On Sep 4, 2010, at 6:53 PM, st...@wittongilbert.free-online.co.uk wrote:
Hi I know asking which test to use is frowned upon on this list...
so please do read on for at least a couple on sentences...
I have some multivariate data slit as follows
Tumour Site (one of 5 categories) #
Chemo
Hi,
Sorry for a basic questions on linear models.
I am trying to adjust raw data for both fixed and mixed effects. The data that
I
output should account for these effects, so that I can use the adjusted data
for
further analysis.
For example, if I have the blood sugar levels for 30
Why should height be a random effect?
I suspect you may need a tutorial on what a random effect in a mixed
model is. I see no obvious reason why one would cluster on height.
Perhaps if you clarify, it may become obvious either what your
concerns are (and that your model is correct) or that you
, September 2, 2010 12:46:13 PM
Subject: Re: [R] Linear models (lme4) - basic question
Why should height be a random effect?
I suspect you may need a tutorial on what a random effect in a mixed
model is. I see no obvious reason why one would cluster on height.
Perhaps if you clarify, it may become
, September 2, 2010 12:46:13 PM
Subject: Re: [R] Linear models (lme4) - basic question
Why should height be a random effect?
I suspect you may need a tutorial on what a random effect in a mixed
model is. I see no obvious reason why one would cluster on height.
Perhaps if you clarify, it may become
James Nead james_nead at yahoo.com writes:
Sorry, forgot to mention that the processed data will be used as input for a
classification algorithm. So, I need to adjust for known effects before I can
use the data.
I am trying to adjust raw data for both fixed and mixed effects.
The
Bolker bbol...@gmail.com
To: r-h...@stat.math.ethz.ch
Sent: Thu, September 2, 2010 2:06:47 PM
Subject: Re: [R] Linear models (lme4) - basic question
James Nead james_nead at yahoo.com writes:
Sorry, forgot to mention that the processed data will be used as input for a
classification algorithm. So
.
*From:* Ben Bolker bbol...@gmail.com
*To:* r-h...@stat.math.ethz.ch
*Sent:* Thu, September 2, 2010 2:06:47 PM
*Subject:* Re: [R] Linear models (lme4) - basic question
James Nead james_nead at yahoo.com http://yahoo.com writes
:* Ben Bolker bbol...@gmail.com
*To:* r-h...@stat.math.ethz.ch
*Sent:* Thu, September 2, 2010 2:06:47 PM
*Subject:* Re: [R] Linear models (lme4) - basic question
James Nead james_nead at yahoo.com http://yahoo.com writes:
Sorry, forgot to mention that the processed data will be used
Dear Greg,
Thanks for the tip. As I am new in R can you please provide me a script how
do to so. It will help my learning process.
Thanks,
Asher
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Sent from
] On Behalf Of ashz
Sent: Thursday, August 19, 2010 3:02 AM
To: r-help@r-project.org
Subject: Re: [R] Linear regression equation and coefficient matrix
Dear Greg,
Thanks for the tip. As I am new in R can you please provide me a script
how
do to so. It will help my learning process.
Thanks
Hi,
I have 20*60 data matrix (with some NAs) and I wish to perfom a Pearson
correlation coefficient matrix as well as simple linear regression equation
and coefficient of determination (R2) for every possible combination. Any
tip/idea/library/script how do to so.
Thanks,
As hz
--
View
On Wed, Aug 18, 2010 at 6:09 AM, ashz a...@walla.co.il wrote:
Hi,
I have 20*60 data matrix (with some NAs) and I wish to perfom a Pearson
correlation coefficient matrix as well as simple linear regression equation
The correlation matrix can be readily obtained by calling cor() on the
entire
Hmm, after reading one of your other posts, I am thinking you may
*just* want all pairwise combinations. This worked for me:
# Create a sample data frame with 60 named columns
x - data.frame(matrix(rnorm(1200), ncol = 60, dimnames = list(NULL,
paste(Col, 1:60, sep=''
# calculate the
Hi,
Thanks, the cor() works.
