similar to: Extrapolation of rarefaction curve

Hi List, I’m using the vegan function specaccum to produce a rarefaction curve. In the function’s help it says: “Function ‘specaccum’ finds species accumulation curves or the number of species for a certain number of sampled sites or individuals”. Well, I would like to finds this curve for individuals, but when I compute it the function (using the ‘rarefaction’ method) gives me Sites, Richness

Hi everybody, i'm trying to draw rarefaction curves to estimate a population size from genotyped faeces. I used the Gimlet software which gave me a script and a "rarefaction.txt" file. I've copied both files in the work directory of R. I changed library(nls) by library(stats) in the script. But now, i'm still unable to run it. If i ask to show error messages, the software

Dear R-experts, I write to you to know if somebody is aware of a R package (or function) able to plot graphs for extrapolation. I need to be clear on what extrapolation really is to me. It is when we use the model for X variables outside the range of X variables that were used to construct the model and estimates.? What I am really looking for is that beyond confidence intervals for predictions

Dear all, I'm trying to fit a curve to some 1 year failure-time data, so that I can extrapolate and predict failure rates up to 3 years. The data is in the general form: Treatment Time Status Treatment A 28 0 Treatment B 28 0 Treatment B 28 0 Treatment A 28

Dear R-Experts, Here below my R code working but I don't know how to complete/finish my R code to get the final plot with the extrapolation for the10 more years. Indeed, I try to extrapolate my data with a linear fit over the next 10 years. So I create a date sequence for the next 10 years and store as a dataframe to make the prediction possible. Now, I am trying to get the plot with the

Here's a question for ecologists on the r-help list-- I'm addressing this to ecologists in particular because they're most likely to be familiar with the equation in question but I'll be happy to discuss the problem with anyone who's willing to take a whack at it. I'm trying to write a function to calculate the large sample variance of species richness estimates by

Hi, I am facing a problem in extrapolation of data series. It is a series of Bond yields, I am having the yield for 1 year to 30 years. I want to find the yield for 0.5 year and 30.5 years. I used the Langrange's Extrapolation but the extrapolation deviates from the normal trend ( as we can see in theoritical yield curves) very sharply, as go on increasing my years from 30 years to 35 years

Dear?R experts, ? I?am trying to do linear extrapolation on a dataset?like the attached document. I looked at the approx and approxfun function that seem to do this function, but not fully understand them. I was wondering if someone could help with writing commands to do the following based on the attached file's example: ? ID#1 and ID?#2 both have response parameters ("MEASUREMENT"

Readers, Data was imported using the read csv command: dataimport<-read.csv("/path/to/dataimport.csv") 10,2000 12,2001 13,2002 15,2003 17,2004 Using the help contents for 'predict.lm' (i.e. ?predict.lm) a new data frame was created dataimportextra<-data.frame(x=seq(1990,2010,1)) predict(lm(dataimport),dataimportextra[,2],se.fit=TRUE)

Dear list members, I have a data.frame so shaped: Sector Quadrants Plot Sic Time Species1 Species2 Species3 .. Species-n 1 1 1 1 5 0 0 1 . 0 2 1 1 1 12 1 1 1 . 0 3 1 1 1 34 0 1 0 . 0 4 1 1 1 23 1 1 0 . 0 5 2 1 1 22 1 1 1 . 1 6 2 1 1 10 1 1 1 . 1 7 2 1 1 2 1 0 0 . 0 8 2 1 1 2 0 0 1 . 0 9 3 1 1 12 0 0 0 . 1 . . . . . . . . . .

Hi all, I'm fitting a line to my dataset. Later I want to predict missing values that exceed the [min,max] interval of my empirical data, therefore I choose surface="direct" for extrapolation. l1<-loess(y1~x1,span=0.1,data.frame(x=x1,y=y1),control=loess.control(surface="direct")) In my application it is highly important that the fitted line intercepts at the point of

Dear list members, I am analysing my microarray data using limma package. Now I encounter several problems. Looking forward to your suggestions! Question 1: During the process of background correction using method="normexp", four warning messages appeared as "NaNs produced in: log(x)" (as you can see in the program posted below). What does that mean? How will it effect the

Hi, We have an experiment where individuals responses were measured over 5 days. Some responses were not obtained because we only allowed individuals to respond within a limited time-frame. These individuals are given the maximum response time as they did not respond, yet we feel they may have done if given time (and by looking at the rest of their responses over time, the non-response days stand

DeaR list, I'm running an external program that computes some electromagnetic response of a scattering body. The numerical scheme is based on a discretization with a characteristic mesh size "y". The smaller y is, the better the result (but obviously the computation will take longer). A convergence study showed the error between the computed values and the exact solution

I am new to R and am trying to fit a survival curve with a weibull hazard function to a set of data giving the probability of survival to age x, given the year of birth, in the form: Probability of survival: Birth year 1980 1981 ... 2003 .2 0.90 0.89 ... 0.87 1 0.80 0.81 ... 0.79 age 2 0.75 0.74 ... 0.73 3 0.70 0.69 ... 0.68 5 0.50 0.49 ... 0.43 10 0.30 0.31 ... 0.26 I would like to be

Dear All, I am analysing a dataset on levels of herbivory in seedlings in an experimental setup in a rainforest. I have seven classes/categories of seedling damage/herbivory that I want to analyse, modelling each separately. There are twenty maternal trees, with eight groups of seedlings around each. Each tree has a TreeID, which I use as the random effect (blocking factor). There are two

Dear all I generated predicted risk of death for each subject in the study by means of Cox proportional hazards model at 8 year of follow-up, a time point at which follow-up was more than 90% complete. It is possible to extrapolate to 10-year the predicted risk of each subjet by assuming an exponential distribution? Any help would be greatly appreciated. Thanks for your consideration.

Dear R experts: I need to get this plot, but also with 95% confidence interval bands: hour <- c(1, 2, 3, 4, 5, 6) millivolts <- c(3.5, 5, 7.5, 13, 40, 58) plot(hour, millivolts, xlim=c(1,10), ylim=c(0,1000)) pm <- lm(millivolts ~ poly(hour, 3)) curve(predict(pm, data.frame(hour=x)), add=TRUE) How can the 95% confidence interval band curves be plotted too? Sincerely,

Dear R colleagues: Am trying to fit a simple NL model to determine Economical Optimum Nitrogen Rates. The segmented (quadratic + plateau) model only works with some y's, in some cases I get a "singular gradient" error. I'll appreciate any ideas in how to solve the singular gradient error. Thanks, Jose # The following code works using yield2 in the nls model but not using

Dear R-help, I have fitted a glm logistic function to dichotomous forced choices responses varying according to time interval between two stimulus. x values are time separation in miliseconds, and the y values are proportion responses for one of the stimulus. Now I am trying to extrapolate x values for the y value (proportion) at .25, .5, and .75. I have tried several predict parameters, and they