is something I want to keep for a long time, then it
pays in the long run to use the fewest packages possible.
Tim
-Original Message-
From: R-help On Behalf Of Richard O'Keefe
Sent: Thursday, September 7, 2023 6:52 AM
To: Bert Gunter
Cc: r-help
Subject: Re: [R] Finding combination of states
ient way to do this, of course.
> > >>
> > >> Cheers,
> > >> Bert
> > >>
> > >>
> > >> On Mon, Sep 4, 2023 at 12:57 PM Eric Berger
> > wrote:
> > >>>
> > >>> My initial response was buggy and a
it(s, NULL))
> >>> sum(v[-1]==v[-length(v)]) == 0
> >>> }
> >>>
> >>> f <- function(states, nsteps, first, last, rm_repeat=TRUE) {
> >>> if (nsteps < 3) stop("nsteps must be at least 3")
> >>> out <-
te(first,
>>> myExpand(states, nsteps-2) |>
>>> apply(MAR=1, \(x) paste(x, collapse="")),
>>> last, sep="")
>>> if (rm_repeat) {
>>> ok <- sapply(out, no_repeat)
>>> out &
tates, nsteps-2) |>
>> apply(MAR=1, \(x) paste(x, collapse="")),
>> last, sep="")
>> if (rm_repeat) {
>> ok <- sapply(out, no_repeat)
>> out <- out[ok]
>> }
>> out
>> }
>
nit,expand.grid(rep(list(states), ntimes-2)), final))
>>> }
>>>
>>> e.g.
>>>
>>> > g(4, LETTERS[1:5], "B", "D")
>>> [1] "BAAD" "BBAD" "BCAD" "BDAD" "BEAD" &q
"BEBD" "BACD" "BBCD" "BCCD" "BDCD" "BECD" "BADD"
>> [17] "BBDD" "BCDD" "BDDD" "BEDD" "BAED" "BBED" "BCED" "BDED"
>> [25] "
nk the OP will have to clarify his query.
>
> Cheers,
> Bert
>
> On Mon, Sep 4, 2023 at 9:25 AM Ebert,Timothy Aaron wrote:
>
>> Does this work for you?
>>
>> t0<-t1<-t2<-LETTERS[1:5]
>> al2<-expand.grid(t0, t1, t2)
>> al3<-paste(al2$Var1,
On Mon, Sep 4, 2023 at 9:25 AM Ebert,Timothy Aaron wrote:
> Does this work for you?
>
> t0<-t1<-t2<-LETTERS[1:5]
> al2<-expand.grid(t0, t1, t2)
> al3<-paste(al2$Var1, al2$Var2, al2$Var3)
> al4 <- gsub(" ", "", al3)
> head(al3)
>
> T
10:17 AM
To: Christofer Bogaso
Cc: r-help
Subject: Re: [R] Finding combination of states
[External Email]
The function purrr::cross() can help you with this. For example:
f <- function(states, nsteps, first, last) {
paste(first, unlist(lapply(purrr::cross(rep(list(v),nsteps-2)),
\(x) p
The function purrr::cross() can help you with this. For example:
f <- function(states, nsteps, first, last) {
paste(first, unlist(lapply(purrr::cross(rep(list(v),nsteps-2)),
\(x) paste(unlist(x), collapse=""))), last, sep="")
}
f(LETTERS[1:5], 3, "B", "E")
[1] "BAE" "BBE" "BCE" "BDE" "BEE"
Let say I have 3 time points.as T0, T1, and T2.(number of such time
points can be arbitrary) In each time point, an object can be any of 5
states, A, B, C, D, E (number of such states can be arbitrary)
I need to find all possible ways, how that object starting with state
B (say) at time T0, can
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