1. You should regress Elevation on Volume, no?
2. You are calling lm incorrectly for prediction. Please read ?lm and
related links carefully and/or consult a tutorial. R-Help is really not the
first place you should look for this sort of detailed info.
3. I think this is what you want:
lm1 <- lm
I think you might be looking for
?contrasts
to form the contrast matrix.
Rich
On Mon, May 13, 2019 at 7:31 AM Witold E Wolski wrote:
>
> I am looking for a function to compute contrasts with a interface
> similar to that of
>
> lmerTest::contest
> multcomp::glht
>
> i.e. taking the model and a c
Thanks Brian for all your kind help.
"didn't mean to imply that the different parameterization of the contrasts
would make the lm estimates agree more with the lmer estimates, only that
it might be easier to compare the regression summary output to see how
similar/dissimilar they were ".
Got it no
Utkarsh: I think the differences between the lm and lmer estimates of the
intercept are consistent with the regularization effect expected with
mixed-effects models where the estimates shrink towards the mean slightly.
I don't think there is any reason to expect exact agreement between the lm
and
Hi Brian,
This makes some sense to me theoretically, but doesn't pan out with my
experiment.
The contrasts default was the following as you said:
> options("contrasts")
$contrasts
unordered ordered
"contr.treatment" "contr.poly"
I changed it as follows:
> options(contracts
Your lm() estimates are using the default contrasts of contr.treatment,
providing an intercept corresponding to your subject 308 and the other
subject* estimates are differences from subject 308 intercept. You could
have specified this with contrasts as contr.sum and the estimates would be
more ea
Hello Thierry,
Thank you for your quick response. Sorry, but I am not sure if I follow
what you said. I get the following outputs from the two models:
> coef(lmer(Reaction ~ Days + (1| Subject), sleepstudy))
Subject(Intercept) Days
308292.1888 10.46729
309173.5556 10.46729
3101
The parametrisation is different.
The intercept in model 1 is the effect of the "average" subject at days ==
0.
The intercept in model 2 is the effect of the first subject at days == 0.
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Bio
One technique for dealing with this is called 'multiple imputation'.
Google for 'multiple imputation in R' to find R packages that implement
it (e.g., the 'mi' package).
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Mar 15, 2016 at 8:14 AM, Lorenzo Isella
wrote:
> Dear All,
> A situation
IMHO this is not a question about R... it is a question about statistics
whether R is involved or not. As such, a forum like stats.stackexchange.com
would be better suited to address this.
FWIW I happen to think that expecting R to solve this for you is unreasonable.
--
Sent from my phone. Ple
On 16/11/15 20:49, Ragia Ibrahim wrote:
Dear group IF I had an objective function and some constrains formed
in linear model form. is there a way,..library in R that helps me to
solve such amodel and find the unknown variable in it?
This is a very ill-posed question and is unlikely to provoke
On Jun 13, 2013, at 2:21 PM, Bert Gunter wrote:
> Lorenzo:
>
> 1. This is a statistics question, not an R question.
>
> 2. Your statistical background appears inadequate -- it looks like
> Poisson regression, which would fall under "generalized linear
> models". But it depends on how "discrete
Lorenzo:
1. This is a statistics question, not an R question.
2. Your statistical background appears inadequate -- it looks like
Poisson regression, which would fall under "generalized linear
models". But it depends on how "discrete" discrete is (on some level,
all measurements are discrete, dis
Carmo
Cc: r-help@r-project.org; Adams, Jean
Assunto: Re: [R] linear model coefficients by year and industry, fitted values,
residuals, panel data
On 2013-04-04 02:11, Cecilia Carmo wrote:
> Thank you all. I'm very happy with this solution. Just two questions:
> I use mutate() with package
ter Ehlers
De: Peter Ehlers [ehl...@ucalgary.ca]
Enviado: quarta-feira, 3 de Abril de 2013 19:01
Para: Adams, Jean
Cc: Cecilia Carmo; r-help@r-project.org
Assunto: Re: [R] linear model coefficients by year and industry, fitted values,
residuals, panel data
A
Carmo
De: Peter Ehlers [ehl...@ucalgary.ca]
Enviado: quarta-feira, 3 de Abril de 2013 19:01
Para: Adams, Jean
Cc: Cecilia Carmo; r-help@r-project.org
Assunto: Re: [R] linear model coefficients by year and industry, fitted values,
residuals, panel data
A few minor improvements to Jean's post
On 04/04/2013 07:54 AM, Adams, Jean wrote:
Peter.
