search for: spl

What should have been a routine upgrade to ZFS 0.6.4.1-1.el6.x86_64 somehow failed, and now I've lost access to a 2T raidz2 array. I've tried removing all zfs and related packages (spl*), then starting fresh. For reasons that have have nothing to do with this problem (I don't think), I'm still booted to kernel 2.6.32-431.29.2.el6.x86_64, and there are no other kernels installed. The %post script in the spl-dkms package appears to have a serious problem. Complete captures...

Dear all, After struggling for some time with *apply() and eva() without success, I decided to ask for help. I have 3 lists labeled with, each contains 3 different interpolation functions with identical names: > names(missgp0) [1] "spl.1mb" "spl.2mb" "spl.5mb" > > names(missgp1) [1] "spl.1mb" "spl.2mb" "spl.5mb" > > names(missgp2) [1] "spl.1mb" "spl.2mb" "spl.5mb" > ( In case it matters the functions accept and return one argum...

...1 Jun 2015 04:46, "Robert G. (Doc) Savage" <dsavage at peaknet.net> wrote: > > What should have been a routine upgrade to ZFS 0.6.4.1-1.el6.x86_64 > somehow failed, and now I've lost access to a 2T raidz2 array. I've > tried removing all zfs and related packages (spl*), then starting fresh. > For reasons that have have nothing to do with this problem (I don't > think), I'm still booted to kernel 2.6.32-431.29.2.el6.x86_64, and there > are no other kernels installed. > > The %post script in the spl-dkms package appears to have a serious &g...

...r=c(3,15,0,2)... I get the same margins in all of the plots. That leaves lots of white space in plots where the annotations are short. Here is the code I'm using: library(gplots) opar <- par(mar=c(3,15,0,2), bg="white", cex=1, oma = c(0, 0, 2, 0), mgp=c(3,0.5,0)) for (i in names(spl)) { .order <- order(spl[[i]]$x) barplot2(spl[[i]]$x[.order], names.arg=as.character(spl[[i]]$os[.order]), horiz=TRUE, las=1, cex.names=0.7, cex.main=0.9, cex.axis=0.7,...

I am not certain that this is the right place to post this, but maybe some kind soul can point me in the right direction if not. And I apologize if this is the wrong spot...The reason I posted here is that I recall running across a syslinux changlog that said "..added support for reading .spl..." files. I want to customize the suse boot up splash screens (for a CD install). I have been attacking it from the gfxboot and bootsplash side of things...but I believe that the [resolution].spl files they supply are one of my final links...I have searched and searched but can not seem to fi...

...sfully combined my data frames, and am now on my next hurdle. I had combined the data and quarter, and used tapply to count the entries for each unique date/quarter pair. ar= tapply(ewrgnd$gw, list(ewrgnd$dq), sum) #for each date/quarter combination sums the gw (which are all 1) dq=row.names(ar) spl=strsplit(dq) But I need to split them back into the separate date and quarter. So I used strsplit(), and get > spl [[1]] [1] "2009-01-01" "60" [[2]] [1] "2009-01-01" "61" [[3]] [1] "2009-01-01" "62" [[4]] [1] &qu...

Dear all, I'm trying to use the Pspline add-on package to fit a quintic spline (norder =3), but I keep running into a Singularity error. > traj.spl <- smooth.Pspline(time, x, norder=3 ) Error in smooth.Pspline(time, x, norder = 3) : Singularity error in solving equations > Playing around with the other parameters produce...

...ame[30:40], paste(state.name[30:40], "University", sep=" "), paste("Th University of", state.name[30:40], sep=" "),paste("University o", state.name[30:40], sep=" "))) da <- sapply(d, as.character) # factor to character transformation spl <- strsplit(da, " ") # spliting components dd <- character(dim(da)[1]) # initializing empty vector for (i in 1:dim(da)[1]) { if (sum(c("New", "Jersey", "University") %in% spl[[i]]) >= 3) dd[i] <- "The University of New Jersey"...

...model whish was denoted by a smooth function?but?unfortunately?don't manage to?do?it. I presume that such characteristics from a smooth curve can be determined by the first and second derivative operators. The following are the example of fitting a nonparametric regression model via smoothing spline function from the Help file in R. ####################################################### attach(cars) plot(speed, dist, main = "data(cars)? &? smoothing splines") cars.spl <- smooth.spline(speed, dist) lines(cars.spl, col = "blue") lines(smooth.spline(speed, dist, df...

Some wavelet analysis experts have implemented periodic boundary conditions for signals. I need to implement a circular buffer. Something like: "12345abcdefgh12345abcdefgh" so that at each step the riightmost element is moved to the leftmost index and everything else is properly shifted: "h12345abcdefgh12345abcdefg", "gh12345abcdefgh12345abcdef", .... My

Dear R helpers Suppose following is an output due to some R process. I wish to save it as a table in 'temp.db' df <- data.frame(x = c(5, 4, 3, 11), y = c(25, 16, 9, 121))     library(RSQLite)     write('** Initializing','')     drv <- dbDriver("SQLite", shared.cache = TRUE)     con <- dbConnect(drv, dbname = "temp.db",

hola soy nuevo usuario de R y necesito crear un objeto p tal que, p=lambda*integral[f´´(x)^2 dx ] donde "lambda" es uno de los parametros que resultan de la funcion "smooth.spline()" y la integral es sobre la derivada 2 de esa misma funcion... dos cuestiones: 1) como extraigo lambda de los resultados de smooth.spline() para usarlo como objeto cuando lo requiera y 2) como puedo obtener con R la integral del segundo termino? gracias -- Lic. Nahuel E. Farias Laborato...

