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Hi Say I want to add manually an intercept in the function lm. Even if almost all results will be identical, few stats are different as DF counting will be different as intercept will not be included in "automatic" case, while it will be in "manual" case. See: ###usual lm on freeny fr<-lm(freeny.y~freeny.x) ###manual lm on freeny man<-cbind(1,freeny.x)

Full thread here: https://github.com/tidyverse/broom/issues/287 Reproducible example: is.object(freeny$y) # [1] TRUE attr(freeny$y, 'class') # [1] "ts" class(freeny$y) # [1] "ts" # ts attribute wiped by model.frame class(model.frame(y ~ ., data = freeny)$y) # [1] "numeric" attr(model.frame(y ~ ., data = freeny)$y, 'class') # NULL # but still:

Hi My goal is to do a (multiple) regression, just knowing that my Y variables will be the say k first variables of a matrix/data frame. I thought I should do it with eval(parse)) but encounter a strange problem. See: lm(y~.-y, data=freeny) #that's what I want to do in the one equation case #Problem is I don't know name of the variable... only that it is the first one... #so idea is to

Hello! I would love to be able to include an external variable to a lm call, I mean something: if(TRUE) a<-freeny.x[,4] else a<-NULL lm(freeny.y~freeny.x[,-4] +a) but it does not work with a<-NULL, whereas lm(freeny.y~freeny.x[,-4] +NULL) I don't understand why and did not find an answer in the manuals... do you see it? Any idea? Thanks!!

Hi, A recent thread provided a (working) construct for lm: lm(as.matrix(freeny[ix]) ~., freeny[-ix]) Can someone explain what is meant by the formula in that expression, that is, what does "mymatrix~." do? I couldn't find any such example in the lm() or formula() help pages. thanks Carl

This one showed up while looking at one of Ripley's other reports: > data(freeny) > model.frame(y~1,data=freeny,subset=1:10) y 1962.25 8.79236 1962.5 8.79137 1962.75 8.81486 1963 8.81301 1963.25 8.90751 1963.5 8.93673 1963.75 8.96161 1964 8.96044 1964.25 9.00868 1964.5 9.03049 > model.frame(y~1,data=freeny,subset=1:10)$y Warning: Replacement length not a

Dear all, I have discovered an object named 'freeny.x' which seems in-built. Can somebody tell me why it is there and what purpose it is serving? 'ls()' syntax does not seem to catch this object in the environment. Thanks,

Hi, I’d like to fit one GLM model for each possible combination of inputs (i.e. exhaustive search). The package leaps can help me to generate all possible variable subsets, but I’ll appreciate your guidance as of how to generate one model for each of those possible subsets. I’m new in R! Thanks in advance for your help. library(leaps) subsets <- regsubsets( y ~ . , data=freeny, nbest=100,

I noticed this in the NEWS: o Datasets 'sunspot.month' and 'sunspot.year' are available separately but not via data(sunspot) (which is used by package lattice to retrieve a dataset 'sunspot'). Now, the 'sunspot' dataset included in lattice was actually a mistake, I realized later that I instead wanted the 'sunspots' data from

Hi all, I need to (quickly!grrr) update IP addresses for which a given machine will answer. I did this: /sbin/ifconfig eth0:0 xx.xx.xx.10 netmask 255.255.255.0 /sbin/ifconfig eth0:1 xx.xx.xx.11 netmask 255.255.255.0 /sbin/ifconfig eth0:2 xx.xx.xx.30 netmask 255.255.255.0 /sbin/ifconfig eth0:3 xx.xx.xx.31 netmask 255.255.255.0 /sbin/ifconfig eth0:4 xx.xx.xx.13 netmask 255.255.255.0

### First, is there a way to access a univariate time series as a matrix instead of a vector? # For example: data(UKLungDeaths) # If I do apply(mdeaths,1,cumsum) # Gives an error as mdeaths is not a matrix but a vector, although when I look at it : mdeaths # the ts object has a matrix like "appearance" # The only way of doing it I've found is: mdeaths2<-as.matrix(mdeaths)

Hi Everyone, I am pretty new to Rails. I have very little idea about everything. But for my new project i have been asked to do some research on Solr search and implement it on Rails. Please provide me some link on tutorials and guides about Solr on Rails. Trouble is... have to do it by tomorrow. -- Posted via http://www.ruby-forum.com/. -- You received this message because you are

Dear helpers, I am trying to write a function to return prediction values using package ts. I have written three different versions since I am not sure what's wrong with my func2. func and func1 return the same results.But func1 and func2 don't. In particular, the only difference between "func1" and "func2" is the function variable name being y and data, respectively.

Problem: How can I format tickmark labels of log scaled axes of lattice graphics in the usual `xxx'-Format (and not in the scientific format). Example: (according to the help-page of xyplot): In the first plot I get the xxx-Format, in the second plot I get the scientific format (10^xxx): data(sunspot) plot( 1:37, sunspot, log='y',type='l') xyplot( sunspot ~ 1:37,

Hi folks, I've put my new DNS server in place, told the primary that this IP would be pulling zones and restarted BIND on the primary. I configured my named.conf file to pull zones from the primary, started BIND on the new box. Here's where I get confused ++++++++++++++++++++++++ [root at sunspot etc]# /etc/init.d/named status number of zones: 169 debug level: 0 xfers running: 0 xfers

Hi I have a few problems with tail/head when applied to multiple time series. I'm not sure as whether I did not understand the function or whether it correspond to an unexpected behavior. When head(a,n) is applied on data.frame or matrix, it returns a data-frame or matrix with first n obs of *each* variable. When applied to a mts object, it returns first n obs of *first* variable only,

Any help with this problem would be greatly appreciated: I need to produce a custom plot i haven't come across in R. Basically, I want to show means, 1st standard deviation and 5th and 95th percentiles visually, using something resembling a boxplot. Is it possible to completely customize a boxplot so that it shows means as the bar (instead of, not as well as medians), standard deviations at

How do I add to a trellis plot the best fit line from a robust fit? I can use panel.lm to add a least squares fit, but there is no panel.rlm function. -- Alan S Barnett <asb at mail.nih.gov> NIMH/CBDB

Is there a way of getting a list of indices or something indicating which set of points lqs when doing a regression? The $bestone, seems to return only 2 pts. no matter how good or bad the fit. I checked the archive for the past few months, but didn't find anything. thanks in advance, Mark Hall -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help

Hello all, I'm using R2.7.0 (on Windows 2000) and I'm trying do run a robust regression on following model structure: model = "Y ~ x1*x2 / (x3 + x4 + x5 +x6)" where x1 and x2 are both factors (either 1 or 0) and x3.....x6 are numeric. The error code I get when running rlm(as.formula(model), data=daymean) is: error in rlm.default(x, y, weights, method = method, wt.method =