For suggestion 1, what advantages are there to using coef() rather than
$coef?
Just thought I'd chip in: It is considered, uh, politically correct to use
extractor functions rather than digging out components of objects
in a "direct" manner.
Peter.
For suggestion 1, what advantages are there to using coef() rather than
$coef?
For suggestion 2, thanks! I'm new to the plyr package and wasn't aware of
the mutate() function.
Jean
On Wed, Apr 3, 2013 at 1:01 PM, Peter Ehlers wrote:
> A few minor improvements to Jean's post suggested
A few minor improvements to Jean's post suggested inline below.
On 2013-04-03 05:41, Adams, Jean wrote:
Cecilia,
Thanks for providing a reproducible example. Excellent.
You could use the ddply() function in the plyr package to fit the model for
each industry and year, keep the coefficients, a
Cecilia,
Thanks for providing a reproducible example. Excellent.
You could use the ddply() function in the plyr package to fit the model for
each industry and year, keep the coefficients, and then estimate the fitted
and residual values.
Jean
library(plyr)
coef <- ddply(final3, .(industry, yea
Hello,
The example below gives you some information about using predict(). This was
originally used to predict NAs.
dat1<-data.frame(Year=1962:1986,Discharge=c(rnorm(15,25),rep(NA,5),1:5))
lm1<-lm(Discharge~Year,dat1)
dat2<-data.frame(Year=dat1[,1])
dat1$fit<-predict(lm1,newdata=dat2)
#The c
On Tue, Jul 24, 2012 at 2:06 PM, wrote:
> Yes, why wouldn't I? It's a linear model between two sets of data: x and y.
Conventionally, one predicts y based on x -- which is specified y ~ x,
not x ~ y. (Predictors on the RHS, predicted on the LHS)
>
> Also, what would the new data be if i want to
Yes, why wouldn't I? It's a linear model between two sets of data: x and y.
Also, what would the new data be if i want to predict into the future? So,
for example, the data goes from a month ago to today. I want to predict what
tomorrow's data would be. So what is "newdata"?
--
View this mes
On Jul 24, 2012, at 1:38 PM, cm wrote:
> How do I set it up? Because when I do predict(model) I get a ton of points,
> not just one.
You need to supply newdata= . predict() without new data gives predicted
values for the predictors you for the model to.
Incidentally, repeating Uwe -- are
How do I set it up? Because when I do predict(model) I get a ton of points, not
just one.
- Original Message -
From: "Uwe Ligges-3 [via R]"
Date: Tuesday, July 24, 2012 2:28 pm
Subject: Re: Linear Model Prediction
To: cm
>
>
>
>
> On 24.07.2012 20:20, cm wrote:
> > I have data X a
On 24.07.2012 20:20, cm wrote:
I have data X and Y, and I want to predict what the very next point would be
based off the model. This is what I have:
model=lm(x~y)
Hmmm, are you sure about the above code?
I think I want to use the predict function, but I'm not exactly sure what to
do.
Y
Hi Weidong,
thank you very much. It really works fine.
Robert
2011/6/12 Weidong Gu :
> this may work.
> X<-data.frame(sapply(X,function(x) as.factor(x)))
> reg3=lm(Y~.,data=X)
> dummy.coef(reg3)
>
> Weidong Gu
>
> On Sun, Jun 12, 2011 at 4:55 PM, Robert Ruser wrote:
>> Hi,
>> but I want to get t
this may work.
X<-data.frame(sapply(X,function(x) as.factor(x)))
reg3=lm(Y~.,data=X)
dummy.coef(reg3)
Weidong Gu
On Sun, Jun 12, 2011 at 4:55 PM, Robert Ruser wrote:
> Hi,
> but I want to get the coefficients for every variables from x1 to x5.
> (x1 was an example)
>
> Robert
>
> 2011/6/12 Jorge
Hi,
but I want to get the coefficients for every variables from x1 to x5.
