You really really need to go through one of the many R (web) tutorials
where this is discussed in detail. Or see the Intro to R tutorial that
ships with R. This is not the proper venue for learning how R handles
linear models and how its formula interface works. A terse treatment
can be found in
Hi there,
I have a experimental design related question. I have done a experiment
with 3 factors. The design matrix is similar to:
data.frame(Factor.1 = rep(rep(0:2, each = 3),3), Factor.2 =
rep(c("U","S","N"), 9), Factor.3 = rep(0:2, each = 9))
Factor.1 Factor.2 Factor.3
1 0
I tried your suggestion on the toy example that was posted by David Carlson
a while back:
set.seed(42)
x - 1:15
y - x/(1+x) + rnorm(15,0,0.02)
I found that I had to:
(a) Wrap the 1/x inside I().
(b) Set the offset term to be rep(1,length(x)).
Bottom line:
fit - glm(y ~
Rolf Turner rolf.turner at xtra.co.nz writes:
On 21/08/13 11:23, Ye Lin wrote:
T
hanks for your insights Rolf! The model I want to fit is y=x/a+x with
no intercept, so I transformed it to 1/y=1+a/x as they are the same.
For crying out loud, they are ***NOT*** the same. The equations
On 13-08-21 05:17 PM, Rolf Turner wrote:
Thott about this a bit more and have now decided that I don't understand
after all.
Doesn't
glm(1/y~x,family=gaussian(link=inverse))
fit the model
1/y ~ N(1/(a+bx), sigma^2)
whereas what the OP wanted was
y ~
Hey All,
I wanna to fit a model y~x/(a+x) to my data, here is the code I use now:
lm((1/y-1)~I(1/x)+0, data=b)
and it will return the coefficient which is value of a
however, if I use the code above, I am not able to draw a curve the
presents this equation. How can I do this?
Thanks for your
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Ye Lin
Sent: Tuesday, August 20, 2013 4:40 PM
To: R help
Subject: [R] how to code y~x/(x+a) in lm() function
Hey All,
I wanna to fit a model y~x/(a+x) to my data, here is the code I
use now:
lm((1
(1) It is not acceptable to use wanna in written English. You should say
I want to fit a model .
(2) The model you have fitted is *not* equivalent to the model you first
state.
If you write y ~ x/(a+x) you are tacitly implying that
y = x/(a+x) + E
where the errors E are
T
hanks for your insights Rolf! The model I want to fit is y=x/a+x with no
intercept, so I transformed it to 1/y=1+a/x as they are the same. but i
will look up nls() and see how to fit the model without transformation.
On Tue, Aug 20, 2013 at 2:45 PM, Rolf Turner rolf.tur...@xtra.co.nz wrote:
On 21/08/13 11:23, Ye Lin wrote:
T
hanks for your insights Rolf! The model I want to fit is y=x/a+x with
no intercept, so I transformed it to 1/y=1+a/x as they are the same.
For crying out loud, they are ***NOT*** the same. The equations y =
x/(a+x) and
1/y = 1 + a/x are indeed
Rolf:
Thanks for this.
It nicely illustrates what to me is a fundamental problem: For many
scientists, it is not math (stats) that is the stumbling block, but
rather the failure to understand how variability (noise) affects the
experimental/observational process. One does tend to get more
Dave, your situation is clearer now. You wrote (see the full context at
the end of the message):
From this, you will see that I have 4 control sites and 7 treatment
sites that are measured each week. All 13 locations have different
names, and Location is a random varaible. Is Location
Hi All,
I am frustrated by mixed-effects model! I have searched the web for
hours, and found lots on the nested anova, but nothing useful on my
specific case, which is: a random factor (C) is nested within one of the
fixed-factors (A), and a second fixed factor (B) is crossed with the
first
Dear Dave, there are some inconsistencies in your explanation of the
problem. You said your variables are:
CO is a continuous response variable,
Week is a fixed categorical factor,
Habitat is a fixed categorical factor, and
Location is a random categorical factor nested within Habitat.
