search for: y10

...ts and other characters. I know that this is a vague idea, and that sorting character strings is a complex topic, but perhaps a couple of examples will clarify what I mean: > s <- c("x1b", "x1a", "x02b", "x02a", "x02", "y1a1", "y10a2", + "y10a10", "y10a1", "y2", "var10a2", "var2", "y10") > sort(s) [1] "var10a2" "var2" "x02" "x02a" "x02b" "x1a" [7] "x1b" "y10&...

...erits(x, "data.frame") : Argument "xdata" is missing, with no default" Please note that the same is happening when I create a matrix from the data.frame with the function cbind (i.e. names(data) [1] "shortrate" "y1" "y5" "y10" "y15" "y20" [7] "y25" xdata_cbind(data$y10,data$shortrate) is.matrix(x.data) TRUE) Thanks in advance, Chiara

Hi, I'm a novice R user, much more used to SAS. My problem is pretty simple - basically, in a data frame, I have variables named x1,....,x10 and y1,...,y10; and I would like to create r1 = x1 / y1 etc Apologies if this is way too rudimentary - but I couldn't find any posts online which solve this exact issue. Cheers, Andre ********************************************************************** This email and any attachments are confidential, prot...

...,0.93,1.01,0.86) y2=c(0.05,0.16,0,0.26,0.15,0.61,0.33,0.31,0.33) y3=c(-0.23,0,0.16,-0.03,0.22,-0.03,-0.17,-0.05) y4=c(0.04,-0.21,-0.04,-0.26,-0.03,-0.04,-0.05) y5=c(-0.02,-0.34,0.06,-0.22,-0.11,-0.31) y6=c(0.20,0.01,0.01,-0.26,0.01) y7=c(-0.06,-0.14,0.39,0.23) y8=c(0.21,0.10,0.09) y9=c(-0.24,-0.23) y10=c(0.13) y=c(y1,y2,y3,y4,y5,y6,y7,y8,y9,y10) ## autoregressive paramters ,table 1 biometrika 1999,page 685 om2=matrix(0,nrow=55,ncol=3) om1=poly(c(1:10),degree=3) ##polynomial design matirx with level =11,cubic om2[1:10,]=t(matrix(rep(om1[1,],10),ncol=10)) om2[11:19,]=t(matrix(rep(om1...

...t graphic parameter, but I tried plot a graph with 10 intervals in x axis with par(xaxp = c(0, 150, 10) but to no avail: par(xaxp =c (0, 150, 10)) plot(age, y18, type="n", ylab="Height (m)", xlab=" age (Yrs)", font = 2) lines(age, y6, lty = 4, lwd = 2) lnes(age, y10, lty = 1, lwd =2) lines(age, y14, lty = 2, lwd =2) lines(age, y18, lty = 3, lwd = 2) abline(v=50) The plot shown only 3 intervals with tick at 0, 50, 100, and 150. and > par()$xaxp [1] 0 150 3 I use current version Lattice and Grid. > version _ platform i38...

...uot;A5" "A6" "A7" "A8" "A9" "A10" Now assign each element as name for a data frame: (d <- data.frame(cbind(X1=1,Y1=1:10,X2=1,Y2=1:10,X3=1,Y3=1:10,X4=1,Y4=1:10,X5=1,Y5 =1:10, X6=1,Y6=1:10,X7=1,Y7=1:10,X8=1,Y8=1:10,X9=1,Y9=1:10,X10=1,Y10=1:10))) x <- paste("X",1,sep="") y <- paste("Y",1,sep="") for (i in 2:10) {x[i] <- paste("X",i,sep="") y[i] <- paste("Y",i,sep="")} for (k in 1:10) {assign(char[k],d[,c(x[k],y[k])])}...

...110 X111 X112 X113 X114 1 2 1 2 1 2 1 2 1 2 1 2 1 2 X115 X116 X21 X22 X23 X24 X25 X26 X27 X28 X29 X210 X211 X212 1 2 0 1 0 1 1 1 0 1 0 1 0 1 X213 X214 X215 X216 Y1 Y2 Y3 Y4 Y5 Y6 Y7 Y8 Y9 Y10 1 1 0 1 1 2 3 4 5 6 7 8 1 2 Y11 Y12 Y13 Y14 Y15 Y16 3 4 5 6 7 8 So, e.g. for a 3 column matrix: > matrix(do.call(c,imp), ncol=3) [,1] [,2] [,3] [1,] 1 0 1 [2,] 2 1 2 [3,] 1 0 3 [4,] 2 1...

