Re: [R] lm fails on some large input

The final goal is to make two lines and find the intersection point. I don't want to argue more about the reason. The tol suggestion is reasonable, and I'll take that. 2019/4/19 4:12, Jeff Newmiller: The fact that you think x~y is interchangeable with y~x suggests to me that you will have a

Re: [R] lm fails on some large input

gt; Wang > Sent: Thursday, April 18, 2019 12:36 PM > To: Michael Dewey ; r-help@r-project.org > Subject: Re: [R] lm fails on some large input > > I just want to make a line out of timestamps vs some coordinates, so y~x or > x~y doesn't matter. > > Yes, I know the answer. Wh

Re: [R] lm fails on some large input

The fact that you think x~y is interchangeable with y~x suggests to me that you will have a difficult time convincing R Core that this is a bug. I recommend that you take at leastan upper division college course in linear regression first. On April 18, 2019 9:35:55 AM PDT, Dingyuan Wang

Re: [R] lm fails on some large input

I just want to make a line out of timestamps vs some coordinates, so y~x or x~y doesn't matter. Yes, I know the answer. When trying R, I'm surprised that R can't solve that either. I first noticed that PostgreSQL can't solve it, and found that they fixed that in pg 12.

Re: [R] lm fails on some large input

I make a general rule not to stick time values into numerical analysis algorithms without first subtracting a reasonable epoch (to obtain difftime) and then using as.numeric.POSIXt with the units argument set explicitly so the analysis uses numeric values that I can interpret. While the

Re: [R] lm fails on some large input

Dear Peter, > -Original Message- > From: peter dalgaard [mailto:pda...@gmail.com] > Sent: Thursday, April 18, 2019 12:23 PM > To: Fox, John > Cc: Michael Dewey ; Dingyuan Wang > ; r-help@r-project.org > Subject: Re: [R] lm fails on some large input > > Um,

Re: [R] lm fails on some large input

> On Apr 18, 2019, at 8:24 AM, Michael Dewey wrote: > > Perhaps subtract 1506705766 from y? Good advice. Some further notes follow. One can specify `tol` to have a smaller than default value e.g. m2 <- lm(x ~ y, tol=1e-12) which is accurate: plot(y,x) abline(coef=coef(m2)) Users

Re: [R] lm fails on some large input

This sort of data arises quite easily if you deal with time/dates around now. E.g., > d <- data.frame( + when = seq(as.POSIXct("2017-09-29 18:22:01"), by="secs", len=10), + measurement = log2(1:10)) > coef(lm(data=d, measurement ~ when)) (Intercept) when

Re: [R] lm fails on some large input

d Dingyuan Wang, > >> -Original Message- >> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Michael >> Dewey >> Sent: Thursday, April 18, 2019 11:25 AM >> To: Dingyuan Wang ; r-help@r-project.org >> Subject: Re: [R] lm fails on some large inp

Re: [R] lm fails on some large input

Dear Michael and Dingyuan Wang, > -Original Message- > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Michael > Dewey > Sent: Thursday, April 18, 2019 11:25 AM > To: Dingyuan Wang ; r-help@r-project.org > Subject: Re: [R] lm fails on some large input >

Re: [R] lm fails on some large input

Perhaps subtract 1506705766 from y? Saying some other software does it well implies you know what the _correct_ answer is here but I would question what that means with this sort of data-set. On 17/04/2019 07:26, Dingyuan Wang wrote: Hi, This input doesn't have any interesting properties

[R] lm fails on some large input

Hi, This input doesn't have any interesting properties except y is unix time. Spreadsheets can do this well. Is this a bug that lm can't do x ~ y? R version 3.5.2 (2018-12-20) -- "Eggshell Igloo" Copyright (C) 2018 The R Foundation for Statistical Computing Platform: x86_64-pc-linux-gnu

Re: [R] lm equivalent of Welch-corrected t-test?

> On 13 Nov 2018, at 16:19 , Paul Johnson wrote: > > Long ago, when R's t.test had var.equal=TRUE by default, I wrote some > class notes showing that the result was equivalent to a one predictor > regression model. Because t.test does not default to var.equal=TRUE > these days, I'm curious

[R] lm equivalent of Welch-corrected t-test?

