similar to: Help with pexp( )

I wonder if something like this has been reported before: > pexp(85:86,0.438) [1] 1 -Inf --please do not edit the information below-- Version: platform = i386-pc-mingw32 arch = i386 os = mingw32 system = i386, mingw32 status = major = 1 minor = 5.0 year = 2002 month = 04 day = 29 language = R Windows NT 4.0 (build 1381) Service Pack 6 Search Path: .GlobalEnv,

Hi R-users: [R 2.2 on OSX 10.4.3] I have a (sparse) vegetation data frame with 500 rows (sampling units) and 177 columns (plant species) where the data represent % cover. I need to summarize the cover data by returning the names of the most dominant and the second most dominant species per plot. I reduced the data frame to omit cover below 5%; this is what it looks like stacked. I have

Full_Name: M Welinder Version: 1.4 OS: (src) Submission from: (NULL) (192.5.35.38) It seems to me that pexp can be improved in the lower_tail=TRUE and log_p=FALSE case by using expm1. Something like -expm1 (-x / scale); I think. -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send

Hi All, I have a function that is used with data frames having multiple id's per row and it aggregates the data down to 1 id per row. It also randomly selects one of the within-id values of a variable (mod), which often differ within-id. Assume this data frame (below) is much larger and I want to repeat this function, say 100 times, and then derive the mean values of r over those 100

Hi everybody, while performing ks.test for a standard exponential distribution on samples of dimension 2500, generated everytime as new, i had this strange behaviour: >data<-rexp(2500,0.4) >ks.test(data,"pexp",0.4) One-sample Kolmogorov-Smirnov test data: data D = 0.0147, p-value = 0.6549 alternative hypothesis: two.sided >data<-rexp(2500,0.4)

Dear R People: Hope you're having a nice day. Here is a character vector: > yz [1] "pexp(3.2,rate=1)" > str(yz) chr "pexp(3.2,rate=1)" > And I would like to evaluate that vector. I tried: > eval(as.expression(yz)) [1] "pexp(3.2,rate=1)" > But that doesn't work. Any suggestions would be most welcome. I have a feeling that it's quite

Does anyone know if R have any problems with the exponential random number generation (function rexp)? I comment it because I executed data<-sort(rexp(100)) plot(data,dexp(data)/(1-pexp(data)),type="l") and the graphic isn't constant. (Note: exponential distribution have a constant hazard failure rate). Thank you, Juan

Hello R-Users, I'm new to R so I apologize in advance for any big mistake I might be doing. I'm trying to fit a set of samples with some probabilistic curve, and I have an important question to ask; in particular I have some data, from which I calculate manually the CDF, and then I import them into R and try to fit: I have the x values (my original samples) and the y values

fitz_ra wrote > I know this is posted a lot, I've been through about 40 messages reading > how to do this so let me apologize in advance because I can't get this > operation to work unlike the many examples shown. > > I have a 2 row matrix >> temp > [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] > [,9] [,10] > [1,] 17.000

Hi, is there a way I can calculate a summary statistics for a columns matrix let say we have this matrix x <- matrix(sample(1:8000),nrow=100) colnames(x)<- paste("Col",1:ncol(x),sep="") if I used summary summary(x) i get the output for each column but I need the output to be in matrix with rownames and all the columns beside it this how I want it

Hi all, Using the exponential distribution to exemplify: The dexp function is the PDF (1) and pexp is the CDF (2), that is obtained integrating the PDF. How can I get the qexp and the rexp? Considering that I have the PDF, how this two are mathematically related to the PDF? (1) ke^{-kx} (2) 1-e^{kx} Thanks in advance.

Dear R People: I am trying to build R-2.5.0 from source on a Windows machinee. I downloaded the tools and such, and started things off. I received some odd errors messages, looked on the R Search and found a similar problem. The suggestion was to re-load "ld.exe". I downloaded and installed everything again. Same problem. I am including my output.

I have many text files in the format below and in certain rare instances such as below there can be nothing in one of the fields so a double comma is written but I won't know this because I am reading in many,many files sequentially. # TEXT FILE 2004-02-10 00:01:31.00000,,105.60000000 2004-02-10 00:01:32.00001,,105.60000000 2004-02-10 00:01:45.00000,,105.60000000 2004-02-10

dear list I use runif to generate a ramdom number between min and max runif(n, min=0, max=1) however , the syntaxe of rexp does not allow that rexp(n, rate = 1) and it generate a number with the corresponding rate. The question is: how to generate a number between min and max using rexp(). Regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000

(1) ...need to speed up a monte-carlo sampling...any suggestions about how I can get R to use all 8 cores of a mac pro would be most useful and very appreciated... (2) spent the last few hours trying to get pnmath to compile under os- x 10.5.4... using gcc version 4.2.1 (Apple Inc. build 5553) as downloaded from CRAN, xcode 3.0... ...xcode 3.1 installed over top of above after

Hi list, I have a group of data. It looks like they follow a exponential distribution. In R, how can I esimate lamda, that is the rate in pexp, of the distribution and can I use Kolmogorov-Smirnov for hypothesis testing in such a situation? I have read the "8.2 Examing the distribution of a set of data" of "An Introduction to R" but I did not find any clues on this issue.

Hi, I am new in R and I have problem with reading this data file 0 TITLE Title 0 XLEGEND Legend -1 LABEL x 1 1 12 1 2 30 1 3 34 I want to read only lines starting with 1 (it indicates 1st plotting line) and create data set from second and third value on this row. Thank for advice Jakub Zlamal -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list

Dear R-users, is there a function for the Chi-square Goodness of Fit Test in R (compare function chisq.gof in S/S-PLUS). Thanks for help. Matthias Seitzinger -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the

Dear All, Apologies if my questions are too basic for this list. I am given a set of data corresponding to list of detection times (real, non-integer numbers in general) for some events, let us say nuclear decays to fix the ideas. It is a small dataset, corresponding to about 400 nuclear decay times. I would like to test the hypothesis that these decay times are Poissonian-distributed. What is

Hi guys Given a exponential curve, is there any function on r that can generate exponential distributed random numbers? in General I want an function that can transform one probability distribution into another?? Regards ****************************************************************** Bander ************************************* [[alternative HTML version deleted]]