search for: time3

Greetings listeRs - Given a data frame such as times time1 time2 time3 time4 1 70.408543 48.92378 7.399605 95.93050 2 17.231940 27.48530 82.962916 10.20619 3 20.279220 10.33575 66.209290 30.71846 4 NA 53.31993 12.398237 35.65782 5 9.295965 NA 48.929201 NA 6 63.966518 42.16304 1.777342 NA one can use "transform"...

...have seen much more sophisticated examples using lattice (e.g each person has a separate panel/plot). I am open to alternative code as to how I could present this data. Time1 <- c(9.0,6.0,1.0,5.0,7.0,9.0,5.0,7.5,6.0,8.0,5.0,5.0,9.0,4.0,5.0,5.0) Time2 <- c (10,5,3,3,3,6,7,8,5,8,7,7,9,8,5,3) Time3 <- c (10,0,3,0,0,6,0,0,0,0,0,0,0,0,0,0) df <- rbind (Time1, Time2, Time3) dft <- (t(df)) dft barplot(dft, beside = TRUE, main= "Risk score by assessment", xlab = " Score", ylab = "frequency", col="blue") Any assistance is much appreciated, rega...

...ot;%Y %m %d %H %M") > time2=strptime("2009 05 31 23 07 00",format="%Y %m %d %H %M") > > diff(c(time1,time2),units="mins") Time difference of 10 mins .. I get the correct response in minutes. But if I try the same thing with different values, say.. > time3=strptime("2009 06 01 00 47 00",format="%Y %m %d %H %M") > time4=strptime("2009 06 01 00 57 00",format="%Y %m %d %H %M") > > diff(c(time3,time4)) Time difference of NA secs ...which is not what I´m looking for. The difference should also be 10 minute...

...ollowing code   Dose<-1000 Tinf <-0.5 INTERVAL <-8 TIME8 <-matrix(c((0*INTERVAL):(1*INTERVAL))) TIME7 <-matrix(c((0*INTERVAL):(2*INTERVAL))) TIME6 <-matrix(c((0*INTERVAL):(3*INTERVAL))) TIME5 <-matrix(c((0*INTERVAL):(4*INTERVAL))) TIME4 <-matrix(c((0*INTERVAL):(5*INTERVAL))) TIME3 <-matrix(c((0*INTERVAL):(6*INTERVAL))) TIME2 <-matrix(c((0*INTERVAL):(7*INTERVAL))) TIME1 <-matrix(c((0*INTERVAL):(8*INTERVAL))) CDURINF1 <-((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME1)) CAFTINF1 <-((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME1))) CONC1...

Dear list-members I am new to R and a statistics beginner. I really like the ease with which I can extract and manipulate data in R, and would like to use it primarily. I've been learning by checking analyses that have already been run in SAS. In an experiment with Y being a response variable, and group a 2-level between-subject factor, and time a 5-level within-subject factor. 2

...px>px1){ #nested if stmt to see if subsequent times are higher than base, thus replacing base with new max (I am seeking max in the time period in this instance) px2<-px #px2 would be the new higher maximum time2<-time #time associated with px2 px3<-px+1 #price immediately following max time3<-time+1 #time immed. follows max out<-(cur_date,px2,time2,px3,time3) #output the high price/time and the immediately following price/time cat(out,"\n")} This code however, does nothing. Any help would be apreciated. The manuals seem to only take one so far. Thanks, George [[a...

Probably a stupidly simple question, but I wouldn't know how to google it: xyplot(neuro ~ time | UserID, data=data_sub) creates a proper plot. However, if I add type = "l" the lines do not go first through time1, then time2, then time3 etc but in about 50% of all subjects the lines go through points seemingly random (e.g. from 1 to 4 to 2 to 5 to 3). The lines always start at time point 1, though. Defining "time" as factor or ordered doesn't change this. "neuro" is a numerical variable. It's probably...

...a slightly different mind-set behind their experiments than, say, an ecologist, who is interested in the effects of time per se, and not just the effects of a treatment. For example, here's my dataset, say, looking at plant height in cm with and without fertilizer Treatment, Time1, Time2, Time3, Time4, Time5 Fertilizer, 1, 4, 8, 10, 12 Control,1,2,3,4,5 Fertilizer,1,8,10,12,20 Control,1,3,5,6,6 Fertilizer,2,5,10,20,25 Control,1,2,4,4,4 Clearly there is a time*treatment interaction (just eyeballing the dataset) My question is, how does one set this up using the anova.mlm approach so t...

...260.48540 16.48991 15.796653 0.0000 Tra0 -9.38703 23.74893 -0.395261 0.6935 LugarChixchulub -0.15377 19.60260 -0.007845 0.9938 Lugar Hobonil -173.21354 15.89736 -10.895741 0.0000 Time2 -14.74999 14.55909 -1.013112 0.3135 Time3 14.42177 15.64594 0.921758 0.3589 Time4 14.77803 16.72367 0.883659 0.3790 Tra0:Time2 17.93859 20.78257 0.863156 0.3901 Tra0:Time3 -48.77118 22.17628 -2.199250 0.0302 Tra0:Time4 -52.63611 23.20192 -2.268610 0...

