dear Sumi,
I am not familiar with the dplyr package (%>%..), however if you want to
fit the model for each subject times freq interaction, a simple for loop
will suffice.
Here possible code:
Assuming d is the dataframe, something like
subj<-levels(d$subject)
fr<-unique(d$freq)
#new
dear all,
I am stuck on the following problem. Give a string like
ss1<- "z:f(5, a=3, b=4, c='1:4', d=2)"
or
ss2<- "f(5, a=3, b=4, c=\"1:4\", d=2)*z"
I would like to remove all entries within parentheses.. Namely, I aim to
obtain respectively
"z:f()" or "f()*z"
I played with sub() and
dear all,
Is the following intentional? Am I missing anything in documentation?
d<-data.frame(y=rnorm(10,5,.5),exp=rnorm(10), age=rnorm(10))
formula(lm(exp(y)~exp+age, data=d))
#--> exp(y) ~ exp + age
formula(lm(exp(y)~., data=d))
#--> exp(y) ~ age
variable 'exp' (maybe indicating
Hi Witold,
use do.call()
list.args<-list(...)
#modify 'list.args' (add/delete/modify)
do.call(image, list.args)
best,
vito
Il 11/05/2016 10.45, Witold E Wolski ha scritto:
Hi,
I am looking for a documentation describing how to manipulate the
"..." . Searching R-intro.html gives to many
dear all,
I don't know if that problem is related to the Rcmdr package itself..
(Sekhar try to install any other packages..)
I am experiencing the same problem, in that when typing
> install.packages("_ANY_PACKAGE_")
I get the message
Warning message:
package ‘_ANY_PACKAGE_’ is not available
McMaster University
Hamilton, Ontario
Canada L8S 4M4
web: socserv.mcmaster.ca/jfox
From: R-help [r-help-boun...@r-project.org] on behalf of Vito M. R. Muggeo
[vito.mug...@unipa.it]
Sent: February 9, 2016 10:15 AM
To: Sekhar Venkatesan; Duncan Murdoch; R-help@r
dear Milan,
I think you should consult a (local) statistician for your analyses.
R (and packages) are only software, and you need theoretical background.
best,
vito
Il 05/05/2015 14.25, Milan Cisty ha scritto:
Dear list members,
Is it possible to compute in R AIC, when model was fitted by
dear jpm,
segmented can't deal with I(1) regression.. However the segmented
default method could be used on objects fitted by any function which
fits I(1) *linear* regression,
Please contact me off list for details,
best,
vito
Il 18/03/2015 2.16, jpm miao ha scritto:
Hi,
If the
dear Stanislav,
Your data show two slopes with a kink at around 0. Thus, yet another
approach would be to use segmented regression to fit a piecewise linear
relationship with unknown breakpoint (being estimated as part of model
fitting). While the resulting fitting is likely to be (slightly)
dear all,
a student of mine brought to my attention the following, somewhat odd,
behaviour of summary.lm() when the response variance is zero (yes,
possibly meaningless from a practical viewpoint). Namely something like
n=10;k=1;summary(lm(rep(k,n)~rnorm(n)))
The values of k, n and the
()
R version 3.0.2 (2013-09-25)
Platform: x86_64-w64-mingw32/x64 (64-bit)
-Original Message-
From: Vito M. R. Muggeo [mailto:vito.mug...@unipa.it]
Sent: Wednesday, March 12, 2014 6:27 AM
To: r-help@r-project.org
Subject: [R] summary.lm() for zero variance response
dear all,
a student
dear Katie,
Since you are looking for exactly 1 breakpoint (namely you know the
number of breakpoints), I suggest to use bootstrap restarting (default
in segmented) with the rough value from davies.test() as a starting
value, namely
o-davies.test(reg1.2,~lagBYmean)
dear Daniel,
yet another package performing growth modelling is quantregGrowth. It
uses quantile regressions with B-splines and quadratic penalties to
ensure flexible estimation with additional noncrossing and monotonicity
(optional) constraints.
The paper underlying the package is here:
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