Regarding the simple linear regression equation (mainly, the slope
parameter) and r2. I think I was not writing it well. I need to do it just
for the columns. If I have a, b, c, d columns I wish to compute the relation
of there data, e.g., between a-b, a-c, a-d,
: Wednesday, August 18, 2010 8:43 AM
To: r-help@r-project.org
Subject: Re: [R] Linear regression equation and coefficient matrix
Hi,
Thanks, the cor() works.
Regarding the simple linear regression equation (mainly, the slope
parameter) and r2. I think I was not writing it well. I need to do
Please read the posting guide and include a standalone example.
Maybe you want something like the results from
lm(weight ~ Time, data = ChickWeight, subset = Diet==1)
lm(weight ~ Time, data = ChickWeight, subset = Diet==2)
## ... etc ...
Then you could do
(m - lm(weight ~ Time*Diet, data =
hi,
maybe an ANCOVA is what you want, which is also done by lm in R
lm(y~x*z)
Am 12.08.2010 17:11, schrieb JesperHybel:
I have a simple dataset of a numerical dependent Y, a numerical independent X
and a categorial variable Z with three levels. I want to do linear
regression Y~X for each
Example is spot on - sr for not providing one myself.
The results you calculate are what I'm looking for.
Would like a function F where I could type:
F(weight ~ Time, data = ChickWeight, SOME ARGUMENT = Diet))
Resulting in
for (i in 1:4){
print(
lm(weight ~ Time, data = ChickWeight, subset =
Have a look at lmList() in the nlme package.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of JesperHybel
Sent: Friday, August 13, 2010 7:56 AM
To: r-help@r-project.org
Subject: Re: [R] Linear regression on several groups
I have a simple dataset of a numerical dependent Y, a numerical independent X
and a categorial variable Z with three levels. I want to do linear
regression Y~X for each level of Z. How can I do this in a single command
that is without using lm() applied three isolated times?
--
View this message
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 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
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
Hi R experts,
I have the following timeseries data:
#example data structure
a - c(NA,1,NA,5,NA,NA,NA,10,NA,NA)
c - c(1:10)
df - data.frame(timestamp=a, sequence=c)
print(df)
where i would like to linearly interpolate between the points 1,5, and
10 in 'timestamp'. Original timestamps should not
On Thu, Jul 29, 2010 at 5:16 PM, Ralf B ralf.bie...@gmail.com wrote:
Hi R experts,
I have the following timeseries data:
#example data structure
a - c(NA,1,NA,5,NA,NA,NA,10,NA,NA)
c - c(1:10)
df - data.frame(timestamp=a, sequence=c)
print(df)
where i would like to linearly interpolate
On Jun 28, 2010, at 10:58 AM, Martin Spindler wrote:
Hello,
currently I am estimating an ordered probit model with the function
polr
(MASS package).
Is there a simple way to obtain values for the prediction of the index
function ($X*\hat{\beta}$)?
(E..g. in the GLM function there is
Hello,
currently I am estimating an ordered probit model with the function polr
(MASS package).
Is there a simple way to obtain values for the prediction of the index
function ($X*\hat{\beta}$)?
(E..g. in the GLM function there is the linear.prediction value for this
purpose).
If not,
Joris,
Thank you, I have corrected my mistakes. I very much appreciate your time
and patience.
All my best,
Cobbler.
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I checked your data. Now I have to get some sense out of your code. You do :
G - vowel_features[15]
cvc_lda - lda(G~ vowel_features[15], data=mask_features,
na.action=na.omit, CV=TRUE)
Firstly, as I suspected, you need to select a column by using
vowel_features[,15] . Mind the comma!