Hi again,
Thank you very much for taking the time to respond to my question. I am
sorry that my explanation was confusing. Please allow me to try to clarify.
First, please ignore my attempts to define a lmer model. By putting
forward my best first guess, which was clearly wrong, I have
Dear list members,
I need to fit a regression with two inequality constraints. I look up in
the literature and R archive and found some ways to do it, for example
using the ic.infer package.
However I do not know how to write in R my constraints. In the package
help, it is told that I
Dear list members,
I need to fit a regression with two inequality constraints. I look up in
the literature and R archive and found some ways to do it, for example
using the ic.infer package.
However I do not know how to write in R my constraints. In the package
help, it is told that
Hi
I have say a large vector of 3500 digits. Initially the digits are 0s and
1s. I need to check for a rule to change some of the 0s to -1s in this
vector. But once I change a 0 to -1 then I need to start applying the rule
to change the next 0 only after I see the next 1 in the vector.
Say for
If I take your meaning correctly, you want something like this.
x - c(0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0,
+ 1)
easy - function(x) {
+ state - 0
+ for (i in 1:length(x)) {
+ if (x[i] == 0)
+ x[i] - state
+ state - 0
+ if (x[i] ==
You've tried:
diff(c(0, x)) ?
On Wed, Jul 28, 2010 at 3:10 PM, Raghu r.raghura...@gmail.com wrote:
Hi
I have say a large vector of 3500 digits. Initially the digits are 0s and
1s. I need to check for a rule to change some of the 0s to -1s in this
vector. But once I change a 0 to -1 then I
On Wed, Jul 28, 2010 at 2:10 PM, Raghu r.raghura...@gmail.com wrote:
Hi
I have say a large vector of 3500 digits. Initially the digits are 0s and
1s. I need to check for a rule to change some of the 0s to -1s in this
vector. But once I change a 0 to -1 then I need to start applying the rule
Thanks Henrique.
On Wed, Jul 28, 2010 at 9:20 PM, Henrique Dallazuanna [via R]
ml-node+2305552-1908682012-309...@n4.nabble.comml-node%2b2305552-1908682012-309...@n4.nabble.com
wrote:
You've tried:
diff(c(0, x)) ?
On Wed, Jul 28, 2010 at 3:10 PM, Raghu [hidden
Thanks Matt, I will try putting the other rule into this.
On Wed, Jul 28, 2010 at 9:06 PM, Matt Shotwell shotw...@musc.edu wrote:
If I take your meaning correctly, you want something like this.
x - c(0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0,
+ 1)
easy - function(x) {
+
Thanks Gabor.
On Wed, Jul 28, 2010 at 9:24 PM, Gabor Grothendieck [via R]
ml-node+2305561-610338075-309...@n4.nabble.comml-node%2b2305561-610338075-309...@n4.nabble.com
wrote:
On Wed, Jul 28, 2010 at 2:10 PM, Raghu [hidden
email]http://user/SendEmail.jtp?type=nodenode=2305561i=0
wrote:
On Wed, Jul 28, 2010 at 2:18 PM, Henrique Dallazuanna www...@gmail.comwrote:
You've tried:
diff(c(0, x)) ?
This is clever, but not quite what he's asking for--it converts a sequence
of 1's into a 1 followed by zeroes.
Jonathan
On Wed, Jul 28, 2010 at 3:10 PM, Raghu
Hi,
I have coding question on mixed model in R. I am using R2.11.0 in
windows. I have an experiment with 2 fixed effect factors - A and B. The
levels of B are within the levels of A factor. The model is very similar
to a split plot design except the nesting relationship between the 2
fixed
Dear Everybody:
I want to test an interaction of two repeated measures in lmer()
I've response times (Y) from N subjects with two within-subject-factors.
aov(Y ~ A*B + Error(subjects/A*B), ...) shows an insigificant
interaction p(A:B) = .35
now lmer()
lm1 = lmer(Y ~ 1+A+B+(1+A*B|subjects),
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