Hi
I have the following problem :
I have 3 vectors xx, yy, zz :
xx - c(5479, 6209, 6940, 7670, 8766, 9496, 10227, 11048, 11778,
12509, 13239, 13970,
14700, 15340, 15948)
yy - c( 267, 275, 281, 287, 296, 306, 316, 325, 334, 351, 365, 377,
389, 419, 419)
zz - c( 3, 3, 3, 3, 4, 4, 4, 4,
How can a person in a controlled environment install additional R
packages..
Charles.
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
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PLEASE do read the posting guide
Hi Michel,
lines(xx,yy,col=zz-2,type=s)
If you use a color vector, say cols, then you can also do
lines(xx,yy,col=cols[zz-2],type=s)
Hope it helps,
Tsjerk
On Mon, Sep 16, 2013 at 8:42 AM, Arnaud Michel michel.arn...@cirad.frwrote:
Hi
I have the following problem :
I have 3 vectors xx,
Hi Tsjerk
Thank you but the color always remains black !
I would want that the color changes on the same graph (color = 3 on the
4 first steps, col = 4 on 5 following steps
Michel
Le 16/09/2013 09:01, Tsjerk Wassenaar a écrit :
Hi Michel,
lines(xx,yy,col=zz-2,type=s)
If you use a
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Evan Sticca
Sent: Friday, September 13, 2013 6:48 PM
To: r-help@r-project.org
Subject: [R] Splitting data into two camps
Hello R-help,
I have recently generated some
Hi Michel,
In that case, you need to use segments:
?segments
For a line, it works like:
plot(xx,yy,type=n)
segments(xx[-1],yy[-1],xx[-length(xx)],yy[-length(yy)],col=zz,lwd=2)
For a step function, you'll have to do a bit more work :)
Cheers,
Tsjerk
On Mon, Sep 16, 2013 at 9:20 AM, Arnaud
Hi,
Maybe the following might help you:
s - seq(length(xx)-1)
plot(xx, yy, type=n)
segments(xx[s], yy[s], xx[s+1], yy[s], col=zz, lwd=2)
segments(xx[s+1], yy[s], xx[s+1], yy[s+1], col='grey')
Regards,
Pascal
On 16/09/2013 15:42, Arnaud Michel wrote:
Hi
I have the following problem :
I
Thanks Pascal and Tsjerk
Michel
Le 16/09/2013 09:42, Pascal Oettli a écrit :
Hi,
Maybe the following might help you:
s - seq(length(xx)-1)
plot(xx, yy, type=n)
segments(xx[s], yy[s], xx[s+1], yy[s], col=zz, lwd=2)
segments(xx[s+1], yy[s], xx[s+1], yy[s+1], col='grey')
Regards,
Pascal
On
Le vendredi 13 septembre 2013 à 23:38 +0400, Maxim Linchits a écrit :
This is a condensed version of the same question on stackexchange here:
http://stackoverflow.com/questions/18789330/r-on-windows-character-encoding-hell
If you've already stumbled upon it feel free to ignore.
My problem is
On 09/16/13 19:01, Charles Thuo wrote:
How can a person in a controlled environment install additional R
packages...
Use a local library in your user space. Create a directory --- I
called mine Rlib,
and it is located in my home directory. I.e. my local library is
/home/rolf/Rlib.
Then
On Monday, September 16, 2013 09:01:03 Charles Thuo wrote:
How can a person in a controlled environment install additional R
packages..
Charles.
Hello Charles,
a slight variation of what Rolf wrote, this is my setup:
in my ~/.Rprofile (which is read at R's startup), I set the R_LIBS_USER
On Sun, 15 Sep 2013, Andrew Crane-Droesch wrote:
The c2d4u PPA is the main search result when googling upgrade R 3.0.1
ubuntu.
And it should be, because it is more likely that a PPA re-distribution works
better for Ubuntu than a general distribution, even if it is an exceptional
case with
Josh,
A couple of things:
1) It would be helpful if you can provide some reproducible data and the
code you have developed thus far.
