search for: regularment

Hello all, I'm sorry if my question seems basic. Im studying a responses (Yes,No) in a survey and, thanks to GLM I obtain the following relation with my variables : (Yes,No)~ ?0 + Age We note this this certain type of (Yes,No) response is linked to age (p<0.05 in glm) . After that we calculated : model1=glm(cbind(Yes,No) ~ Age + Times + Type, family=binomial) summary(model1)

Dear Lenny, You can do this by using Age as an offset factor. dataset$wAge <- dataset$Age * 1.02 glm(cbind(Yes,No) ~ offset(wAge) + Times + Type, family=binomial, data = dataset) Best regards, ir. Thierry Onkelinx Statisticus / Statistician Vlaamse Overheid / Government of Flanders INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR NATURE AND FOREST Team Biometrie &

Dear Thierry, Thanks a lot for this answer, I mean i want to obtain such model *Behavior1 = ?0+?1*Age* , the purpose is to obtain *?1*. I want to be sure that the odds value could be the ?1. Or how to calculate it ? Thanks again for your precious help. Lenny <http://www.avg.com/email-signature?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail>

days=Sys.Date()-1:70 price=abs(rnorm(70)) regular=rep(c(0,0,0,0,1,0,1,0,0,1),c(7,7,7,7,7,7,7,7,7,7)) y=data.frame(cbind(days,price,regular)) y is like days price regular 1 14990 0.16149463 0 2 14989 1.69519358 0 3 14988 1.57821998 0 4 14987 0.47614311 0 5 14986 0.87016180 0 6 14985 2.55679229 0 7 14984 0.89753533 0 the output I want:

??? Hi, I'm facing a challenge with selinux and because I don't got an explanation elsewhere, I'm trying to explain here. I have decided to mount /var/spool/cron on a separate partition? and apply quota for regular users. But quotacheck replyes with a "permission denied" . quotacheck: Cannot create new quotafile /var/spool/cron/aquota.user.new: Permission denied

Hello, Could someone recommend a good book on regular expressions with focus on applications/use as it might relate to R. I remember there was a mention of such a reference book recently, but I could not locate that message on the archive. Thanks. -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830

Hi all, I am hoping that someone can help me with a problem I am having with column headings. I have read a table into R using read.table: the rows are documents, and the columns are counts of regular expression matches (so that the column heading is the given regular expression). My problem is that read.table seems to be trying to interpret the regular expressions, or has trouble with the

Hello all, I have a question. I have an asterisk box setup with an x100p card installed. I have about 3 VoIP phones connected to it. The x100p is connected to my pots line. I also have regular phones connected the the pots line as well like most normal people would. When I get an incomming call all the regular phones ring and my VoIP phones ring. This is good. My question is this. Say

On Tue, Nov 2, 2010 at 3:36 PM, Salomon Brys <salomon.brys at gmail.com> wrote: > In my case, I am working on a posix regular expression library that uses > LLVM to execute a regular expression. > Each regular expression is a very small sets of code and is faster to > interpret than to JIT then execute. > I wanted to JIT only when the regular expression is executed more than

I''m looking to write a regular expression that will match valid URLs. My problem is that it almost works, except it accepts URLs with / in the middle of them, suchs as: http://www.ruby/rails.com It looks (to me) like my regular expression should not match strings like that, but it does. Here is the regular expression:

Hi R, Again struck with regular expressions... Suppose, S=c("World_is_beautiful", "one_two_three_four","My_book") I need to extract the last but one element of the strings. So, my output should look like: Ans=c("is","three","My") gsub() can do this...but wondering how do I give the regular expression....

The interpretation of regular expressions with repetition quantifiers in the 'gregexpr' function seems to have changed between R Version 2.2.0 and 2.3.0. The 'gsub' function, however, gives the same results in R Versions 2.2.0 and 2.3.0. Below is an example that demonstrates the version differences of the 'gregexpr' function. I am not sure whether this new behavior is

Hi, I was recently misfortunate enough to have to use regular expressions to sort out some data in R. I'm working on a data file which contains taxonomical data of bacteria in hierarchical order. A sample of this file can be generated using: tax.data <- read.table(header=F, con <- textConnection(' G9SS7BA01D15EC Bacteria(100) Cyanobacteria(84) unclassified G9SS7BA01C9UIR

Dear R People: Could someone recommend a good reference on regular expressions, please? Thanks in advance, Sincerely, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodgess at gmail.com

Dear Everyone, I try to automatically manipulate the data of a variable (class = factor) like x 220 220a 221 221b B221 Into two variables (class = numeric) like x y 220 0 220 1 221 0 221 1 221 1 y has to carry the information about the class (number or string) of the former x-Variable. I could do it by hand like x[x == "220a"] <- 220

In my case, I am working on a posix regular expression library that uses LLVM to execute a regular expression. Each regular expression is a very small sets of code and is faster to interpret than to JIT then execute. I wanted to JIT only when the regular expression is executed more than 20 times. The benchmarks I have run show that my library is amzingly fast to execute a regular expression

Hi, Is there a way to use regular expressions to capture two or more words in a sentence? For example, I wish to to find all the lines that have the words "thomas", "perl", and "program", such as "thomas uses a program called perl", or "perl is a program that thomas uses", etc. I'm sure this is a very easy task, I would greatly appreciate

Dear R developpers, I am a user of Splus since many years and I have developped lots of functions to plot graph of data and model results of irregular or regular times series. In Splus regular times series are created using the rts function and irregular time series using the its functions. In both cases, times is given as a numeric vector. There is no problem in plotting a regular and an

Hi Experts, This might be silly question that I am asking, but no way as I am new to R. I want to list the files in a directory using regular expression like A_B*_C*.csv etc. How to make this possible in R ? I tried like this list.files(dir=".", pattern="A_B*_C*.csv") but this gives no output, whereas list.files(.... pattern="*.csv") giving all the .csv files in

Hello, this is my first post. I have a large CSV file where I need to fill in the 1st and 2nd column with a Loan # and Account name that would be found in a line of text : like this: ,,Loan #:,ML-113-07,Account Name:, Quilting Boutique,,,,,,,,,,, I would like to place the Loan #: ML-113-07 in the first column and the account name quilting boutique in the second column. If possible I would also