Long ago, when R's t.test had var.equal=TRUE by default, I wrote some class notes showing that the result was equivalent to a one predictor regression model. Because t.test does not default to var.equal=TRUE these days, I'm curious to know if there is a way to specify weights in an lm to obtain

Re: [R] lm model with many categorical variables

You need statistical help, which is generally off topic here. I suggest you post to a statistcal site like stats.stackexchange.com instead. Better yet, find a local statistical expert with whom you can consult. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep

Re: [R] lm model with many categorical variables

> On 20 Sep 2016, at 11:34, Michael Haenlein wrote: > > Dear all, > > I am trying to estimate a lm model with one continuous dependent variable > and 11 independent variables that are all categorical, some of which have > many categories (several dozens in some cases).

[R] lm model with many categorical variables

Dear all, I am trying to estimate a lm model with one continuous dependent variable and 11 independent variables that are all categorical, some of which have many categories (several dozens in some cases). I am not interested in statistical inference to a larger population. The objective of my

Re: [R] lm() silently drops NAs

> peter dalgaard > on Tue, 26 Jul 2016 23:30:31 +0200 writes: >> On 26 Jul 2016, at 22:26 , Hadley Wickham >> wrote: >> >> On Tue, Jul 26, 2016 at 3:24 AM, Martin Maechler >> wrote: >>>

Re: [R] lm() silently drops NAs

Agh. I've argued elsewhere that the default behaviour should be to fail, and the user should take the responsibility to explicitly handle the missing values, even if that simply be by changing the argument. Probably Peter and I have different experiences with the completeness of datasets, but

Re: [R] lm() silently drops NAs

> On 26 Jul 2016, at 22:26 , Hadley Wickham wrote: > > On Tue, Jul 26, 2016 at 3:24 AM, Martin Maechler > wrote: >> ... > To me, this would be the most sensible default behaviour, but I > realise it's too late to change without breaking many

Re: [R] lm() silently drops NAs

> I think that's a bit too strict for me, so I wrote my own: > > na.warn <- function(object, ...) { > missing <- complete.cases(object) > if (any(missing)) { > warning("Dropping ", sum(missing), " rows with missing values", > call. = FALSE) > } > > na.exclude(object, ...) > } That

Re: [R] lm() silently drops NAs

On Tue, Jul 26, 2016 at 3:24 AM, Martin Maechler wrote: > I have been asked (in private) Martin was very polite to not share my name, but it was me :) > > Hi Martin, > > y <- c(1, 2, 3, NA, 4) > x <- c(1, 2, 2, 1, 1) > > t.test(y ~ x) > lm(y ~ x)

Re: [R] lm() silently drops NAs

I have been asked (in private) > Hi Martin, y <- c(1, 2, 3, NA, 4) x <- c(1, 2, 2, 1, 1) t.test(y ~ x) lm(y ~ x) > Normally, most R functions follow the principle that > "missings should never silently go missing". Do you have > any background on why these

Re: [R] lm() with spearman corr option ?

I think you would just need to replace the lm() function call with cor(x,y,method="spearman". It would probably be more informative to actually plot by the magnitude of the correlation coefficient (all |r| >= 0.20 or something similar) rather than just by those with P <=0.05. Brian Brian S.

Re: [R] lm() with spearman corr option ?

Please read ?lm! -- where it says: method: the method to be used; for fitting, currently only method = "qr" is supported; method = "model.frame" returns the model frame (the same as with model = TRUE, see below). More to the point, your request for a "spearman" method for lm() makes little or

[R] lm() with spearman corr option ?

Hi, A following function was kindly provided by GGally’s maintainer, Barret Schloerke. function(data, mapping, ...) { p <- ggplot(data = data, mapping = mapping) + geom_point(color = I("blue")) + geom_smooth(method = "lm", color = I("black"), ...) + theme_blank() +

[R] lm-step: use train-data model to validate test-data

I have model-data named as: model that is split as model.T(train) and model.V(test or validation). The least square model (from lm to step) is built withmodel.T and I like to see how model.T is robust by comparing predicted model.V toactual model.V. How do I get score for model.V based on

Re: [R] lm

Your first predict sets up a newdata with a column name that is not the same as the one that you used in the lm formula. The failure to match causes R to look in the global environment, where it finds AT. You really should make a habit of always putting your regression data into a data.frame

[R] lm

Hello,I used l  to draw a figure, but I got different result (26 vs 301) when I input the same parameters. > length(predict(lm(A$Counts ~ AT),list(ATE=timevalues))) [1] 26 > length(predict(lm(A$Counts ~ ATE),list(ATE=timevalues))) [1] 301 all variables are initialized as below: >A <-