2) This is more of a stackexchange.com or crossvalidated.com question.
That said...without seeing the data...
Dendrograms/hclust are generated by using a distance
Le lundi 16 septembre 2013 à 10:40 +0200, Milan Bouchet-Valat a écrit :
Le vendredi 13 septembre 2013 à 23:38 +0400, Maxim Linchits a écrit :
This is a condensed version of the same question on stackexchange here:
Dear R list,
I want to aggregate the number of individuals 'IND' of the same ORDER,
within each site and season CAMP,TRANS... but I also want to keep record of
the habitat HAB and LOTE
For example I have this:
CAMP LOTE HAB TRANS IND ORDEN
1765 C1 B1 BB1 7 HEMIPTERA
Hi,
Try:
aggregate(IND~.,data=net1,sum)
CAMP LOTE HAB TRANS ORDEN IND
1 C1 B1 C C1 0
2 C1 B1 B B3 ACARI 3
3 C1 B1 B B1 ARANEAE 1
4 C1 B1 B B3 ARANEAE 2
5 C1 B1 B B3 COLEOPTERA 2
6 C1 B1 B
Neal,
I like this answer. Simple and clean. Don't know why I didn't think of that
before.
Thanks!
--
Noah Silverman, M.S., C.Phil
UCLA Department of Statistics
8117 Math Sciences Building
Los Angeles, CA 90095
On Sep 4, 2013, at 3:12 PM, Neal Fultz nfu...@gmail.com wrote:
print(1:100)
What question (or questions) are you trying to answer? Any advice we may
give will depend on what you are trying to accomplish.
On Sat, Sep 14, 2013 at 2:12 PM, Saumya Gupta saumya.gu...@outlook.comwrote:
I have a dataset which has several predictor variables and a dependent
variable, score
Hello,
In R 3.0.1, I get the following warning that I do not get in R 2.15.3:
data(mtcars)
mycov.rob(mtcars[,1:3], method=mcd)
Error in .C(mve_fitlots, as.double(x), as.integer(n), as.integer(p), :
mve_fitlots not available for .C() for package MASS
It seems like there was a change in MASS?
Here is an example using grid functions, based on an example from Deepayan
(https://stat.ethz.ch/pipermail/r-help/2005-April/069459.html)
I hope this helps.
library(grid)
library(lattice)
ft -
grid.layout(nrow = 2, ncol = 4,
heights = unit(rep(1, 2), lines),
widths
Hi the list,
I am using the function microbenchmark to measure the performance of some code. But I notice that
the first execution of the code takes much longueur than the next executions.
I compare it to several executions of the code :
--- 8 --
A - matrix(1:9,3)
nbReroll - 1000
Hi all,
I wonder if there is a way to draw two separate legends in xyplot as I
would like to separate the legend for data and the legend for reference
lines I add. I can use key argument to draw one legend with everything
together. What I really want is to put one legend at the bottom and the
Hello R Team,
Thanks
for this gigantic software Called R. I am new to R software. My name is Amit
Khatri and currently I am working as a Research Student in
Department of Economics University of Mumbai, Mumbai, India. I need your
suggestion
on How to use
HI,
Not sure how you wanted the results with the rows having NAs.
net1[,sapply(net1,is.factor)]-lapply(net1[,sapply(net1,is.factor)],as.character)
with(net1,aggregate(IND,list(CAMP,LOTE,HAB,TRANS,ORDEN),FUN=sum))
#or
with(net1,aggregate(IND,list(CAMP,LOTE,HAB,TRANS,ORDEN),FUN=sum,na.rm=TRUE))
UTF-8 on windows is a huge pain, this bites me often. Usually I give
up and do the analysis on a Linux server. In previous struggles with
this I've found this blog post enlightening:
https://tomizonor.wordpress.com/2013/04/17/file-utf8-windows/
Best,
Ista
On Mon, Sep 16, 2013 at 10:38 AM, Milan
Hi All,
I need some help regarding how to set up a breakpoint in debug. For example, I
have a very simple/naïve function (a useless function just for demo)
f = function()
{
x = 10;
len = 100;
a = 1;
for(i in 1:len)
{
a = a * i;
}
y = x + a;
y;
}
If I need to
You could just use debug(f) and then when the Browser opens and the loop begins
type 'c', that jumps over the loop to next line after the loop.