(x1 was an example)
Robert
2011/6/12 Jorge Ivan Velez :
> Hi Robert,
>
> Try this:
> reg2 <- lm( Y ~ factor(x1) + factor(x2) + factor(x3) + factor(x4) +
> factor(x5) - 1, data = X )
> cof(ref2)
> HTH,
> Jorge
>
> On Sun, J
Hi Robert,
Try this:
reg2 <- lm( Y ~ factor(x1) + factor(x2) + factor(x3) + factor(x4) +
factor(x5) - 1, data = X )
cof(ref2)
HTH,
Jorge
On Sun, Jun 12, 2011 at 4:40 PM, Robert Ruser <> wrote:
> Prof. Ripley, thank you very much for the answer but wanted to get
> something else. There is an
Prof. Ripley, thank you very much for the answer but wanted to get
something else. There is an example and an explanation:
options(contrasts=c("contr.sum","contr.poly")) # contr.sum uses ‘sum
to zero contrasts’
Y <- c(6,3,5,2,3,1,1,6,6,6,7,4,1,6,6,6,6,1)
X <- structure(list(x1 = c(2L, 3L, 1L, 3L,
?dummy.coef
(NB: 'R' does as you tell it, and if you ask for the default contrasts
you get coefficients a2 and a3, not a1 and a2. So perhaps you did
something else and failed to tell us? And see the comment in
?dummy.coef about treatment contrasts.)
On Sun, 12 Jun 2011, Robert Ruser wrote
Thank you very much for all of your help.
On Mon, Apr 4, 2011 at 6:10 PM, Steven McKinney wrote:
>
>
>> -Original Message-
>> From: stephen sefick [mailto:ssef...@gmail.com]
>> Sent: April-04-11 2:49 PM
>> To: Steven McKinney
>> Subject: Re: [R] Linea
> -Original Message-
> From: stephen sefick [mailto:ssef...@gmail.com]
> Sent: April-04-11 2:49 PM
> To: Steven McKinney
> Subject: Re: [R] Linear Model with curve fitting parameter?
>
> Steven:
>
> I am really sorry for my confusion. I hope this now makes s
> -Original Message-
> From: stephen sefick [mailto:ssef...@gmail.com]
> Sent: April-03-11 5:35 PM
> To: Steven McKinney
> Cc: R help
> Subject: Re: [R] Linear Model with curve fitting parameter?
>
> Steven:
>
> You are exac
McKinney wrote:
>
>> -Original Message-
>> From: stephen sefick [mailto:ssef...@gmail.com]
>> Sent: April-01-11 5:44 AM
>> To: Steven McKinney
>> Cc: R help
>> Subject: Re: [R] Linear Model with curve fitting parameter?
>>
>> Setting Z=Q-A would
> -Original Message-
> From: stephen sefick [mailto:ssef...@gmail.com]
> Sent: April-01-11 5:44 AM
> To: Steven McKinney
> Cc: R help
> Subject: Re: [R] Linear Model with curve fitting parameter?
>
> Setting Z=Q-A would be the incorrect dimensions. I could Z=Q
On 2011-04-01 05:44, stephen sefick wrote:
Setting Z=Q-A would be the incorrect dimensions. I could Z=Q/A. Is
fitting a nls model the same as fitting an ols? These data are
hydraulic data from ~47 sites. To access predictive ability I am
removing one site fitting a new model and then accessin
Setting Z=Q-A would be the incorrect dimensions. I could Z=Q/A. Is
fitting a nls model the same as fitting an ols? These data are
hydraulic data from ~47 sites. To access predictive ability I am
removing one site fitting a new model and then accessing the fit with
a myriad of model assessment c
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of stephen sefick
> Sent: March-31-11 3:38 PM
> To: R help
> Subject: [R] Linear Model with curve fitting parameter?
>
> I have a model Q=K*A*(R^r)*(S^s)
>
> A, R, and S are data
See:
http://en.wikipedia.org/wiki/Coefficient_of_determination#Adjusted_R2
and the implementation in summary.lm :
ans$adj.r.squared <- 1 - (1 - ans$r.squared) * ((n -
df.int)/rdf)
Brian Smith wrote:
Hi,
Sorry for the naive question, but what exactly does the 'Adjusted R
No, as the error states, you need random effects in lmer. But, you don't for
lm() and that is what you're running with no random effects. However, some
caution is warranted on the comparison.
