similar to: how to output profile plots for groups using lattice package

I've got 4 Cisco 7960's and they're behind a firewall (sonic wall). The asterisk box is on a WAN connection on the other end of a DS3, the phones connect fine to the Asterisk server as you can see from the output of show sip peers below. tp3/tp3 <firewall-ip> D N 255.255.255.255 60665 Unmonitored tp2/tp2 <firewall-ip> D N

Hi all, I have 2 questions: 1)How do I calculate the mean on an imported txt file? I've imported the file below and that's what it looks like imported. How do I then calcuate the mean, median, or mode on the column LeafArea using the desktop R package? Any help would be greatly appreciated!! Thanks, Nat LeafType Leaflets LeafArea ShapeRatio LeafWeight LeafThickness 1 1

Hi, I have a 3d array as below, I want to make this array to a matrix of p=50(rows) and n=20(columns) with the coverage values . The code before the array is: library(binom) Loading required package: lattice pi.seq<-seq(from = 0.01, to = 0.5, by = 0.01) no.seq<-seq(from = 5, to = 100, by = 5) cp.all = binom.coverage( p = pi.seq, n = no.seq , conf.level = 0.95, method = "exact")

Hi all I have created a lattice graph but the text appears too big. Just wondering if there was a way I could change the font size. tp1.nDai=xyplot(value ~ yr | Name, data = nDai,type = 'o') update(tp1.nDai, aspect = c(0.8), layout = c(4, 4)) I would appreciate your help. Kind regards andy Andrew McFadden MVS BVSc Incursion Investigator Investigation & Diagnostic Centres -

On 12/1/2012 7:10 PM, Nick Lewycky wrote: > > I don't know of any, but I was wondering if you could point me to the > paper that describes predictive commoning? I could only find the > second-order predictive commoning paper, which if I understand correctly > is a much newer and different algorithm. I think the original paper was some internal IBM publication. The idea is

Dear Group, I have the following data matrix which is a timeseries. > dput(tData) structure(list(A = c(0.2, 0.13, 0.05, 0.1, 0.02, 0.18, 0.09, 0.06, 0.13), B = c(0.15, 0.06, 0.09, 0.02, 0.03, 0.12, 0.01, 0.15, 0.06), C = c(-0.1, 0, -0.07, -0.06, -0.05, -0.05, -0.06, -0.08, -0.07), D = c(-0.15, -0.05, -0.1, -0.03, -0.13, -0.04, -0.1, -0.04, -0.15), E = c(-0.17, -0.16, -0.08, -0.07, -0.09,

Hi There, I'm facing one problem to construct a vector using the "for" command: I have one matrix named 'dados' (same as /data/ from portuguese), for example: > dados[140:150,] [,1] [,2] [,3] [1,] 212.7298 0.14 0.11 [2,] 213.3778 0.14 0.11 [3,] 214.0257 0.15 0.11 [4,] 214.6737 0.15 0.12 [5,] 215.3217 0.15 0.12 [6,] 215.9696 0.15 0.12 [7,] 216.6176 0.16

Hi R Helpers, Is it possible to interpret the top level of the list as a date after downloading all the option chain data for a ticker? For example, after I run aapl_total<-getOptionChain("AAPL", NULL) the top descriptor of the lists is a date (Mar.09.2018, Mar.23.2018, etc.). So if want to subset down to those parts of the list that correspond to say, (expiration)

Hey Guys, I know it's only TP1 and maybe? this isn't the best place to ask but where do I start? I've download the debian experimental Samba 4 TP1 package and installed it. The debian installer did the provisioning and I've set up dns. I set up a share and it all seems to work fine with the built-in administrator account. But what now? how do I join computers to the domaind? Do

Well, here's one way to do it: (dat is your example data frame) Cutoff <- seq(0, .15, .01) Pop <- with(dat, sapply(Cutoff, \(p)sum(Totpop[Pct >= p]))) I think there must be a more efficient way to do it with cumsum(), though. Cheers, Bert On Sat, Oct 14, 2023 at 12:53?AM Jason Stout, M.D. <jason.stout at duke.edu> wrote: > > This seems like it should be simple but I

