Dear expeRts,
These days i and some of my friends often encount the same problem when
installing packages, e.g. when i try to install.packages(stringi), i will get
:
package ‘stringi’ successfully unpacked and MD5 sums checked
Warning in install.packages :
unable to move temporary
Hi
You still do not disclose important info about details of your functions.
However, when you want to perform indexing like you show, you maybe can get rid
of NULL and use zero instead.
a-1:5
a[-c(1,3)]
[1] 2 4 5
a[-c(0,1,3)]
[1] 2 4 5
a[-c(1,0,3)]
[1] 2 4 5
a[-c(0,1,0,3,0)]
[1] 2 4 5
On 11/09/2014 07:19, PO SU wrote:
Dear expeRts,
These days i and some of my friends often encount the same problem when installing
packages, e.g. when i try to install.packages(stringi), i will get :
package ‘stringi’ successfully unpacked and MD5 sums checked
Warning in install.packages
It really suprise me that a[-c(0,1,2)] works as a[-c(1,2)].
But, unfortunately, a[-0] still can't work . That 's mean:
i want:
a-1:3
a[-0]
1 2 3
or
a[-NULL]
1 2 3
--
PO SU
mail: desolato...@163.com
Majored in Statistics from SJTU
At 2014-09-11 02:24:25, PIKAL Petr
Hi
If you insist on this behaviour you need to redefine [ function to suit your
needs. You can see there is already plenty of methods for this operator.
methods([)
[1] [.acf* [.arrow* [.AsIs [.bibentry*
[5] [.data.frame [.Date [.difftime
Hi,
I have two vector of margins. Now I want to create fill matrix that
reflects the margins.
seats - c(17,24,28,30,34,36,40,44,46,50)
mandates - c(107,23,24,19,112,19,25,20)
Both vectors adds up to 349. So I want a 10x8 matrix with row sums
corresponding to seats and column sums
I got the benchmark script, which I've attached, from Texas Advanced
Computing Center. Here are my results (elapsed times, in secs):
Where can we get the benchmark script?
__
R-help@r-project.org mailing list
Hi,
My R Studio version is 3.1.0 but JRI package is not getting installed on
it.It is giving the following error.
package ‘JRI’ is not available (for R version 3.1.0)
Is there any another way to do this?
Thanks
__
R-help@r-project.org mailing list
Sorry for that, Glenn.
http://statweb.stanford.edu/~tibs/ElemStatLearn/
Have also a look at this that I've run into:
http://codecondo.com/9-free-books-for-learning-data-mining-data-analysis/
Best regards,
Angel
De: Glenn Doherty [mailto:glennrdohe...@gmail.com]
Enviado el: miércoles, 10 de
Hi,
As far as I know, there is any package JRI. You need to install rJava.
Regards,
Pascal
On Thu, Sep 11, 2014 at 6:08 PM, madhvi.gupta madhvi.gu...@orkash.com wrote:
Hi,
My R Studio version is 3.1.0 but JRI package is not getting installed on
it.It is giving the following error.
package
On 10/09/2014, 9:53 PM, PO SU wrote:
Tks, i think using logical index is a way, but to do that, i have to keep a
vector as long as the original vector. that's, to exclude position 1 and 3
from
a-1:5
I have to let b-c(F,T,F,T,T) and exec a[b], not a[-c(1,3)]. which c(1,3) is
much
Please check this link, particularly the red line: http://rforge.net/JRI/
Regards,
Pascal
On Thu, Sep 11, 2014 at 6:44 PM, madhvi.gupta madhvi.gu...@orkash.com wrote:
I already have rjava.JRI package is for r java interface
On 09/11/2014 02:44 PM, Pascal Oettli wrote:
Hi,
As far as I
Orignally i don't want to do the if ( length) check because i know that in a
1 loops, after may be 10 or 20 or 100 loops , i will not be empty.
so i mean , in the left loops, i would always check something not needed to
check which i would not like to do.
--
PO SU
mail:
Actually, i thought the way:
a-1:3
b-NULL or 2
a[-b] will not work if b is NULL
A-c(a,1)
B-c(b,length(A))
A[-B] will get the same result as if b is NULL get a, if b is 2 get a[-2]
I think it works well in considering memory use or efficiency or code tidy.
