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
Dear all;
I have a dataframe with several columns. The columns are the elevation,
volume and the area of the cells (which were placed inside a polygon). I
have extracted them from DEM raster to calculate the volume under polygon
and the elevation for a specific volume of the reservoir.
> head(x6
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
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 contrast vector or matrix as an argument,
but for linear models, and without the multiple testing adjusted made
by multcomp::glht.
Thank you
--
Wit
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
Hi experts,
While the slope is coming out to be identical in the two methods below, the
intercepts are not. As far as I understand, both are formulations are
identical in the sense that these are asking for a slope corresponding to
'Days' and a separate intercept term for each Subject.
# Model-1
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
Dear All,
A situation that for sure happens very often: suppose you are in the
following situation
set.seed(1235)
x1 <- seq(30)
x2 <- c(rep(NA, 9), rnorm(19)+9, c(NA, NA))
x3 <- c(rnorm(17)-2, rep(NA, 13))
y <- exp(seq(1,5, length=30))
mm<-lm(y~x1+x2+x3)
i.e. you try a simple linear regressio
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
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?
thanks in advance
Ragia
__
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
Dear All,
I am struggling with a linear model and an allegedly trivial data set.
The data set does not consist of categorical variables, but rather of
numerical discrete variables (essentially, they count the number of times
that something happened).
Can I still use a standard linear regressi
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
Hi R-helpers,
My real data is a panel (unbalanced and with gaps in years) of thousands of
firms, by year and industry, and with financial information (variables X, Y, Z,
for example), the number of firms by year and industry is not always equal, the
number of years by industry is not always e
Dear R-users,
in the last days I have been trying to estimate a normal linear model with
equality and inequality constraints.
Please find below a simple example of my problem.
Of course, one could easily see that, though the constraints are consistent,
there is some redundancy in the specific
a=dat2)
#The code below is to replace the NA values with predicted.
#dat1<-within(dat1,{Dischargenew<-ifelse(is.na(Discharge)==T,fit,Discharge)})
#dat1new<-dat1[,c(1:2,4)]
A.K.
- Original Message -
From: cm
To: r-help@r-project.org
Cc:
Sent: Tuesday, July 24, 2012 2:20 P
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
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)
I think I want to use the predict function, but I'm not exactly sure what to
do.
Thank you!
--
View this message in context:
http://r.789695.n4.nabble.com/Line
I cleaned up my old benchmarking code and added checks for missing
data to compare various ways of finding OLS regression coefficients.
I thought I would share this for others. the long and short of it is
that I would recommend
ols.crossprod = function (y, x) {
x <- as.ma
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
Dear R Users,
Using lm() function with categorical variable R use contrasts. Let
assume that I have one X independent variable with 3-levels. Because R
estimate only 2 parameters ( e.g. a1, a2) the coef function returns
only 2 estimators. Is there any function or trick to get another a3
values. I
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
McKinney wrote:
-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 I have a
alf 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 I have and K is a curve fitting parameter. I
>> have linearized a
> -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)
&
I have a model Q=K*A*(R^r)*(S^s)
A, R, and S are data I have and K is a curve fitting parameter. I
have linearized as
log(Q)=log(K)+log(A)+r*log(R)+s*log(S)
I have taken the log of the data that I have and this is the model
formula without the K part
lm(Q~offset(A)+R+S, data=x)
What is the fo
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
Hi,
Sorry for the naive question, but what exactly does the 'Adjusted R-squared'
coefficient in the summary of linear model adjust for?
Sample code:
> x <- rnorm(15)
> y <- rnorm(15)
> lmr <- lm(y~x)
> summary(lmr)
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-1
..@r-project.org] On
> Behalf Of Brian Smith
> Sent: Friday, February 25, 2011 10:06 AM
> To: r-help@r-project.org
> Subject: [R] linear model lme4
>
> Hi,
>
>
> I wanted to check the difference in results (using lme4) , if I treated a
> particular variable (beadch
Hi,
I wanted to check the difference in results (using lme4) , if I treated a
particular variable (beadchip) as a random effect vs if I treated it as a
fixed effect.
