On Tue, Jan 25, 2011 at 08:58:31AM +, Prof Brian Ripley wrote:
[...]
> >If k may be 0, then it is better to use
> >
> > for (n in seq(length=k))
> >
> >since seq(length=0) has length 0.
>
> Since you keep mentioning that, it is actually much better to use
> seq_len(k) (and seq_along(x) instea
Now I understand what the difference between a primitive and a
non-primitive!
Thanks for the clarification!
Ivan
Le 1/25/2011 18:03, Bert Gunter a écrit :
Well, I'm not Prof. Ripley, but the answer is: Look at the code.
seq_len, seq.int, and seq_along call Primitives, which are implemented
in C
Well, I'm not Prof. Ripley, but the answer is: Look at the code.
seq_len, seq.int, and seq_along call Primitives, which are implemented
in C, and therefore MUCH faster than seq(), which is implemented as
pure R code (and is also a generic, so requires method dispatch).
Though for small n (up to a f
Hi
r-help-boun...@r-project.org napsal dne 25.01.2011 10:58:36:
> ooh.. I have another question.
> What if I want to add the value in the vector a to the hello each time
it
> prints.
> Here is your output
>
> a <- c(2,3,5,5,5,6,6,7)
> mapply(rep, "hello", rle(a)$lengths, USE.NAMES = FALSE)
>
>
ooh.. I have another question.
What if I want to add the value in the vector a to the hello each time it
prints.
Here is your output
a <- c(2,3,5,5,5,6,6,7)
mapply(rep, "hello", rle(a)$lengths, USE.NAMES = FALSE)
[[1]]
[1] "hello"
[[2]]
[1] "hello"
[[3]]
[1] "hello" "hello" "hello"
[[4]]
[1] "
Mr Ripley,
May I ask why seq_len() and seq_along() are better than seq()?
Thanks,
Ivan
Le 1/25/2011 09:58, Prof Brian Ripley a écrit :
On Tue, 25 Jan 2011, Petr Savicky wrote:
On Mon, Jan 24, 2011 at 11:18:35PM +0100, Roy Mathew wrote:
Thanks for the reply Erik, As you mentioned, grouping co
Dear Erik,
Thanks for the mapply idea. I never got around to understand all those apply
functions.
I am still curious as to why the other loop didnt work. I even tried the
debug but doesnt help.
Anyway I will leave that for now.
Thanks a lot for your help.
Regards,
Roy
On Mon, Jan 24, 2011 at 11:
On Tue, Jan 25, 2011 at 09:05:03AM +0100, Petr Savicky wrote:
[...]
> to foreach loop in Perl. If v is a vector, then
>
> for (n in v)
>
> first creates the vector v and then always performs length(v) iterations.
I forgot that ‘break’ may stop the loop. See ?"for" for further
information. In p
On Tue, 25 Jan 2011, Petr Savicky wrote:
On Mon, Jan 24, 2011 at 11:18:35PM +0100, Roy Mathew wrote:
Thanks for the reply Erik, As you mentioned, grouping consecutive elements
of 'a' was my idea.
I am unaware of any R'ish way to do it. It would be nice if someone in the
community knows this.
T
Hi
r-help-boun...@r-project.org napsal dne 24.01.2011 23:18:35:
> Thanks for the reply Erik, As you mentioned, grouping consecutive
elements
> of 'a' was my idea.
> I am unaware of any R'ish way to do it. It would be nice if someone in
the
> community knows this.
>
> The error resulting in th
On Mon, Jan 24, 2011 at 11:18:35PM +0100, Roy Mathew wrote:
> Thanks for the reply Erik, As you mentioned, grouping consecutive elements
> of 'a' was my idea.
> I am unaware of any R'ish way to do it. It would be nice if someone in the
> community knows this.
>
> The error resulting in the NA was
Thanks for the reply Erik, As you mentioned, grouping consecutive elements
of 'a' was my idea.
I am unaware of any R'ish way to do it. It would be nice if someone in the
community knows this.
The error resulting in the NA was pretty easy to fix, and my loop works, but
the results are still wrong (
Roy Mathew wrote:
Thanks for the reply Erik, As you mentioned, grouping consecutive
elements of 'a' was my idea.
I am unaware of any R'ish way to do it. It would be nice if someone in
the community knows this.
Is this the idea you're trying to execute? It uses ?rle and ?mapply.
a <- c(2,3
On Mon, Jan 24, 2011 at 07:16:58PM +0100, Roy Mathew wrote:
> Dear R-users,
> This is a loop which is part of a bigger script. I managed to isolate the
> error in this loop and simplified it to the bare minimum and made it
> self-contained.
>
> a<-c(2,3,4,5,5,5,6,6,6,7)
>
> for(n in 1:10)
> {
> p
Roy,
I have no idea what you're actually trying to do here, but
it looks like there would be a more natural R'ish way if
you're concerned about grouping consecutive elements of 'a'.
At any rate, within your while loop, you're incrementing n by
1, and eventually n will be 10, which will be transf
Dear R-users,
This is a loop which is part of a bigger script. I managed to isolate the
error in this loop and simplified it to the bare minimum and made it
self-contained.
a<-c(2,3,4,5,5,5,6,6,6,7)
for(n in 1:10)
{
print(paste("n: ",n))
z1<-a[n]
#make a list container
ldata<-list()
t=1
while(z1
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