similar to: zipfR help

Using R, I plotted a log-log plot of the frequencies in the Brown Corpus using plot(sort(file.tfl$f, decreasing=TRUE), xlab="rank", ylab="frequency", log="x,y") However, I would also like to add lines showing the curves for a Zipfian distribution and for Zipf-Mandelbrot. I have seen these in many articles that used R in creating graphs. Thank you! [[alternative HTML

Dear R-users, I am new to R and would like to use it for fitting the zipf distribution to some numeric data that I have. Here's the snippet that I use: library(VGAM) X <- read.table(file("~\\mydata.txt", encoding="latin1")) w <- as.vector(t((X[2]))) w <- w/sum(w) y <- (1:length(w)) fit = vglm (y ~ 1, zipf, tra=TRUE, weight=w) zipf(N=NULL,

Dear all, Using: library(vegan) data(BCI) mod <- radfit(BCI[1,]) mod RAD models, family poisson No. of species 93, total abundance 448 par1 par2 par3 Deviance AIC BIC Null 39.5261 315.4362 315.4362 Preemption 0.042797 21.8939 299.8041 302.3367 Lognormal 1.0687 1.0186 25.1528 305.0629 310.1281

Hi everyone. Newbie to statistics. I have 40 matrices of ~400 values. how may I determine whether the distribution follows zips law? response <-sample (1:20,400*4, replace= TRUE) Thank you vry much. -- View this message in context: http://r.789695.n4.nabble.com/A-introductory-question-about-Zips-law-Newbie-to-statistics-tp4521190p4521190.html Sent from the R help mailing list archive at

Hola, no me funciona: Te contesto lo mismo que le contesté a otro compañero de la lista, espero ser más claro. Les comparto la carpeta con la capa y los datos que estoy usando. https://www.dropbox.com/s/umy0evt3qm1wr4d/dissolve.zip Esto es lo que estoy haciendo: library(rgdal) library(maptools) library(sp) library(classInt) library(RColorBrewer) zm <- readOGR(?.?, ?zmdis?) data <-

Hola, El problema con la propuesta de Olivier es que los intervalos son diferentes para cada variable. La forma sencilla y rápida es: spplot(zm["part88"], col.regions=plotclr, at=class$brks) Pero para que quede más elegante hay que dar algunos pasos más: ## Intervalos en forma character op <- options(digits=4) tab <- print(class) options(op) intChar <- names(tab) ## Indice

Hola, Intento representar en un mapa participaciones porcentuales de los sectores económicos y no logro hacerlo con spplot. He intentado con spplot(zm, c(?part88?, ?part93?), cuts=4, col.regions=brewer.pal(4, ?Set3?)) Lo pude hacer utilizando el base graphics de R, definiendo: > plotvar88 <- zm$part88 > nclr <- 8 > plotclr <- brewer.pal(nclr, "PuOr") > plotclr

Hi We have a couple of FSC Computers here which seem to have a very strange bios. I am trying to boot a MS-DOS disk to flash an Scsi Raid controller. Since the computer doesn't have a floppy anymore I tried syslinux from an usb stick. The menu loads fine but I cannot boot disc images with memdisk on this computer. It works fine on others, of course. But now comes the weird part, if I use the

Dear R user, I want to estimate the parameter of ZETA or/and ZIPF distributions using R, given a series of integer values. Do you know a package (similar to MASS) or a function (similar to fitdistr) I can use to estimate the parameter of these distributions using MLE method? Otherwise do you know a function (which use MLE method to estimate distribution parameters) that allow me to specify a

Hola, Estoy tratando de sobreponer un mapa sobre otro, uno que sirva de ?background? y otro con datos. Mi primer intento lo hice usando el paquete PBSmapping. Aquí lo que hice: library(PBSmapping) estados <- importShapefile(?Entidades_2013.shp") #Aquí cargamos las entidades federativas (32 entidades) zonas <- importShapefile(?zm.shp?) #Aquí cargamos las zonas metropolitanas del país

Hi, Is there a Zipf-like distribution RNG in R? Thanks, Weiguang

I have the following daily exchange rate series (from january 1st 1996 to december 31st 2008) and I want to obtain them monthly series from it. I've read about the 'zoo' library but I'm not getting it how to do it. These are the data (left column day-month-year, right column the index) 31/12/1993 1,12509 03/01/1994 1,12509 04/01/1994 1,12558 05/01/1994 1,1258 06/01/1994 1,12596

Hello R-list members, I’m trying to make a min-avg-max plot (like a boxplot, without box and outliers, i.e., for each class of "x" a vertical line indicating the range (max-min) and an horizontal line or point indicating the mean) with data like: Year Avg Min Max 98 10 7 13 99 12 10 15 01 11 6 14 Which command should be used? Many thanks, Antonio Olinto

I am simulating an ANOVA model and get a strange behavior from the summary function. To be more specific: please run the following code and see for yourself: the summary()[["r.squared"]] values of two identical models are quite different!! ## 3 x 3 ANOVA of two factors x and z on outcome y s.size <- 300 # the sample size p.z <- c(0.25, 0.5, 0.25) # the probabilities of factor z ##

Inthe code below I was trying to to obtain the GMM estimates for CAPM (REGRESSION) for 36 stocks each have 180 observations,however it only gives me one output rather than 36. In SAS i would just put in a *By statement*. I have a variable TICKER that categorize them into 36 groups. *How can I obtain all 36 output instead of just one.* **

Dear all, How can I fit a truncated power law to a vector? I can't find a function to do that. If the function provides an AIC, even better. John

I am working on a simple pilot project comparing the capability of SQL, SAS and R to perform a rolling mean per the following instructions. I have completed the SQL and SAS analysis, so now it's R's turn. Calculate mean values of x (x=count) for each date in the dataset where mean = the average count of days [t-9] through day [t-3] for each date/illness combination. Dataset aggpilot

Hi all R users, I want to create a matrix having n columns and 2^n rows, and all its entries are only 0 or 1. In each row, column i is 0 means dimension i is chosen, 0 means not. The matrix will contains all the possible combination of those n dimensions. Here is an example, if n = 3, the matrix will look like: 0 0 0 0 0 1 0 1 0 1 0 0 0 1 1 1 0 1 1 1 1 I know I can use n "for" loops

Hi, today i wrote a function in R of the type: index.refraction <- function(Temp,Press, RH, CO2) When i try to plug a negative number in Temp, i got this type of error: " n <- index.refraction(Temp= -40,100,80,CO2) Messages d'avis : 1: In Ops.ordered(left, right) : '-' is not meaningful for ordered factors 2: In Ops.factor(left, right) : - not meaningful for factors 3:

Dear all, I have a dataset. I want to make barplot from this data. Zero1 <- " V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 0 0 0 1 0 0 0 Positive 2 0 0 1 0 1 0 1 1 Negative 3 0 0 1 0 0 0 1 1 Positive 4 0 1 0 1 1 1 0 1 Negative 5 0 0 1 0 1 1 0 0 Positive 6 0 1 0 0 1 1 1 1 Negative 7 1 0 1 1 1 1 1 1 Negative 8 0 0 0 0 1 0 0 1