search for: day1

I have the following function defined as below match.trace <- function(dfobj, distance, day1, day2) { day1 <- substitute(dfobj$day1); day1 day2 <- substitute(dfobj$day2) distance <- substitute(dfobj$distance) xx <- NULL for (i in 0:10) xx[i+1] <- with(dfobj, cor(Lag((day1-day1[1]),i), (day2-day2[1]), use='pair')) i <- match(max(xx), xx) with(dfobj...

...order of the factors in the fitted model object, and this has me baffled. I see this dependency with the data.frame below but not with an example (table 6.4) from Montgomery's DOE book. This is with R 1.3.0 on Debian GNU-Linux. Where have I gone wrong? > centerpts run sample CH50mg 1 day1 dev126 0.56 2 day1 dev126 0.70 3 day1 dev126 0.82 4 day1 dev126 0.72 5 day2 dev126 0.57 6 day2 dev126 0.60 7 day3 dev126 0.61 8 day3 dev126 0.64 9 day3 dev126 0.68 10 day3 dev126 0.68 11 day1 dev118 0.77 12 day1 dev118 0.80 13 day1 dev118 0.86 14 day2 dev118 0.71...

...", self.id, self.child_id, self.startdate, self.enddate]) @days_inside_new_day.collect {|d| d.destroy} # Split days that have startdate <= self.startdate and enddate >= self.enddate # # before after # |==| # |---------| |---|==|--| # @day1 = Day.find(:all, :conditions => [ "id != ? and child_id = ? and startdate <= ? and enddate >= ?", self.id, self.child_id, self.startdate, self.enddate])[0] if (@day1) @day2 = @day1.clone if (@day1.startdate < self.startdate) @day1.enddate = self.star...

Hi all, I have been bumbling around with r for years now and still havent come up with a solution for plotting reliable graphs of relationships from a linear regression. Here is an example illustrating my problem 1.I do a linear regression as follows summary(lm(n.day13~n.day1+ffemale.yell+fmale.yell+fmale.chroma,data=surv)) which gives some nice sig. results Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -0.73917 0.43742 -1.690 0.093069 . n.day1 1.00460 0.05369 18.711 < 2e-16 *** ffemale.yell 0.22419 0.06...

I would like to include this in a package. The S3 methods on R CMD check says * checking S3 generic/method consistency ... WARNING window: function(x, ...) window.chron: function(data, day1, hour1, day2, hour2, ...) See section 'Generic functions and methods' of the 'Writing R Extensions' manual. I have looked and can not figure it out. This function is for convience. What can I do to fix this? This is the first time I have tried this, so be easy, and explain it to...

Hi, I have a data.frame SAMPLES with columns: Site Site# Season Day1 Day2 Day3 Day1, Day2, Day3 are class "Date", the other columns are numeric or factor. I have a date "mydate" that may or may not be listed in my data.frame and I need to find that out. If "mydate" is there, I want to get the number of the data.frame row whe...

Can someone help me with the proper usage of the reshape function? Let's say I have a dataset with columns like this (wide format): Id Sex Group Test Day1 Day 2 Day 3... And I want to transpose this into something like this (thin): Id Sex Group Test Time Where the new column labeled time contains all the time variables (Day 1, Day 2, Day 3...) that were in the wide format. I just cannot get it to work right using the reshape fun...

...lp I am closer to solving my problem but I have some small problem. So let's say I start with >data number day hour 1 17 10 2 17 11 3 17 6 4 18 4 5 18 10 6 19 8 7 19 8 I want to split to odd days, which I am able to do, I call this object frames, which looks like: > frames $`1` c1 day1 hour1 1 1 17 10 2 2 17 11 3 3 17 6 $`2` c1 day1 hour1 4 6 19 8 5 7 19 8 Now I want to make plot of the hours of both days, but not by hand. I need some sort of loop for this. How is this done? So par(mfrow=c(1,2)) for(?) plot(?hours?) thanks for the help --...

...lity.analysis\\urine\\mz.spot.sam.dat.new", header = TRUE ) pvalue = read.table("Z:\\bruce.9.3.09.sample.stability.analysis\\urine\\all.urine.features.t.test.result", header = TRUE ) library(compositions) p = function(a,b){ y = pvalue[,a] if(y<0.01){ index = which(y, arr.ind=TRUE) day1 = raw_urine[index,3:7] day2 = raw_urine[index,b] graph = {matplot(raw_urine[index,1],day1,lwd = 3) matpoints(raw_urine[index,1],day2,lwd = 3, col="red")} print(graph) }} the above is my sample code in which i'm trying to output some graphs. but however, I can get return() to return v...

