100,ncol(x))
a2[upper.tri(a2)] <- t(a2)[upper.tri(a2)]
> identical(a, a2)
[1] TRUE
Best,
Nello
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Kulupp
Sent: Montag, 24. Juni 2013 11:02
To: r-help@r-project.org
Subject: [R] Avoidi
Dear R-experts,
I'd like to avoid the use of very slow 'for'-loops but I don't know how.
My data look as follows (the original data has 1600 rows and 30 columns):
# data example
c1 <- c(1,1,1,0.25,0,1,1,1,0,1)
c2 <- c(0,0,1,1,0,1,0,1,0.5,1)
c3 <- c(0,1,1,1,0,0.75,1,1,0.5,0)
x <- data.frame(c1,
Hi Trying,
It would be helpful if you provided reproducible examples. It would
also be polite to sign a name so that we have something by which to
address you.
On Tue, Jul 12, 2011 at 8:00 AM, Trying To learn again
wrote:
> Hi all,
>
> I have this information on a file ht.txt, imagine it is a da
Hi all,
I have this information on a file ht.txt, imagine it is a data frame without
labels:
1 1 1 8 1 1 6 4 1 3 1 3 3
And on other table called "pru.txt" I have sequences similar this
4 1 1 8 1 1 6 4 1 3 1 3 3
1 6 1 8 1 1 6 4 1 3 1 3 3
1 1 1 8 1 1 6 4 1 3 1 3 3
6 6 6 8 1 1 6 4 1 3 1 3 3
I want
Hi, I am working on a dataset in which a number of venture capitalists invest
in a number of firms. What I am creating is an asymmetric matrix M in which
m(ij) is the volume (sum) of coinvestments of VC i with VC j (i.e., how much
has VC i invested in companies that VC j also has investments in). T
On 30.05.2010 19:23, Alan Lue wrote:
Is there a performance advantage to doing this, as opposed to growing
the vector within the loop? I suppose R could have to dynamically
reallocate memory at some point?
Right, but that takes time since memory management is always expensive
(and this way
Is there a performance advantage to doing this, as opposed to growing
the vector within the loop? I suppose R could have to dynamically
reallocate memory at some point?
Alan
2010/5/30 Uwe Ligges :
>
>
> On 26.05.2010 08:52, Alan Lue wrote:
>>
>> Come to think of it, we can't save the output of
On 26.05.2010 08:52, Alan Lue wrote:
Come to think of it, we can't save the output of each invocation and
concatenate it later, since we need the output as input for the next
iteration.
Yes, but you can do it a bit cleverer than before by initializing to the
fill length as in:
r.seq <- nu
Hi:
On Tue, May 25, 2010 at 11:43 PM, Alan Lue wrote:
> Since `for' loops are slow in R, and since `apply' functions are
> faster, I was wondering whether there were a way to use an apply
> functionor to otherwise avoid using a loopwhen iterating over a
> statement that updates its input.
>
T
Come to think of it, we can't save the output of each invocation and
concatenate it later, since we need the output as input for the next
iteration.
Alan
On Tue, May 25, 2010 at 11:43 PM, Alan Lue wrote:
> Since `for' loops are slow in R, and since `apply' functions are
> faster, I was wonderin
Since `for' loops are slow in R, and since `apply' functions are
faster, I was wondering whether there were a way to use an apply
function—or to otherwise avoid using a loop—when iterating over a
statement that updates its input.
For example, here's some such code:
r.seq <- 2 * (1 / d$Dt[1] - 1)
thank you very much
--- On Sat, 10/17/09, Kenn Konstabel wrote:
> From: Kenn Konstabel
> Subject: Re: [R] avoiding loops in equation
Thank God for R-help mailing list. Thanks..
> To: "Julius Tesoro"
> Date: Saturday, October 17, 2009, 1:51 PM
> a3 <- sapply(acc,
To illustrate my problem, I have a complex code of several matrices and a
vector. To simplify I only used two matrices and a vector as an example:
pex<-function(acc, pga, std){
(acc-pga)/std
}
acc<-seq(.1,1,.1)
pga<-matrix(rnorm(9,4,.1),3,3)
std<-matrix(rnorm(9,4,.5),3,3)
I tried calculating
bco.com
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Martin Morgan
> Sent: Wednesday, September 02, 2009 9:17 AM
> To: Alexander Shenkin
> Cc: r-help@r-project.org; spec...@stat.berkeley.edu;
> cbe
Alexander Shenkin wrote:
> Though, from my limited understanding, the 'apply' family of functions
> are actually just loops. Please correct me if I'm wrong. So, while
> more readable (which is important), they're not necessarily more
> efficient than explicit 'for' loops.
