The parameter is different because the model without intercept assumes that
effect of f1 is independent on the effect of f2. So you force f1b:f2ll to
be 0.
The interpretation is the same. The fit is conditional on the model
(interaction or no interaction).
ir. Thierry Onkelinx
Instituut voor
Hi all,
I am doing anova multi factor and I found different Intercept when model
has interaction term.
I have the follow data:
set.seed(42)
dt - data.frame(f1=c(rep(a,5),rep(b,5)),
f2=rep(c(I,II),5),
y=rnorm(10))
When I run
summary.lm(aov(y ~ f1 * f2, data =
Dear Mario,
The interpretation is the same: the average at the reference situation
which is the group that has f1 == f1 level1 and f2 == f2 level1.
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie Kwaliteitszorg
Dear Thierry,
That is the problem. I read that interpretation is the same, but the
Intercept value of summary is different:
The mean of level a of f1 and level I of f2 (first level of each
factor) is 0.7127851.
When I run model with interaction term:
summary.lm(aov(y~f1*f2,data=dt))
On 26/04/15 10:52, Uwe Ligges wrote:
On 26.04.2015 00:26, Rolf Turner wrote:
SNIP
If you insist on using an antiquated version you will have to install
from source, which can be a little bit of a challenge --- particularly
if you are using Windoze, which I presume (Psigh!!!) that you are.
On 26 Apr 2015, at 17:30 , Thierry Onkelinx thierry.onkel...@inbo.be wrote:
Dear Mario,
The interpretation is the same: the average at the reference situation
which is the group that has f1 == f1 level1 and f2 == f2 level1.
A little more precisely: It is the estimate of the expected
On 27.04.2015 00:04, Rolf Turner wrote:
I meant Thanks Uwe That was a fumble-fingered typo; I wasn't
trying to be funny.
Sorry 'bout that!
Don't worry, when quickly typing mails, I typically generate lots of
worse typos, Ralf. ;-)
[Sorry, but I could not resist.]
Best wishes from
We need some idea of what you were actually doing (i.e code, data, ... )
See http://adv-r.had.co.nz/Reproducibility.html for some suggestions.
John Kane
Kingston ON Canada
-Original Message-
From: m...@markdrummond.ca
Sent: Sun, 26 Apr 2015 16:41:32 -0400
To:
I was getting pretty much the same error recently, when I joined this
list (I'm new to R), and the solution turned out to be that I needed
to call as.integer() on a certain value, before passing it to a
function. In my case it was a value read/returned from commandArgs(),
but I suppose there are
I meant Thanks Uwe That was a fumble-fingered typo; I wasn't
trying to be funny.
Sorry 'bout that!
cheers,
Rolf
--
Rolf Turner
Technical Editor ANZJS
Department of Statistics
University of Auckland
Phone: +64-9-373-7599 ext. 88276
Home phone: +64-9-480-4619
knittr is giving me the above error. The code it is failing on is
multiplying two numeric features of a data frame. I can run the code
by hand and it works fine, but when I try to knit my document, knittr
chokes on the same line.
When kitting:
Quitting from lines 161-175
I'd say what you did in the past was definitely a hack that made too
strong assumptions on the internal implementation of lapply() etc. It
basically relied on *apply() to loop over the elements using an index
variable. There any many ways to do this and it seems like in R 3.2.0
there was change.
On Sun, Apr 26, 2015 at 1:39 PM, Adams, Jean jvad...@usgs.gov wrote:
Nick,
I don't know of a way to do what you want ... tell R to ignore all errors
... but, I do have a suggestion.
Since you regard these errors as non-essential, why not edit your code
to reflect that? For example,
Dear list-members,
I find a annoying difference between R 3.1.3 and R 3.2.
To get the element name of a list within lapply() or mclapply() call, I
used the trick below:
For example:
essai - list(T2345=c(5, 6, 7), T5664=c(9, 12, 17, 16))
lapply(essai, function(x) plot(x, main=
Not reproducible [1], so any response likely to be a guess. However, you likely
have not put everything that is in your interactive environment into the knitr
document, so you are not working with the same data in those two environments.
[1]
Nick,
I don't know of a way to do what you want ... tell R to ignore all errors
... but, I do have a suggestion.
Since you regard these errors as non-essential, why not edit your code to
reflect that? For example, instead of writing
plot(df$x1, df$y1)
write
if (x1 %in% names(df) y1
Thanks all for the responses. As Murphy would have it, after posting
my query I found the problem. I had a function defined that did some
value mapping and I had a stray line of code in the function. Actually
a legitimate line of code that was just in the wrong place.
Cheers,
Mark
On Sun, Apr
Kumar and Jim,
The phi coefficient is identical to the Pearson coefficient in the case of a
2 x 2 data set.
As it says in the help file for phi:
Since the phi coefficient is just a Pearson correlation applied to
dichotomous data, to find a matrix of phis from a data set involves
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