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Aug 5, 2010, at 6:50 AM, Giuseppe Amatulli wrote:
Hi,
can somebody tell me why R is not able to calculate a linear model
written in this way?
lm (seq(1:100)~seq(1:100))
Call:
lm(formula = seq(1:100) ~ seq(1:100))
Coefficients:
(Intercept)
50.5
Warning messages:
1: In model.matrix
On 08/05/2010 05:50 AM, Giuseppe Amatulli wrote:
Hi,
can somebody tell me why R is not able to calculate a linear model
written in this way?
lm (seq(1:100)~seq(1:100))
Call:
lm(formula = seq(1:100) ~ seq(1:100))
Coefficients:
(Intercept)
50.5
Warning messages:
1: In model.matrix.def
On Nov 13, 2009, at 11:49 AM, Sam Albers wrote:
Hello R list,
snipped answered question
Sorry to not use your data but it's not in a form that lends itself
very well to quick testing. If you had included the input commands I
might have tried it.
No problem not use my data. For futur
On Fri, Nov 13, 2009 at 11:49 AM, Sam Albers wrote:
> No problem not use my data. For future reference, would it have been easier
> to attach a .csv file and then include the appropriate read.csv command? I
> realized that the easier one makes it to help, the easier it is to get a
> response.
>
>
> Hello R list,
>>
>> This is a question for anyone who has used the by() command. I would like
>> to
>> perform a regression on a data frame by several factors. Using by() I
>> think
>> that I have able to perform this using the following:
>>
>> lm.r <- by(master, list(Sectionf=Sectionf, startd=s
On Nov 12, 2009, at 8:26 PM, Sam Albers wrote:
Hello R list,
This is a question for anyone who has used the by() command. I would
like to
perform a regression on a data frame by several factors. Using by()
I think
that I have able to perform this using the following:
lm.r <- by(master, l
Hi,
You have not given us all the data needed to reproduce your analysis
(what is SectionF?), but the issue is probably that lm.r is a list and
you're not treating it that way. Try
srt(lm.r)
and
summary(lm.r[[1]])
You may also want to look at the the lmList() function in the lme4 package.
-Ist
Rnewb wrote:
>
> I would like to perform a regression like the one below:
>
> lm(x ~ 0 + a1 + a2 + a3 + b1 + b2 + b3 + c1 + c2 + c3, data=data)
>
> However, the data has the property that a1+a2+a3 = A, b1+b2+b3 = B, and
> c1+c2+c3 = C, where A, B,
>
Ravi Varadhan has an example how this co
On 21-Sep-09 20:38:25, Douglas M. Hultstrand wrote:
> Hello,
>
> I am deriving near real-time liner relationships based on 5-min
> precipitation data, sometimes the non-qced data result in a
> slope of NA. I am trying to read the coefficient (in this example x)
> to see if it is equal to NA, if it
On Sep 21, 2009, at 4:50 PM, David Winsemius wrote:
On Sep 21, 2009, at 4:38 PM, Douglas M. Hultstrand wrote:
Hello,
I am deriving near real-time liner relationships based on 5-min
precipitation data, sometimes the non-qced data result in a slope
of NA. I
am trying to read the coefficien
On Sep 21, 2009, at 4:38 PM, Douglas M. Hultstrand wrote:
Hello,
I am deriving near real-time liner relationships based on 5-min
precipitation data, sometimes the non-qced data result in a slope of
NA. I
am trying to read the coefficient (in this example x) to see if it
is equal
to NA, if
> if("fit$coef[[2]]" == "NA") {.cw = 1}
See ?is.na
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-conta
On Sat, 2009-08-29 at 12:56 +0100, Markus Gesmann wrote:
> Dear R-help,
>
> Suppose I have the following data:
>
> df=data.frame(x=1:10, y=c(1,2,3,4,5,12,14,16,18,20))
> plot(y~x, df, t="b")
>
> How can I fit a model which estimates the slopes between x = 1-5, 5-6,
> and 6-10?
Does the segmen
On Aug 29, 2009, at 7:56 AM, Markus Gesmann wrote:
Dear R-help,
Suppose I have the following data:
df=data.frame(x=1:10, y=c(1,2,3,4,5,12,14,16,18,20))
plot(y~x, df, t="b")
How can I fit a model which estimates the slopes between x = 1-5,
5-6, and 6-10?
Adding the factor f:
df$f <- gl(
On Sat, 8 Aug 2009, Katharina May wrote:
Thanks to somebody I got the hint to use offset for the purpose of
validating if there's
a difference between the intercept and slope of a model and some
provided values for
the coefficients intercept and slope.