Re: [R] lm model exported from R to excel

. Cheers Petr -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Livia Maria Vestergaard Sent: Wednesday, May 06, 2015 11:37 AM To: r-help Subject: [R] lm model exported from R to excel Hi all I all. I am wondering whether anybody know how to export

Re: [R] lm model exported from R to excel

Hi Livia, One way is to use the delim.table function in the prettyR package. See the examples, in particular the final one. The resulting TAB delimited file will usually import directly into Excel. Jim On Wed, May 6, 2015 at 7:36 PM, Livia Maria Vestergaard lves...@student.sdu.dk wrote: Hi all

Re: [R] lm model exported from R to excel

-Original Message- From: Livia Maria Vestergaard [mailto:lves...@student.sdu.dk] Sent: Wednesday, 6 May 2015 22:37 To: Duncan Mackay; R Subject: SV: [R] lm model exported from R to excel Hi Duncan Thank you so much - it worked :) Best Livia Fra

[R] lm model exported from R to excel

Hi all I all. I am wondering whether anybody know how to export an output of an lm model from R to excel in order to have excel recognize the table that comes and divide the numbers in the table into columns and rows? I really hope it is possible? :) Best Livia

Re: [R] lm model exported from R to excel

and Soil Science University of New England Armidale NSW 2351 Email: home: mac...@northnet.com.au -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Livia Maria Vestergaard Sent: Wednesday, 6 May 2015 19:37 To: r-help Subject: [R] lm model exported from R

Re: [R] lm model exported from R to excel

; R Subject: Re: [R] lm model exported from R to excel Hi Duncan Thank you so much - it worked :) Best Livia Fra: Duncan Mackay [dulca...@bigpond.com] Sendt: 6. maj 2015 14:26 Til: R; Livia Maria Vestergaard Emne: RE: [R] lm model exported from R to excel

Re: [R] lm model exported from R to excel

Hi Duncan Thank you so much - it worked :) Best Livia Fra: Duncan Mackay [dulca...@bigpond.com] Sendt: 6. maj 2015 14:26 Til: R; Livia Maria Vestergaard Emne: RE: [R] lm model exported from R to excel Hi Livia There are several html packages

Re: [R] lm model exported from R to excel

it Duncan -Original Message- From: John Kane [mailto:jrkrid...@inbox.com] Sent: Thursday, 7 May 2015 01:34 To: Duncan Mackay; R Subject: Re: [R] lm model exported from R to excel And for those of us who know close to nothing about HTML I found just now that under a basic print.xtable commmand

Re: [R] lm model exported from R to excel

to LaTeX where the output looks beautiful. I like booktabs :) John Kane Kingston ON Canada -Original Message- From: dulca...@bigpond.com Sent: Thu, 7 May 2015 00:32:48 +1000 To: r-help@r-project.org Subject: Re: [R] lm model exported from R to excel If you know some basic html

Re: [R] LM() and time in R

a regression about to improve the model? Cheers Livia :) Fra: Boris Steipe [boris.ste...@utoronto.ca] Sendt: 29. april 2015 20:25 Til: Livia Maria Vestergaard Cc: r-help mailing list Emne: Re: [R] LM() and time in R Have a look at the help page

Re: [R] LM() and time in R

Vestergaard Cc: r-help mailing list Emne: Re: [R] LM() and time in R Have a look at the help page for the function ts() (type ?ts at the R prompt). Other than that, you haven't provided nearly enough information for us to diagnose your issue. Please see here for some hints on how to ask

Re: [R] LM() and time in R

Have a look at the help page for the function ts() (type ?ts at the R prompt). Other than that, you haven't provided nearly enough information for us to diagnose your issue. Please see here for some hints on how to ask questions productively: http://adv-r.had.co.nz/Reproducibility.html

[R] LM() and time in R

Hello, I need some help with a project that I’m working one. Im trying to make a l regression model (lm) in r with time as independant variable and gas prices as the depended. But It seems like everything im trying to run it, R freeze, I think that I need to tell R somehow that my time is

Re: [R] LM() and time in R

Hi Livia, From your description, it sounds like the time variable is actually a vector of date strings like 30/04/2015. These will usually be input as factors in R, meaning that you probably had a very large number of categorical values instead of numeric. To find out, do this:

Re: [R] lm() funtion

try lm.ridge from MASS package. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented,