Hello everyone, I want to transform a data.frame into an array (lets call it mydata), where: mydata[[1]] is the first imputed dataset...and for each mydata[[d]], the first p columns are covariates X, and the last one is the outcome Y. Lets assume a simple data.frame: Imputed = data.frame( X1 = c(1,2,1,2,1,2,1,2, 1,2,1,2,1,2,1,2), X2 =

...se the method ?confint? will return a wrong evaluation of the confidence interval without throwing any warning. I suggest to use the fixed values that are already contained within the mle object: > fit0 at fullcoef[grep('y', names(fit0 at fullcoef))] y1 y2 y3 y4 y5 y6 y7 y8 y9 y10 y11 26 17 13 12 20 5 9 8 5 4 8 Additionally, the method ?coef? returns both the coefficient and the fixed parameters - while should return only the parameters - , while the ?fullcoef? method is not implemented - while present in the documentation of 'mle-class?. Best, Stefa...

...110 X111 X112 X113 X114 1 2 1 2 1 2 1 2 1 2 1 2 1 2 X115 X116 X21 X22 X23 X24 X25 X26 X27 X28 X29 X210 X211 X212 1 2 0 1 0 1 1 1 0 1 0 1 0 1 X213 X214 X215 X216 Y1 Y2 Y3 Y4 Y5 Y6 Y7 Y8 Y9 Y10 1 1 0 1 1 2 3 4 5 6 7 8 1 2 Y11 Y12 Y13 Y14 Y15 Y16 3 4 5 6 7 8 So, e.g. for a 3 column matrix: > matrix(do.call(c,imp), ncol=3) [,1] [,2] [,3] [1,] 1 0 1 [2,] 2 1 2 [3,] 1 0 3 [4,] 2 1...

...h stands for gene expression divergence and 14 matrices which stands for gene sequence divergence. I have tried joining them by using the concatanation function giving SequenceDivergence <- c(X1,X2,X3,X4,X5,X6,X7,X8,X9,X10,X11,X12,X13,X14) ExpressionDivergence <- c(Y1,Y2,Y3,Y4,Y5,Y6,Y7,Y8,Y9,Y10,Y11,Y12,Y13,Y14) where X1,X2..X14 are the expression matrices containing r-values and Y1,Y2..Y14 are the ones with patristic distances Now, I want to plot SequenceDivergence vs. Expression Divergence Tried doing that using plot (Sequence Divergence vs. Expression Divergence) But then getting...

...gt; with 10 intervals in x axis with par(xaxp = c(0, 150, 10) > but to no avail: > > par(xaxp =c (0, 150, 10)) > plot(age, y18, type="n", ylab="Height (m)", xlab=" age > (Yrs)", font > = 2) > lines(age, y6, lty = 4, lwd = 2) > lnes(age, y10, lty = 1, lwd =2) > lines(age, y14, lty = 2, lwd =2) > lines(age, y18, lty = 3, lwd = 2) > abline(v=50) > > The plot shown only 3 intervals with tick at 0, 50, 100, and 150. and > > > par()$xaxp > [1] 0 150 3 > > I use current version Lattice and Grid. &...

...h stands for gene expression divergence and 14 matrices which stands for gene sequence divergence. I have tried joining them by using the concatanation function giving SequenceDivergence <- c(X1,X2,X3,X4,X5,X6,X7,X8,X9,X10,X11,X12,X13,X14) ExpressionDivergence <- c(Y1,Y2,Y3,Y4,Y5,Y6,Y7,Y8,Y9,Y10,Y11,Y12,Y13,Y14) where X1,X2..X14 are the expression matrices containing r-values and Y1,Y2..Y14 are the ones with patristic distances Now, I want to plot SequenceDivergence vs. Expression Divergence Tried doing that using plot (Sequence Divergence vs. Expression Divergence) But then getting...