...all functions Pos100415,Pos100416,Trad100415...are all data frames I have one dataframe each day. For example, Pos100415 is "Pos" for day 15/04/2010. Now I need to access data.frames in my environment for a function, according to this following scheme: Pos100415 and Pos100416 (i.e. Pos(day1) and Pos(day2)), and Trad100416 (i.e. trad(day2)). I have absolutely no idea how to deal with this indexing problem in a generic manner. Note that all numbers will belong to the list "select". Not sur to be clear enough, but any help would be appreciated. TY

...s should be fairly simple, but I can't quite figure it out and I must not be using the right words when I search for answers. I have a dataset with a number of individuals and observations for each day (7 possible codes plus missing data) So it looks something like this Individual A, B, C, D Day1 1,1,1,1 Day 2 1,3,4,2 Day3 3,,6,4 (I've also tried transposing it so that individuals are rows and days are columns) I want to summarize the total observation codes by individual so that I end up with a table something like this: ,  1, 2 ,3 ,4, 5, 6,7, missing A  2,0,1,0,0,0,0,0 B  1,0,1,0,0,...

...ay in a hospital should be illustrated with a dark colour a if there is a stay at home between 2 hopital stays, it should be illustrated with a bright colour. e.g. P1 |//////| |///| |////////| P2 |//////////| P3 |//| |////| |///| |//| Day1------------Day100---------------------------------Day365 legend: Px Patient with id x |///| Hospital stay (length of bar symbolizes length of stay) Place between 2 hospital stays symbolizes stay at home (not in hospital) Are there any package which can do that? If not, how can I...

...over the three days. In other stats packages I have used, I can just select this data and run the ANOVA function and get the F and p values. However in R, the anova function seems to only work with a fitted model, eg. Linear regression. This function seems to assume there is a relationship such as day1~ day 4 + day 8, but in my case there isn't - I just want to perform an ANOVA without regression. If anyone could point me in the right direction I'd greatly appreciate it, Thanks [[alternative HTML version deleted]]

...yntax. The use of dat and dat2 is not an error. I'm pulling data from 2 sources for the model. for (i in 1:dim(dat2)[[1]]) { assign("modelb",i) <- lm(log(dat$flux) ~ dat$Tsoil_flux, subset = dat$chamber == dat2$chamber[i] & dat$year == dat2$year[i] & dat$doy >= dat2$day1[i] & dat$doy <= dat2$day2[i]) dat2$coef[i] <- coef(assign("modelb",i, sep = ""))[[2]] dat2$Rsq[i] <- summary(assign("modelb",i, sep = ""))[[9]] } I have also tried assign("modelb",1:i) #following the ?assign example paste("modelb&...

Hi, If I have a data frame A with the following format: Day1 Day2 Day3 Day4 1 1979-11-02 1979-11-03 1979-11-04 <NA> 2 1979-12-06 <NA> <NA> <NA> 3 1979-12-13 1979-12-14 1979-12-15 1979-12-16 4 1979-12-20 <NA> <NA> <NA> And a date &q...

Dear all, I have the following data in excel: day1 y 1 2 3 2 3 x1 0.2 0.3 0.4 0.3 0.2 x2 7 3.4 2 8 6 day2 y 2 4 3 2 2 x1 0.4 0.5 0.3 0.3 0.2 x2 7 8 9.1 6 5 I have the following problems: first of all, when I ask R to read the file (with the package xlsReadWrite and the command read.xls) it has a problem with the fact that the left most corner...

I would like to write some data to different files. I can create the filename Day1.txt like this: filen <- paste("Day", l, ".txt", sep="") and then I'm using a For loop to write out one row of a matrix, something like this: For (j in 1:10) { cat(mat[1,j], ",", file=filen, append=TRUE) } cat("\n", file=filen, append=TRU...

Hi all Have recently watched Matt Jordan's session on Kamailio World 2014 On slides 26-29 of his presentation (http://www.kamailio.org/events/2014-KamailioWorld/day1/09-Matt.Jordan-Asterisk12-And-PJSIP.pdf) he speaks about a (completely new, for me at least) approach to build scalable telephony systems, using N instances of Kamailio and N instances of Asterisk Are there any whitepapers, howtos, "implementation experience reports", whatever, avail...

This patch adds support for symbol versions. It is based on what libvirt does. The generated syms file looks like: LIBNBD_1.0 { global: nbd_...; nbd_...; local: *; }; In a future stable 1.2 release, new symbols would go into a new section which would look like this: LIBNBD_1.2 { global: nbd_new_symbol; nbd_another_new_symbol; local: *; } LIBNBD_1.0; In my testing the

Hi, i did a mistake with my first post. I have to reshape data from this matrix: id x1 x2 x3 x4 day1 day2 day3 day4 day5 day6 day7 day8 day9 1 0.129 0.797 0.231 0.615 4 4 1 1 1 1 3 3 3 2 0.420 0.376 0.501 0.282 4 4 4 4 5 4 2 5 5 3 0.377 0.486 0.897 0.699 4 4 4...