[R] lm() funtion

Hi, Currently i am working with the lm() function for some regressions required for my project. suppose the formula parameter in that is given by response ~ terms,after some testing i found out that when the number of observations under terms is less than the number of columns or features

Re: [R] lm() funtion

You really really really need to work with a local statistical expert, as your post indicates fundamental confusion. Furthermore, statistical issues are off topic here. Cheers, Bert On Friday, April 24, 2015, Praveen kr singh pcubesi...@gmail.com wrote: Hi, Currently i am working with the

[R] lm weights argument within function

Hi all, I need to loop over lm within a function using weights. For example: mydata = data.frame(y=rnorm(100, 500, 100), x= rnorm(100), group=rep(c(0,1), 50), myweight=1/runif(100)) reg.by.wt - function(formula, wt, by, data) { if(missing(by)) { summary(lm(formula=formula, data=data,

Re: [R] lm weights argument within function

One way to accomplish this is to assign a new environment to the formula, an environment which inherits from the formula's original environment but one that you can add things to without affecting the original environment. Also, since you do this in a function only the copy of the formula in the

[R] lm and 450k data

Hi, I'm quite new to R and currently trying to use lm to fit linear models but I am currently stuck my code is as follows: Model1 = function(meth_matrix,exposure, X1, X2, X3, batch) { mod = lm(meth_matrix[, methcol]~exposure+X1+X2+X3+batch) res = summary(mod)$coef[2,]

[R] lm models over all possible pairwise combinations of the columns of two matrices

Dear all, I am working through a problem at the moment and have got stuck. I have searched around on the help list for assistance but could not find anything - but apologies if I have missed something. A dummy example of my problem is below. I will continue to work on it, but any help would be

Re: [R] lm models over all possible pairwise combinations of the columns of two matrices

Well... If my arithmetic and understanding is correct, that's 32 billion combinations, which, to put it politely, is nuts. As all you'll be doing is generating random numbers anyway, the fastest way to do this is just to use a random number generator. Cheers, Bert Bert Gunter Genentech

[R] lm predictions for rows with missing y values

Hi. I'm trying to produce lm fitted values and standard errors for cases with missing y values. I know how to compute these myself with matrix algebra, but I'm wondering if there is an appropriate na.action in the lm function to do this. Here is some simple code where I use na.action=NULL with

Re: [R] lm predictions for rows with missing y values

Yes. I believe what you're looking for is: See ?predict.lm and what it has to say about the na.action=na.exclude argument to lm. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not

Re: [R] lm predictions for rows with missing y values

Hello, I believe you want na.action = na.exclude. lmnew - lm(newy ~ newx,newdata,na.action=na.exclude) na.action can not be set to TRUE or FALSE. From the help page ?lm na.action a function which indicates what should happen when the data contain NAs. The default is set by the

[R] lm(y ~ group/x ) + predict.lm(...,type=terms)

Hi, all I am trying to figure out the computation result for predict.lm(...,type=terms) when the original fitting model has a nesting term, lm(y ~ group/x ). A example, set.seed(731) group - factor(rep(1:2, 200)) x - rnorm(400) fun1 - function(x) -3*x+8 fun2 - function(x) 15*x-18 y -

[R] lm(y ~ group/x ) + predict.lm(...,type=terms)

Hi, all I am trying to figure out the computation result for predict.lm(...,type=terms) when the original fitting model has a nesting term, lm(y ~ group/x ). A example, set.seed(731) group - factor(rep(1:2, 200)) x - rnorm(400) fun1 - function(x) -3*x+8 fun2 - function(x) 15*x-18 y -

[R] lm(y ~ group/x ) + predict.lm(...,type=terms)

Hi, all I am trying to figure out the computation result for predict.lm(...,type=terms) when the original fitting model has a nesting term, lm(y ~ group/x ). A example, set.seed(731) group - factor(rep(1:2, 200)) x - rnorm(400) fun1 - function(x) -3*x+8 fun2 - function(x) 15*x-18 y -

Re: [R] lm(y ~ group/x ) + predict.lm(...,type=terms)

Please see the posting guide: posting 3 times only reduces your chance of an informative answer (as does posting in HTML). Your posting is too vague for others to know what it is you do not understand. R is Open Source: please read the sources for the definitive answer to the 'know the

