Better ask for local help if you can't reduce your code to some minimal
examples so that we can understand easily what you are looking for.
On 02.03.2015 10:38, Rami Alzebdieh wrote:
Dear Sir,
I start using (R) 3 months ago, and I am still learning,
Same for me after more than 16
Dear Sir,
I start using (R) 3 months ago, and I am still learning, I have a project and I
am using R in this project, my friend helped me to build a code for this
project and it's working perfect, but I need to make a small change in, it
looks very simple but for me it's very complicated. I
Hola David,
¿Seguro que buscas las combinaciones?
Creo que lo que buscas es esto...
#
MuestraS - c(1 ,1, 1, 1, 0, 1, 1, 0, 1, 1)
library(combinat)
resPer - permn(MuestraS)
matresPer - matrix(unlist(resPer), nrow=factorial(length(MuestraS)),
ncol=length(MuestraS))
Buena tarde amigos,
En días pasados hice algunas consultas y ya pude salir de las dudas que
tenia en ese momento, ahora requiero de su colaboración con lo siguiente:
Tengo un vector dicotomico (Binario) con la siguiente información que me
surgio de algunos procesos anteriores:
MuestraS
[1] 1
Angela:
These are statistical, not R, issues I believe, and you appear to be
out of your depth statistically here. I suggest you talk to a local
statistical resource or, if you can't find such help, post on a
statistical site like stats.stackexchange.com.
Cheers,
Bert
Bert Gunter
Genentech
Hi,
I would like to know how to decide the weight in a WLS model in R?
For example, In the pipeline data from faraway, I try to fit a
regression model Lab ~ Field (non-constant variance). I wish to use weights
to account for the non-constant variance. So how to decide the weight in
the WLS
On 02/03/2015 11:53 AM, Mello Cavallo, Alice wrote:
I copied the file into the bin folder of R ...
perf_data - read.csv(PerfResultsCSv.csv)
Error in file(file, rt) : cannot open the connection
In addition: Warning message:
In file(file, rt) :
cannot open file 'PerfResultsCSv.csv': No
Hello,
I apologies for bringing up next and break in loops given that there is so
much on the net about it, but I've tried numerous examples found using
Google and just can't seem to get this to work.
This is a simple version of what I am doing with matrices but it shows the
issue. I need to
Muchas gracias por sus respuestas tan oportunas y precisas,
Use partes de los dos códigos y hasta el momento todo va perfecto.
Nuevamente gracias y estaré en contacto ante cualquier inconveniente que se
me presente.
Saludos,
DC.
El 2 de marzo de 2015, 17:50, Carlos Ortega
On 03/03/15 15:04, Jeff Newmiller wrote:
Your example is decidedly not expressed in R, though it looks like
you tried. Can you provide the hand-computed result that you are
trying to obtain?
Note that the reason you cannot find anything about next or break in
R is that they don't exist.
Point
Sigh. To be positive is to be wrong at the top of one's lungs. Next I will be
told R has a goto statement.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.us
Here is an implementation.
(x -
c(23,35,22,11,10,1,14,15,13,15,17,16,154,13,24,25,25,25,25,25,22,11,15,15))
[1] 23 35 22 11 10 1 14 15 13 15 17 16 154 13 24 25 25 25
25 25 22 11 15 15
(y - c(0,cumsum(x)))
[1] 0 23 58 80 91 101 102 116 131 144 159 176 192 346 359
Your example is decidedly not expressed in R, though it looks like you tried.
Can you provide the hand-computed result that you are trying to obtain?
Note that the reason you cannot find anything about next or break in R is that
they don't exist. There are generally alternative ways to
On 03/03/15 16:08, Jeff Newmiller wrote:
Sigh. To be positive is to be wrong at the top of one's lungs. Next I
will be told R has a goto statement.
I am ***positive*** that it hasn't! :-) Well, 99.999% confident.
Although I guess it's not inconceivable that some misguided nerd might
Hello All,
I have to admit that I am not that good when it comes to vectorizing a
function. I need some insight. Is the below a case where vectorization can be
accomplished to improve speed?
Below the function a sample data - as you can see it is not delimited. However, the
record length
Here is an implementation.
t -
data.frame(x=c(1,1,1,1,1,2,2,2,2,2),y=c(a,a,a,b,b,a,a,b,b,b))
t
x y
1 1 a
2 1 a
3 1 a
4 1 b
5 1 b
6 2 a
7 2 a
8 2 b
9 2 b
10 2 b
assignSeq
function(test)
{
temp - test[order(test$x),]
InC - numeric(length(test))
inD - unique(test$x)
countAll
Hello List,
I am trying to obtain a table containing absolute and relative
frequencies but it must be done by strata. Each strata have to contain
totals and subtotals being the sum of the subtotals equal to the total
in upper strata in same column. As this could be some vague I am
Dear R-help,
I'm getting an error with Glm (from the rms package) when the
equivalent model using glm does not give an error. This is using rms
4.3-0 in R 3.1.1.