#this is a daily series of precipitation data. I would like to condense it into weekly means. How can I do this #as a side note I would like to do this same thing to two years worth of fifteen minute interval data and make it into #a series of daily averages (there are 96 readings per day) #is aggregate the right function? or... y <- c(1.23, 0, 0, 0, 0, 0, 0, 0, 0, 0.27, 0, 0.29, 0, 0, 0,

That's very helpful and instructive, thank you! Jason Stout, MD, MHS Box 102359-DUMC Durham, NC 27710 FAX 919-681-7494 ________________________________ From: John Fox <jfox at mcmaster.ca> Sent: Saturday, October 14, 2023 10:13 AM To: Jason Stout, M.D. <jason.stout at duke.edu> Cc: r-help at r-project.org <r-help at r-project.org> Subject: Re: [R] Create new data frame with

This seems like it should be simple but I can't get it to work properly. I'm starting with a data frame like this: Tract Pct Totpop 1 0.05 4000 2 0.03 3500 3 0.01 4500 4 0.12 4100 5 0.21 3900 6 0.04 4250 7 0.07 5100 8 0.09

Package? The **names** of the top levels of your lists, "Mar.09.2018", "Mar.23.2018" certainly look like dates and if they are -- I have no idea what package/context is -- they certainly could be formatted as such. See e.g. "date-time" . There are also several package that provide date tools. Cheers, Bert Bert Gunter "The trouble with having an open mind is

I'm trying to collapse the following table along the sub-group factor. In this case, collapsing means taking the average ages within a subgroup and creating a new table. I seem to be running into trouble trying to create this new data frame. I can use the ave() function to find averages within a subgroup but how do I maintain the Group/Gender factors after collapsing? (see bottom table) Can

hello, i'm pete ,how can i order rows of matrix by max to min value? I have a matrix of membership degrees, with 82 (i) rows and K coloumns, K are clusters. I need first and second largest elements of the i-th row. for example 1 0.66 0.04 0.01 0.30 2 0.02 0.89 0.09 0.00 3 0.06 0.92 0.01 0.01 4 0.07 0.71 0.21 0.01 5 0.10 0.85 0.04 0.01 6 0.91 0.04 0.02 0.02 7 0.00 0.01 0.98 0.00 8 0.02

Under the hood, sapply() is also a loop (at the interpreted level). As is lapply(), etc. -- Bert On Sun, Oct 15, 2023 at 2:34?AM Jason Stout, M.D. <jason.stout at duke.edu> wrote: > > That's very helpful and instructive, thank you! > > Jason Stout, MD, MHS > Box 102359-DUMC > Durham, NC 27710 > FAX 919-681-7494 > ________________________________ > From: John

Hi, R-users, If I have a data frame like this: >x<-data.frame(g=c("g1","g2","g1","g1","g2"),v=c(1,7,3,2,8)) g v 1 g1 1 2 g2 7 3 g1 3 4 g1 2 5 g2 8 It contains two groups, g1 and g2. Now for each group I want the max v: > aggregate(x$v,list(g=x$g),max) g x 1 g1 3 2 g2 8 Beautiful. But what if I want to keep the row index of (g1

Hi, I wonder how to pass several functions and their arguments as arguments to a function. For example, the main function is f = function(X ) { process1(X) ... process2(X) } I have a few functions that operate on X, e.g. g1(X, par1), g2(X, par2), g3(X, par3). par1, par2 and par3 are parameters and of different types. I would like to pass g1, g2, g3 and their arguments to f and g1,

Dear R users, I used to "OPTIM" to minimize the obj. function below. Even though I used the true parameter values as initial values, the results are not very good. How could I improve my results? Any suggestion will be greatly appreciated. Regards, Kathryn Lord #------------------------------------------------------------------------------------------ x = c(0.35938587,