Just install Older R from the project R archives in a separate directory
Best Regards
Please excuse the brevity of the message. This message was sent from a mobile
device.
From: R-help on behalf of Marc Girondot via
R-help
Sent: Friday, December 6, 2019 1:09
I use R 3.6.1 in macOSX 10.15.1 (Catalina).
I cannot install the last version (3.0-7) of the package raster from
source or from binary. In both cases I get
??*** caught segfault ***
address 0x31, cause 'memory not mapped'
when I try to load it.
Same occurs when I use the development version
Here is the modified code of the reproducible example I sent previously.
Notice the .self$ prefix to funcB which is what has changed
foo <- setRefClass("foo",
fields=list(x="numeric"),
methods=list(
initialize=function(a) {
This particular task is not a problem about R.
It is a problem n combinatorics.
Start with the obvious brute force algorithm
(1) Let S be the union of all the sets
(2) For each K in 0 .. |S|
(3) Enumerate all |S| choose K subsets C of S
(4) If C satisfies the condition, report it and stop
The best advice that anyone could give:
See fortunes::fortune("Friends") .
cheers,
Rolf Turner
On 6/12/19 4:39 am, Thomas Subia wrote:
Colleagues,
I'm trying to extract a cell from all Excel files in a directory.
library(readxl)
files <- list.files(pattern="*.xls", full.names = FALSE)
Here is the code of a reproducible example:
foo <- setRefClass("foo",
fields=list(x="numeric"),
methods=list(
initialize=function(a) {
x <<- a
},
funcA=function(f)
On Thu, 5 Dec 2019 15:39:56 +
Thomas Subia wrote:
> date <- lapply(files, read_excel, sheet="Sheet1", range=("B5"))
> date_df <- as.data.frame(date)
> trans_date <-t(date_df)
> mydates <- list(trans_date)
This feels a bit excessive for what looks like a one-dimensional string
vector. Why is
It might be easier to implement in R if you employ the base functions
that take arrays and operate on them as if they represented sets. See
the help() for "union", "intersect", "setdiff", "setequal" and the
operator "%in%".
--
Best regards,
Ivan
__
On 04/12/2019 2:07 p.m., Nestor Toledo wrote:
Hello everyone, even I'm not fluent in coding, R has become a
fundamental part of my daily work as a researcher and I'm very much
grateful for such a wonderful, open tool. However, I have faced in many
opportunities the problems associated with
Colleagues,
I'm trying to extract a cell from all Excel files in a directory.
library(readxl)
files <- list.files(pattern="*.xls", full.names = FALSE)
date <- lapply(files, read_excel, sheet="Sheet1", range=("B5"))
date_df <- as.data.frame(date)
trans_date <-t(date_df)
mydates <-
Hello,
A ggplot graph follows almost exactly my previous code. The *only*
difference is in facet_wrap(). See below.
library(ggplot2)
idv <- grep("part", names(DB)[-(3:4)], ignore.case = TRUE, value = TRUE)
dblong <- reshape2::melt(DB[-(3:4)], id.vars = idv)
dblong <- reshape2::melt(dblong,
I would second Rui's suggestion. However, as a package developer and
maintainer, I think
it is important to note that users need to be encouraged to use good tools. I
work with optimization
codes. My software was incorporated into the optim() function a LONG time ago.
I have updated
and
On December 5, 2019 3:39:07 AM PST, "Александр Дубровский"
wrote:
>Task:
>A family of sets of letters is given. Find K for which one can
>construct a
>set consisting of K letters, each of them belonging to exactly K sets
>of a
>given family.
...
>
> [[alternative HTML version
Hello,
If you are talking about CRAN packages, like it seems you are, then
there is no general purpose solution. What is deprecated depends on each
package's team of developers/maintainer.
Since R is open source and so must be all CRAN packages, a possible
solution is to have your own
Hello,
in some R sessions, method dispatch for objects of the (S4) class “lavaan"
fail. An example from such a “bad” session:
> library(lavaan)
> HS.model <- ' visual =~ x1 + x2 + x3
+ textual =~ x4 + x5 + x6
+ speed =~ x7 + x8 + x9 '
> fit <- cfa(HS.model, data =
Task:
A family of sets of letters is given. Find K for which one can construct a
set consisting of K letters, each of them belonging to exactly K sets of a
given family.
Possible solution:
For each letter, we will have a separate 'scoop', in which we will' put '
the letter. This can be done using
Hello everyone, even I'm not fluent in coding, R has become a
fundamental part of my daily work as a researcher and I'm very much
grateful for such a wonderful, open tool. However, I have faced in many
opportunities the problems associated with updates/upgrades of packages.
Frequently packages
This is a consolation, because I cannot get it in ggplot either!
Thanks for the code!
F.
> Il giorno 5 dic 2019, alle ore 11:17, Jim Lemon ha
> scritto:
>
> Sorry it's not ggplot, I couldn't work that one out.
>
> # using the data frame structure that Rui
Hello there,
Yes, I'd tried scale as well. I mean, I could do my preprocessing
separately and it was working fine.
I was just wondering how preProcess argument in train function works. As
far as I know, when preProcess argument is set, it normalizes inputs but
not outputs.
Then I've figured we
Exactly. I was trying to remelt data in the right way, but I could not get
there yet. Can you suggest me this code?
Thanks a lot
F.
--
> Il giorno 5 dic 2019, alle ore 11:11, Jim Lemon ha
> scritto:
>
> Hi Francesca,
> Do you want something like this?
>
> Jim
Hi Francesca,
Do you want something like this?
Jim
On Thu, Dec 5, 2019 at 6:58 PM Francesca wrote:
>
> Hi, sorry for bothering again.
> I was wondering how I can reshape the data, if in your code,
> I would like to have only two panels, where in the panel with Participation
> =0, I represent
(He hecho todo con R, analisis y graficos)
El jue., 5 dic. 2019 a las 9:51, IRIS ALVES MOTA ()
escribió:
> Buenos dias,
> Soy Iris, soy portuguesa, y hice un Master relacionado con agronomia y
> metabolomica de plantas en la Universidad de Salamanca. He estado
> trabajando en CIALE, en
Buenos dias,
Soy Iris, soy portuguesa, y hice un Master relacionado con agronomia y
metabolomica de plantas en la Universidad de Salamanca. He estado
trabajando en CIALE, en Salamanca, estos ultimos meses.
Entretanto he mandado un artículo a una revista (de mi thesis del Master) y
he recibido
Hola Jaume:
Los datos que pones en la fórmula son diferentes de los que introduces
en el newdata en la función predict(). Además, el Intercept en la
fórmula lo lo has metido cambiado de signo. Si lo haces correctamente,
coinciden el resultado del predict y el resultado de la fórmula:
> prob<-
Hola, Jaume:
En la asignación de 'prob' te falta un signo menos (al no dejar espacios
es algo difícil de ver) y no evalúas con los mismos datos que pones en
el predict (tem = 25, pot = 0, time = 3). Cambiando eso ahora sí sale lo
mismo, como debería.
> prob <- -1.89521331 - 0.02303313*25 +
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