subexpression). ;extract dialed number

Hello, I have a vector of dates and I would like to grep the year component from this vector (= all digits after the last punctuation character) dates <- c("28.7.08","28.7.2008","28/7/08", "28/7/2008", "28/07/2008", "28-07-2008", "28-07-08") the resulting vector should look like "08" "2008"

Hello, What about an `invert` argument in grep, to return elements that are *not* matching a regular expression : R> grep("pink", colors(), invert = TRUE, value = TRUE) would essentially return the same as : R> colors() [ - grep("pink", colors()) ] I'm attaching the files that I modified (against today's tarball) for that purpose. Cheers, Romain --

completion is semi-broken in today's r-devel, and the reason seems to be some regular expression changes: > sessionInfo() R version 2.6.0 Under development (unstable) (2007-05-22 r41673) i686-pc-linux-gnu locale: [...] attached base packages: [1] "stats" "graphics" "grDevices" "utils" "datasets" "methods" [7]

Hi all, Is there any existing pass that performs fast local common subexpression elimination? The reason I'm asking is because global common subexpression elimination (GVN or GVN-PRE) is too slow for my JIT purposes, but I see lots of value in local CSE. Thanks, Nicolas -------------- next part -------------- An HTML attachment was scrubbed... URL:

On Mon, 29 Jan 2007, Gil Dogon wrote: > Now the problem with this code , is that the calculation of the address > mat[i][j] which is done by the (two) getelementptr instructions > is quite expensive (involving at least two multiplications and one > addition) hence it actualy should have been moved out of the inner loop. Right. > and not twice. Anyway this is just a syptom of a

Hello. I have a problem which is quite basic for array optimization, amd I wonder whether I am missing something, but I could not find the LLVM pass that does it. Consider the following code snippet: int test() { int mat[7][7][7]; int i,j,k,sum=0; for(i=0;i<7;i++){ for(j=0;j<7;j++){ for(k=0;k<7;k++){ sum+=mat[i][j][k]^mat[i][j][k^1]; } } } return

Hi Does anyone know if there is a function similar to as.numeric that will extract a numeric prefix from a string as in the following examples? x<-c(3, "abc", 5.67, "2.4a", "6a", "6b", "2.4.a", 3, "4.2a") df.x<-data.frame(Code=x) x.str<-levels(df.x[,1]) # required function result 2.40 3.00 4.20 5.67 6.00 NA Thanks Mike White

Hello, I have build a syntax to find out if a given substring is included in a larger string that works like this: d1$V1[regexpr("some text = 9",d1$V2)>0] <- 9 and this works all right till "some text" contains standard ASCII set. However, it does not work when accents are included as the following: d1$V1[regexpr("some t?xt = 9",d1$V2)>0] <- 9 I have

I'm trying to use the regexpr function to locate the decimal in a character string. Regardless of the position of the decimal, the function returns 1. For example, > regexpr(".", "Female.Alabama") [1] 1 attr(,"match.length") [1] 1 In trying to figure out what was going on here, I tried the below command: > gsub(".", ",",

Dear R Helpers, I am running R 2.6.2 on a Windows XP machine. I am trying to use regexpr to locate full stops in strings, but, without success. Here an example:- f="a,b.c at d:" #define an arbitrary test string regexpr(',',f) #find the occurrences of ',' in f - should be one at location 2 # and this is what regexpr finds #[1] 2