similar to: Generalized Logistic and Richards Curve

CRAN et al., I would like to add an extented introduction or other arbitrary sections to my package lmomco. I have been shipping inst/doc/Introduction.Rd. I would like to have this content inserted to the front of the PDF build for the CRAN. The R-exts.pdf seems to be a little silent on this subject? For my purposes, I have been doing this R CMD Rd2dvi --pdf

Dear R forum >From the given data, I have estimated the parameters of Wakeby distribution using lmomco package as library(lmomco) (amounts <- read.csv("input_S.csv")$amount) # ___________________________________________________________ # Wakeby distribution - Parameter estimation N                      = length(amounts) lmr                    = lmom.ub(amounts)

I'm trying to plot an extrapolated logistic sigmoid curve using glm(..., family=binomial) as follows, but neither the fitted() points or the predict()ed curve are plotting correctly: > year <- c(2003+(6/12), 2004+(2/12), 2004+(10/12), 2005+(4/12)) > percent <- c(0.31, 0.43, 0.47, 0.50) > plot(year, percent, xlim=c(2003, 2007), ylim=c(0, 1)) > lm <- lm(percent ~ year)

Dear R helpers, xx = data.frame(country = c("USA", "UK", "Canada"), x = c(10, 50, 20), y = c(40, 80, 35), z = c(70, 62, 10)) > xx        country      x     y    z 1      USA        10    40  70 2      UK          50   80   62 3     Canada    20   35   10 I need to arrange this as a new data.frame as follows - country       type     values USA           

Hi guys, I'm trying to use lmomco package. first I did the manual calculation on what is the estimates scale and location parameter given L-CV=0.2, L1=1000 L-moments and k (shape parameter) =- 0.1. so what i get is: location: 821.0445 scale: 260.7590 shape: -0.1000 #I assign this as GEV vectors using vec2par GEVpara2<-vec2par(c( 821.0445 , 260.7590 ,-0.1),'gev') #then I

Hi, I have a data set which is assumed to follow weibull distr'. How can I find of cdf for this data. For example, for normal data I used (package - lmomco) >cdfnor(15,parnor(lmom.ub(c(df$V1)))) Also, lmomco package does not have functions for finding cdf for some of the distributions like lognormal. Is there any other package, which can handle these distributions?

Dear R helpers I have following input f = c(257, 520, 110). I need to generate a decreasing sequence (decreasing by 100) which will give me an input (in a tabular form) like 257, 157, 57 520, 420, 320, 220, 120, 20 110, 10 I tried the following R code f = c(257, 520, 110) yy = matrix(data = NA, nrow = 3, ncol = 6) for (i in 1:3)      {      value = NULL      for (j in 1 : 6)           {

Dear R helpers Suppose I have a vector as vect1 = as.character(c("ABC", "XYZ", "LMN", "DEF")) > vect1 [1] "ABC" "XYZ" "LMN" "DEF" I want to reverse the order of this vector as vect2 = c("DEF", "LMN", "XYZ", "ABC") Kindly guide Regards Vincy [[alternative HTML

Dear R helpers, I have one basic doubt about the value of pi. In school, we have learned that pi = 22/7 (which is = 3.142857). However, if I type pi in R, I get pi = 3.141593. So which value of pi should be considered? Regards Vincy [[alternative HTML version deleted]]

Hello, I am using the lmomco package (lmom.ub and pargev) to compute the GEV parameters (location, scale, and shape), which are used to estimate return values. I was wondering how/if I can calculate upper and lower confidence (CI_u, CI_l) intervals for each return frequency using the GEV parameters to fill-in the table below? Xi (location) = 35.396 Alpha (scale) = 1.726 Kappa (shape) =

Dear R helpers I have a data which gives Month-wise and Rating-wise Rates. So the input file is something like month           rating           rate January        AAA             9.04 February      AAA             9.07 .......................................... .......................................... Decemeber     AAA            8.97  January           BBB           11.15 February        

Dear R helpers Wish you all a very Happy and Prosperous New Year 2011. I have following query. country = c("US", "France", "UK", "NewZealand", "Germany", "Austria", "Italy", "Canada") Through some other R process, the result.csv file is generated as result.csv      var1   var2  var3  var4    var5    var6   var7  

Dear R helpers I have a dataframe as df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189)) > df    x   y 1  1 102 2 14 500 3  3  40 4 21 101 5 11 189 # Actually I am having dataframe having multiple columns. I am just giving an example. I need to subtract all the rows of df by the first row of df i.e. I need to subtract each element of 'x' column by 1. Likewise I

Dear R list, I have one very elementary question regrading correlation between two variables. x = c(44,46,46,47,45,43,45,44) y = c(44,43,41,41,46,48,44,43) > cov(x, y) [1] -2.428571 However, if I try to calculate the covariance using the formula as covariance = sum((x-mean(x))*(y-mean(y)))/8       # no of of paired obs. = 8 or     covariance = sum(x*y)/8-(mean(x)*mean(y)) gives

Dear R list As suggested by Prof Brian Ripley, I have tried to read acf literature. The main problem is I am not the statistician and hence have some problem in understanding the concepts immediately. I came across one literature (http://www.stat.nus.edu.sg/~staxyc/REG32.pdf) on auto-correlation giving the methodology. As per that literature, the auto-correlation is arrived at as per following.

Hi, I am using parnor function of lmomco package. I believe it provides mean and std. dev for the set of data. But the std. dev provided does not match with the actual std. dev of the data which is 247.9193 (using sd function). Am I missing something here? > lmr <- lmom.ub(c(123,34,4,654,37,78)) > parnor(lmr) $type [1] "nor" $para [1] 155.0000 210.2130 >

Dear R helpers, (At the outset I sincerely apologize if I have not put forward my following query properly, though I have tried to do so.) Following is a curtailed part of my R - code where I am trying to generate say 100 random no.s for each of the products under consideration. library(plyr) n = 100 my_code = function(product, output_avg, output_stdev)     { BUR_mc = rnorm(n, output_avg,

Dear R helpers Suppose I have a vector as vect_1 = c("AAA", "AA", "A", "BBB", "BB", "B", "CCC") vect_1_id = c(1:length(vect_1)) Through some process I obtain vect_2_id = c(2, 3, 7), then I need a new vector say vect_2 which will give me vect2 = ("AA", "A", "CCC")  i.e. I need the subset of

Hi, I have one basic doubt. Suppose X ~ N(50,10). I need to calculate Probability X = 50. dnorm(50, 50, 10) gives me [1] 0.03989423 My understanding is (which is bit statistical or may be mathematical) on a continuous scale, Probability of the type P(X = .....) are nothing but 1/Infinity i.e. = 0. So as per my understanding P(X = 50) should be 0, but even excel also gives 0.03989422. Obviously

Hi R-users, I'm trying to repeat the same procedure to 1000 data set. I know this is very easy, but I got stuck finding the right and fastest way in running it. IID50=Riidf[1:50,1:1000] #where IID50 is a dataframe consist of 1000 time series(as column) and 50 time scales (row). #what I tried to do: estIID50=rep(NA,1000) for (i in 1:1000) estIID50[i]=pargev(lmom.ub(IID50[1:50,i])) #warning