Well, sort of...
aggregate() is basically a wrapper for lapply(), which ultimately must loop
over the function call at the R interpreter level, as opposed to vectorized
functions that loop at the C level and hence can be orders of magnitude
faster. As a result, there is often little difference in
The small example below works lighting-fast; however, when I run the same
script on my real problem, a 1Gb text file, the for loops have been running
for over 24 hrs and I have no idea if the processing is 10% done or 90%
done. I have not been able to figure out a betteR way to code up the
You are certainly in Circle 2 of 'The R Inferno',
which I suspect is where almost all of the
computation time is coming from.
Instead of doing:
divChng - rbind(divChng,c(datTS$ts[1], SEG[j], DC, GRW,
max(datTS$iter)))
it would be much better to create 'divChng' to
be the final length and then
You don't need loops at all.
grw - aggregate(gw ~ ts + ISEG + iter, data = dat, FUN = sum)
GRW - aggregate(gw ~ ts + ISEG, data = grw, FUN = function(x){max(x) -
min(x)})
DC - aggregate(div ~ ts + ISEG, data = subset(dat, IRCH == 1), FUN =
function(x){max(x) - min(x)})
iter -
Hi,
Try ?duplicated()
apply(x,2,function(x) {x[duplicated(x)]-;x})
A.K.
Hi all,
I have a dataset of around a thousand column and a few thousands
of rows. I'm trying to get all the possible combinations (without
repetition) of the data columns and process them in parallel. Here's a
I don't think you answered the OP's query, although I confess that I
am not so sure I understand it either (see below). In any case, I
believe the R level loop (i.e. apply()) is unnecessary. There is a
unique (and a duplicated()) method for data frames, so simply
unique(x)
returns a data frame
Hi everyone,
I'm using the following code to go over every element of a data frame (row
wise). The problem I am facing is that the outer 'x' variable is not
incrementing itself, thus, only one row of values is obtained, and the
program does not proceed to the next row.
This is the code:
Hi, sorry that got sent without the output :
Please ignore the aligning in the input, I am re-adding that here as well :
1GENEACDEFGHIKLMNPQRSTVWY2amt:Amet_00012902334171612422939635201325342732312
3amt:Amet_000219315421218835254372613914212030084
Output:
On 25-03-2013, at 18:43, Sahana Srinivasan sahanasrinivasan...@gmail.com
wrote:
Hi everyone,
I'm using the following code to go over every element of a data frame (row
wise). The problem I am facing is that the outer 'x' variable is not
incrementing itself, thus, only one row of values is
while(x=21)
{
while(y=rown)
{
n-as.numeric(df[[y]][x]);
if(n0)
{
while(k=lim)
{
k-k+1;
} # while loop for k closes
opdf[[y]][x]-sum;
} # if statement closes
y-y+1;
} #while for y closes
x-x+1;
} #while with x
Your immediate problem is that 'y' is
not reset to 1.
Easier code to write would be to use
'for' loops rather than 'while' loops.
Better still would be to use neither if
possible. I suspect that you are in
Circle 3 of 'The R Inferno'.
http://www.burns-stat.com/documents/books/the-r-inferno/
I sent a post yesterday that I found out why my function didn't work. It's
ok now it works.
Thank you all.
--
View this message in context:
http://r.789695.n4.nabble.com/nested-for-loops-tp3992089p3994996.html
Sent from the R help mailing list archive at Nabble.com.
Thank you , this works but I have to do it with nested for loops...
Could you suggest me a way ?
--
View this message in context:
http://r.789695.n4.nabble.com/nested-for-loops-tp3992089p3992324.html
Sent from the R help mailing list archive at Nabble.com.
Why do you need to do it with nested for loops? It is of course possible - and
I hinted how to do it in my first email - but there's no reason as far as I can
see to do so, particularly as a means of MLE. Sounds suspiciously like
homework...
Michael
On Nov 4, 2011, at 10:14 PM, nick_pan
If in fact this is homework, you will do yourself, your classmates, and
possibly your teacher if you let them know that, at least in R, almost
anything you can do in a for() loop can be done easier and faster with
vectorization. If you teacher can't comprehend this, get him fired.
a-c(4,6,3)
You need to define l as a dimensioned object , either a vector or an array,
and then assign the value of your calculation to the correctly indexed
location in that object. Otherwise you are just overwriting the value each
time through the loop. Use these help pages (and review Introduction to R
Carl: Almost anything you can do in a for() loop can be done easier and
faster with vectorization.--
That is false: while this is certainly true for a great many basic
vectorized operations, it is certainly false for most other things --
simulations are a typical example. Note that __ply type
I found the way out - it was because the borders of the vectors was close
enough thats why I had the same result while I was adding points to the
sequence. The example I gave was irrelevant but I made in order to find out
that the problem was.
Thank you all for your answers.
--
View this message
Bert, this is not helpful. Since for loops and apply functions are not
vectorized, why are you admonishing Carl that vectorizing doesn't always speed
up algorithms? He didn't reference apply functions as being vectorized. But you
seem to be doing so.
I would assert that vectorizing DOES always
No idea how this relates to what you said originally but glad you got
it all worked out.
And let us all reiterate: really, don't use nested for loops...there's
a better way: promise!
