interval between specific characters in a string...
On 04/12/2022 00:25, Hadley Wickham wrote:
On Sun, Dec 4, 2022 at 12:50 PM Herv? Pag?s <hpages.on.github at gmail.com> wrote:
On 03/12/2022 07:21, Bert Gunter wrote:
Perhaps it is worth pointing out that looping constructs like lapply() can
be avoided and the procedure vectorized by mimicking Martin Morgan's
solution:
## s is the string to be searched.
diff(c(0,grep('b',strsplit(s,'')[[1]])))
However, Martin's solution is simpler and likely even faster as the regex
engine is unneeded:
diff(c(0, which(strsplit(s, "")[[1]] == "b"))) ## completely vectorized
This seems much preferable to me.
Of all the proposed solutions, Andrew Hart's solution seems the most
efficient:
big_string <- strrep("abaaabbaaaaabaaabaaaaaaaaaaaaaaaaaaab", 500000)
system.time(nchar(strsplit(big_string, split="b", fixed=TRUE)[[1]]) + 1)
# user system elapsed
# 0.736 0.028 0.764
system.time(diff(c(0, which(strsplit(big_string, "", fixed=TRUE)[[1]]
== "b"))))
# user system elapsed
# 2.100 0.356 2.455
The bigger the string, the bigger the gap in performance.
Also, the bigger the average gap between 2 successive b's, the bigger
the gap in performance.
Finally: always use fixed=TRUE in strsplit() if you don't need to use
the regex engine.
You can do a bit better if you are willing to use stringr:
library(stringr)
big_string <- strrep("abaaabbaaaaabaaabaaaaaaaaaaaaaaaaaaab", 500000)
system.time(nchar(strsplit(big_string, split="b", fixed=TRUE)[[1]]) + 1)
#> user system elapsed
#> 0.126 0.002 0.128
system.time(str_length(str_split(big_string, fixed("b"))[[1]]))
#> user system elapsed
#> 0.103 0.004 0.107
(And my timings also suggest that it's time for Herv? to get a new computer :P)
LOL
Actually my timings were for
big_string <- strrep("abaaabbaaaaabaaabaaaaaaaaaaaaaaaaaaab", 1500000)
but I mixed up things when I copy-pasted them in my email.
Even though I still need a new laptop and I'm in the process of getting
a new one ;-)
H.
Herv? Pag?s Bioconductor Core Team hpages.on.github at gmail.com