similar to: problem using identify() after plot()

I can't seem to find just what I'm looking for in R help, Everitt and Hothorn HSAUR, Murrell's book, or the R graphics gallery at http://addictedtor.free.fr/graphiques/. Probably not looking efficiently, but anyway, If my data look like this: > head(data) cat startyear studentid 1 other 2001 12 2 UHS 2001 17 3 Lourdes 2001 10 4

I am trying the following lattice graphics code: > library(lattice) > attach(junk) > summary(junk) mean.barr variable value Min. : 3.50 item15 :15 Min. :0.0000 1st Qu.: 9.50 item14 :14 1st Qu.:0.0000 Median :11.00 item16 :14 Median :0.0000 Mean :11.64 item1 : 0 Mean :0.6744 3rd Qu.:16.00 item2 : 0 3rd Qu.:1.0000 Max. :20.00 item3 :

The ts method for plot() is quirky. You can use the default method: plot(as.vector(time(ttt)), as.vector(ttt), type = "p", col=ifelse(ttt<8, "black", "red")) Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, May 1, 2018 at 1:17 PM, Christopher W Ryan <cryan at binghamton.edu> wrote: > How would I color points conditional on their value in a plot

Excellent! Worked like a charm. Thanks. --Chris Ryan On Tue, May 1, 2018 at 4:33 PM, William Dunlap <wdunlap at tibco.com> wrote: > The ts method for plot() is quirky. You can use the default method: > > plot(as.vector(time(ttt)), as.vector(ttt), type = "p", col=ifelse(ttt<8, > "black", "red")) > > > Bill Dunlap > TIBCO Software

You may also want to check this out: plot(ttt, type = "p") points(ttt, col = ifelse(ttt < 8, "black", "red")) Eivind K. Dovik Bergen, NO On Tue, 1 May 2018, Christopher W Ryan wrote: > Excellent! Worked like a charm. Thanks. > > --Chris Ryan > > On Tue, May 1, 2018 at 4:33 PM, William Dunlap <wdunlap at tibco.com> wrote: > >> The

Dear R users, A have a dataframe (matrix) with two collumns (plot, and diameter (d)). I want all diameters values for different combination of plots. For example I want all d values for all posible combination, 100C2 (all d values for plot 1 with all d values in the plot 2.......with all d values from plot 1 with all d values from plot 100, ...... with all d values from plot 99 with all d values

Dear R-User, Appreciate any helps. It looks simple, but I don't have a clue. Given that I have a dataframe of tree population with three variables: sp=species , d0=initial_size grow=growth increment from initial size per year How can I calculate the future growth increment of each tree for the next 3 years. The following Rscript was written, #---------- a0 <-

Running R 2.13.1 on Windows XP. I would like to get week of the year (1-52) for each date. library(chron) dts <- dates(c("02/27/92", "02/27/92", "01/14/92","02/28/92", "02/01/92")) dts dts.chron <- as.chron(dts) dts.chron class(dts.chron) # all of these component extractions work: months(dts.chron) weekdays(dts.chron) years(dts.chron)

How would I color points conditional on their value in a plot of a time series. Something like this: ## demonstration data ttt <- ts(rpois(12, lambda = 8), start = c(2000, 1), freq = 4) ttt plot(ttt, type = "p") ## doesn't work--all points the same color plot(ttt, type = "p", col = ifelse(ttt < 8, "black", "red")) ## also doesn't work--all

Hello, It's an alternative to use SAS by function in R? I want to plot d histograms by plot.from example bellow: Thank you! plot d 1 1 16.3 2 1 25.0 3 1 57.8 4 1 17.0 5 2 10.8 13 2 96.4 17 3 76.0 18 3 32.0 19 3 11.0 20 3 11.0 24 3 106.0 25 3 12.5 21 4 19.3 22 4 12.0 26 4 15.0 27 5 99.3 32 7 11.0 36

Is there a way to automate fitting and assessing loglinear models for several nominal variables . . . something akin to step or drop1 or add1 for linear or logistic regression? Thanks. --Chris -- Christopher W. Ryan, MD SUNY Upstate Medical University Clinical Campus at Binghamton 40 Arch Street, Johnson City, NY 13790 cryanatbinghamtondotedu "If you want to build a ship, don't drum

Hello, I have several strings where I am trying to eliminate the period and everything after the period, using a regular expression. However, I am having trouble getting this to work. > x = "wa.w" > gsub(x, "\..*", "", perl=TRUE) [1] "" Warning messages: 1: '\.' is an unrecognized escape in a character string 2: unrecognized escape removed

Hi Tim This is brilliant - thank you!! I've had to tweak the basePath line a bit (I am on a Linux machine), but having done that, the code works as intended. This is a truly helpful contribution that gives me ideas about how to work it through for the missing fields, which is one of the major sticking points I kept bumping up against. Thank you so much for this. All the best Andy On

Hi R folks, I have a massive array (object name "points") in the following form [,1] [,2] [1,] 1369 22 [2,] 1370 22 [3,] 1368 23 [4,] 1369 23 [5,] 1370 23 [6,] 1371 23 (10080 rows truncated) These represent pixel coordinates of interest in a jpeg image. I need to find the distance from each point to all other points of interest. The only way I can see to do this

Hi, I'm having a problem with black boxes and black widgets appearing in Wine programs. See these two screenshots[0], [1] for examples. One program is a Windows application, but the other is Winecfg. Any clues what is wrong here? This is Wine 1.3.0 compiled on Debian squeeze. Thanks, Andrew Barr [0] http://img843.imageshack.us/img843/5373/screenshotho.png [1]

Hi all, I want to get the index numbers of all elements of a vector which match any of a long series of possible values. Say x <- c(1,2,3,4) and I want to know which values are equal to 1, 2 or 4. I could do which(x == 1 | x==2 | x==4) [1] 1 2 4 This gets really ugly though, when the list of values of interest is really long. Is there a nicer way to do this? Something akin to the MySQL

Hi R-developers In the package Parallel, the function parLapply(cl, x, f) seems to allow transmission of only one parameter (x) to the function f. Hence in order to compute f(x, y) parallelly, I had to define f(x, y) as f(x) and tried to access y within the function, whereas y was defined outside of f(x). Script: library(parallel) f <- function(x) { z <- 2 * x + .GlobalEnv$y # Try to

I have two large datasets (156K and 2.06M records). Each row has the hour that an event happened, represented by an integer from 0 to 23. R's histogram is combining some data. Here's the command I ran to get the histogram: > histinfo <- hist(crashes$hour, right=FALSE) Here's histinfo: > histinfo $breaks ?[1] ?0 ?1 ?2 ?3 ?4 ?5 ?6 ?7 ?8 ?9 10 11 12 13 14 15 16 17 18 19 20 21

suppose I have some logical vector x <- as.logical(c(0,0,0,1,0,0,1,1,0)) x How would I make the words TRUE appear on the screen in a different color from the words FALSE? Thanks. --Chris -- Christopher W. Ryan, MD SUNY Upstate Medical University Clinical Campus at Binghamton 40 Arch Street, Johnson City, NY 13790 cryanatbinghamtondotedu "If you want to build a ship, don't drum

Hi all, I am having trouble assigning a value within a vector using a variable defined within a function. I suspect this has something to do with namespaces but I cannot figure it out. Please reply directly to me if you can help. ###begin code #simple vector test<-c(4,5,6) #simple function to get a value from a vector #works just like I would expect, has no problem indexing using the local