similar to: fit an exponential curve

Dear all, I am attempting to explain patterns of arthropod family richness (count data) using a regression model. It seems to be able to do a pretty good job as an explanatory model (i.e. demonstrating relationships between dependent and independent variables), but it has systematic problems as a predictive model: It is biased high at low observed values of family richness and biased low at

I have a set of time-series climate data with missing entries. I need to add rows for these missing entries to this data set. The only way I know to do this is unsing a foreloop, but this won't work on a list. I've tried to convert the list to a data frame, but that won't happen, either. I want to fill rows in this table: > newtest[10:15,] yrmos yearmo snow.sum snow.mean

I am trying to run the following code in R on a Linux cluster. I would like to use the full processing power (specifying cores/nodes/memory). The code essentially runs predictions based on a GAM regression and saves the results as a CSV file for multiple sets of data (here I only show two). Is it possible to run this code using HPC packages such as Rmpi/snow/doParallel? Thank you!

Hello everyone, I am working for a few days already on a basic algorithm, very common in applied agronomy, that aims to determine the degree-days necessary for a given individual to reach a given growth stade. The algorithm (and context) is explained here: http://www.oardc.ohio-state.edu/gdd/glossary.htm , and so I implemented my function in R as follows: DD <- function(Tmin, Tmax, Tseuil,

Dear all, I have weekly data by city (variable citycode). I would like to take the average of the previous two, three, four weeks (without the current week) of the variable called value. This is what I have tried to compute the average of the two previous weeks; df = df %>% mutate(value.lag1 = lag(value, n = 1)) %>% mutate(value .2.previous = rollapply(data = value.lag1,

Estimados erreros, Estoy intentando entender como calcula el paquete dismo ( https://cran.r-project.org/web/packages/dismo/index.html) un coeficiente de variación. Os pongo un ejemplo: tmin <- c(10,12,14,16,18,20,22,21,19,17,15,12) # temperatura mínima media mensual de un año tmax <- tmin + 5 # temperatura máxima media mensual de un año prec <- c(0,2,10,30,80,160,80,20,40,60,20,0)

Hola, en la misma definici?n de la funci?n: # P15. Precipitation Seasonality(Coefficient of Variation) # the "1 +" is to avoid strange CVs for areas where mean rainfaill is < 1) p[,15] <- apply(prec+1, 1, cv) Un saludo, Jorge On Martes, 19 de Junio de 2018 13:07:27 Marcelino de la Cruz Rot escribi?: > Hola Jaume: > > Si miras el c?digo de biovars() ver?s que la

Hi, I am using mgcv:gam and have developed a model with 5 smoothed predictors and one factor. gam1 <- gam(log.sp~ s(Spr.precip,bs="ts") + s(Win.precip,bs="ts") + s( Spr.Tmin,bs="ts") + s(P.sum.Tmin,bs="ts") + s( Win.Tmax,bs="ts") +factor(site),data=dat3) The total deviance explained = 70.4%. I would like to extract the variance explained

Hope these help for alternatives to lm()? I show the use of a 2nd order polynomial as an example to generalize a bit. Sometimes from the subject line two separate responses can appear as reposts when in fact they are not... (though there are identical reposts too). I should probably figure a way around that. --- Stephen Tucker <brown_emu at yahoo.com> wrote: > ## Data input > input

Hola!! tengo un data.frame donde cada fila corresponde a un año y cada columna a un mes (De enero a diciembre) > head(valT) V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 1941 18.0 16.3 15.2 10.1 8.1 8.3 8.8 9.2 7.9 12.2 11.9 14.6 1942 17.2 15.9 13.6 11.6 8.7 6.2 6.4 7.2 9.7 12.0 14.1 16.7 1943 17.6 17.3 13.5 12.5 10.5 7.0 8.2 7.9 -999.9 -999.9

Hi folks, I've looked through the list archives and online resources, but I haven't really found an answer to this -- it's pretty basic, but I'm (very much) not a statistician, and I just want to check that my solution is statistically sound. Basically, I have a data file containing two columns of data, call it data.tsv: year count 1999 3 2000 5 2001 9 2002 30 2003 62 2004 154

I have a 45 x 16 data frame consisting of dissimilarities among 10 colors, giving in each column the 45 = 10*9/2 pairwise judgments for one of 16 subjects. The rownames identify each pair of colors, e.g, "AC" = ("A","C"), and the pairs are ordered by columns in the lower triangle of each distance matrix. > helm.raw <-

I have one data frame with a column of dates and I want to fill another data frame with one column of dates, one of years, one of months, one of a unique combination of year and month, and one of days, but R seems to have some problems with this. My initial data frame looks like this (ignore the NAs in the other fields): > mans[1:10,] date loc snow.new prcp tmin snow.dep tmax 1

I am sure that this sort of thing has been asked and answered before, so in case my suggestions don't work for you, just search the archives a bit more. I am also sure that it can be handled directly by numerous functions in numerous packages, e.g. via time series methods or by calculating running means of suitably shifted series. However, as it seems to be a straightforward task, I'll

Hi R bayesians, I need an advise how to resolve the two different estimates applying a traditional glm (TG) and a bayes glm (BG), and different results depending on the data formats of response data and the prior specs using bayesglm in R. I'm not familiar with bayes estimate and my colleague asked me to look into this because the EPA from France reported a quite different estimates for

Hi R Users, I am going to run a multiple linear regression with around 57 independent variables. Each time I run the model with just 11 variables, the results are reasonable. With increasing the number of independent variables more than 11, the coefficients will get ?NA? in the output. Is there any limitation for the number of independent variables in multiple linear regressions in R? I attached

This is an example from chapter 11 of the 6th edition of Devore's engineering statistics text. It happens to be a balanced data set in two factors but the calculations will also work for unbalanced data. I create a factor called 'cell' from the text representation of the Variety level and the Density level using '/' as the separator character. The coefficients for the linear

Hello, Is there anyone who could give me an example of how to import an ascii grid (i.e. ArcGIS exported raster) into R. I want to use it with gstat but don't know the appropriate import routine. Thanks very much for your help. Regards, femke Femke Reitsma Graduate Student (ABD) Geography Department 2181 LeFrak Hall University of Maryland College Park, MD 20742 Phone: 301-405-4121

Dear All: One more thing. I want to add the normal curve to the histogram. Is there away to stretch the peak of the curve to the top of the histogram or at least near to the top of the histogram. Please see the code below. Lizard.tail.lengths <- c(6.2, 6.6, 7.1, 7.4, 7.6, 7.9, 8, 8.3, 8.4, 8.5, 8.6,8.8, 8.8, 9.1, 9.2, 9.4, 9.4, 9.7, 9.9, 10.2, 10.4, 10.8,11.3, 11.9) x<-seq(5,12, 0.001)

Hi everyone, I'm trying to perform a bi exponential Fit with the package NLS. the plinear algorithm seems to be a good choice see: p<-3000 q<-1000 a<--0.03 b<--0.02 t<-seq(0:144);t y<-p*exp(a*t) + q*exp(b*t)+rnorm(t,sd=0.3*(p* exp(a*t) + q*exp(b*t))) fittA <- nls(y~cbind(exp(a*t), exp(b*t)), algorithm="plinear",start=list(a=-.1, b=-0.2), data=list(y=y, t=t),