Best
Simon
On Sep 16, 2013, at 9:05 PM, Hui Du hui...@dataventures.com wrote:
Hi All,
I need some help regarding how to set up a breakpoint in
Hi,
library(MASS)
Couldn't find the function ?mycov.rob()
mycov.rob(mtcars[,1:3],method=mcd)
#Error: could not find function mycov.rob
??mycov.rob
No vignettes or demos or help files found with alias or concept or
title matching ‘mycov.rob’ using regular expression matching.
Though,
On 16/09/2013 3:05 PM, Hui Du wrote:
Hi All,
I need some help regarding how to set up a breakpoint in debug. For example, I
have a very simple/naïve function (a useless function just for demo)
f = function()
{
x = 10;
len = 100;
a = 1;
for(i in 1:len)
{
a =
On Sep 16, 2013, at 10:53 AM, Saumya Gupta wrote:
I have a training dataset which contains statistics of football players for
the year 2009, and their ranks for the year 2010. For example:
RHelp is not the place to ask for help on homework or Kaggle challenges.
Read:
Hi Duncan,
I've put an example file online at
https://docs.google.com/file/d/0B73Ve8vxnjR6QnRESXBQTHRUME0/edit?usp=sharing,
with a screenshot showing the expected contents of the file at
https://docs.google.com/file/d/0B73Ve8vxnjR6b1ZSQmtsRXdadVU/edit?usp=sharing
Hopefully you'll find this easy
Le lundi 16 septembre 2013 à 13:39 -0400, Duncan Murdoch a écrit :
On 16/09/2013 12:04 PM, Maxim Linchits wrote:
Here is that old post:
http://r.789695.n4.nabble.com/read-csv-and-FileEncoding-in-Windows-version-of-R-2-13-0-td3567177.html
In that post, you'll see I asked for a sample file.
Hi,
I had an example like this:
iduseraction
1 12 login
2 12 view
3 12 view
4 12 view
5 12 login
6 12 view
7 12 view
8 12 login
which I used to split using split(dat1,cumsum(dat1$action==login)).
If I had a
Hello,
That's an even simpler case for ?split.
dat - read.table(text =
iduserIP
1 12 ip1
2 12 ip1
3 12 ip2
4 12 ip2
5 12 ip2
6 12 ip3
7 12 ip3
8 12 ip3
, header = TRUE)
split(dat, dat$IP)
Hope this
Thanks... I don't know why I didn't try... guess was in hurry...
I apologize for posting such a simple question
On Mon, Sep 16, 2013 at 3:44 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
That's an even simpler case for ?split.
dat - read.table(text =
iduserIP
1 12
I missed that, thank you.
This is from a function someone else wrote, and they modified
cov.rob(); I will look through what they have done.
Thank you.
On Mon, Sep 16, 2013 at 2:05 PM, arun smartpink...@yahoo.com wrote:
Hi,
library(MASS)
Couldn't find the function ?mycov.rob()
Hello everyone,
I have been struggling quite a bit with R whenever I try to fit ARMA or
ARIMA models and produce forecasts for datasets containing daily
observations.
Can somebody tell me whether R can handle daily time series or not? I have
the impresion that R cannot fit nor produce forecasts
Joshua,
I'm not sure I understand your aim correctly, but if I do, here's my
advice: If you are able to find the clusters according to rows or
columns using clustering, you must be using some kind of a distance
matrix that encodes whether two antibodies should be in one bin for
rows, and a
Hi,
Try:
sum(sapply(1:100,function(i) i^3+ 4*(i^2)))
#[1] 26855900
169551560477066118158651749177/7963268583173904
#[1] 2129170437
sum(sapply(1:25,function(i) ((2^i)/i)+ ((3^i)/(i^2
#[1] 2129170437
A.K.
