similar to: Best Fit line trouble with rsruby

Hi All, Could anyone please tell the installation steps of RSruby gem on Windows XP. I have latest version of ruby & R installed on Windows. Thanks -- View this message in context: http://r.789695.n4.nabble.com/R-and-Ruby-integration-using-RSruby-gem-tp4633020.html Sent from the R help mailing list archive at Nabble.com.

Readers, Data set: t,c 0,100 40,78 80,59 120,38 160,25 200,21 240,16 280,12 320,10 360,9 400,7 graphdata<-read.csv('~/tmp/data.csv') graphmodeld<-lm(log(graphdata[,2])~graphdata[,1]) graphmodelp<-exp(predict(graphmodeld)) plot(graphdata[,2]~graphdata[,1]) lines(graphdata[,1],graphmodelp) Please what is the function and syntax to obtain gradient values for the model curve at

Hello everyone, I''m new here. I''ve been working with prototype and plotr for about a month now, off and on, and I have pretty much hit the wall on using the data returned by Ajax.Request. I''m using some php code to return a string: {''foo'': [[0,0.0865334429075127], [1,0.0828179861705063], [2,0.0828173042602942], [3,0.0841707718624196]]} But I keep

Hello, I am searching for the best method to plot two variables with points whose output color depends on the size of a third variable. For example, the darkness of the x-y point would increase incrementally based on the size of the z value, similar to the colramp parameter in geneplotter. This would be analagous to symbols(), except changing the selection from the color gradient rather than the

When I installed RSRuby (Ruby interface to the R programming language), an error occured : cannot find the R library. I defined Rlibrary : R\library in PATH on Windows but it didn't work with the same error. RSRuby ask R to permit the option 'shared library' but when I install R, it didn't ask me for this option. I want to ask you how to share le library in R? And when I can find

Hi, I am trying to install RSRuby (http://rubyforge.org/projects/rsruby/) which somewhat analogous to RPy. Installation fails using the cran R 2.8.0 packages for Ubuntu Intrepid because RDevices.h does not appear to be included in the r-cran-core package as it was previously. It is in the rbase_2.8.0.orig.tar.gz in the same directory meaning it hasn't been taken out of R altogether, and I

Hey guys, I'd really appreciate any help. I have a multivariate analysis done, the output of which is: > GraphData <-read.table("eigen.coa") > GraphData V1 V2 V3 V4 1 1 0.371970 0.8552 0.8552 2 2 0.061785 0.1420 0.9972 3 3 0.001211 0.0028 1.0000 4 4 0.000000 0.0000 1.0000 > summary(GraphData) V1 V2 V3

> On Apr 6, 2018, at 8:03 AM, David Winsemius <dwinsemius at comcast.net> wrote: > > >> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com> wrote: >> >>> Sent: Friday, April 06, 2018 at 5:55 AM >>> From: "David Winsemius" <dwinsemius at comcast.net> >>> >>> >>> Not correct. You already have

I have never found the R symbolic differentiation helpful because my functions are typically quite complicated, but was prompted by Steve Ellison's suggestion to try it out in this case: ################# reprex (see reprex package) graphdta <- read.csv( text = "t,c 0,100 40,78 80,59 120,38 160,25 200,21 240,16 280,12 320,10 360,9 400,7 ", header = TRUE ) nd <- c( 100, 250,

Hello,   I am trying to calculate predicted values derived from one dataset into a hypothetical dataset. I tried this line of code:   graphdata$fmgpredvalues <- predict(Acs250.3.4, graphdata)   and received the following error message:   ERROR: ZXend[1], drop = FALSE] %*%lmeFit$beta   I have made sure all variable names are the same between the two datasets and all factors are appropriately

> Sent: Friday, April 06, 2018 at 1:44 PM > From: "Jeff Newmiller" <jdnewmil at dcn.davis.ca.us> > > You did not try my suggestion. You tried David's, which has a leftover mistake from your guesses about what the argument to coef should be. Yes, sorry for the mistake. coef(graphmodeld) (Intercept) graphdata[, 1] 4.513544204 -0.006820623 This corresponds

This smells like homework, which the Posting Guide indicates is off topic. I am not aware of "the function" that will solve this, but if you know what a gradient is analytically then you should be able to put together a solution very similar to the code you already have with the addition of using the coef function. -- Sent from my phone. Please excuse my brevity. On April 5, 2018

> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com> wrote: > >> Sent: Friday, April 06, 2018 at 5:55 AM >> From: "David Winsemius" <dwinsemius at comcast.net> >> >> >> Not correct. You already have `predict`. It is capale of using the `newdata` values to do interpolation with the values of the coefficients in the model. See:

> Sent: Friday, April 06, 2018 at 5:55 AM > From: "David Winsemius" <dwinsemius at comcast.net> > > > Not correct. You already have `predict`. It is capale of using the `newdata` values to do interpolation with the values of the coefficients in the model. See: > > ?predict > The ? details did not mention interpolation explicity; thanks. > The

Hello, I am trying to fit my Elisa results (absorbance readings) to a standard curve. To create the standard curve model, I performed a 4-parameter logistic fit using the 'drc' package (ExpectedConc~Absorbance). This gave me the following: > FourP A 'drc' model. Call: drm(formula = Response ~ Expected, data = SC, fct = LL.4()) Coefficients: b:(Intercept) c:(Intercept)

> > We're developing a software tool that does statistical analysis and forecasting using a number of historical data sets (time series). We're looking for a statistician with experience in R to help us build out the statistical backbone of the software and provide ongoing analysis for our clients. This position will be 50% focused on research and development of the statistical

Hi I have measured the UV absorbance (abs) of 10 solutions of a substance at known concentrations (conc) and have used a linear model to plot a calibration graph with confidence limits. I now want to predict the concentration of solutions with UV absorbance results given in the new.abs data.frame, however predict.lm only appears to work for new "conc" variables not new "abs"

Hi everyone, I would like to fit a predictive model to my data in order to compare absorbance readings to a toxin standard. This data was obtained by exposing red blood cells to different toxin concentrations, which lead to the lysis of the red blood cells, increasing the absorbance (hemoglobin is freed). The data has a sigmoid shape (see below), so I thought about fitting a logistic model to the

Hello, I'm new here, but will try to be as specific and complete as possible. I'm trying to use “lm“ to first estimate parameter values from a set of calibration measurements, and then later to use those estimates to calculate another set of values with “predict.lm”. First I have a calibration dataset of absorbance values measured from standard solutions with known concentration of

> Sent: Friday, April 06, 2018 at 4:53 AM > From: "Jeff Newmiller" <jdnewmil at dcn.davis.ca.us> > To: "g l" <gnulinux at gmx.com> > coef( graphmodeld ) > coef(graphmodelp) Error: $ operator is invalid for atomic vectors A quick search engine query revealed primarily references to the dollar sign ($) operator which does not seem relevant to this