similar to: lognorm

Full_Name: Anthony Gichangi Version: 1.90 OS: Windows XP Pro Submission from: (NULL) (130.225.131.206) The function rlnorm generates negative values for lognormal distribution. x- rlnorm(1000, meanlog = 0.6931472, sdlog = 1) Regards Anthony

I am trying to fit a lognormal distribution to a set of data and test its goodness of fit with regard to predicted values. I managed to get so far: > y <- c(2,6,2,3,6,7,6,10,11,6,12,9,15,11,15,8,9,12,6,5) > library(MASS) > fitdistr(y,"lognormal",start=list(meanlog=0.1,sdlog=0.1)) meanlog sdlog 1.94810515 0.57091032 (0.12765945) (0.09034437) But I would

Hello, I tried to fit a lognormal distribution by using optim. But sadly the output seems to be incorrect. Who can tell me where the "bug" is? test = rlnorm(100,5,3) logL = function(parm, x,...) -sum(log(dlnorm(x,parm,...))) start = list(meanlog=5, sdlog=3) optim(start,logL,x=test)$par Carsten. [[alternative HTML version deleted]]

Dear all, Is there a way to generate K numbers of integer (K = 10^6). The maximum value of the integer is 200,000 and minimum is 1. And the occurrences of this integer follows a lognormal distribution. - Gundala Viswanath Jakarta - Indonesia

Hi list, Does anyone know the code behind rlnorm(n, meanlog = 0, sdlog = 1)? I am going to write it in c#. thanks Alireza [[alternative HTML version deleted]]

rlnorm takes two 'shaping' parameters: meanlog and sdlog. meanlog would appear from the documentation to be the log of the mean. eg if the desired mean is 1 then meanlog=0. So to generate random values that fit a lognormal distribution I would do this: rlnorm(N , meanlog = log(mean) , sdlog = log(sd)) But when I check the mean I don't get it when sdlog>0. Interestingly I

Dear R-helpers, I was testing the truncgof CRAN package, found something that looked like a bug, and did my job: contacted the maintainer. But he did not reply, so I am resending my query here. I installed package truncgof and run the example for function ad.test. I got the following output: set.seed(123) treshold <- 10 xc <- rlnorm(100, 2, 2) # complete sample xt <- xc[xc >=

Hi R lovers! I'd like to know how to use the parameter 'start' in the function fitdistr() obviously I have to provide the initial value of the parameter to optimize except in the case of a certain set of given distribution Indeed according to the help file for fitdistr " For the following named distributions, reasonable starting values will be computed if `start'

Hola Ruben, Sí precisamente es lo que comentas, en matemáticas no se suele llamar bucketización (este término se emplea más en informática) sino datos agrupados. Pero la idea es la que tu mismo dices. Respecto a las gráficas que has puesto, me han aclarado mucho sobre el tema, gracias. Si realizo lo mismo, por ejemplo con nbucket=1000 sigo obteniendo un p-valor de 1. Es decir, que casi le

Hello, I have a question regarding the use of an offset term with survreg(), in the Survival library. In particular, I am trying to figure out on what scale the offset term should be. Here's a simple example with no censoring and no coefficients: --------- y = rlnorm(1000, meanlog = 10, sdlog = 2) delta = rep(1, 1000) int = rep(1, 1000) survreg(Surv(y,delta)~offset(10*int), dist =

Dear all, I have a code that generates data vectors within R. For example assume: z <- rlnorm(1000, meanlog = 0, sdlog = 1) Every time a vector has been generated I would like to export it into a csv file. So my idea is something as follows: for (i in 1:100) { z <- rlnorm(1000, meanlog = 0, sdlog = 1) write.csv(z, "c:/z_i.csv") Where "z_i.csv" is a filename that is

Dear All I am trying to replicate a numerical application (not computed on R) from an article. Using, ks.test() I computed the exact D value shown in the article but the p-values I obtain are quite different from the one shown in the article. The tests are performed on a sample of 37 values (please see "[0] DATA" below) for truncated Exponential, Pareto and truncated LogNormal

I am trying to plot a lognormal density curve on top of an existing histogram. Can anybody suggest a simple way to do this? Even if someone could just explain how to plot a regular normal density curve on top of an existing histogram, it would be a big help. Also, is there some way to search through the R-help archives other than simple browsing? Thank you so much. Your help and time is greatly

Dear all I am struggling to fit data which form something like CDF by lognorm. Here are my data: proc <- c(0.9, 0.84, 0.5, 0.16, 0.1) size <- c(0.144, 0.172, 0.272, 0.481, 0.583) plot(size, proc, xlim=c(0,1), ylim=c(0,1)) fit<-nls(proc~SSfpl(size, 1, 0, xmid, scal), start=list(xmid=0.2, scal=.1)) lines(seq(0,1,.01), predict(fit, newdata=data.frame(sito=seq(0,1,.01))), col=2) I tried

* fitdistr? * it seems unusual (to me) to fit directly to the data with lognormal... fitting a normal to the log of the data seems more in keeping with the assumptions associated with that distribution. -- Sent from my phone. Please excuse my brevity. On July 10, 2017 7:27:47 AM PDT, PIKAL Petr <petr.pikal at precheza.cz> wrote: >Dear all > >I am struggling to fit data which form

Dear R Expert   allow me to ask a quick qestion: I have a mean value of 6 and a SD of 3 describing my distribution. I would like to "convert" this distribution into a log normal distribution that would best describe it when resimulated using log normal distribution. Currently I am using another software to estimate the respective mean and SD on the log scale and the results are: 1.6667

How about proc <- c(0.9, 0.84, 0.5, 0.16, 0.1) size <- c(0.144, 0.172, 0.272, 0.481, 0.583) plot(size, proc, xlim=c(0,1), ylim=c(0,1)) fit<-nls(proc~plnorm(size, log(xmid), sdlog, lower=FALSE), start=list(xmid=0.2, sdlog=.1)) summary(fit) lines(fitted(fit)~size) -pd > On 10 Jul 2017, at 16:27 , PIKAL Petr <petr.pikal at precheza.cz> wrote: > > Dear all > > I am

Hi R lovers Here is a numerical vector test > test [1] 206 53 124 112 92 77 118 75 48 176 90 74 107 126 99 84 114 147 99 114 99 84 99 99 99 99 99 104 1 159 100 53 [33] 132 82 85 106 136 99 110 82 99 99 89 107 99 68 130 99 99 110 99 95 153 93 136 51 103 95 99 72 99 50 110 37 [65] 102 104 92 90 94 99 76 81 109 91 98 96 104 104 93 99 125 89

Hi Great. I did not think that such combination is posssible. Thanks. Petr > -----Original Message----- > From: peter dalgaard [mailto:pdalgd at gmail.com] > Sent: Tuesday, July 11, 2017 1:11 AM > To: PIKAL Petr <petr.pikal at precheza.cz> > Cc: r-help at r-project.org > Subject: Re: [R] fit lognorm to cdf data > > How about > > proc <- c(0.9, 0.84, 0.5,

Hi R users, I am a new user, still learning basics of R. Is there anyway to extract y (or x) value for a known x (or y) value from ecdf (empirical cumulative distribution function) curve? Thanks in advance. Mano. [[alternative HTML version deleted]]