--
PO SU
mail:
On 2014-09-10 10:45, PIKAL Petr wrote:
On 2014-09-08 15:47, William Dunlap wrote:
d - data.frame(Choices=c(One,Two,One,Three), X=1:4)
i - 1 # possible output of menu(unique(d$Choices))
d[ d$Choices[i] == d$Choices, ]
# Choices X
#1 One 1
#3 One 3
testd -
On 11/09/2014, 6:04 AM, PO SU wrote:
Actually, i thought the way:
a-1:3
b-NULL or 2
a[-b] will not work if b is NULL
A-c(a,1)
If a is a long vector, that is a very expensive operation, at least as
expensive as constructing a logical vector of the same length as a.
B-c(b,length(A))
On Tue, 9 Sep 2014 04:42:11 PM Norbert Dörrer wrote:
Dear Experts
I'm quite new to the world of R; so maybe my question is kind of
beginners...
I am right now trying to analyse questionnaire data (CAWI) for
producing a
image/positioning map of the competitors of my company; when
On Thu, 11 Sep 2014 08:58:54 PM Jim Lemon wrote:
Sorry, copying fumble. The last line in the loop should read:
sum(nddf[subject,seq(2,10,by=2)]))
Jim
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Hi,
I want to generate a LD plot that looks like the triangle in the lower part
of the output of snp.plotter()
As I would not use the values of the original LD measure (instead I use a
more general measure) I have calulated pairwise correlation on my own. I
could arrange the results in a data
On 11 Sep 2014, at 08:24, PIKAL Petr petr.pi...@precheza.cz wrote:
Hi
You still do not disclose important info about details of your functions.
However, when you want to perform indexing like you show, you maybe can get
rid of NULL and use zero instead.
a-1:5
a[-c(1,3)]
[1] 2 4 5
Do you have an example of what you would like your output to look like? It
is a little difficult to fully understand what you are looking for. You
only have 18 values but are looking to fill at 10x8 matrix (i.e. 80
values). If you can clarify better we may be better able to help you.
Charles
I have :
rs - c(3, 2, 3, 4)
cs - c(4, 5, 3)
And want:
matrix
[,1] [,2] [,3]
[1,] 120
[2,] 101
[3,] 111
[4,] 121
The rowSums in the above matrix is equal to sum(rs) and colSums is
equal to sum(cs). It's sort of a matrix expansion where the margins
are known
I'm out of the office today, but will resend it tomorrow.
Jonathan Anspach
Intel Corp.
Sent from my mobile phone.
On Sep 11, 2014, at 3:49 AM, arnaud gaboury arnaud.gabo...@gmail.com wrote:
I got the benchmark script, which I've attached, from Texas Advanced
Computing Center. Here are my
You'll find R-benchmark-25.R, which I assume is the same and the proper
pointer to use, at http:// http://r.research.att.com/benchmarks/
r.research.att.com http://r.research.att.com/benchmarks//benchmarks/
http://r.research.att.com/benchmarks/
Henrik
I'm out of the office today, but will resend
On 11/09/2014 3:29 AM, Philippe GROSJEAN wrote:
On 11 Sep 2014, at 08:24, PIKAL Petr petr.pi...@precheza.cz wrote:
Hi
You still do not disclose important info about details of your functions.
However, when you want to perform indexing like you show, you maybe can get rid of
NULL and use
..°}))
) ) ) ) )
( ( ( ( (Prof. Philippe Grosjean
) ) ) ) )
( ( ( ( (Numerical Ecology of Aquatic Systems
) ) ) ) ) Mons University, Belgium
( ( ( ( (
..
On 11 Sep 2014,
Hi,
I got the following data frame:
dat1 - read.table(text=a,b
1,A1
2,A1
3,A1
4,A1
5,A1
6,A2
7,A2
8,A2
9,A2
10,A2
11,B1
12,B1
13,B1
14,B1
15,B1,sep=,,header=T)
I would like to add a new column dat1$new based on column b (dat$b) in
which values will be substituted according to their unique
Note that in the data you sent, b is a factor:
str(dat1)
'data.frame': 15 obs. of 2 variables:
$ a: int 1 2 3 4 5 6 7 8 9 10 ...
$ b: Factor w/ 3 levels A1,A2,B1: 1 1 1 1 1 2 2 2 2 2 ...
So all you need is
dat1$new - as.numeric(dat1$b)
table(dat1$new)
table(dat1$new)
1 2 3
5 5 5
On Thu, Sep 11, 2014 at 10:49 AM, raz barvazd...@gmail.com wrote:
Hi,
I got the following data frame:
dat1 - read.table(text=a,b
1,A1
2,A1
3,A1
4,A1
5,A1
6,A2
7,A2
8,A2
9,A2
10,A2
11,B1
12,B1
13,B1
14,B1
15,B1,sep=,,header=T)
I would like to add a new column dat1$new based
This too.
