similar to: Panel.abline would not show beyond a certain slope value

Hi, I have a data frame like this: > dput(df) structure(list(ID = 1:8, Type = c("gmx mdrun -ntmpi 8 -ntomp 1 -s benchPEP.tpr -nsteps 10000 -resethway", "gmx mdrun -ntmpi 8 -ntomp 1 -s benchPEP.tpr -nsteps 10000 -resethway", "gmx mdrun -ntmpi 8 -s benchPEP.tpr -nsteps 4000 -resetstep 3000", "gmx mdrun -ntmpi 8 -s benchPEP.tpr -nsteps 4000 -resetstep

I managed to solve the problem myself without using this code. thx 2011-11-24 12:26 keltezéssel, Kehl Dániel írta: > Dear Community, > > I am trying to write code for the following problem. > Lets assume we have a beta distribution. > I know one quantile, lets say, 10% of the mass lies above .8, that is > between .8 and 1. > In addition, I know that the average of this

Hi all, Here is my problem: I performed bioassays using a unique insecticide on 9 different genotypes and got their mortality depending on the dose of insecticide used. Now, I want to see wether some genotypes are different or not in their responses to insecticide. My problem is that I have up to four replicates for some genotypes, but only one for other... Due to this unbalanced design, I

Consider I have the following data: AgeRange AgeOfPerson PersonNo FriendsAtYear0 FriendsAtYear1 FriendsAtYear2 FriendsAtYear3 FriendsAtYear4 FriendsAtYear5 10 - 12 11 1 0 1 2 2 3 3 10 - 12 12 2 0 1 2 2 3 3 15 - 18 13 3 1 2 3 4 6 7 15 - 18 14 4 1 3 4 5 7 7 30 - 40 33 5 3 5 5 6 8 9 30 - 40 36 6 4 4 4 4 4 4 I want to plot the number of friends against number of years, as to show how friendships

Hello, I am trying to figure out how to make a bode plot (a.k.a. bode diagram) in ggplot2. An example of such a diagram can be found here: http://meweb.ecn.purdue.edu/~me475/ctm/freq/bode2.GIF The example above shows the gain and phase characteristic of a linear system. In my case, I would like to compare visually several systems on the same diagram. The data in the data frame is arranged

Hi, Use ?split() #dat1 is the dataset: lst1<- split(dat1,dat1$BaseProd) lst1 #$`2231` ?# BaseProd? CF OSA #1???? 2231 0.5 0.7 #2???? 2231 0.8 0.6 #3???? 2231 0.4 0.8 # #$`2232` ?# BaseProd CF OSA #4???? 2232? 1?? 2 #5???? 2232? 3?? 1 # #$`2233` ?# BaseProd? CF OSA #6???? 2233 0.9 0.5 #7???? 2233 0.7 0.5 #8???? 2233 4.0 5.0 #9???? 2233 5.0 7.0 lst1[[1]] #? BaseProd? CF OSA #1???? 2231 0.5 0.7

Hi!! Can anyone help me, i have problems to converge the following data with 5 nonlinears models that i evaluated. Firtly, i send my data (totalsinatipicos) that i just try to fit with the nonlinear models. Next, i have the following script where i called the data as totalsinatipicos. I made selfstarting each nonlinear model. ###Library library(NRAIA) ###Data d<-totalsinatipicos

Hi, I am trying to add text to the bottom of a lattice bwplot with multiple panels. I would like to add a label below each boxplot, but the labels do not come from the data. I've tried the following, code: f1 <- c(rep(c(rep("a", 3), rep("b", 3), rep("c", 3)), 2)) f2 <- c(rep("A", 9), rep("B", 9)) dv <- c(0.9, 0.8, 0.85, 0.6, 0.65,

Folks, The contour() function wants x and y to be in increasing order. I have a situation where I have a grid in x and y, and associated z values, which looks like this: x y z [1,] 0.00 20 1.000 [2,] 0.00 30 1.000 [3,] 0.00 40 1.000 [4,] 0.00 50 1.000 [5,] 0.00 60 1.000 [6,] 0.00 70 1.000 [7,] 0.00 80 0.000 [8,] 0.00 90