Hi Janis,
As you have suggested below is the output for the following:
test.vowel - vowel_features[,1:10]
test.mask - mask_features[,1:10]
dput(test.vowel)
dput(test.mask)
--- NOTE: outputs are limited
test_vowel first 12 columns are all zero (total of 26 columns)
V1 V2 V3 V4
Thanks for being patient with me.
I guess my problem is with understand how grouping in this particular case
is used:
one of the sample codes I found online
(http://www.statmethods.net/advstats/discriminant.html)
library(MASS)
fit - lda(G ~ x1 + x2 + x3, data=mydata, na.action=na.omit, CV=TRUE)
It's not your questions, Cobbler, but could you PLEASE just do what we asked
for?
Copy-paste the following in R and copy-paste ALL output you get in your next
mail.
test.vowel - vowel_features[,1:10]
test.mask - mask_features[,1:10]
dput(test.vowel)
dput(test.mask)
I don't know whether your
Could you provide us with data to test the code? use dput (and limit the
size!)
eg:
dput(vowel_features)
dput(mask_features)
Without this information, it's impossible to say what's going wrong. It
looks like you're doing something wrong in the selection. What should
vowel_features[15]
28, 2010 8:50 AM
To: cobbler_squad
Cc: r-help@r-project.org
Subject: Re: [R] Linear Discriminant Analysis in R
Could you provide us with data to test the code? use dput
(and limit the
size!)
eg:
dput(vowel_features)
dput(mask_features)
Without this information, it's impossible
Joris,
You are a life saver. Based on two sample files above, I think lda should go
something like this:
vowel_features - read.table(file = mappings_for_vowels.txt)
mask_features - data.frame(as.matrix(read.table(file =
3dmaskdump_ICA_37_Combined.txt)))
G - vowel_features[15]
cvc_lda - lda(G~
Dear R gurus,
Thank you all for continuous support and guidance -- learning without you
would not be efficient.
I have a question regarding LD analysis and how to best code it up in R.
I have a file of (V52 and 671 time points across all columns) and another
file of phonetic features (each
Why exactly do you need lda and not another method? For lda to be
applicable, you should check :
1) whether the regressors are normally distributed within the classes
2) whether the variance-covariance matrices are equal for all classes
Essentially, this means that the boundary between both
Daniel Barton daniel.barton at umontana.edu writes:
Hello All,
I am looking for an R library/function that allows the specification
of a custom link function in a linear mixed model.
You might want to try on the r-sig-mixed-mod...@r-project.org list.
I would probably call this a
Hi,
I was wondering if you can show me how to plot the discriminant boundary lines
for an LD analysis in R. I am curious as to perform this on a larger scale, so
I was wondering if you can provide me an example on the infamous 'iris' data.
Here's what I have so far:
iris.lda=lda(Species ~.,
My data looks like following:
cera3[i, ] batch lcl29 pdt
Untreated 3.185867 1 0 0
Untreated.4 3.185867 0 0 0
LCL29 4.357552 1 1 0
LCL29.6 3.446256 0 1 0
PDT 2.765535 1 0 1
PDT.5 3.584963
gauravbhatti wrote:
My data looks like following:
cera3[i, ] batch lcl29 pdt
Untreated 3.185867 1 0 0
Untreated.4 3.185867 0 0 0
LCL29 4.357552 1 1 0
LCL29.6 3.446256 0 1 0
PDT 2.765535 1
Hi,
I calculated the linear predictors derived from weibull model using ovarian
data sets. I calculated the linear predictors as the sum of covariates weighted
by the weibull coefficients and compared to the linear predictors generated by
survreg function. Why are they different? note that the
Note that the first set of coefficients minus the second set of coefficients
is constant (12.78262)
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Dear list,
I need to fit a multiple regression in which individual data point in the
independent and the dependent variables have an associated variance or
standard deviation. To make things clear, instead of having just a table of
dependent (X) and independent (Y1, Y2) variable values like
Hello,
I was doing a linear regression with the following formula:
*lm(y~x+0)*, so it passes through the origin. But when I called the summary
of the regression i saw that R squared is abnormally high (it's a lot lower
in other programs such as SigmaPlot and MS Excel).The manual explained the
Dear Csanad,
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
Reproducible code/problem are welcome.