Hi Allie -- This uses a
On Tue, 1 Sep 2009, dolar wrote:
Would like some tips on how to avoid loops as I know they are slow in R
If I understand your criterion (and calling your data.frame 'dat'):
criterion <- as.matrix(dist(dat$a)) <= 5 & outer(dat$a,dat$a,">=")
criterion %*% as.matrix(dat[, c("b","c")])
b
Another advantage of the apply family of functions is that
they determine the size and type of their output in an
efficient way, which is sometimes tricky when you write
the loop yourself.
- Phil Spector
Statistica
If you can do it- try a for loop and another solution to prove this to
yourself. A for loop can get a little unwieldy for a novice like me
to understand the code, but doable. The simpler the better, but they
are not terribly slow. I have run into a couple of situations where a
vectorized solutio
> Would like some tips on how to avoid loops as I know they are slow in R
They are not slow. They are slower than vectorised equivalents, but
not slower than apply and friends.
Hadley
--
http://had.co.nz/
__
R-help@r-project.org mailing list
https:/
Though, from my limited understanding, the 'apply' family of functions
are actually just loops. Please correct me if I'm wrong. So, while
more readable (which is important), they're not necessarily more
efficient than explicit 'for' loops.
allie
On 9/2/2009 3:13 AM, Phil Spector wrote:
> Here's
Here's one way (assuming your data frame is named dat):
with(dat,
data.frame(a,t(sapply(a,function(x){
apply(dat[a - x >= -5 & a - x <=
0,c('b','c')],2,sum)}
- Phil Spector
St
Would like some tips on how to avoid loops as I know they are slow in R
i've got a data frame :
a b c
1 5 2
4 6 9
5 2 3
8 3 2
What i'd like is to sum for each value of a, the sum of b and the sum of c
where a equal to or less than (with a distance of 5)
i.e. for row three
we have a=
"DM" == David Masson <[EMAIL PROTECTED]>
on Wed, 05 Nov 2008 15:13:37 +0100
DM> I have a question concerning avoiding loops.
DM> I know the function "apply" and I have used it several times, but I feel
DM> blocked
DM> with this situation :
DM>
DM> E <- array(X, dim =
I have a question concerning avoiding loops.
I know the function "apply" and I have used it several times, but I feel
blocked
with this situation :
E <- array(X, dim = c(L,nlon,nlat) )
data <- matrix(Y, nrow=nlon, ncol=nlat )
G <- vector(length=L)
for (l in 1:L)
{
G[l] <- function.F(
ill.venables/
> >
> > -Original Message-
> > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
> > project.org]
> > On Behalf Of Ingmar Visser
> > Sent: Thursday, 27 March 2008 7:58 AM
> > To: R-help@r-project.org
> > Subject: [R] avoiding loops
&
OTECTED] [mailto:[EMAIL PROTECTED]
> project.org]
> On Behalf Of Ingmar Visser
> Sent: Thursday, 27 March 2008 7:58 AM
> To: R-help@r-project.org
> Subject: [R] avoiding loops
>
> Hi,
> I need to compute an array from a matrix and an array:
>
> A <- array(1:20,
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Ingmar Visser
Sent: Thursday, 27 March 2008 7:58 AM
To: R-help@r-project.org
Subject: [R] avoiding loops
Hi,
I need to compute an array from a matrix and an array:
A <- array(1:20,c(2,2,5))
B <- matrix(1:10,
Hi,
I need to compute an array from a matrix and an array:
A <- array(1:20,c(2,2,5))
B <- matrix(1:10,5)
And I would like the result to be an array consisting of the following:
rbind(A[1,,1]*B[1,],
A[2,,1]*B[1,])
rbind(A[1,,2]*B[2,],
A[2,,2]*B[2,])
rbind(A[1,,3]*B[2,],
A[2,,3]*B[2,])
etc.
He
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