...eader every 121 records This function works well for smallish records getcsv=function(fname){ ff=file(description = fname) x <- readLines(ff) closeAllConnections() x <- x[x != ""] # REMOVE BLANKS x=x[grep("^[-0-9]", x)] # REMOVE ALL TEXT spl=strsplit(x,',') # THIS PART IS SLOW, BUT MANAGABLE xx=t(sapply(1:length(spl),function(temp)as.vector(na.omit(as.numeric(spl[[temp]]))))) return(xx) } It's not elegant, but it works. For 121,000 records it completes in 2.3 seconds For 121,000*5 records it completes in 63 second...

...ments to plot.effect which default to row=1, col=1, nrow=1, ncol=1, more=FALSE so that I can plot multiple effect plots on one layout? I was hoping that a call like last <- rep(c(TRUE, FALSE), c(3, 1)) lay <-matrix(c(1:4,), 2, 2, byrow = T) for(i in 1:4) { tmp.lm <- lm(y ~ a + b, data) spl <- which(lay == i, arr.ind = TRUE) rowN = spl[1] colN = spl[2] plot(all.effects(tmp.lm), row = rowN, col = colN, nrow = 2, ncol = 2, more = last[i]) } would work (my data are complicated, and have not had the time to set up a replicable example). Here is what I ran, and what R told me:...

...> y <- fano(p1, x) + fano(p2, x) + fano(p3, x) > jitter(y, a=0.005*max(y)) > } > > y <- exampleFunction(x) > > sample.df <- data.frame(x = x, y = y) > > with(sample.df, plot(x, y, t="l")) # there are three peaks, located > around x=2, 4 ,5 > y.spl <- smooth.spline(x, y) # smooth the noise > lines(y.spl, col="red") > I wish to obtain the zeros of the first and second derivatives of the smoothed spline y.spl. I can use uniroot() or optim() to find one root, but I cannot find a reliable way to iterate and find the desire...

Hi, How can I extract part of an array? I would like to extract table Supported from this array. If this is not possible, how do I convert array to list? I'm sorry this is not an reproducible example. > spl <- tapply(temp$var1, list(temp$var2, temp$var3, temp$var3), mean) > spl , , Supported 07 08 A 68.38710 71.48387 B 21.67742 20.83871 C 55.74194 61.12903 AL L 26.19816 27.39631 , , Not_supported 07 08 A NA 82.38710 B...

hi all I have R object look like this: > spl $SB012XSB044 DPW Cross 1 66.6 SB012XSB044 2 96.5 SB012XSB044 3 78.8 SB012XSB044 4 68.6 SB012XSB044 5 62.0 SB012XSB044 6 72.1 SB044XSB012 7 72.2 SB044XSB012 8 69.6 SB044XSB012 9 87.9 SB044XSB012 10 84.4 SB044XSB012 11 51.9 SB044XSB012 12 65.5 SB044XSB012 $SB012XSB099 DPW...

...0 Jan 14 15:45 ./baz/glup drwxr-xr-x 3 jh jh 4096 Jan 14 15:50 ./foo drwxr-xr-x 2 jh jh 4096 Jan 14 15:31 ./foo/bar -rw-r--r-- 1 jh jh 0 Jan 14 15:44 ./foo/quuz lrwxrwxrwx 1 jh jh 3 Jan 14 15:50 ./foo/wrt -> spl rsync -aHSvn --delete foo/ baz/ building file list ... done deleting glup quuz wrt -> spl wrote 133 bytes read 24 bytes 314.00 bytes/sec total size is 3 speedup is 0.02 Here it should have said it would copy bar. rsync -aHSvn --delete baz/ foo/ building file list ... done deleting wrt dele...

...del and weighting type (1/y) as in the instrument to see if I'll get similar results. However, in some instances I don't get any results - just errors. Here is an example calibration data, representative of my experiment. Instrument soft had no problem fitting it: x <- structure(list(SPL = structure(c(1L, 1L, 2L, 2L, 3L, 3L, 4L, 4L, 5L, 5L, 6L, 6L, 7L, 7L), .Label = c("St1", "St2", "St3", "St4", "St5", "St6", "St7"), class = "factor"), MFI = c(10755.5, 9839, 5142.5, 4857, 1510.5, 1505, 502.5, 451, 215,...

...pplicable to R users). How can I fit separate regression lines for every subgroup? So far, I can only plot the overall fitted line. The code looks like this: trellis.device() sps<-trellis.par.get("superpose.symbol") sps$pch<-1:7 trellis.par.set("superpose.symbol",sps) spl <- trellis.par.get("superpose.line") ps$lty <- 1:7 trellis.par.set("superpose.line",spl) xyplot(a~b|factor+treatment, groups=external,data=ownframe,layout=c(2,2), panel=function(x,y,subscripts,...){panel.superpose(x,y,subscripts,...);panel.lmline(x,y)}) Than...