> On Dec 11, 2016, at 5:35 PM, frede...@ofb.net wrote:
>
> Dear R-Help,
>
> I was going to ask Jeff to read the entire works of William
> Shakespeare to learn why his reply was not helpful to me...
>From a typical US 10th grade English assignment:
Cassius:
"The fault, dear Brutus, is not in ou
Dear R-Help,
I was going to ask Jeff to read the entire works of William
Shakespeare to learn why his reply was not helpful to me...
Then I realized that the answer, as always, lies within...
desub <- function(y) {
e1=substitute(y, environment())
e2=do.call(substitute,list(e1), e
Hi Tan Dai,
It looks like you have little or no variance in your measurements, and
so the result is degenerate. The functions you are using are probably
working properly, but there is nothing to display.
Jim
On Sat, Dec 10, 2016 at 6:04 PM, Tan Dai Chuan via R-help
wrote:
> Hi, i face the probl
There are limits to how much people will do your debugging for you, but it
looks like you are unpacking beta,gamma from par, but packing gamma,beta into
start. Otherwise, print some of the values computed in the function and check.
-pd
> On 10 Dec 2016, at 20:01 , Fernando de Souza Bastos wrot
Hi
Great to hear you have it working.
Figuring out the names of grobs takes two things:
1. someone has to name the grobs
2. grid.ls()
The reason why I did my example using 'lattice' is because 'lattice'
names all of its grobs. There is a document ...
http://lattice.r-forge.r-project.org/Vi
Dear Bert,
This is awesome, thanks a lot!
Best,
Marlin
On Sun, 2016-12-11 at 06:52 -0800, Bert Gunter wrote:
> Use list indexing, "[[" not "[" .
>
> > df <- data.frame(a=1:3,b=letters[1:3])
> > x <- "new"
> > df[[x]]<- I(list(1:5,g = "foo", abb = matrix(runif(6),nr=3)))
> > df
>
> a b
Thank you Duncan, it really was that!
Fernando de Souza Bastos
Professor Assistente
Universidade Federal de Viçosa (UFV)
Campus UFV - Florestal
Doutorando em Estatística
Universidade Federal de Minas Gerais (UFMG)
Cel: (31) 99751-6586
2016-12-11 11:25 GMT-02:00 Duncan Murdoch :
> On 10/12/2
If you see my previous example, I have tried something like
> df[,n3] <- I(mylist)
However, in my case, the name of the new column is in a variable (by
user input) which can not be directly used by the dollar assign. On the
other hand, "[<-" does not work correctly even if I wrap the list into
"
Glad to help.
However, I need to publicly correct my misstatement. Both "[[" and "["
can be used and are useful for list indexing. As ?"[" clearly states,
the former selects only a single column, while the latter can select
several.
Also:
"Both [[ and $ select a single element of the list. The m
Use list indexing, "[[" not "[" .
> df <- data.frame(a=1:3,b=letters[1:3])
> x <- "new"
> df[[x]]<- I(list(1:5,g = "foo", abb = matrix(runif(6),nr=3)))
> df
a b new
1 1 a 1, 2, 3,
2 2 b foo
3 3 c 0.248115
> df$new
[[1]]
[1] 1 2 3 4 5
$g
[1] "foo"
$abb
[,1]
On 10/12/2016 2:01 PM, Fernando de Souza Bastos wrote:
The Log.lik function below returns the value '-INF' when it should return
the value -5836.219. I can not figure out the error, does anyone have any
suggestions?
I haven't read it carefully, but a likely problem is that you are using
constr
?data.frame says:
"If a list or data frame or matrix is passed to data.frame it is as if
each component or column had been passed as a separate argument
(except for matrices of class "model.matrix" and those protected by
I). "
So doing what Help says to do seems to do what you asked:
> df <- da
Hi Francesca,
I'm not sure what you are doing here, but try this:
regnames<-paste("r",letters[1:8],sep="")
for(i in 1:8) {
response<-rnorm(20)
coef1<-rnorm(20)
coef2<-rnorm(20)
age<-sample(20:50,20)
gender<-sample(c("M","F"),20,TRUE)
assign(regnames[i],lm(response~coef1+coef2+age+gender))
}
No. Read Hadley Wickham's "Advanced R" to learn why not.
--
Sent from my phone. Please excuse my brevity.
On December 10, 2016 10:24:49 PM PST, frede...@ofb.net wrote:
>Dear R-Help,
>
>I asked this question on StackOverflow,
>
>http://stackoverflow.com/questions/41083293/in-r-how-do-i-define-a-f
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