By the way, who you believe is right?
bests
milton
On Sat, Jan 9, 2010 at 5:40 PM, Csanád Bertók
Dear R-gurus,
Here is what I need to do..
I have two .txt files that are in a matrix form (each looks something like
this:
0.0334820.02238 0.026677
0.0345530.0232260.028855
0.0350170.0232620.02941
0.0362620.023306
Hi everybody,
I'm trying to construct contrasts for an ANOVA to determine if there is a
significant trend in the means of my groups.
In the following example, based on the type of 2x3 ANOVA I'm trying to perform,
does the linear polynomial contrast generated by contr.poly allow me to test
for
...@stat.math.ethz.ch
Onderwerp: [R] linear contrasts for trends in an anova
Hi everybody,
I'm trying to construct contrasts for an ANOVA to determine if there is a
significant trend in the means of my groups.
In the following example, based on the type of 2x3 ANOVA I'm trying to perform,
does the linear
I want to perform linear regression on groups of consecutive rows--say 5 to
10 such--of two matrices. There are many such potential groups because the
matrices have thousands of rows. The matrices are both of the form:
shp[1:5,16:20]
SL495B SL004C SL005C SL005A SL017A
-2649 1.06 0.56
Perhaps along these lines:
1st #need to decide what your group width is , so the second number
inside the extraction call will be that number minus 1:
for (x in seq(1:1000, by=6) {
temp - na,omit( shp[x:(x+5), ] ) # Need the parens in x:(x+5)
lm( formula, data=temp)
}
Or
But I do feel compelled to ask: Do you really get meaningful
information from lm applied to 5 cases? Especially when the predictors
used may not be the same from subset to subset???
Thanks again for your help David. Your question is a good one. It's a bit
complicated but here's the
Dear all,
I am trying to recreate a discriminant analysis in R based on the article from
Dong,J.-J.,etal.,Discriminant analysis of the geomorphic characteristics and
stability of landslide dams, Geomorphology (2009).
I used lda (MASS) to determine the discriminant functions but I noticed that
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
Douglas Bates-2 wrote:
On Thu, Nov 12, 2009 at 10:14 AM, milton ruser milton.ru...@gmail.com
wrote:
Hi Ana,
I am not quite sure if it is the problem, but if you call your data.frame
as
exp,
you will crash exp() function... try use another name for your
data.frame.
By the way, I
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),
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
Hi Ana, you did again :-)
require(fortunes)
fortune(dog)
Firstly, don't call your matrix 'matrix'. Would you call your dog 'dog'?
Anyway, it might clash with the function 'matrix'.
-- Barry Rowlingson
R-help (October 2004)
if you call your data data you will crach data() function.
On Fri, 13 Nov 2009, milton ruser wrote:
Hi Ana, you did again :-)
require(fortunes)
fortune(dog)
Firstly, don't call your matrix 'matrix'. Would you call your dog 'dog'?
Anyway, it might clash with the function 'matrix'.
-- Barry Rowlingson
R-help (October 2004)
if you call your data
Milton's point is dead-on, and I would highly encourage you to give the
posting guide a look.
That said... you might try na.action = na.omit in your call to...
actually, we don't know what function you are using (see first point).
Regardless, sounds like you have missing data and na.action
On Thu, Nov 12, 2009 at 10:14 AM, milton ruser milton.ru...@gmail.com wrote:
Hi Ana,
I am not quite sure if it is the problem, but if you call your data.frame as
exp,
you will crash exp() function... try use another name for your data.frame.
By the way, I suggest you not use attach().
Try
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
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.
CAN ANYONE PLEASE HELP ME WITH THIS
i HAVE TO DO A MIXED EFFECT LINEAR MODEL WITH MY DATA DUE TO THE FACT THAT I
have pseudoreplication!