An example is shown below. I have set the seed value, but the error is
not specific to this seed value.
Thanks for any help anyone
Hi
I do not see much logic in your removal of outliers.
you can easily find which values differ from previous one by more than 15
myts[c(FALSE,abs(diff(myts$x))15),]
but I did not understand why do you keep values from row 8 and 10.
Your example can be solved by
myts$y[myts$x15]-1
Dear R-users,
I would like to speed up a double-loop I developed for detecting and
removing outliers in my whole data.frame. The idea is to remove data with a
too big difference with the previous value. If detected, this test must be
done here on maximum the next 10 values following the last
Hi,
Maybe a beginning of solution with this?
test -
data.frame(x=c(1,1,1,1,1,1,2,2,2,2,2,2),y=c(a,a,a,b,b,b,a,a,b,b,b,a))
test[order(test$x),]
out - split(test,test$x)
for (i in 1:length(out)) {
foo - unique(out[[i]][,2])
out[[i]][,2] - rep(foo,(nrow(out[[i]])/(length(foo }
On 01/03/2015 17:21, Duncan Murdoch wrote:
On 01/03/2015 10:13 AM, Ista Zahn wrote:
I think this might be due to the removal of the -remote option in
firefox. Some discussion and details are available at
https://lists.gnu.org/archive/html/emacs-orgmode/2015-02/msg00946.html
Yes, that's it. R
Is this a very recent version of Firefox? Have a look at the thread
Re: [R] Firefox not showing R help.
It could be related, in which case you might want to try the up-to-the-minute
latest version of R-patched (aka 3.1.3RC).
Otherwise, as usual, you may need to talk with the relevant package
Have a look at the caschrono package.
There's an excellent associated book by the author of the package -Yves Aragon-
but it's in French; if you don't read French, the package documentation is very
clear.
José
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On
I'm having trouble pulling down data from a website with my code below as I
keep encountering the same error, but the error occurs on different pages.
My code below loops through a wensite and grabs data from the html table. The
error appears on different pages at different times and I'm not
Using this dataset:
dat - read.table(textConnection(daynoRes.QwRes.Q
1 237074.41 215409.41
2 2336240.20 164835.16
3 84855.42 357062.72
4 76993.48 386326.78
5 73489.47 307144.09
6 70246.96 75885.75
7 69630.09 74054.33
8 66714.78 70071.80
9 122296.90 66579.08
10
On Mar 2, 2015, at 11:43 AM, Morway, Eric emor...@usgs.gov wrote:
Using this dataset:
dat - read.table(textConnection(daynoRes.QwRes.Q
1 237074.41 215409.41
2 2336240.20 164835.16
3 84855.42 357062.72
4 76993.48 386326.78
5 73489.47 307144.09
6 70246.96 75885.75
7
Dear Eric,
Here is a solution using the plyr package.
library(plyr)
dat$flg - dat$wRes.Qdat$noRes.Q
dat$group - cumsum(c(0, abs(diff(dat$flg
ddply(dat, group, function(x){
if(x$flg[1] nrow(x) = 5){
x$plygn - seq_along(x$group)
} else {
x$plygn - NA
}
x
})
Best regards,
ir.
I copied the file into the bin folder of R ...
perf_data - read.csv(PerfResultsCSv.csv)
Error in file(file, rt) : cannot open the connection
In addition: Warning message:
In file(file, rt) :
cannot open file 'PerfResultsCSv.csv': No such file or directory
I also installed and load RWeka
This somewhat simpler rvest code does the trick for me:
library(rvest)
library(dplyr)
i - 1:10
urls - paste0('http://games.crossfit.com/scores/leaderboard.php?stage=5',
'sort=0division=1region=0numberperpage=100competition=0frontpage=0',
Hi Petr,
Thanks for your reply,
Actually it's not what I'm looking for. The aim is not simply to remove each
value 15.
In my loop, I consider the first numeric value of my column as correct.
Then, I want to test the second value. If the absolute difference with the
previous correct one is 15,
Hi,
This is my first post to R-help. I'm having trouble getting rgeos to work.
Info on the server and packages I'm using:
$ *uname -a*
Linux some-server.somewhere.com 2.6.32-431.11.2.el6.x86_64 #1 SMP Tue Mar
25 19:59:55 UTC 2014 x86_64 x86_64 x86_64 GNU/Linux
$ *R --version*
R version 3.0.2
Is this file in your working directory? (To know your working directory
use: getwd() )
If not, put it in there.
2015-03-02 11:53 GMT-05:00 Mello Cavallo, Alice mel...@wit.edu:
I copied the file into the bin folder of R ...
perf_data - read.csv(PerfResultsCSv.csv)
Error in file(file,
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