similar to: [OT]Web-Based Data Brushing

Is there a package (other than xgobi which requires an X server) that will do scatterplot brushing? I see a mention in the mail archive of R-orca by Anthony Rossini but it is not in the current list of packages. My OS is Windows XP version 5.1, service pack 2 R version 2.4.1 (2006-12-18) Thanks [[alternative HTML version deleted]]

I apologize in advance if I picked the wrong list to post this to. I have made an effort to find the answers to these questions on CRAN, but if they are there, I couldn't find them, and I was going to email the developer of StructTS directly but could not find who that is. I have 2 interrelated questions about StructTS 1. Where can I obtain the source code for StructTS if I wanted to

Hi: I am working with someone remotely to allow them access to our data. The follow command using "download.file" works perfectly on my Mac: > > download.file(url="http://oceanwatch.pfeg.noaa.gov:8081/thredds/ > wcs/satellite/AG/ssta/14day? > request=GetCoverage&version=1.0.0&service=WCS&format=NetCDF3&coverage= >

Hi All: We are developing (actually updating) a piece of code that runs in R. Due to some limitations of some the libraries we use, at the present time we would be limited to the 32-bit version of R on Windows. Does anyone have a feel (or even real knowledge) if most people these days run R on Windows in 32-bit or 64-bit mode? This would affect some development decisions. Thanks -Roy M.

From the r-help posting guidelines: > Good manners: Remember that customs differ. Some people are very direct. Others surround everything they say with hedges and apologies. Be tolerant. Rudeness is never warranted, but sometimes `read the manual' is the appropriate response. Don't waste time discussing such matters on the list. Ad hominem comments are absolutely out of place. It

Let me take an artifical matrix : dat = matrix(rnorm(200*200), 200, 200) My goal is to visualize this matrix according to the procedure, described in previous mails. I took Mendelssohn's advice and got following advice : ?plot.im Z <- setcov(owin()) plot(Z) .................... etc However I can not reproduce this example in my problem. How I can change my data

The min_height = -0.25 is there to make it show cycle values down to -1/4. You may want to change it to -1 so it shows more of the cycle values. Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, Oct 17, 2017 at 1:26 PM, Roy Mendelssohn - NOAA Federal < roy.mendelssohn at noaa.gov> wrote: > yes, thanks, and I was getting close to that. One thing I found is the > manual says the

yes, thanks, and I was getting close to that. One thing I found is the manual says the height is the distance above the y-line, which should be, but doesn't have to be positive. In fact, the time series are estimates of a cycle, and has negative values, which unfortunately are not included in my sub-sample. And the negative values are not handled properly (the series disappears for

Thanks to all that replied. I had just looked through the httr code and sure enough for a .csv mime time it calls readr::read_csv(). The httr::content docs suggest not using automatic parsing in a package, rather to determine mime type and parse yourself and Ben's suggestion also works if I do: junk <- readr::read_csv(r1$content, col_types = cols()) Perfect. Using httr rather than

Does the following work for you? ggplot2::ggplot(plotFrame, aes(x = time, y = depth, height = cycle, group = depth)) + ggridges::geom_ridgeline(fill="red", min_height=-0.25) Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, Oct 17, 2017 at 12:43 PM, Roy Mendelssohn - NOAA Federal < roy.mendelssohn at noaa.gov> wrote: > I have tried: > > ggplot(plotFrame, aes(x =

Thanks again. I am going to try the different versions. But I probably won't be able to get to it till next week. This is probably at the point where anything further should be sent to me privately. -Roy > On Jun 1, 2017, at 1:56 PM, David L Carlson <dcarlson at tamu.edu> wrote: > > On the off chance that anyone is still interested, here is the corrected function using

I have tried: ggplot(plotFrame, aes(x = time, y = cycle, height = cycle, group = depth)) + geom_ridgeline() ggplot(plotFrame, aes(x = time, y = depth, height = cycle, group = depth)) + geom_ridgeline() ggplot(plotFrame, aes(x = time, y = depth, group = depth)) + geom_density_ridges() none are producing a plot that was a ridgeline for each depth showing the time series at that depth. The plot

Ahoy! That's a message generated by the readr::read_table() function (or it's friends). You can suppress it a number of ways, but this should work as httr::content() will pass through arguments, like col_types = cols(), to the file reader. junk <- httr::content(r1, col_types = cols()) See more here... https://blog.rstudio.com/2016/08/05/readr-1-0-0/

Although it is not an elegant solution, but if your output format is HTML, you can add an arbitrary empty HTML element like <span id="foo"></span> before your code chunk. Then you can jump to this <span> via a link like "see [this code chunk](#foo)". Regards, Yihui -- https://yihui.name On Tue, Aug 29, 2017 at 1:30 PM, Roy Mendelssohn - NOAA Federal

On the off chance that anyone is still interested, here is the corrected function using aperm(): z <- array(1:120,dim=2:5) f2 <- function(a, wh) { idx <- seq_len(length(dim(a))) dims <- setdiff(idx, wh) idx <- append(idx[-1], idx[1], wh-1) aperm(apply(a, dims, rev), idx) } all.equal(f(z, 1), f2(z, 1)) # [1] TRUE all.equal(f(z, 2), f2(z, 2)) # [1] TRUE

Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, >>> junk1 <- junk[, rev(seq_len(10), ] so that junk[1,1,1 ] = junk1[1,10,1] junk[1,2,1] = junk1[1,9,1] etc. The genesis of this is the program is downloading data from a variety of sources on (time, altitude, lat, lon) coordinates, but all

Thanks David, It is working. Holtman's also gave me a solution but, I wanted to have a color pallet for description of colors, that was not in his solution. However I need one small modification. If I want to plot only lower diagonal elements of 'dat' then how should I proceed? What I want is, to visualize only lower diagonal elements and having the color pallet on them only. Also

> On 1 Jun 2017, at 22:42, Roy Mendelssohn - NOAA Federal <roy.mendelssohn at noaa.gov> wrote: > > Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, > >>>> junk1 <- junk[, rev(seq_len(10), ] > > > so that > > junk[1,1,1 ] = junk1[1,10,1] > junk[1,2,1] =

How about this: f <- function(a,wh){ ## a is the array; wh is the index to be reversed l<- lapply(dim(a),seq_len) l[[wh]]<- rev(l[[wh]]) do.call(`[`,c(list(a),l)) } ## test z <- array(1:120,dim=2:5) ## I omit the printouts f(z,2) f(z,3) Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into

Hi All: I am using httr to download files form a service, in this case a .csv file. When I use httr::content on the result, I get a message. Since this will be in a package. I want to suppress the message, but haven't figured out how to do so. The following should reproduce the result: myURL <-