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Hola!! tengo un data.frame donde cada fila corresponde a un año y cada columna a un mes (De enero a diciembre) > head(valT) V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 1941 18.0 16.3 15.2 10.1 8.1 8.3 8.8 9.2 7.9 12.2 11.9 14.6 1942 17.2 15.9 13.6 11.6 8.7 6.2 6.4 7.2 9.7 12.0 14.1 16.7 1943 17.6 17.3 13.5 12.5 10.5 7.0 8.2 7.9 -999.9 -999.9

Hi all; I have a data set with the format below: Year, Day, Hour, Value 2010, 001, 0, 15.9 2010, 001, 1, 7.3 2010, 001, 2, 5.2 2010, 001, 3, 8.0 2010, 001, 4, 0.0 2010, 001, 5, 12.1 2010, 001, 6, 11.6 2010, 001, 7, 13.9 2010, 001, 8, 11.9 2010, 001, 9, 13.6 2010, 001, 10, 16.1 2010, 001, 11, 18.5 That should

Hello, I have to add "Age (bar(x)=14.3) as a title on a chart. I am unable to get this to working. I have tried bquote, substitute and expression, but they are only doing a part of the job. new<- c(14.3, 18.5, 18.1, 17.7, 18, 15.9, 19.6, 17.3, 17.8, 17.5, 15.4, 16.3, 15, 17.1, 17.1, 16.4, 15.2, 16.7, 16.7, 16.9, 14.5, 16.6, 15.8, 15.2, 16.2, 15.6, 15, 17.1, 16.7, 15.6, 15, 15.8, 16.8,

Dear R gurus: Here is the following from Montgomery's Design and Analysis of Experiments, 5th edition. > str(rout1.df) 'data.frame': 16 obs. of 3 variables: $ resp: num 18.2 18.9 12.9 14.4 27.2 24 22.4 22.5 15.9 14.5 ... $ A : Factor w/ 2 levels "-1","1": 1 1 1 1 2 2 2 2 1 1 ... $ B : Factor w/ 2 levels "-1","1": 1 1 1 1 1 1 1 1 2

Dear All: I have some questions about the output of coxph. Below is the input and output: ---------------------------------------- > coxph(formula = Surv(futime, fustat) ~ age + rx + ecog.ps, data = + ovarian, x = TRUE) Call: coxph(formula = Surv(futime, fustat) ~ age + rx + ecog.ps, data = ovarian, x = TRUE) coef exp(coef) se(coef) z p age 0.147 1.158

is there a way to use the actual index value for plotting zoo objects this is the way that the index is set up and a sample range of what I would like to plot 01/01/06 00:00:00 - 01/01/06 23:45:00 { library(zoo) # chron library(chron) fmt.chron <- function(x) { chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x)) }} x <- structure(c(15.57, 15.5,

for learning purposes and also to help someone, i used roger peng's document to get the mle's of the gamma where the gamma is defined as f(y_i) = (1/gammafunction(shape)) * (scale^shape) * (y_i^(shape-1)) * exp(-scale*y_i) ( i'm defining the scale as lambda rather than 1/lambda. various books define it differently ). i found the likelihood to be n*shape*log(scale) +

Hi, I am trying to do simple linear regression using dates in R but receiving error messages. With the data shown below, I would like to regress x on y. x y 11/12/1999 56.8 11/29/1999 17.9 01/04/2000 27.4 1/14/2000 96.8 1/31/2000 49.5 R gives the following error messages after reading the linear regression command: Error in storage.mode(y) <-

Hi, I''m running Opensolaris 2009.06, and I''m facing a serious performance loss with ZFS ! It''s a raidz1 pool, made of 4 x 1TB SATA disks : zfs_raid ONLINE 0 0 0 raidz1 ONLINE 0 0 0 c7t2d0 ONLINE 0 0 0 c7t3d0 ONLINE 0 0 0 c7t4d0 ONLINE 0 0

Hello there, I am not quite sure how to interpret the output of Rprof (in the following the output I was staring at). I was poking around the web a little bit for documentation but without much success. I guess if I want to figure out what takes so long in my code the 2nd table $by.total and the total.pct column (pct = percent) is the most helpful. What does it mean that [ or [.data.frame is

