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 dataframe
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 gsub(
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 "experience")
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 not
John Fox, Professor
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; Dunc
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 (f
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 me
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 relati
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) wo
ted R-squared:NaN
F-statistic: NaN on 1 and 8 DF, p-value: NA
sessionInfo()
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
Subje
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 covari
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)
start.psi<-as.numeric(o$st
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:
htt
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