On 04/12/2022 00:25, Hadley Wickham wrote:
On Sun, Dec 4, 2022 at 12:50 PM Hervé Pagès 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:
##
On 04/12/2022 10:25, Hadley Wickham wrote:
On Sun, Dec 4, 2022 at 1:22 PM wrote:
This may be a fairly dumb and often asked question about some functions like
strsplit() that return a list of things, often a list of ONE thing that be
another list or a vector and needs to be made into someth
On Sun, Dec 4, 2022 at 1:22 PM wrote:
>
> This may be a fairly dumb and often asked question about some functions like
> strsplit() that return a list of things, often a list of ONE thing that be
> another list or a vector and needs to be made into something simpler..
>
> The examples shown bel
On Sun, Dec 4, 2022 at 12:50 PM Hervé Pagès 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 sear
Thanks. Very informative.
I certainly missed this.
-- Bert
On Sat, Dec 3, 2022 at 3:49 PM Hervé Pagès
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
022 6:50 PM
To: Bert Gunter ; Rui Barradas
Cc: r-help@r-project.org; Evan Cooch
Subject: Re: [R] interval between specific characters in a string...
On 03/12/2022 07:21, Bert Gunter wrote:
> Perhaps it is worth pointing out that looping constructs like lapply()
> can be avoided and t
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 solutio
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
Às 17:18 de 02/12/2022, Evan Cooch escreveu:
Was wondering if there is an 'efficient/elegant' way to do the following
(without tidyverse). Take a string
abaaabbabaaab
Its easy enough to count the number of times the character 'b' shows up
in the string, but...what I'm looking for is outpu
Evan, there are oodles of ways to do many things in R, and mcu of what the
tidyverse supplies can often be done as easily, or easier, outside it.
Before presenting a solution, I need to make sure I am answering the same
question or problem you intend.
Here is the string you have as an example:
s
Here's a function that can get the interval sizes for you.
getStringSegmentLengths <- function(s, delim, ...) {
nchar(unlist(strsplit(s, delim, ...))) + 1L
}
It uses strsplit to return a list of all the segments of the string
separated by delim. delim can be a regular expression and with ...,
You could split the string into letters and figure out which ones are �b�
which(strsplit(x, "")[[1]] == "b")
and then find the difference between each position, �anchoring� at position 0
> diff(c(0, which(strsplit(x, "")[[1]] == "b")))
[1] 2 4 1 6 4
From: R-help on behalf of Evan Cooch
Date:
try
gregexpr('b+', target_string)
which looks for one or more b characters, then get the attribute
"match.length"
On Fri, Dec 2, 2022, 18:56 Evan Cooch wrote:
> Was wondering if there is an 'efficient/elegant' way to do the following
> (without tidyverse). Take a string
>
> abaaabbabaaab
>
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