similar to: case conversion and/or string comparison

Dear R users, Could somebody tell me why the grey color ramp is repeated in this graph, eventhough the ramp values go from 0 to 1? I must be missing something obvious, but I can not see it! z <- c(0.064329041,0.117243316,0.161565116,0.19923015,0.231642175,0.259835539,0.284571226, 0.038507288,0.094184749,0.140959431,0.180803984,0.215159105,0.245096084,0.271412845,

Full_Name: Oliver Niggemann Version: Windows 1.4 patched OS: Windows NT 4.0 Submission from: (NULL) (141.169.250.14) Hello, I am working with the persp command. I want to change to color of the facets according to their z-value. The colors seems to be transposed, i.e. the color of a facet A seems to be used for a facet B which is a (defined?) way off.... An example: nx <- 21 ny <- 21

Hello, Before I consider purchasing it, can someone comment on the general applicability of John Chambers' book titled "Programming with Data" for R programming. I am currently using R-1.3.1 with the Omega SLanguage and RSMethod packages. Many thanks, Randall -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Hi all, Appologies for the rather basic IO question but I am rather new to R... Migrated from IDL/Matlab recently. I have a rather simple Fortran control file (sigh...) that I am trying to parse and read using R. My problem is that the file's format is somewhat flexible. Imagine: --- 1> 39 1901 2> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 3> 22 24 26 28 30 32 34 36

Hi all I am analyzing a data set containing information about the behaviour of marine molluscs on a vertical wall. Since I have replicate observations on the same individuals I was thinking to use the geepack library. The data are organised in a dataframe with the following variables Date = date of sampling, Size = dimensions (mm) Activity duration of activity (min) Water = duration of

Thorsten is right. There is a direct formula for computing the Moore-Penrose inverse using the singular value composition of a matrix. This is incorporated in the following: mpinv <- function(A, eps = 1e-13) { s <- svd(A) e <- s$d e[e > eps] <- 1/e[e > eps] return(s$v %*% diag(e) %*% t(s$u)) } Hope it helps. Dietrich

Hi all, I've tried installing NUT from both the Debian packages and the 2.0.4-pre1 source but I'm unable to sort out the USB permissions when using the newhidups driver and an APC XS series UPS. In each case, I'm able to run newhidups as user root but running it as user nut fails: --- as 'nut' # /lib/nut/newhidups -u nut -a apc1 Network UPS Tools: New USB/HID UPS driver 0.28

Hi all, I analysed my data with lme and after that I spent a lot of time for mean separation of treatments (post hoc). But still I couldn’t make through it. This is my data set and R scripts I tried. replication fertilizer variety plot height 1 level1 var1 1504 52 1 level1 var3 1506 59 1 level1 var4 1509 54 1 level1 var2 1510 48 2 level1 var1 2604 47 2 level1 var4 2606 51 2 level1 var3

Surely I am confused about something... still. Having read parts of S-programming, MASS, and Programming with Data, I am no closer to understanding how to make a simple vector containing list objects. If I have somehow skipped over a relevant section in these books, please let me know what sections I should have read. This is a simpler case of what I posted on Friday. I am trying to build

Hi all, I found the web site for R and decided to install it on my freebsd system; however, I have run into 'permission problems' during execution. I do the standard .configure, make, make check, make install and install into the default location of /usr/local/... but when a normal user attempts to run 'R' I get the following: --- [tulip ~]% R R : Copyright 2001, The R

Hi there I thought this would be of particular interest to people using 'optim' functions and perhaps people involved with R development. I've been beaten down by R trying to get it to perform an optimization on a mass-balance model. I've written the same program in excel, and using the 'solver' function, it comes up with an answer for my variables (p, ACT, which

Hi all, I am writing relatively large data files from some simulation code that runs on a variety of different platforms. I just migrated from my own (broken) file format that was not particularly platform independent to the XDR format. Is it possible to read these XDR files directly into R? Cheers, Randall -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help

Hi, there I am a graduate student new to coding in S who is hitting a bit of a wall at present using an "optim" function. I am running into some troubles, and was hoping someone might be able to recognize where I am going wrong. As background: I have constructed a loop that carries out a 365-day calculation for a mass-balance model. Basically, the model depends on 2 variables (p,

# Dear R list, # # I'm needing help with lattice, regression and respective lines. # My data is below: bra = gl(2, 24, label = c('c', 's')) em = rep(gl(3, 8, label = c('po', 'pov', 'ce')), 2) tem = rep(c(0, 0, 30, 30, 60, 60, 90, 90), 6) tem2 = tem^2 r = rep(1:2, 24) y = c(40.58, 44.85, 32.55, 35.68, 64.86, 51.95, 42.52, 52.21,

Hola, El problema que tienes es en esta sentencia: ymd_hms(TEM$TIMESTAMP) No la estás asignando a nada. Lo suyo es que hagas algo así: TEM$myTimeStamp <- ymd_hms(TEM$TIMESTAMP) Y crees el intervalo (seq...) on esta nueva variable. El error que obtienes es que "TEM$TIMESTAMP" al leerla del fichero es un string, pero en la lectura automáticamente (salvo que le digas que no lo

Un ejemplo con un modelo más simple: He especificado este modelo: >formula(m2.pile) ger ~ tem + pot + time Si hago predict me da: >predict(m2.pile,newdata=data.frame(tem=25,pot=0,time=3),type="response") 0.08243262 Extraigo los coeficientes: > coef(m2.pile) (Intercept) tem pot time -1.89521331 -0.02303313 4.74499714 0.02043222 Ahora calculo la

Dear All, I have been trying to to do this for a few days now. I can generate persp plot OK and can generate it in colour using the command persp(x,y,z,col=terrain.colors(20)) However, I was wondering if it is possible to shade the 3d surface like a contour plot. i.e. black for large z, white for small z, say Thanks in advance Colin

Estimad en s errer en s He hecho este modelo glm m1.pile<-glm(ger~tem+pot+time+I(tem^2)+I(tem^2):pot ,family="binomial" ,data=long.PILE ) Que nos da la probabilidad de germinación de una semilla en función de tem (Temperatura), pot (Humedad del suelo) y time (Tiempo que la semilla pasa en esas condiciones). Ahora quiero, para diferentes tem, pot

As you can see: time is from 1 to 1460, 1 means they the measure Tem for the globe at the first 6 hours , 2=after 12 hours .....; and so on 1460 *6 = 8760 hours which equals 1 year first of all I want to convert all data from kelvin to degree second I want to convert from 6 hourly to daily and make it like this raw data what i want to make time

Ross posted this code: persp(x, y, z, col = cm.colors(10)[floor(9 * hgt + 1)], theta = 35) I hadn't seen the cm.colors function before; it's very nice! I wanted to modify it a bit to allow the start and end colors as well as the max saturation to be specified, and I think I may have turned up a bug in the hsv() function. Here's the evidence of the bug: Look at the results of