For the first case, my formula is:
lmer.result <- lmer(expression ~ cancerClass + (1|beadchip))
For the second case, I want t
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
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.default(mt, mf, contrasts) :
the response appeared on
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
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=startd), function(x) lm
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, and C are positive constants. So there are two
extra degrees of freedom, a
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
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 is equal to NA assign it a value of 1. I am having trouble
w
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
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 is equal to NA assign it a value of 1. I am having trouble
w
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$f <- gl(
Dear R-help,
Suppose I have the following data:
df=data.frame(x=1:10, y=c(1,2,3,4,5,12,14,16,18,20))
plot(y~x, df, t="b")
How can I fit a model which estimates the slopes between x = 1-5, 5-6,
and 6-10?
Adding the factor f:
df$f <- gl(2,5)
Allows me to fit a linear model with interaction
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
Hi there,
I've got a question which is really trivial for sure but still I have
to ask as I'm not
making any progress solving it by myself (please be patient with an
undergraduate
student):
I've got a linear model (lm and lmer fitted with method="ML").
Now I want to compare the coefficients (slop
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
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. In the model print, I want to
show the p-values of all my coefficients.
Thank you in adv
Hi all,
I have a question about linear model with interaction:
I created a data frame df like this:
>df
V1 V2 V3 V4 V5
1 6.414094 c t a g
2 6.117286 t a g t
3 5.756922 a g t g
4 6.090402 g t g t
...
which holds the response in the first column and letters (a,c,g,t) in th
4-455-3265
http://www.StatisticalEngineering.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Melissa2k9
Sent: Friday, April 03, 2009 5:51 AM
To: r-help@r-project.org
Subject: [R] Linear model, finding the slope
Hi
for some data I working
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
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.
The variable names I hav
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
Behalf Of kayj
Sent: Thursday, 12 February 2009 6:36 AM
To: r-help@r-project.org
Subject: [R] Linear model
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
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 intercept is 48.402.
I tried to calculate the p-value in R
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
Hey,
I am modelling a linear regression Y=X*B+E. To compute the effect of “group”
the B-values of the regressors/columns that code the interaction effects (col.
5-8 and col. 11-14, see below) have to be weighted with non-zero elements
within the contrast "Group 1" minus "Group 2" (see below). My
> 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
Dear All,
I have a question which seems trivial, but I reached a dead end.
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 dev
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$coefficents,it came to
nothing,
such as I need to extract the estimates,how to do it by using
OTECTED] [mailto:[EMAIL PROTECTED]
> On Behalf Of Manli Yan
> Sent: Tuesday, June 03, 2008 9:07 PM
> To: r-help@r-project.org
> Subject: [R] linear model in the repeated data type~
>
> here is the data:
> y<-c(5,2,3,7,9,0,1,4,5)
> id<-c(1,1,6,6,7,8,15,15,19)
> t<-c
Try this:
f <- function(x)any(is.na(coefficients(x)))
models <- by(table1[c("y", "t")], table1$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
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(50,56,50,56,50,50,50,60,50)
table1<-data.frame(y,id,t)//longitudinal data
the above is only part of data.
what I want to do is to use the linear model for each id ,then get the
estimate value,like:
fit1<-lm(y~t,data=tabl
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]>
> Subje
[EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Manli Yan
Sent: Tuesday, June 03, 2008 9:07 PM
To: r-help@r-project.org
Subject: [R] linear model in the repeated data type~
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(50,56,50,56,50,50,5
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(50,56,50,56,50,50,50,60,50)
table1<-data.frame(y,id,t)//longitudinal data
what I want to do is to use the linear model for each id ,then get the
estimate value,like:
fit1<-lm(y~t,data=table1,subset=(id==1))
but ,you c
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