similar to: 'Fitting' a model at predefined points

Hi, Does any one know if it is possible to create an R script that can use command line parameters. I can execute an R script from the command line, but I cannot figure out how to pass parameters to the script. The only resources I have found seem somewhat involved or incomplete. Any help is appreciated. Akintayo [[alternative HTML version deleted]]

Consider the following example (based on an example in Pat Altham's GLM notes) pyears <- scan() 18793 52407 10673 43248 5710 28612 2585 12663 1462 5317 deaths <- scan() 2 32 12 104 28 206 28 186 31 102 Smoke <- gl(2,1,10,labels=c("No","Yes")) Age <- gl(5,2,10,labels=c("35-44","45-54","55-64","65-74","75-84"),

?? If I haven't misunderstood, they are completely different! 1) NIR must be a matrix, or poly(NIR,...) will fail. 2) Due to the previously identified bug in poly, degree must be explicitly given as poly(NIR, degree =2,raw = TRUE). Now consider the following example: > df <-matrix(runif(60),ncol=3) > y <- runif(20) > mdl1 <-lm(y~df*I(df^2)) > mdl2

Marc: 1. I am aware of the need to explicitly name arguments after ... -- see the R Language definition where this can be inferred from the argument matching rules. 2. I am aware of the stated exception for poly(). However: > x1 <- runif(20) > x2 <- runif(20) > mx <- cbind(x1,x2) > poly(mx,2) Error in poly(dots[[i]], degree, raw = raw, simple = raw) : 'degree'

> On Jul 13, 2017, at 10:43 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > poly(NIR, degree = 2) will work if NIR is a matrix, not a data.frame. > The degree argument apparently *must* be explicitly named if NIR is > not a numeric vector. AFAICS, this is unclear or unstated in ?poly. I still get the same error with: library(pld) data(gasoline) gasTrain <-

Hi, I can not comprehend the linear fitting results of polynoms. For example, given the following data (representing y = x^2): > x <- 1:3 > y <- c(1, 4, 9) performing a linear fit > f <- lm(y ~ poly(x, 2)) gives weird coefficients: > coefficients(f) (Intercept) poly(x, 2)1 poly(x, 2)2 4.6666667 5.6568542 0.8164966 However the fitted() result makes sense: >

Bert, The 'degree' argument follows the "..." argument in the function declaration: poly(x, ..., degree = 1, coefs = NULL, raw = FALSE, simple = FALSE) Generally, any arguments after the "..." must be explicitly named, but as per the Details section of ?poly: "Although formally degree should be named (as it follows ...), an unnamed second argument of length 1

I would like to use samp as a part of a filename that I can change. My source files are .csv files with date as the file name, and I would like to be able to type in the date (later perhaps automate this using list.files) and then read the csv and write the pdf automatically. I have tried different combinations with "" and () around samp, but I keep getting the error "object

Hello guys, I would like to ask for help to understand what is going on in "func2". My plan is to generalize "func1", so that are expected same results in "func2" as in "func1". Executing "func1" returns... 0.25 with absolute error < 8.4e-05 But for "func2" I get... Error in dpois(1, 0.1, 23.3065168689948, 0.000429064542600244,

my reproducible example test<-structure(list(site = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L), .Label = c("A", "B", "C", "D", "E"), class = "factor"),

> On Jul 13, 2017, at 7:43 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > Below. > > -- Bert > Bert Gunter > > > > On Thu, Jul 13, 2017 at 3:07 AM, Luigi Biagini <luigi.biagini at gmail.com> wrote: >> I have two ideas about it. >> >> 1- >> i) Entering variables in quadratic form is done with the command I >>

(To the list moderator: I just subscribed to the list. Apologies for not having done so longer before trying to post.) Hi all, I am currently using lmer to analyze data from an experiment with a single fixed factor (treatment, 6 levels) and a single random factor (block). I've been trying to follow the online guidance for estimating p-values for parameter estimates on these and other

I'm new to R (and to S) and am wondering about code from pages 72 and 83 of MASS (Venables+Ripley, 3rd edition), to draw an envelope on a QQ plot. Copying from the book, I've got: #... code whose gist is "a.fit <- nls(..." num.points <- length(resid(a.fit)) qqnorm(residuals(a.fit)) # illustrate data-model residuals qqline(residuals(a.fit)) samp <-

Hi, I am facing the problem that I want to plot a histogram chart set freq to true and overlay with normal or weibull or exponential distribution curve. The sample code is shown as below: >samp<-c(-8.2262,-8.2262,-8.2262,-8.20209,-8.09294,-8.07321,-8.07321, -8.07321,-8.07175,-8.04948,-8.04948,-8.04948,-8.03848,-8.03848,

Dear all, I am currently analyzing count data from a hierarchical design, and I?ve tried to follow the suggestions for a correct estimation of p-values as discusssed at R-Wiki (http://wiki.r-project.org/rwiki/doku.php?id=guides:lmer-tests&s=lme%20and%20aov). However, I have the problem that my model only consists of parameters with just 1 d.f. (intercepts, slopes), so that the

Hi, I would like to fit my data with a 4th order polynomial. Now I have only 5 data point, I should have a polynomial that exactly pass the five point Then I would like to compute the "fitted" or "predict" value with a relatively large x dataset. How can I do it? BTW, I thought the model "prodfn" should pass by (0,0), but I just wonder why the const is

Dear R users I am trying to obtain p-values for (quasi)poisson lmer models, including Markov-chain Monte Carlo sampling and the command summary. > > My problems is that p values derived from both these methods are totally different. My question is (1) there a bug in my code and > (2) How can I proceed, left with these uncertainties in the estimations of > the p-values? > > Below

Hi Bert, Ok, to your initial point, the key nuance is that if 'x' is a vector, you can leave the 'degree' argument unnamed, however, if 'x' is a matrix, you cannot. That aspect of the behavior does not seem to change if poly() is called stand alone or, as suggested in ?poly, within a formula to be parsed. Working on tracing through the code using debug(), the error is

G'day all, looking at http://cran.at.r-project.org/web/checks/check_results_Sleuth2.html I noticed that r-prerel-* and r-devel-* issue notes. Apparently, now examples are more thoroughly checked by R CMD check and the note pointed out that a package used in the examples was not declared. This was easily fixed by adding 'Suggests: lattice' to the DESCRIPTION file (now this change has

Ccing the maintainer if the tuneR package. Looks to me like sawtooth (and square) don't behave as expected when using xunit="samples". Workaround is to use xunit="time" instead: sawtooth(110,duration=1/100,samp.rate=sample_rate,xunit="time") I looked at the code but found it to be opaque. -- Sent from my phone. Please excuse my brevity. On November 12, 2017