[R] lm Regression takes 24+ GB RAM - Error message

Hello, I am a rather unexperienced r-user (learned the language 1 month ago) and run into the following problem using a local computer with 6 cores 24 GB RAM and R 2.15 64-bit. I didn't install any additional packages 1. Via the read.table command I load a data table (with different data types)

Re: [R] lm and Formula tutorial

Dear Alex, Here you have some url's: http://data.princeton.edu/R/linearModels.html http://www.r-bloggers.com/r-tutorial-series-simple-linear-regression/ Regards, Eva --- El mié, 6/3/13, Alaios ala...@yahoo.com escribió: De: Alaios ala...@yahoo.com Asunto: [R] lm and Formula tutorial Para: R

Re: [R] lm Regression takes 24+ GB RAM - Error message

On Wed, Mar 6, 2013 at 9:51 AM, Jonas125 schleeberge...@pg.com wrote: Hello, I am a rather unexperienced r-user (learned the language 1 month ago) and run into the following problem using a local computer with 6 cores 24 GB RAM and R 2.15 64-bit. I didn't install any additional packages 1.

Re: [R] lm and Formula tutorial

I may be wrong, but I believe what makes it difficult is that the Help file assumes some linear model statistics that you may not have. I suggest that you look for a tutorial on linear models first and then re-read the Help. Incidentally, the provenance of the syntax is GLIM (correction requested

Re: [R] lm Regression takes 24+ GB RAM - Error message

The datatable (and the split obviously) only contain characters and numeric data. I found that 4 regression in a row work if I don't use the calculated columns as variables but 2 of the original columns. RAM usage stays below 3GB! -- Why does R has such problems with the calculated columns?

Re: [R] lm Regression takes 24+ GB RAM - Error message

Le mercredi 06 mars 2013 à 08:31 -0800, Jonas125 a écrit : The datatable (and the split obviously) only contain characters and numeric data. I found that 4 regression in a row work if I don't use the calculated columns as variables but 2 of the original columns. RAM usage stays below 3GB!

Re: [R] lm Regression takes 24+ GB RAM - Error message

Length(Datasplit) = 7100 I did a regression for Datasplit[[1]] and the calculated columns -- the object size is 70 MB. Quite large Assuming that R cannot handle inf values in regressions (didn't have the time to google it) How can I avoid the calculation of infinite values? Like If the

Re: [R] lm Regression takes 24+ GB RAM - Error message

Le mercredi 06 mars 2013 à 09:18 -0800, Jonas125 a écrit : Length(Datasplit) = 7100 I did a regression for Datasplit[[1]] and the calculated columns -- the object size is 70 MB. Quite large 7100*70/1024 = 485 (GB) No wonder why you run out of memory quite fast. You probably do not need

[R] lm and Formula tutorial

Dear all, I was reading last night the lm and the Formula manual page, and 'I have to admit that I had tough time to understand their syntax. Is there a simpler guide for the dummies like me to start with? I would like to thank you in advance for your help Regards Alex [[alternative

Re: [R] lm regression query

Smells like homework to me. If so, we don't do homework on this list. -- Bert On Thu, Feb 14, 2013 at 3:55 PM, email email8...@gmail.com wrote: Hello: I have a 4-column dataset: Crime, Education, Urbanization, Age. I want to construct a multiple linear regression to find the effect of

[R] lm regression query

Hello: I have a 4-column dataset: Crime, Education, Urbanization, Age. I want to construct a multiple linear regression to find the effect of Education, Urbanization, and Age on Crime lm(Crime ~ Education + Urbanization + Age) If I use + in above statement, does it mean it will build a model to

Re: [R] lm regression query

Hi, Did you read the help file? Particularly the section Details. ?lm Regards, Pascal Le 15/02/2013 08:55, email a écrit : Hello: I have a 4-column dataset: Crime, Education, Urbanization, Age. I want to construct a multiple linear regression to find the effect of Education, Urbanization,

Re: [R] lm regression query

You will also need to specify/ name your model: Mod - lm(Crime~. ~Nicole Ford Graduate Instructor Department of Government and International Affairs University of South Florida office: SOC 012M e: nmhi...@mail.usf.edu http://gia.usf.edu/student/nford/ Sent from my iPhone On Feb 14,

Re: [R] lm regression query

You could also run it and find out. There are many R tutorials free online. I presume crime is continuous, as well...? ~Nicole Ford Graduate Instructor Department of Government and International Affairs University of South Florida office: SOC 012M e: nmhi...@mail.usf.edu