Although after reading and trying it for several times can get around due to
Error in na.fail.default(list(date = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, :
missing
Dear Ana Golveia,
It is completelly impossible someone realise what kind or help you need or
what is happening. I suggest you give a look on the posting guide, mainly
that part about a minimum reproducible code with self explaining
information, etc.
Cheers
milton
On Wed, Nov 11, 2009 at 7:22
Milton's point is dead-on, and I would highly encourage you to give the
posting guide a look.
That said... you might try na.action = na.omit in your call to...
actually, we don't know what function you are using (see first point).
Regardless, sounds like you have missing data and na.action is
-project.org
Onderwerp: [R] linear trend line and a quadratic trend line.
Dear list users
How is it possible to visualise both a linear trend line and a quadratic
trend line on a plot of two variables?
Here my almost working exsample.
data(Duncan)
attach(Duncan)
plot(prestige ~ income
Eric Fail wrote:
Dear list users
How is it possible to visualise both a linear trend line and a quadratic trend
line on a plot
of two variables?
Here my almost working exsample.
data(Duncan)
attach(Duncan)
plot(prestige ~ income)
abline(lm(prestige ~ income), col=2, lwd=2)
Now
Hello
I was wondering if there is any possibility of estimating a linear
regression model with GARCH errors in R? I tried using fGarch library, but I
could only find for instance how to estimate arma-garch models or similar
combinations.
Thank you,
Ioana
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Hi R users
I used R to get the results of a linear regression
reg-lm(y~x)
here are the results:
# Call:
# lm(formula = donnees$txi7098 ~ donnees$txs7098)
#
# Residuals:
# Min
Hi CE.KA,
Take a look at the following:
# Data
set.seed(123)
x - rnorm(100)
y - 2 + 1.5*x + rnorm(100)
# Regression model
reg - lm(y ~ x)
# The summary
summary(reg)
# Objects present in the summary()
names(summary(reg))
# Extracting the coefficients
summary(reg)$coeff
HTH,
Jorge
On Wed,
Dear Sir or Madam,
I am a student at MSc Probability and Finance at Paris 6 University/
Ecole Polytechnique. I am using R and I can't find an answer to the
following question. I will be very thankful if you can answer it.
I have two vectors rendements_CAC40 and rendements_AlcatelLucent.
I use
On Tue, Oct 13, 2009 at 11:17:11PM +0200, Alexandre Cohen wrote:
I have two vectors rendements_CAC40 and rendements_AlcatelLucent.
I use the lm function as follows, and then the sumarry function:
regression=lm(rendements_CAC40 ~ rendements_AlcatelLucent);
sum=summarry(regression);
[...]
I
On 13-Oct-09 21:17:11, Alexandre Cohen wrote:
Dear Sir or Madam,
I am a student at MSc Probability and Finance at Paris 6 University/
Ecole Polytechnique. I am using R and I can't find an answer to the
following question. I will be very thankful if you can answer it.
I have two vectors
Alexandre,
Let me add two small points to Ted's exposition:
1. you can use the extractor function coefficients(),
or just coef() on the summary:
coef(summary(regression))
which will also give you the matrix of estimates, etc.
2. You will find that using the function str() often is
I am looking for some advice on an appropriate statistical analysis in R
*Experimental question* : We are interested in how communities from
different streams may vary in their response to experimental
temperature. Specifically we are interested to test for differences in
the slope and
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
Daniel Perkins wrote:
Ideally we would do an ANCOVA to test for differences in slope or
intercepts for the different streams. However as there were repeated
measures and unequal n and unbalanced design, I have used a linear mixed
effect model (from nlme package in R) in the form:
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,
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
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
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,
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,
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
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
I am new to R and I want to solve this following problem using R.