Michael
On Sat, Nov 5, 2011 at 2:20 PM, nick_pan nick_pa...@yahoo.gr wrote:
I found the way out - it was
Hi all , I have written a code with nested for loops .
The aim is to estimate the maximum likelihood by creating 3 vectors with the
same length( sequence )
and then to utilize 3 for loops to make combinations among the 3 vectors ,
which are (length)^3 in number , and find the one that maximize
Your problem is that you redefine l each time through the loops and
don't record old values; you could do so by using c() for
concatenation, but perhaps this is what you are looking for:
exp(rowSums(log(expand.grid(V1, V2, V3
Hope this helps,
Michael
On Fri, Nov 4, 2011 at 7:49 PM,
Dear Admin,
I will appreciate if you advise me an effective way to write the following R
code including nested for loops. I cannot do it by using expand.grid
function because it results with memory allocation problems.
Thanks for your time and consideration.
for(d1 in 0:n){
for(d2 in 0:n){
for(d3
Le 05/07/10 23:06, Senay ASMA a écrit :
Dear Admin,
I will appreciate if you advise me an effective way to write the following R
code including nested for loops. I cannot do it by using expand.grid
function because it results with memory allocation problems.
Thanks for your time and
What do you want to do with the data being genereated? In the loop
you have, it will just return the last value generated. Let me ask my
favorite question: What is the problem you are trying to solve. If
you get a memory problem with expand.grid, then if you are trying to
store the values in
bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens
baloo mia
Verzonden: dinsdag 22 december 2009 2:07
Aan: r-help@r-project.org
Onderwerp: [R] Nested For loops
Dear R experts,
Might be very simple question to ask but would be insightful. As the same story
] Nested For loops
Dear R experts,
Might be very simple question to ask but would be insightful. As the same story
of nested for loops. following is the code that I am using to get the
autocorrelation function of the sample data. I have tried to get rid of for
loops but since I am touching R
is the lag.
Baloo
--- On Tue, 12/22/09, ONKELINX, Thierry thierry.onkel...@inbo.be wrote:
From: ONKELINX, Thierry thierry.onkel...@inbo.be
Subject: RE: [R] Nested For loops
To: baloo mia baloo_...@yahoo.com, r-help@r-project.org
Date: Tuesday, December 22, 2009, 1:00 AM
Baloo,
Why don't you use
Dear R experts,
Might be very simple question to ask but would be insightful. As the same story
of nested for loops. following is the code that I am using to get the
autocorrelation function of the sample data. I have tried to get rid of for
loops but since I am touching R after such a long
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Michael Knudsen
Sent: Monday, July 13, 2009 10:39 PM
To: r-help@r-project.org
Subject: [R] Nested for loops
Hi,
I have spent some time locating a quite subtle (at least
Make it
for (i in 1:9)
This is not the general solution, but in your case when i=10 you do not want to
do anything.
--- On Tue, 14/7/09, Michael Knudsen micknud...@gmail.com wrote:
From: Michael Knudsen micknud...@gmail.com
Subject: [R] Nested for loops
To: r-help@r-project.org
Received
On Tue, Jul 14, 2009 at 8:03 AM, Moshe Olshanskym_olshan...@yahoo.com wrote:
Make it
for (i in 1:9)
Thanks. That's also how I solved the problem myself. I just somehow
think it makes my code look rather clumsy and opaque. Maybe I just
have to get used to this kind of nasty tricks.
This is
On Tue, Jul 14, 2009 at 8:20 AM, Michael Knudsenmicknud...@gmail.com wrote:
What do you mean? It looks a like a very general solution to me.
Just got an email suggesting using the functions col and row. For example
temp = matrix(c(1:36),nrow=6)
which(col(temp)row(temp))
This gives the indices
On Jul 14, 2009, at 2:25 AM, Michael Knudsen wrote:
On Tue, Jul 14, 2009 at 8:20 AM, Michael
Knudsenmicknud...@gmail.com wrote:
What do you mean? It looks a like a very general solution to me.
Just got an email suggesting using the functions col and row. For
example
temp =
Try this:
seq. - function(from, to) seq(from = from, length = max(0, to - from + 1))
seq.(11, 10)
On Tue, Jul 14, 2009 at 1:38 AM, Michael Knudsenmicknud...@gmail.com wrote:
Hi,
I have spent some time locating a quite subtle (at least in my
opinion) bug in my code. I want two nested for
On Tue, Jul 14, 2009 at 1:56 PM, David Winsemiusdwinsem...@comcast.net wrote:
temp[ upper.tri(temp) ]
[1] 7 13 14 19 20 21 25 26 27 28 31 32 33 34 35
Thanks! I didn't know about that function; it certainly makes things a
lot easier. For example, until now I have used the following, homemade
On Tue, Jul 14, 2009 at 2:29 PM, Gabor
Grothendieckggrothendi...@gmail.com wrote:
seq. - function(from, to) seq(from = from, length = max(0, to - from + 1))
Really nice! Thank you!
--
Michael Knudsen
micknud...@gmail.com
http://lifeofknudsen.blogspot.com/
Hi,
I have spent some time locating a quite subtle (at least in my
opinion) bug in my code. I want two nested for loops traversing the
above-diagonal part of a square matrix. In pseudo code it would
something like
for i = 1 to 10
{
for j = i+1 to 10
{
// do something
}
}
However,
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