similar to: lm output

Hi All Apologies if this is not the correct list for this question. The Rglpk package offers the following example in its documentation library(Rglpk) ## Simple mixed integer linear program. ## maximize: 3 x_1 + 1 x_2 + 3 x_3 ## subject to: -1 x_1 + 2 x_2 + x_3 <= 4 ## 4 x_2 - 3 x_3 <= 2 ## x_1 - 3 x_2 + 2 x_3 <= 3 ## x_1, x_3 are non-negative integers ## x_2 is a non-negative real

I have a zoo object that contains 2 time series named "A-B" and "V1". When I create a third series "V2", the name of the "A-B" series is changed to "A.B". Although I could recreate the names for the 3 series I am wondering if there is a way of preventing the name change from happening ( ... maybe an equivalent of the keep.names=TRUE statement

I am running the following line to download data from the US Energy Information Administration. This function has worked successfully for me in the past but yesterday gave the error/warning messages below. If I simply type "http://ir.eia.gov/wpsr/psw09.xls" (no quotes) into a browser, the file is available to download. I am running R 2.13.0 on Windows XP. # Download File Attempt

Hello all, I have 3 time series (tt) that I've fitted segmented regression models to, with 3 breakpoints that are common to all, using code below (requires segmented package). However I wish to specifiy a zero coefficient, a priori, for the last segment of the KW series (green) only. Is this possible to do with segmented? If not, could someone point in a direction? The final goal is to

kiotoqq wrote > I want to add a shorter column to my dataset with the function "merge", > it > should be filled with NAs wo be as long as the other colums, like this: > > id age > 9 46 > 8 56 > 6 52 > 5 NA > 4 NA > 3 NA > 1 NA > > i did this: > pa1 <- merge(pa1, an1, by="mergeid") > > and it says

I am trying to rename column names in a dataframe within a function. I am seeing an error (listed below) that I don't understand. Would be grateful of an explanation of what I am doing wrong and how I should rewrite the function to allow me to be able to rename my variables. Thanks. # Test Function myfunc <-function(var){ d = c(1,2,3,4,5) dts =

Hi All I am trying to use the round function on some columns of a dataframe while leaving others unchanged. I wish to specify those columns to leave unchanged. My attempt is below - here, I would like the column d3 to be left but columns d1, d2 and d4 to be rounded to 0 decimal places. I would welcome any suggestions for a nicer way of doing this. d1= rnorm(10,10) d2= rnorm(10,6) d3=

Hi All Apologies if this is not the correct mailing list for this question. I have installed RExcel and RCom from CRAN. R is indicating that I need an additional file/package "statconnDCOM" from rcom.univie.ac.at I have been trying to access this website over the last 2 days be it appears to be down. Does anyone know when this site is likely to be back up and/or know of an

David Stoffer describes some challenges with R's output when fitting ARIMA models for different orders (see Issue 2 at http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm). R doesn't fit an intercept in the model if there is any differencing. David describes a workaround using the xreg parameter to force R to calculate an intercept. Assume I have a variable y and 3 explanatory variables a,

Hi I am using R 2.2.0 under SuSE 10 I want to use lm() to get the slope and intercept for several daatasets and store them in a database. So far so good - but how do I extract the slope and the intercept from the result from lm()? my code looks like this: lmNNDens <- lm(log(DensNN$MeanNN) ~ log(DensNN$MeanDensity)) anovaLM <- anova(lmNNDens) Results$slope[No] <- ???lmNNDens???

Is there a way to do a linear regression with lm (having one predictor variable) and constrain the slope of the line to equal 1? Tom -- View this message in context: http://www.nabble.com/Linear-Regression-with-lm-Forcing-the-Slope-to-Equal-1-tf4828804.html#a13815432 Sent from the R help mailing list archive at Nabble.com.

Hello, I would like to fit data with the following formula : y=V*(1+alpha*(x-25)) where y and x are my data, V is a constant and alpha is the slope I'm looking for. How to translate this into R-language ? At the moment, I only know : lm(y ~ x) Sorry for such a basic question. I thought I could find the solution in a post but I have to confess that, up to know, I'm not able to understand

Is there an easy way to do a linear model with an a priori known slope? In essence, I want to minimize the residuals around a line of known slope. -R -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the

Dear R users, experimenting with the lm function in R, I've encountered some behaviour I don't understand with my limited knowledge of regression. I made a data-'set' of three measurements (see syntax below). Using lm (linear model) to fit the regression-line, I expected to find an intercept of 2.0 and a slope of 0, but in fact the slope is slightly below zero. Amazed by

Dear help-list, I would like to perform a linear regression on the log10 of the two vectors ov.mag.min and res.600nm. The slope and intercept of the regression I use to plot a wider range of ov.mag.min in a double log plot. For a linear regression on only tow points, wouldn't I expect the results for two.points.min to match pretty exactly res.600nm? It does not seem to be the case here.

Hello I want to compare the slope (let's say 'b') of a linear model obtained with lm() to a theoretical value (let's say 'th'). To do so, I think I should compute a 't value' using something like : (b - 'th')/standard.deviation(b) and then look at the p-value of this computed t. I don't understand how to do this in a simple way, just using lm()

Hi All I have a vector x containing 2 levels x = c(1,1,1,0,0,0,0,1,1,0,0,0,1,1,0) I would like to derive the following summarization Level Count 1 3 0 4 1 2 0 3 1 2 0 1 I have generated an inelegant solution using lags and loops but feel sure that there must be a better approach. If anyone has any thoughts I would be very grateful if you would share them. Thanks and best regards Pete --

I have not really used lm before and I was hoping for some help on a simple problem. Here is a toy version of the problem I want to solve. y x grp -.9 1 a -.8 2 a -.7 3 a -.7 1.5 b -.5 2.5 b -.3 3.5 b -.19 2.7 c -.11 3.7 c -.41 4.7 c I want to fit a model that has one y-intercept and three slopes, one for

Hello R-Users, I have I simple question that I could not solve: How to extract the elevation and the slope values from a linear model (lm) separately? coef(model.lm) gives both of them. Thanks, Antonio Olinto -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info",

Dear all, I am not fluent in R and am struggling to 1) apply a lm to a weight-size dataset, thus the model has to run separately for each species, each year; 2) extract coefs, r-squared, n, etc. The data look like this: year sps cm w 2009 50 16 22 2009 50 17 42 2009 50 18 45 2009 51 15 45 2009 51 16 53 2009 51 17 73 2010 50 15 22 2010 50 16 41 2010 50 16 21 2010