My Objective function is a linear function with Quadratic constraints.I want
to know how to solve this problem and which package will be helpful for me
for solving such type of problems.Moreover my one constraint is linear and
On 14/09/2009, Steve Lianoglou mailinglist.honey...@gmail.com wrote:
Hi,
On Sep 14, 2009, at 9:47 AM, e-letter wrote:
Readers,
I have been reading the r book (Crawley) and tried to use the
influence measures function for linear regression, as described. I
have one datum that I wish to
e-letter wrote:
On 14/09/2009, Steve Lianoglou mailinglist.honey...@gmail.com wrote:
Hi,
On Sep 14, 2009, at 9:47 AM, e-letter wrote:
Readers,
I have been reading the r book (Crawley) and tried to use the
influence measures function for linear regression, as described. I
have one datum
On 15/09/2009, Uwe Ligges lig...@statistik.tu-dortmund.de wrote:
e-letter wrote:
On 14/09/2009, Steve Lianoglou mailinglist.honey...@gmail.com wrote:
Hi,
On Sep 14, 2009, at 9:47 AM, e-letter wrote:
Readers,
I have been reading the r book (Crawley) and tried to use the
influence
e-letter wrote:
On 15/09/2009, Uwe Ligges lig...@statistik.tu-dortmund.de wrote:
e-letter wrote:
On 14/09/2009, Steve Lianoglou mailinglist.honey...@gmail.com wrote:
Hi,
On Sep 14, 2009, at 9:47 AM, e-letter wrote:
Readers,
I have been reading the r book (Crawley) and tried to use the
Readers,
I have been reading the r book (Crawley) and tried to use the
influence measures function for linear regression, as described. I
have one datum that I wish to show in the graph but exclude from the
regression and ab line.
x y
0 5
10 9
20 10
30 19
40 4
Hi,
On Sep 14, 2009, at 9:47 AM, e-letter wrote:
Readers,
I have been reading the r book (Crawley) and tried to use the
influence measures function for linear regression, as described. I
have one datum that I wish to show in the graph but exclude from the
regression and ab line.
x y
0
Wen Huang
Verzonden: zondag 6 september 2009 17:49
Aan: r-help@r-project.org
Onderwerp: [R] linear mixed model question
Hello,
I wanted to fit a linear mixed model to a data that is similar in terms
of design to the 'Machines' data in 'nlme' package except that each
worker (with triplicates) only
Hello Wen:
On 2009.09.06 10:49:03, Wen Huang wrote:
Hello,
I wanted to fit a linear mixed model to a data that is similar in
terms of design to the 'Machines' data in 'nlme' package except that
each worker (with triplicates) only operates one machine. I created a
subset of
Wen, to follow up on Thierry, your workers are nested in machines (since each
worker only works one machine). Consider fitting a nested model. Though,
with few observations, you might run into the same problem. Further, if you
have observation triplets, and you expect systematic differences
The workers=as.factor(workers) codeline in my post dropped below my name. It
should be in the code before the command line for the linear model.
Daniel Malter wrote:
Wen, to follow up on Thierry, your workers are nested in machines (since
each worker only works one machine). Consider
Hello,
I wanted to fit a linear mixed model to a data that is similar in
terms of design to the 'Machines' data in 'nlme' package except that
each worker (with triplicates) only operates one machine. I created a
subset of observations from 'Machines' data such that it looks the
same as
Hi,
I am not familiar enough with statistics yet. Please excuse me if my
question is wrong.
In the simplest form of linear regression, the data points are two
list of scalars x_1, ..., x_n and y_1, ..., y_n. I am wondering if
there is a linear regression for two lists of vectors X_1, ..., X_n
Hi,
Is there a way I can specify linear contrasts in glm? I'm looking for
something equivalent to SAS' contrast statement.
I'd like to do the following, suppose I have a categorical input with 4
levels (a,b,c,d), I'd like to test something like: (i) a+b=c+d, (ii) a=b,
(iii) a=b+d, etc...
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