Re: [R] Linear model and approx function

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 <-

[R] Linear model and approx function

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. >

Re: [R] linear model contrast in R

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

[R] linear model contrast in R

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 --

Re: [R] Linear model vs Mixed model

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

Re: [R] Linear model vs Mixed model

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

Re: [R] Linear model vs Mixed model

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: >

Re: [R] Linear model vs Mixed model

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

Re: [R] Linear model vs Mixed model

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 310

Re: [R] Linear model vs Mixed model

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

[R] Linear model vs Mixed model

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

Re: [R] Linear Model and Missing Data in Predictors

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:

Re: [R] Linear Model and Missing Data in Predictors

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.

[R] Linear Model and Missing Data in Predictors

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

Re: [R] linear model solving

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

[R] linear model solving

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

[R] Linear Model with Discrete Data

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

Re: [R] Linear Model with Discrete Data

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,

Re: [R] Linear Model with Discrete Data

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 discrete

Re: [R] linear model coefficients by year and industry, fitted values, residuals, panel data

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 suggested

Re: [R] linear model coefficients by year and industry, fitted values, residuals, panel data

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

Re: [R] linear model coefficients by year and industry, fitted values, residuals, panel data

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 plyr

[R] linear model coefficients by year and industry, fitted values, residuals, panel data

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

Re: [R] linear model coefficients by year and industry, fitted values, residuals, panel data

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,

Re: [R] linear model coefficients by year and industry, fitted values, residuals, panel data

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,

Re: [R] linear model coefficients by year and industry, fitted values, residuals, panel data

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 ehl...@ucalgary.ca wrote: A few minor improvements to

Re: [R] linear model coefficients by year and industry, fitted values, residuals, panel data

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.

[R] linear model with equality and inequality (redundant) constraints

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

[R] Linear Model Prediction

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:

Re: [R] Linear Model Prediction

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.

Re: [R] Linear Model Prediction

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 and Y,

Re: [R] Linear Model Prediction

On Jul 24, 2012, at 1:38 PM, cm bunnylove...@optonline.net 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.

Re: [R] Linear Model Prediction

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

Re: [R] Linear Model Prediction

On Tue, Jul 24, 2012 at 2:06 PM, bunnylove...@optonline.net 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

Re: [R] Linear Model Prediction

Subject: [R] Linear Model Prediction 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

[R] linear model benchmarking

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 -

[R] Linear model - coefficients

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

Re: [R] Linear model - coefficients

?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

Re: [R] Linear model - coefficients

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, 3L,

Re: [R] Linear model - coefficients

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

Re: [R] Linear model - coefficients

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 jorgeivanve...@gmail.com: Hi Robert, Try this: reg2 - lm( Y ~ factor(x1) + factor(x2) + factor(x3) + factor(x4) + factor(x5) - 1, data = X  ) cof(ref2) HTH,

Re: [R] Linear model - coefficients

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 robert.ru...@gmail.com wrote: Hi, but I want to get the coefficients for every variables from x1 to x5. (x1 was an example) Robert

Re: [R] Linear model - coefficients

Hi Weidong, thank you very much. It really works fine. Robert 2011/6/12 Weidong Gu anopheles...@gmail.com: 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 robert.ru...@gmail.com

Re: [R] Linear Model with curve fitting parameter?

-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 exactly right sorry I was confused

Re: [R] Linear Model with curve fitting parameter?

-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 sense. b0 == y intercept == y

Re: [R] Linear Model with curve fitting parameter?

Thank you very much for all of your help. On Mon, Apr 4, 2011 at 6:10 PM, Steven McKinney smckin...@bccrc.ca wrote: -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

Re: [R] Linear Model with curve fitting parameter?

McKinney smckin...@bccrc.ca 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 be the incorrect dimensions.  I could Z=Q/A. I

Re: [R] Linear Model with curve fitting parameter?

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 as log(Q)=log(K)+log(A)+r*log(R)+s*log(S) I have taken the log

Re: [R] Linear Model with curve fitting parameter?

...@bccrc.ca 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

Re: [R] Linear Model with curve fitting parameter?

-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/A. I suspect this is confusion

[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 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

Re: [R] Linear Model with curve fitting parameter?

-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

[R] linear model - lm (Adjusted R-squared)?

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 -1.7828

Re: [R] linear model - lm (Adjusted R-squared)?

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] linear model lme4

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

Re: [R] linear model lme4

] 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 (beadchip) as a random effect vs if I treated it as a fixed effect

[R] linear model with similar response predictor

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

Re: [R] linear model with similar response predictor

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

Re: [R] linear model with similar response predictor

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

Re: [R] linear model and by()

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

Re: [R] linear model and by()

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),

Re: [R] linear model and by()

On Fri, Nov 13, 2009 at 11:49 AM, Sam Albers tonightstheni...@gmail.com wrote: snip 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

[R] linear model and by()

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

Re: [R] linear model and by()

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.

Re: [R] linear model with coefficient constraints

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 could be

[R] linear model with coefficient constraints

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,

[R] Linear Model NA Value Test

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

[R] Linear Model NA Value Test

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

Re: [R] Linear Model NA Value Test

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,

Re: [R] Linear Model NA Value Test

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,

Re: [R] Linear Model NA Value Test

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

Re: [R] Linear Model NA Value Test

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 is

[R] linear model with interaction / segments

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

Re: [R] linear model with interaction / segments

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 -

Re: [R] linear model with interaction / segments

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 segmented

Re: [R] linear model: Test difference between coefficients and given values (t.test?)

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

Re: [R] linear model: Test difference between coefficients and given values (t.test?)

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

[R] linear model: Test difference between coefficients and given values (t.test?)

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

[R] Linear model with coefficient restriction

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

Re: [R] Linear model with coefficient restriction

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

[R] Linear model with interaction

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

[R] Linear model, finding the slope

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

Re: [R] Linear model, finding the slope

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.

Re: [R] Linear model, finding the slope

://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 on I am merely

[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 for the intercept where the t-value for the intercept is 48.402. I tried to calculate the p-value in R

Re: [R] Linear model

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 for the intercept where the t-value

Re: [R] Linear model

On Wed, Feb 11, 2009 at 1:36 PM, kayj kjaj...@yahoo.com 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

[R] Linear model: contrasts

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

Re: [R] Linear model: contrasts

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()

[R] Linear model with one known coordinate

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

Re: [R] Linear model with one known coordinate

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 model,

Re: [R] linear model in the repeated data type~

: 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 list 2008/6/3

Re: [R] linear model in the repeated data type~

: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:length(allFits)) allCoef[[i]] - coef

[R] linear model with 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(50,56,50,56,50,50,50,60,50) 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:

Re: [R] linear model with the repeated data type~

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)

Re: [R] linear model in the repeated data type~

: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(50,56,50,56,50,50,50,60,50) table1-data.frame(y,id,t)//longitudinal data what I want to do is to use the linear model for each id

[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(50,56,50,56,50,50,50,60,50) 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 can

Re: [R] linear model in the repeated data type~

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(50,56,50,56,50,50,50,60,50) table1-data.frame(y,id

Re: [R] linear model in the repeated data type~

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 repeated

[R] Linear model with svyglm

Hi, I am a bit unclear if svyglm with family=gaussian is actually a normal linear model including weighting. The goal is to estimate a normal linear model using sample inflation weights. Can anybody illuminate me a bit on this? Thanks a lot! Werner

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