You could also use a Wald test for a line
Thanks to somebody I got the hint to use offset for the purpose of
validating if there's
a difference between the intercept and slope of a model and some
provided values for
the coefficients intercept and slope.
I read ?model.offset and I'm still struggling to use it for my
purpose. If I understoo
On Jun 5, 2009, at 11:10 AM, Axel Leroix wrote:
Hi every one
I perform a simple linear regression
lm(a b + c + d , data = data1)
How to say to R to perform and print the regression with restricting
the coefficient
of the variable c to be equal to 0.1.
?lm
Examine material on specifyin
I'm not sure what you are doing when you "Normalize." Would you explain?
To see if the slope is significant, look at the model summary, in your
example:
summary(model)
Charles Annis, P.E.
charles.an...@statisticalengineering.com
phone: 561-352-9699
eFax: 614-455-3265
http://www.Statistical
Melissa2k9 wrote:
Hi
for some data I working on I am merely plotting time against temperature for
a variable named filmclip. So for example, I have volunteers who watched
various film clips and have used infared camera to monitor the temperature
on their face at every second of the clip.
Th
On Wed, Feb 11, 2009 at 1:36 PM, kayj wrote:
>
> I want to know how accurate are the p-values when you do linear regression in
> R?
>
> I was looking at the variable x3 and the t=10.843 and the corresponding
> p-value=2e-16 which is the same p-value for the intercept where the t-value
> for the in
If you look carefully you will see that the p-value is actually
< 2e-16
where the "<" sign is telling you this is an upper limit. For many machines
this is the double precision epsilon value:
> .Machine$double.eps
[1] 2.220446e-16
So there is no pretence to accuracy, other than "the value is
The problem comes from mixing up general linear model (LM) theory to compute
B with the classical anova estimators. The two methods use different
approaches to solving the normal equations. LM theory uses any generalized
inverse of X'X to solve the normal equations. Yours comes from ginv() whic
> I have a set of points (measurements) and I used lm() to obtain their linear
> regression model. From the biological background this line must pass through
> a point (100,0). Our dataset is not optimal and it shows a slight deviation
> from that coordinate. How can I add the restraint to the mode
hursday, June 05, 2008 3:03 PM
To: Manli Yan
Cc: r-help@r-project.org
Subject: Re: [R] linear model in the repeated data type~
allFits <- lmList(y ~ t|id, data=table1, pool=FALSE)
allCoefs <- sapply(allFits, coef) ## preferred by me
or
allCoefs <- list(length(allFits))
for(i in 1:le
IL PROTECTED]
Sent: Wednesday, June 04, 2008 9:12 PM
To: Austin, Matt
Cc: r-help@r-project.org
Subject: Re: [R] linear model in the repeated data type~
hi:lot thanks,how to use list to extract,I type allFit$coefficents,it came to
nothing,
such as I need to extract the estimates,how to do it by using
hi:lot thanks,how to use list to extract,I type allFit$coefficents,it came
to nothing,
such as I need to extract the estimates,how to do it by using list
2008/6/3 Austin, Matt <[EMAIL PROTECTED]>:
> How about
>
>
> library(nlme)
> allFits <- lmList(y ~ t|id, data=table1, pool=FALSE)
>
> or
>
> a
Try this:
f <- function(x)any(is.na(coefficients(x)))
models <- by(table1[c("y", "t")], table1$id, FUN=lm)
models[!unlist(lapply(models, f))]
On Wed, Jun 4, 2008 at 6:20 PM, Manli Yan <[EMAIL PROTECTED]> wrote:
> here is the data:
> y<-c(5,2,3,7,9,0,1,4,5)
> id<-c(1,1,6,6,7,8,15,15,19)
> t<-c
Try something like this:
fits <- list(500)
for (i in 1:500)
{
if (sum(table1$id == i) == 0) fits[[i]] <- NA
else fits[[i]] <- lm(y~t,data=table1,subset=(id==i))
}
--- On Wed, 4/6/08, Manli Yan <[EMAIL PROTECTED]> wrote:
> From: Manli Yan <[EMAIL PROTECTED]>
> Subject: [R] linear model in the re
How about
library(nlme)
allFits <- lmList(y ~ t|id, data=table1, pool=FALSE)
or
allFits <- by(table1, table1$id, function(x) lm(y ~ t, data=x))
Both ways store the results as a list, so you can access individual results
using list extraction.
--Matt
-Original Message-
From: [EMAIL
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