[R] lm function - strange error

I am following a document teaching how to use regression and right at the onset I get an R error. I understand that variables conc and signal should have the same length but I am using the what R manual suggests to drop the error and it is not cooperating. What gives? The manual says that the

Re: [R] lm function - strange error

On Fri, Nov 9, 2012 at 4:37 PM, tesssa tesara...@gmail.com wrote: I am following a document teaching how to use regression and right at the onset I get an R error. I understand that variables conc and signal should have the same length but I am using the what R manual suggests to drop the

Re: [R] lm function - strange error

-Original Message- From: tesara...@gmail.com Sent: Fri, 9 Nov 2012 08:37:13 -0800 (PST) To: r-help@r-project.org Subject: [R] lm function - strange error I am following a document teaching how to use regression and right at the onset I get an R error. I understand that variables conc

Re: [R] lm function - strange error

Sorry it is not a comma but a tilda. The R.help message editor Please replace the dash with a tilda as lm function requires -- View this message in context: http://r.789695.n4.nabble.com/lm-function-strange-error-tp4649065p4649074.html Sent from the R help mailing list archive at

[R] lm function - na.action

How do you lm throw away excess data points. I am following the documentation with no success -- View this message in context: http://r.789695.n4.nabble.com/lm-function-na-action-tp4649075.html Sent from the R help mailing list archive at Nabble.com.

Re: [R] lm function - strange error

Here is the document http://www.montefiore.ulg.ac.be/~kvansteen/GBIO0009-1/ac20092010/Class8/Using%20R%20for%20linear%20regression.pdf -- View this message in context: http://r.789695.n4.nabble.com/lm-function-strange-error-tp4649065p4649080.html Sent from the R help mailing list archive at

Re: [R] lm function - strange error

On Nov 9, 2012, at 9:16 AM, tesssa wrote: Sorry it is not a comma but a tilda. The R.help message editor Please replace the dash with a tilda as lm function requires It might be interesting to have such a supernatural entity zooming around the Rhelp Universe answering prayers from

Re: [R] lm function - strange error

hope this helps. John Kane Kingston ON Canada -Original Message- From: tesara...@gmail.com Sent: Fri, 9 Nov 2012 09:36:23 -0800 (PST) To: r-help@r-project.org Subject: Re: [R] lm function - strange error Here is the document http://www.montefiore.ulg.ac.be/~kvansteen/GBIO0009-1

Re: [R] lm function - strange error

-Original Message- From: dwinsem...@comcast.net Sent: Fri, 9 Nov 2012 10:56:48 -0800 To: tesara...@gmail.com Subject: Re: [R] lm function - strange error On Nov 9, 2012, at 9:16 AM, tesssa wrote: Sorry it is not a comma but a tilda. The R.help message editor Please

Re: [R] lm function - strange error

On Fri, Nov 9, 2012 at 2:20 PM, John Kane jrkrid...@inbox.com wrote: -Original Message- From: dwinsem...@comcast.net Sent: Fri, 9 Nov 2012 10:56:48 -0800 To: tesara...@gmail.com Subject: Re: [R] lm function - strange error On Nov 9, 2012, at 9:16 AM, tesssa wrote: Sorry

Re: [R] lm function - strange error

) I hope this helps. John Kane Kingston ON Canada -Original Message- From: tesara...@gmail.com Sent: Fri, 9 Nov 2012 09:36:23 -0800 (PST) To: r-help@r-project.org Subject: Re: [R] lm function - strange error Here is the document http://www.montefiore.ulg.ac.be

Re: [R] lm function - strange error

-Original Message- From: sarah.gos...@gmail.com Sent: Fri, 9 Nov 2012 14:34:57 -0500 To: jrkrid...@inbox.com Subject: Re: [R] lm function - strange error On Fri, Nov 9, 2012 at 2:20 PM, John Kane jrkrid...@inbox.com wrote: -Original Message- From: dwinsem

Re: [R] lm function - strange error

Subject: Re: [R] lm function - strange error Hi Tessa, I agree with John. I think you've made a typo, but looking at your data I think the zero concentration should not be there. Try plotting it. conc = c(10, 20, 30, 40, 50) signal = c (4, 22, 44, 60, 82) plot(signal~conc) abline

Re: [R] lm function - na.action

On Fri, Nov 9, 2012 at 5:21 PM, tesssa tesara...@gmail.com wrote: How do you lm throw away excess data points. I am following the documentation with no success Why would you want to throw away this data? I think that's generally considered sub-optimal model-fitting technique, thought it might

[R] lm and glm

Dear list, I am making a linear regression of the following format. lm(Y~X1+X2+log(X3)+log(X4)) Now I would like to check the linear regression above using generalized linear model. Please kindly if the following format is correct and thank you. (If it is wrong, please indicate why.)

