Hello,
1. When there are systematic errors, use ?try or, better yet, ?tryCatch.
Something like the code below will create a list of errors and read in
the data if none occurred.
The code starts by creating an empty list for tryCatch results. It uses
?file.path instead of noquote/paste0 to assem
This discussion has developed in such a way that it seems a better
subject line would be "problem for the hairsplit function". :-)
cheers,
Rolf Turner
--
Honorary Research Fellow
Department of Statistics
University of Auckland
Phone: +64-9-373-7599 ext. 88276
A bit too fast there, Duncan... x[[c(1,2)]] is illegal.
On July 9, 2021 5:16:13 PM PDT, Duncan Murdoch wrote:
>On 09/07/2021 6:44 p.m., Bert Gunter wrote:
>> OK, I stand somewhat chastised.
>>
>> But my point still is that what you get when you "extract" depends on
>> how you define "extract." D
My mental model for the `[` vs `[[` behavior is that `[` indexes multiple
results while `[[` indexes only one item. If returning multiple items from a
list the result must be a list. For consistency, `[` always returns a list when
applied to a list. The double bracket drops the containing list.
On 09/07/2021 6:44 p.m., Bert Gunter wrote:
OK, I stand somewhat chastised.
But my point still is that what you get when you "extract" depends on
how you define "extract." Do note that ?"[" yields a help file titled
"Extract or Replace Parts of an object"; and afaics, the term "subset"
is not ex
"But it takes me a while to get familiar R."
Of course. That is true for all of us. Just keep on plugging away and
you'll get it. Probably far better than I before too long.
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (
Thanks Bert,
I'm reading some books now. But it takes me a while to get familiar R.
Best,
KaiOn Friday, July 9, 2021, 03:06:11 PM PDT, Duncan Murdoch
wrote:
On 09/07/2021 5:51 p.m., Jeff Newmiller wrote:
> "Strictly speaking", Greg is correct, Bert.
>
> https://cran.r-project.org/doc/
OK, I stand somewhat chastised.
But my point still is that what you get when you "extract" depends on
how you define "extract." Do note that ?"[" yields a help file titled
"Extract or Replace Parts of an object"; and afaics, the term "subset"
is not explicitly used as Duncan prefers. The relevant
On 09/07/2021 5:51 p.m., Jeff Newmiller wrote:
"Strictly speaking", Greg is correct, Bert.
https://cran.r-project.org/doc/manuals/r-release/R-lang.html#List-objects
Lists in R are vectors. What we colloquially refer to as "vectors" are more precisely referred to
as "atomic vectors". And withou
"Strictly speaking", Greg is correct, Bert.
https://cran.r-project.org/doc/manuals/r-release/R-lang.html#List-objects
Lists in R are vectors. What we colloquially refer to as "vectors" are more
precisely referred to as "atomic vectors". And without a doubt, this "vector"
nature of lists is a ke
"1. a column, when extracted from a data frame, *is* a vector."
Strictly speaking, this is false; it depends on exactly what is meant
by "extracted." e.g.:
> d <- data.frame(col1 = 1:3, col2 = letters[1:3])
> v1 <- d[,2] ## a vector
> v2 <- d[[2]] ## the same, i.e
> identical(v1,v2)
[1] TRUE
> v3
On 09/07/2021 3:37 p.m., Laurent Rhelp wrote:
Very effective solution, I hope I remember that for the nex time.
The key things to remember are that in R a "list" object is a vector
whose elements are R objects, and that matrices are just vectors with
dimension attributes.
Those two ideas a
Very effective solution, I hope I remember that for the nex time.
Thank you
Le 09/07/2021 à 19:50, David Winsemius a écrit :
On 7/9/21 10:40 AM, Laurent Rhelp wrote:
Dear R-Help-list,
I have a list init_l containing 16 dataframes and I want to create
a matrix 4 x 4 from this list with a
You are right, this extra level disturbed me.
Very impressive this solution, thank you very much.
Le 09/07/2021 à 19:50, Bill Dunlap a écrit :
> Try
> matrix(init_l, nrow=4, ncol=4,
> dimnames=list(c("X1","X2","X3","X4"),c("X1","X2","X3","X4")))
> It doesn't give exactly what your code does
Hi Migdonio,
I did try your code:
# Initialize the rr variable as a list.
rr <- as.list(rep(NA, nrow(ora)))
# Run the for-loop to store all the CSVs in rr.
for (j in 1:nrow(ora))
{
mycol <- ora[j,"fname"]
mycsv <- paste0(mycol,".csv")
rdcsv <- noquote(paste0("w:/
Kai,
> one more question, how can I know if the function is for column
> manipulations or for vector?
i still stumble around R code. but, i'd say the following (and look
forward to being corrected! :):
1. a column, when extracted from a data frame, *is* a vector.
2. maybe your question is "i
it complained about ASSAY_DEFINITIONS not about RESPONDENTS.
Can you try with the ASSAY_DEFINITIONS file?
On Fri, Jul 9, 2021 at 9:10 PM Kai Yang via R-help
wrote:
> Hello List,
> I use for loop to read csv difference file into data frame rr. The data
> frame rr will be deleted after a comp
Hello List,
I use for loop to read csv difference file into data frame rr. The data frame
rr will be deleted after a comparison and go to the next csv file. Below is my
code:
for (j in 1:nrow(ora))
{
mycol <- ora[j,"fname"]
mycsv <- paste0(mycol,".csv'")
rdcsv <- noquote(paste0("'w:/pr
On 7/9/21 10:40 AM, Laurent Rhelp wrote:
Dear R-Help-list,
I have a list init_l containing 16 dataframes and I want to create a
matrix 4 x 4 from this list with a dataframe in every cell of the
matrix. I succeeded to do that but my loop is very uggly (cf. below).
Could somebody help me to
Try
matrix(init_l, nrow=4, ncol=4,
dimnames=list(c("X1","X2","X3","X4"),c("X1","X2","X3","X4")))
It doesn't give exactly what your code does, but your code introduces an
extra level of "list", which you may not want.
-Bill
On Fri, Jul 9, 2021 at 10:40 AM Laurent Rhelp wrote:
> Dear R-Help-li
Dear R-Help-list,
I have a list init_l containing 16 dataframes and I want to create a
matrix 4 x 4 from this list with a dataframe in every cell of the
matrix. I succeeded to do that but my loop is very uggly (cf. below).
Could somebody help me to write nice R code to do this loop ?
Thank
Dear Rui ang Jim,
Thank you very much.
Thank you Rui Barradas, I already tried using your coding and I'm grateful
I got the answer.
ok now, I have some condition on the location parameter which is cyclic
condition.
So, I will add another 2 variables for the column.
and the condition for locati
Hello,
With the condition for the location it can be estimated like the following.
fit_list2 <- gev_fit_list <- lapply(Ozone_weekly2, gev.fit, ydat = ti,
mul = c(2, 3), show = FALSE)
mle_params2 <- t(sapply(fit_list2, '[[', 'mle'))
# assign column names
colnames(mle_params2) <- c("location",
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