...hand with some of my data. I have a very typical analysis in biology, however it is difficult for me to find the right way to analyse. I had a group of animals, I gave them a treatment, and I measure a variable along time -one??s per day- along 5 days,for example(fake data): Animals Time1 Time2 Time3 Time4 1 1 5 3 1 2 2 7 4 1 3 5 5 3 1 4 1 3 7 2 5 2 7 7 1 Please, notice that all the animals get the same treatment, and there are no...

...he few lines drawn are not drawn on a subject-by-subject basis, (2) not all lines are drawn. I hope someone can help me fix my code Thanks, John # Create a matrix to hold our data. mydata <- matrix(nrow=10,ncol=10) dimnames(mydata) <- list(NULL,c("time1", "time2", "time3", "time4", "time5", "clot1", "clot2", "clot3", "clot4", "clot5")) # load data into the matrix. # The first five columns represent time of measurement # Columns 6-10 have the measured values....

...up + time + group:time, random=~1|subject, na.action=na.omit, data=mydata) ...for a dataset where there are two groups being followed over time. pre is before the treatment, and there are 4 post-treatment times (unequally spaced). The fitted coefficients are: (Intercept) pre group2 time2 time3 time4 group2:time2 group2:time3 group2:time4 2.6442 0.5478 1.0183 0.6085 0.2326 0.5059 -1.9177 -0.3500 -2.2029 of which there are only significant p-values (<0.05) for (Intercept), pre, group2:time2 and group2:time4. anova(fit) gives: numDF denDF F-value p-valu...

Dear All, I regularly want to "apply" some function to an array in a way that the arguments to the user function depend on the index on which the apply is working. A simple example is: A <- array( runif(160), dim=c(5,4,8) ) x <- matrix( runif(32), nrow=4, ncol=8 ) b <- runif(8) f1 <- function( A, x, b ) { sum( A %*% x ) + b } result <- rep(0.0,8) for (i in 1:8) {

I have 100+ .csv files which have the basic format: > test X Substance1 Substance2 Substance3 Substance4 Substance5 1 Time1 10 0 0 0 0 2 Time2 9 5 0 0 0 3 Time3 8 10 1 0 0 4 Time4 7 20 2 1 0 5 Time5 6 25 3 2 1 6 Time6 5 30 4 2 2 7 Time7 4 25 5...

...aron & Li, 2006), something along the lines of: aov(outcome ~ group * time + baseline + Error(subj+subj:time)) (from: http://cran.r-project.org/doc/contrib/Baron-rpsych.pdf, p. 41) C. "Repeated measures" MANOVA as follows (using data in wide format): response <- cbind(time1,time2,time3,time4,time5,time6) mlmfit <- lm(response ~ group) mlmfit1 <- lm(response ~ 1) mlmfit0 <- lm(response ~ 0) # Test time*group effect anova.mlm(mlmfit, mlmfit1, X=~1, test="Spherical") # Test overall group effect anova.mlm(mlmfit, mlmfit1, M=~1) # Test overall time effect anova.mlm(...

...an,corstr="exch",data=dat) I obtain the following error: Error in geese.fit(x, y, id, offset, soffset, w, waves, zsca, zcor, corp, : nrow(zsca) and length(y) not match Could someone tell me what I have done incorrectly. Thanks for your time, Juliet. Data Below: id treat time1 time2 time3 time4 chem1 chem2 chem3 chem4 time score chem 1 1 20 18 15 15 1000 1100 1200 1300 0 20 1000 1 1 20 18 15 15 1000 1100 1200 1300 2 18 1100 1 1 20 18 15 15 1000 1100 1200 1300 3 15 1200 1 1 20 18 15 15 1000 1100 1200 1300 6 15 1300 2 1 22 24 18 22 1000 1000 1055 950 0 22 1000 2 1 22 24 18 22 1000 100...

Hi, I need help in calculating Java method time-stamp in following fashion. Consider following method example. long method3(long stop) { try { Thread.sleep(1500); } catch (Exception e) { } //////////////////// real CPU intensive operation /////////////////////////// for (int i = 1; i < stop; i++) { stop = stop * stop * i; };

Hi, I have been struggling with this problem for some time now. Internet, books haven't been able to help me. ## I have factorial design with counts (fruits) as response variable. > str(stubb) 'data.frame': 334 obs. of 5 variables: $ id : int 6 23 24 25 26 27 28 29 31 34 ... $ infl.treat : Factor w/ 2 levels "0","1": 2 2 2 2 1 1 1 2 1 1 ... $ def.treat :

I am running a server inside of VMWare, and the clock gains ~30 seconds every 1000 seconds or 1.03X. I need to keep the drift under the magic 1000 limit that ntpd kills its self, but despite setting maxpoll really low I get: Dec 11 23:58:14 host ntpd[4909]: kernel time discipline status change 41 Dec 11 23:59:17 host ntpd[4909]: kernel time discipline status change 1 Dec 11 23:59:17 host

Dear list, [cross-posting from Stack Overflow where this question has remained unanswered for two weeks] I'd like to perform a numerical integration in one dimension, I = int_a^b f(x) dx where the integrand f: x in IR -> f(x) in IR^p is vector-valued. integrate() only allows scalar integrands, thus I would need to call it many (p=200 typically) times, which sounds suboptimal. The