Re: [R] lm on matrix data

Baoqiang, Here's an approach that should work: (1) Make sure that the column names of trainx and testx are the same. (2) Combine trainy and trainx into a data frame for fitting the model. (2) Use the newdata= argument in the predict() function. (3) Convert testx from matrix to data frame. # some

[R] lm on matrix data

Hi, I have a question about using lm on matrix, have to admit it is very trivial but I just couldn't find the answer after searched the mailing list and other online tutorial. It would be great if you could help. I have a matrix trainx of 492(rows) by 220(columns) that is my x, and trainy is 492

Re: [R] lm on matrix data

On Wed, Oct 10, 2012 at 3:35 PM, Baoqiang Cao bqcaom...@gmail.com wrote: Hi, I have a question about using lm on matrix, have to admit it is very trivial but I just couldn't find the answer after searched the mailing list and other online tutorial. It would be great if you could help. I

Re: [R] lm without intercept

The cor(mtcars$mpg, fitted(m0))^2 method works great - thanks so much Josh! I've had another instance (also ex intercept) where it actually gave the correct number, odd behavior. Thanks for the other posts also. As an aside, in my application a zero intercept makes economic sense and I'm using

Re: [R] lm with a single X and step with several Xi-s, beta coef. quite different:

Hi, Sounds like suppression -- see e.g., http://www.jstor.org/stable/2988294?seq=1 for a discussion. Since this is not an R question but a statistical one, it may be more appropriate to post this question to a statistics forum such as http://stats.stackexchange.com/ Best, Ista On Tue, Aug 7,

[R] lm with a single X and step with several Xi-s, beta coef. quite different:

Hi, (R version 2.15.0) I am running a pgm with 1 response (earlier standardized Y) and 44 independent vars (Xi) from the same data =a2: When I run the 'lm' function on single Xi at a time, the beta coefficient for let's say X1 is = -0.08 (se=0.03256) But when I run the same Y with 44 Xi-s with

Re: [R] lm without intercept

On Feb 18, 2011, at 14:20 , Jan wrote: Hello Achim, Not quite. Consult your statistics textbook for the correct interpretation of p-values. Under the null hypothesis of a true intercept of zero, it is very likely to observe an intercept as large as 13.52 or larger. thank you for that

Re: [R] lm without intercept

On Feb 18, 2011, at 14:20 , Jan wrote: One of the references you googled suggests that intercepts should never be omitted. Is this true even if I know that the physical reality behind the numbers suggests an intercept of zero? No. That'll be a piece of pragmatic advice caused by the

Re: [R] lm without intercept

I've just picked up R (been using Matlab, Eviews etc) and I'm having the same issue. Running reg=lm(ticker1~ticker2) gives R^2=50% while running reg=lm(ticker1~0+ticker2) gives R^2=99%!! The charts suggest the fit is worse not better and indeed Eviews/Excel/Matlab all say R^2=15% with

Re: [R] lm without intercept

Hi, R actually uses a different formula for calculating the R square depending on whether the intercept is in the model or not. You may also find this discussion helpful: http://stats.stackexchange.com/questions/7948/when-is-it-ok-to-remove-the-intercept-in-lm/ If you conceptualize R^2 as the

[R] lm without intercept, false R-squared

Hi is there a command that calculates the correct adjusted R-squared, when I work without intercept? (The R-squared from lm without intercept is false.) Greetings Chrsitof __ R-help@r-project.org mailing list

Re: [R] lm without intercept, false R-squared

On 27.06.2012 09:33, Christof Kluß wrote: Hi is there a command that calculates the correct adjusted R-squared, when I work without intercept? (The R-squared from lm without intercept is false.) Then we need your definition of your version of correct - we know the definition of your

Re: [R] lm without intercept, false R-squared

On 06/27/2012 10:33 AM, Christof Kluß wrote: Hi is there a command that calculates the correct adjusted R-squared, when I work without intercept? (The R-squared from lm without intercept is false.) Hi! This answer in R-FAQ might help you:

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