Well, I don't think language is as much of a problem as your failure to compose
your messages using plain text format. Your examples are all mushed together
since the mailing list removes formatting. See what we see below for example.
On August 20, 2021 12:27:50 PM PDT, Silvano Cesar da Costa
Thanks, Greg.
Turns out that there's an even faster alternative. Note that the OP
asked whether one could include in the result the counts of each
unordered pair, which I assume could be either 2 or 1. This can be
done easily using table(), and it's quite a bit faster for my 1
million pair example
Bert,
> The efficiency gains are due to vectorization and the use of more
> efficient primitives. None of this may matter of course, but it seemed
> worth mentioning.
thanks very much! the varieties of code, and disparities of
performance, are truly wonderful.
Rui's point that what works better
Hi, thanks you for the answer.
Sorry English is not my native language.
But you got it right.
> As C is first and fourth biggest value, you follow third option and
select 3 highest A, 3B 2C and 2D?
I must select the 10 (not 15) highest values, but which follow a certain
order:
3A - 3B - 2C - 2D
Note that:
1. Your solution returns a matrix, not a data frame (or Tibble)
2. Assuming that the order of the entries in the pairs does not matter
(which your solution also assumes and seems reasonable given the OP's
specification), I think that you'll find the following, which returns
the data.fra
Since you are dealing with graphs you could consider using
the igraph package. This is more involved than needed for
what you are
asking but it might be useful for other follow on calculations.
We first define a 2 column matrix of edges, then convert it to
an igraph and simplify it to remove dupli
Nice Rui.
Here's a version in base R with no apply().
unique(data.frame(V1=pmin(x$Source,x$Target), V2=pmax(x$Source,x$Target)))
On Fri, Aug 20, 2021 at 6:43 PM Rui Barradas wrote:
> Hello,
>
> This seems elegant to me but it's also the slowest, courtesy sort.
>
> apply(x, 1, sort) |> t() |> un
Hello,
This seems elegant to me but it's also the slowest, courtesy sort.
apply(x, 1, sort) |> t() |> unique()
(My tests show that for small inputs Greg's base apply is fastest, with
nrow(x) > 700, Eric's dplyr is fastest)
Hope this helps,
Rui Barradas
Às 15:13 de 20/08/21, Greg Minshall
Eric,
> x %>% transmute( a=pmin(Source,Target), b=pmax(Source,Target)) %>%
> unique() %>% rename(Source=a, Target=b)
ah, very nice. i have trouble remembering, e.g., unique().
fwiw, (hopefully) here's a baser version.
x = data.frame(Source=rep(1:3,4),
Target=c(rep(1,3),rep(2,3),rep(3,
x %>% transmute( a=pmin(Source,Target), b=pmax(Source,Target)) %>%
unique() %>% rename(Source=a, Target=b)
On Fri, Aug 20, 2021 at 2:12 PM Greg Minshall wrote:
> Kimmo,
>
> i'll be curious to see other, maybe more elegant, answers. in the
> meantime, this seems to work.
>
>
> x = dat
Kimmo,
i'll be curious to see other, maybe more elegant, answers. in the
meantime, this seems to work.
x = data.frame(Source=rep(1:3,4),
Target=c(rep(1,3),rep(2,3),rep(3,3),rep(4,3)))
y <- apply(x, 1, function(y) return (c(n=min(y), x=max(y
res <- data.frame()
for (n in unique(
Hi!
I am working with a large network data consisting of source-target
pairs stored in a tibble. Now I need to transform the directed dataset
to an undirected network data. This means, I need to keep only one
instance for pairs with the same "nodes". In other words, if my data
has one row with A (
Agreed. Need the rest of a complete example.
On August 19, 2021 11:27:59 PM PDT, PIKAL Petr wrote:
>Hallo
>
>I am confused, maybe others know what do you want but could you be more
>specific?
>
>Let say you have such data
>set.seed(123)
>Var.1 = rep(LETTERS[1:4], 10)
>Var.2 = sample(1:40, replac
Thanks Dr. Burradas too. i also had the same question.
regards
August 20, 2021 6:02 AM, "bharat rawlley via R-help"
wrote:
> Thank you, Dr. Burradas!
> That resolved my query
> Have a great rest of your day
> On Thursday, 19 August, 2021, 04:47:42 pm GMT-4, Rui Barradas
> wrote:
>
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