I have done this on myself using paper and I know the answer for A is
You might consider Projection Pursuit Regression (ppr function). Since the
ranking is just a monotone transformation of the underlying score, ppr can
estimate the transformation and the contribution of the terms into the
score.
On Mon, Sep 16, 2013 at 11:53 AM, Saumya Gupta
to understand the argument is of length zero message, study these
example:
if (grep('a',c('a','b'))==1) 'a' else 'b'
[1] a
if (grep('a',c('c','b'))==1) 'a' else 'b'
Error in if (grep(a, c(c, b)) == 1) a else b :
argument is of length zero
grep('a',c('c','b'))
integer(0)
On 16/09/2013 12:04 PM, Maxim Linchits wrote:
Here is that old post:
http://r.789695.n4.nabble.com/read-csv-and-FileEncoding-in-Windows-version-of-R-2-13-0-td3567177.html
In that post, you'll see I asked for a sample file. I never received
any reply; presumably some spam filter didn't like
Possibly the text() function.
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 9/15/13 9:43 AM, Ankur Seth ankurset...@gmail.com wrote:
I want to put labels a,b,c,d on the data points
I have a training dataset which contains statistics of football players for the
year 2009, and their ranks for the year 2010. For example:
Player
No. of goals
No. of matches
Age
Rank (in 2010)
A
5
1
35
1
B
2
4
23
2
C
1
7
26
3
Here is one approach that uses the tcltk package to create a button on a
different loop than yours, clicking the button will change a variable
(ShouldIStop) in a temporary environment that both your look and the tk
button can access. The loop then just checks the variable from time to
time and
Yes Mac, text function did it. Thanks All for your help.
text(dfPlot$Date, dfPlot$RECLTD,dfPlot$Labels, col=642)
On Tue, Sep 17, 2013 at 12:00 AM, MacQueen, Don macque...@llnl.gov wrote:
Possibly the text() function.
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave.,
Hello,
As for a general purpose bootstrap routine, at an R prompt type the
following.
library(boot)
?boot
Manu other boot strap functions from other packages are available. Good
luck searching.
Hope this helps,
Rui Barradas
Em 16-09-2013 13:45, amit khatri escreveu:
Hello R Team,
it works, but it eliminates the rows with NA
is there a way to keep those?
On Mon, Sep 16, 2013 at 11:22 AM, arun smartpink...@yahoo.com wrote:
Hi,
Try:
aggregate(IND~.,data=net1,sum)
CAMP LOTE HAB TRANS ORDEN IND
1C1 B1 CC1 0
2C1 B1 BB3
Hi,
No problem.
Please ?dput() your dataset.
dat- read.table(text=
strategy region result
conservative desert 64.68427
moderate mountains 10.880242
moderate desert 48.72387
aggressive desert 34.37877
aggressive mountains 37.43783
moderate grassland 60.572490
aggressive forest 5.193187
Hello,
R is able to handle daily time series. A problem arises when you have to
deal with leap years. zoo is able to manage 365 or 366 days a year,
while ts is not able to do that, . But as far as I read, most of ARMA,
ARIMA or SARIMA works with ts, not with zoo. Maybe some else might
provide you
Hi,
t(a*t(b))
# [,1] [,2]
#[1,] 1 8
#[2,] 2 10
#[3,] 3 12
A.K.
Hello eveybody,
I have a vector a and a matrix b :
a
[1] 1 2
b
[,1] [,2]
[1,] 1 4
[2,] 2 5
[3,] 3 6
With simple multiplication I get :
a * b
[,1] [,2]
[1,] 1 8
[2,] 4 5
[3,] 3 12
I would
Hello,
To complete Arun's response, you also have:
sweep(b,2,a,'*')
[,1] [,2]
[1,]18
[2,]2 10
[3,]3 12
or
b %*% diag(a)
[,1] [,2]
[1,]18
[2,]2 10
[3,]3 12
Regards,
Pascal
2013/9/17 arun smartpink...@yahoo.com
Hi,
t(a*t(b))
# [,1]
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