Hi R Users, I am going to run a multiple linear regression with around 57 independent variables. Each time I run the model with just 11 variables, the results are reasonable. With increasing the number of independent variables more than 11, the coefficients will get ?NA? in the output. Is there any limitation for the number of independent variables in multiple linear regressions in R? I attached

Hi there. I noticed that the speex.org website does not appear to give a table of what bitrates you can expect when using the SPEEX codec in CBR (quality) mode. I have created a table for all modes, giving the bitrate for two cases. a) all coded frames saved consecutively in bitstream b) all coded frames saved with byte alignment You can see there is significant padding loss at some (lower)

CRAN (and crantastic) updates this week New packages ------------ * allan (1.0) Alan Lee http://crantastic.org/packages/allan Automates Large Linear Analysis Model Fitting * andrews (1.0) Jaroslav Myslivec http://crantastic.org/packages/andrews Andrews curves for visualization of multidimensional data * anesrake (0.3) Josh Pasek http://crantastic.org/packages/anesrake This

Dear R Users, Could you please help me on the following issue? I have a real large yearly data set. For each year I have 365 flow values. Some of the flow values are not known and that’s why you will see NA written in those slots. I wanted to know, is there a way that I can estimate those values? I tried approx command but it seems least helpful for the kind of issue I am up against.

Folks, A few days ago, I had asked a question on this mailing list about making a contour plot where a function z(x,y) is evaluated on a grid of (x,y) points, and the data structure at hand is a simple table of (x,y,z) points. As usual, R has wonderful resources (and subtle complexity) in doing this, and the gurus of the list showed me the way. Here's a complete working example. One might

Dear arun and all R users, I will first of all try to simply define my issue.. I have data in the following format Year Discharge dd/mm/yyyy x .. … … … There are some NA values in the discharge which I would like to predict by using “predict command”. I cant figure out the way to write the coding for that. Could you please help me on that??? I have also ,written

I'm getting an error message when using the new version of sqldf, > library(sqldf) > str(kdv) 'data.frame': 71 obs. of 3 variables: $ dpss: num 0.117 0.144 0.164 0.166 0.165 ... $ npdp: num 0.1264 0.0325 0.0109 0.0033 0.0055 ... $ logk: num 1.12 1.29 1.41 1.41 1.42 ... > test=sqldf("select * from kdv") Error in get("fun", env = this, inherits =

Folks, I have a small dataset of counts of recoveries on defaulted loans: recoveries<-structure(c(0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1, 0, 0, 0, 0, 0, 4, 0, 1, 2, 2, 12), .Dim = c(11L, 2L), .Dimnames = list( NULL, c("pcts", "counts"))) Here is the data in columnar form: pcts counts [1,] 0.0 0 [2,] 0.1 0 [3,] 0.2 0 [4,] 0.3

Hello zfs-discuss, One of a disk started to behave strangely. Apr 11 16:07:42 thumper-9.srv sata: [ID 801593 kern.notice] NOTICE: /pci at 1,0/pci1022,7458 at 3/pci11ab,11ab at 1: Apr 11 16:07:42 thumper-9.srv port 6: device reset Apr 11 16:07:42 thumper-9.srv scsi: [ID 107833 kern.warning] WARNING: /pci at 1,0/pci1022,7458 at 3/pci11ab,11ab at 1/disk at 6,0 (sd27): Apr 11 16:07:42 thumper-9.srv

The example here puzzles me. It seems like the < operator doesn't work as expected. > l <- 0.6 > u <- seq(0.4, 0.7, 0.1) > u [1] 0.4 0.5 0.6 0.7 > mygrid <- expand.grid("l" = l, "u" = u) > mygrid l u 1 0.6 0.4 2 0.6 0.5 3 0.6 0.6 4 0.6 0.7 > mygridcollapsed <- mygrid[mygrid$l < mygrid$u, ] > mygridcollapsed l u 3 0.6 0.6 4

Hi, I'm looking the best way to fill irregular polygons with patterns, Something like the function grid.pattern do, but my case is with irregular polygons. Whit this script I can get it, but I'm looking for an "elegant" solution.. library(grid) grid.polygon(x=c(0.2, 0.8, 0.6, 0.6, 0.8, 0.2), y=c(0.2, 0.2, 0.3, 0.5, 0.7,0.7), gp=gpar(fill="grey",