Às 17:24 de 18/10/2023, Leonard Mada escreveu:
Dear Rui,
Thank you for your reply.
I do have actually access to the chemical symbols: I have started to
refactor and enhance the Rpdb package, see Rpdb::elements:
https://github.com/discoleo/Rpdb
However, the regex that you have constructed is quite heavy, as it needs
to iterate through all chemical symbols (in decreasing nchar). Elements
like C, and especially O, P or S, appear late in the regex expression -
but are quite common in chemistry.
The alternative regex is (in this respect) simpler. It actually works
(once you know about the workaround).
Q: My question focused if there is anything like is.numeric, but to
parse each element of a vector.
Sincerely,
Leonard
On 10/18/2023 6:53 PM, Rui Barradas wrote:
Às 15:59 de 18/10/2023, Leonard Mada via R-help escreveu:
Dear List members,
What is the best way to test for numeric digits?
suppressWarnings(as.double(c("Li", "Na", "K", "2", "Rb", "Ca", "3")))
# [1] NA NA NA 2 NA NA 3
The above requires the use of the suppressWarnings function. Are there
any better ways?
I was working to extract chemical elements from a formula, something
like this:
split.symbol.character = function(x, rm.digits = TRUE) {
# Perl is partly broken in R 4.3, but this works:
regex =
"(?<=[A-Z])(?![a-z]|$)|(?<=.)(?=[A-Z])|(?<=[a-z])(?=[^a-z])";
# stringi::stri_split(x, regex = regex);
s = strsplit(x, regex, perl = TRUE);
if(rm.digits) {
s = lapply(s, function(s) {
isNotD = is.na(suppressWarnings(as.numeric(s)));
s = s[isNotD];
});
}
return(s);
}
split.symbol.character(c("CCl3F", "Li4Al4H16", "CCl2CO2AlPO4SiO4Cl"))
Sincerely,
Leonard
Note:
# works:
regex = "(?<=[A-Z])(?![a-z]|$)|(?<=.)(?=[A-Z])|(?<=[a-z])(?=[^a-z])";
strsplit(c("CCl3F", "Li4Al4H16", "CCl2CO2AlPO4SiO4Cl"), regex, perl = T)
# broken in R 4.3.1
# only slightly "erroneous" with stringi::stri_split
regex = "(?<=[A-Z])(?![a-z]|$)|(?=[A-Z])|(?<=[a-z])(?=[^a-z])";
strsplit(c("CCl3F", "Li4Al4H16", "CCl2CO2AlPO4SiO4Cl"), regex, perl = T)
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Hello,
If you want to extract chemical elements symbols, the following might
work.
It uses the periodic table in GitHub package chemr and a package stringr
function.
devtools::install_github("paleolimbot/chemr")
split_chem_elements <- function(x) {
data(pt, package = "chemr", envir = environment())
el <- pt$symbol[order(nchar(pt$symbol), decreasing = TRUE)]
pat <- paste(el, collapse = "|")
stringr::str_extract_all(x, pat)
}
mol <- c("CCl3F", "Li4Al4H16", "CCl2CO2AlPO4SiO4Cl")
split_chem_elements(mol)
#> [[1]]
#> [1] "C" "Cl" "F"
#>
#> [[2]]
#> [1] "Li" "Al" "H"
#>
#> [[3]]
#> [1] "C" "Cl" "C" "O" "Al" "P" "O" "Si" "O" "Cl"
It is also possible to rewrite the function without calls to non base
packages but that will take some more work.
Hope this helps,
Rui Barradas
Hello,
You and Avi are right, my function's performance is terrible. The
following is much faster.
As for how to not have digits throw warnings, the lapply in the version
of your function below solves it by setting grep argument invert = TRUE.
This will get all strings where digits do not occur.
split_chem_elements <- function(x, rm.digits = TRUE) {
regex <- "(?<=[A-Z])(?![a-z]|$)|(?<=.)(?=[A-Z])|(?<=[a-z])(?=[^a-z])"
if(rm.digits) {
stringr::str_replace_all(mol, regex, "#") |>
strsplit("#|[[:digit:]]") |>
lapply(\(x) x[nchar(x) > 0L])
} else {
strsplit(x, regex, perl = TRUE)
}
}
split.symbol.character = function(x, rm.digits = TRUE) {
# Perl is partly broken in R 4.3, but this works:
regex <- "(?<=[A-Z])(?![a-z]|$)|(?<=.)(?=[A-Z])|(?<=[a-z])(?=[^a-z])"
s <- strsplit(x, regex, perl = TRUE)
if(rm.digits) {
s <- lapply(s, \(x) x[grep("[[:digit:]]+", x, invert = TRUE)])
}
s
}
mol <- c("CCl3F", "Li4Al4H16", "CCl2CO2AlPO4SiO4Cl")
split_chem_elements(mol)
#> [[1]]
#> [1] "C" "Cl" "F"
#>
#> [[2]]
#> [1] "Li" "Al" "H"
#>
#> [[3]]
#> [1] "C" "Cl" "C" "O" "Al" "P" "O" "Si" "O" "Cl"
split.symbol.character(mol)
#> [[1]]
#> [1] "C" "Cl" "F"
#>
#> [[2]]
#> [1] "Li" "Al" "H"
#>
#> [[3]]
#> [1] "C" "Cl" "C" "O" "Al" "P" "O" "Si" "O" "Cl"
mol10000 <- rep(mol, 10000)
system.time(
split_chem_elements(mol10000)
)
#> user system elapsed
#> 0.01 0.00 0.02
system.time(
split.symbol.character(mol10000)
)
#> user system elapsed
#> 0.35 0.07 0.47
Hope this helps,
Rui Barradas
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