Hi,
why would this simple procedure not work?
--- snip ---
mydf <- data.frame(id_station = 1234, string_data = c(2024, 12, 1, 0, 0),
rainfall_value= 55)
mydf$string_data <- as.factor(mydf$string_data)
values<-as.integer(levels(mydf$string_data))
for (i in 1:length(values)) {
assign(pa
I would like to use your genoPlotR package
(doi:10.1093/bioinformatics/btq413) to compare the genomes of two
isolates of E. coli K-12 that I have. One is a K-12 that was in my
lab's fridge; the other is a derivative of K-12 bought some time ago,
HB101.
I tried to use genoPlotR, but I could not unde
According to https://cran.r-project.org/web/packages/genoPlotR/index.html
the maintainer of genoPlotR is
Lionel Guy
Send your question also to him.
On Tue, Apr 2, 2024 at 11:27 AM Luigi Marongiu wrote:
>
> I would like to use your genoPlotR package
> (doi:10.1093/bioinformatics/btq413) to comp
Already did...
On Tue, Apr 2, 2024 at 10:45 AM Eric Berger wrote:
>
> According to https://cran.r-project.org/web/packages/genoPlotR/index.html
> the maintainer of genoPlotR is
>
> Lionel Guy
>
> Send your question also to him.
>
> On Tue, Apr 2, 2024 at 11:27 AM Luigi Marongiu
> wrote:
> >
>
Using levels rather than length might cause problems. 2024 1, 1, 0, 0 will have
a different number of levels than 2024, 3, 8, 0, 0 and I cannot assume that the
two tailing zeros are zero for all records. The code can be simplified if you
can assume more. It might require more work if I have assu
Note:
> levels(factor(c(0,0,1))) ## just gives you the levels attribute
[1] "0" "1"
> as.character(factor(c(0,0,1))) ## gives you the level of each value in
the vector
[1] "0" "0" "1"
Does that answer your question or have I misunderstood.
Cheers,
Bert
On Tue, Apr 2, 2024 at 12:00 AM Kimmo El
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