I'm not sure what gave me this idea, but I decided to comment out the swap partition in /etc/fstab and reboot my laptop. I did not run swapoff directly at any time. I'm running more things now than I would ever dream of to hammer my 500MB of memory, but I still notice NO slowdown. This is the best desktop experience I've ever had for any OS on any hardware! Here's the first 12

Hi, Below I have a function mlogl_k, later it's called with "nlm" . __BEGIN__ vsamples<- c(14.7, 18.8, 14, 15.9, 9.7, 12.8) mlogl_k <- function( k_func, x_func, theta_func, samp) { tot_mll <- 0 for (comp in 1:k_func) { curr_mll <- (- sum(dgamma(samp, shape = x_func, scale=theta_func, log = TRUE))) tot_mll <- tot_mll + curr_mll }

Dear all, I have a problem with a loop, if anyone has any knowledge on these things I would appreciate some comments. The code below is designed to allow me to extract the top record of the data frame, and them remove rows from the data frame which have an index close to the extracted top record. topstorm<-subset(rankeddataset[1,]) ## Extracts the top storm

I have a data set like this: if I want to less than 200000 obs from this data set. How can I do these? > str(p1982) 'data.frame': 465979 obs. of 6 variables: $ p : Factor w/ 1982 levels "154l_aa","1A0P_aa",..: 1 1 1 1 1 1 1 1 1 1 ... $ aa : Factor w/ 19 levels "ALA","ARG","ASN",..: 2 16 4 5 18 3 19 3 2 9 ... $ as : num 152.0

I have a data set p1982, its structure is the following Then I take 20 observations from this dataset, and assign to pr. in p1982, p has 1982 levels, in dataset pr, p should have 1 levels. But I do str(pr), it shows that p still has 1982 levels. also for these > pr$aa [1] ARG THR ASP CYS TYR ASN VAL ASN ARG ILE ASP THR THR ALA SER CYS LYS THR ALA LYS Levels: ALA ARG ASN ASP CYS GLN

Some day I may figure out how ggplot2 works. I am trying to plot 5 columns of data on a graph (similar to a simple matplot) =========================================================================== library(ggplot2) bmi <- structure(list(pct = 2:21, P10 = c(14.6, 14.5, 14.2, 13.9, 13.7, 13.7, 13.9, 14.2, 14.5, 14.8, 15.3, 15.9, 16.6, 17.2, 17.8, 18.1, 18.3, 18.4, 18.5, 18.6), P25 =

Hello, I saw that Asterisk don't calcultate fine the ANSWEREDTIME. I want that when ANSWEREDTIME =~ 5.6 become 6 and if =~10.3 become 10 because, now, if ANSEREDTIME =~ 15.9, it become 15! it isn't correct How can I have a rounded ANSWEREDTIME ? Where have I to manipulate the sources? thank you -- Francois -------------- next part -------------- A non-text attachment was scrubbed...

why is that? process in 100% last pid: 22917; load averages: 1.73, 1.89, 1.73 up 2+01:27:52 11:12:00 47 processes: 3 running, 43 sleeping, 1 zombie CPU: 15.9% user, 0.0% nice, 9.1% system, 0.0% interrupt, 75.0% idle Mem: 68M Active, 1229M Inact, 558M Wired, 2932K Cache, 416M Buf, 2063M Free Swap: 8192M Total, 8192M Free PID USERNAME THR PRI NICE SIZE RES STATE C

I have a data set like this. I want to do glm, but I get this error: Error in model.matrix.default(mt, mf, contrasts) : cannot allocate vector of length 932889958 I am wondering if my data set is too large or I did something wrong. Is there some limitation for data size for R? thanks, Aimin > p1982<- read.csv("p_1982_aa.csv") > names(p1982) [1] "p"

Hello, I''m struggling to handle the amount of memory that dom0 gets on some system. It''s a system with 8GB of RAM, running 2.6.38-6 with pv_ops and Xen 4.1 If I want to set the amount of memory for dom0, I get very bogus results. See here: http://pastebin.com/gRgY9ERN If I use the M notation (xl mem-set 0 4096M) the outcome is similarily unpredictable, and sometimes I get the