...am { special_use = \Junk } mailbox Sent { special_use = \Sent } mailbox "Sent Messages" { special_use = \Sent } mailbox Trash { special_use = \Trash } prefix = } I monitor IMAP traffic using the MUA trojita. When I select the mailbox Drafts: >>> y10 SELECT Drafts (QRESYNC (1381746443 1)) <<< * FLAGS (\Answered \Flagged \Deleted \Seen \Draft) <<< * OK [PERMANENTFLAGS (\Answered \Flagged \Deleted \Seen \Draft \*)] Flags permitted. Flag is \Draft instead of \Drafts? When I select Sent >>> y12 SELECT Sent (QRESYNC (...

...7.5?22.9?0.05 Y4?5.9?16.2?24.6?12.7?0.2?24.3?3?8.0?26.8?0.04 Y5?7.2?20.4?40.6?18.2?0.8?28.2?5?9.2?44.3?0.02 Y6?5.9?18.6?37.4?14.5?0.3?36.9?8?9.5?32.9?0.10 Y7?8.0?16.1?88.6?24.1?0.1?34.6?2?8.7?11.1?0.02 Y8?13.6?21.1?56.3?19.0?0.7?33.3?6?9..5?10.8?0.06? Y9?11.2?20.4?40.7?12.9?1.1?40.3?3?12.2?6.5?0.04 Y10?7.6?18.3?27.5?8.1?2.3?41.9?2?5.9?2.9?1.00 Y11?8.8?22.2?33.3?8.8?0.6?44.4?4?6.6?55.5?0.09 Y12?9.4?16.5?35.6?16.2?0.7?50.2?5?8..8?31.4?0.07 ? (These files have been given as attachments as well) The data file has 10 variables in all, divided into?two groups: ?G1 consisting of the first 4 variables a...

Can somebody help me with the following issue (SEM in R), please:   When I run the model (includes second order models) in R, it gives me the following:   1)       In sem.default(ram = ram, S = S, N = N, param.names = pars, var.names = vars,  :   Could not compute QR decomposition of Hessian. Optimization probably did not converge.   2)       I have aliased parameters and NaNS   or sometimes when

DeaR useRs: I would like to assign a values in an object using a loop 'for'. This is a reduce example of my problem, my real problem is a few complicated: for (j in 1:10) { x.j<-rnorm(100) } I want to create 10 objects as "x.1, x.2, ... , x.9, x.10" with values in it. I used the "assign" function but nothing happens. Thank you very much

Dear all, I'm looking for primary help at aggregating table() results and at writing a loop (if useful) My dataset ( http://goo.gl/gEPKW ) is composed of 23k rows, each one representing a point in the space of which we know the land cover over 10 years (column y01 to y10). I need to analyse it with a temporal sliding window of 5 years (y01 to y05, y02 to y06 and so forth) For each period I'm looking for specific sequences (e.g., Maize, -noMaize, -noMaize, -noMaize, -noMaize) to calculate the "return time" of principal land covers: barley (2BC), colza...

Hi All, I am making a serious effort to try to learn R, but one hurdle I am facing is that I need to "see" the data as I walk through the examples in the packages. For instance, many examples on the web start by a command like data("wines"). How can I actually view what the dataset looks like prior to transformations and analysis? I have tried to use edit() , print, and

..."y4",610,45,25,0,16,26,28,0,40,2,8 > "y5",100,10,15,25,5,1,8,30,25,15,6 > "y6",150,3,3,0,11,9,6,0,0,20,10 > "y7",340,12,8,6,26,19,0,0,0,20,25 > "y8",300,5,3,0,16,20,11,0,0,10,8 > "y9",380,8,11,5,10,25,9,2,40,30,10 > "y10",40,5,0,0,2,1,2,0,0,0,0 > "y11",230,12,9,0,11,13,4,0,0,0,10 > "y12",180,0,0,0,0,14,0,6,210,4,2 > > --------------060007010402070602040701-- > > > --__--__-- > > Message: 4 > Date: Sat, 7 Dec 2002 16:01:04 -0500 (EST) > From: Naoki Takebayash...