Applied Predictive Modeling
Max Kuhn • Kjell Johnson
Thanks,
Mohan
On Tue, Sep 9, 2014 at 10:03 PM, Subba Rao raspb...@tanucoo.com wrote:
Hi,
I am interested in R programming in the Big Data space. Are there any
books that would be a good starting point for this career path?
You are only able to search twitter history for a short period of time.
gnip.com and similar companies offer historical tweets for sale.
cn
On Sunday, September 7, 2014 9:21:34 AM UTC-5, Axel Urbiz wrote:
Hello,
The function searchTwitter() with the arguments supplied as below would
Dear all, I'm trying the following experiment simulation, but I'm receiving
this error:
probs()Error in x[j, 4] : incorrect number of dimensions
however, the simulation works fine outside the function statement{}. What am I
doing wrong?
# Create some fake data and call the function: df -
Yes, that's the original. Then TACC increased the matrix sizes for their tests.
Jonathan Anspach
Intel Corp.
Sent from my mobile phone.
On Sep 11, 2014, at 9:18 AM, Henrik Bengtsson
h...@biostat.ucsf.edumailto:h...@biostat.ucsf.edu wrote:
You'll find R-benchmark-25.R, which I assume is the
David,
Thanks a lot for your sugestion, it solved my problem!
Greetings,
--
Thiago V. dos Santos
PhD student
Land and Atmospheric Science
University of Minnesota
http://www.laas.umn.edu/CurrentStudents/MeettheStudents/ThiagodosSantos/index.htm
Phone: (612) 323 9898
On Wednesday, September 10,
Look below to see what happens to your formatting when you use html. Don't use
html.
Why do you use x='df' in defining the function
df is a data frame with 5 observations and 4 variables.
'df' is a character vector of length 1. Your function is looking for a data
frame (or matrix) with at
You want r2dtable():
?r2dtable
set.seed(42)
a - r2dtable(1, seats, mandates)
addmargins(a[[1]])
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,]23106212 17
[2,]8011 11021 24
[3,]80527114
On Windows:
The pdf(file1.pdf) command will throw an error if the file1.pdf is open
in a viewer. For example:
pdf(file1.pdf)
plot(1:10)
dev.off()
shell.exec(file1.pdf)
pdf(file1.pdf) # Causes an error
As suggested by the help page for file.access(), I normally use
try(pdf(file1.pdf)) to test
On Sep 11, 2014, at 5:13 AM, Stefan Petersson wrote:
I have :
rs - c(3, 2, 3, 4)
cs - c(4, 5, 3)
And want:
matrix
[,1] [,2] [,3]
[1,] 120
[2,] 101
[3,] 111
[4,] 121
The rowSums in the above matrix is equal to sum(rs) and colSums is
equal to
Hi,
I am getting an error installing the MASS package. Googling suggests
restarting R, but it didn't help. Anyone has a clue?
Many thanks,
Yuan
install.packages(MASS)
Installing package into ‘/PHShome/yl960/R/3.0’
(as ‘lib’ is unspecified)
trying URL
Yes! That's excactly what I need. Thank You so much!
Den 11 sep 2014 21:20 skrev David L Carlson dcarl...@tamu.edu:
You want r2dtable():
?r2dtable
set.seed(42)
a - r2dtable(1, seats, mandates)
addmargins(a[[1]])
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,]231
A poor mans solution would be to try to temporarily delete or move the
file, depending exactly what you wish to do. You can also try:
library(R.utils)
pathname - Arguments$getWritablePathname(file2.pdf, mustNotExist=FALSE)
which does lots of assertions of write permissions of new and existing
I've got a problem with the mice package that I don't understand.
Here's the code:
library(mice)
d - read.csv(https://dl.dropboxusercontent.com/u/24381951/employment.csv;,
as.is=TRUE, row.names=1)d.imp - mice(data=d, m=1)
Result is:
Error in `[.data.frame`(data, , jj) : undefined columns
Dear expeRts,
When i use the following codes:
pdf(file=1.pdf,width=15)
plot(1:3,main=你好)
dev.off()
#There were 12 warnings (use warnings() to see them)
I find that 你好 can't show correctly in pdf file, but i just
plot(1:3,main=你好) to R plot viewer, it's ok .
Is there anyone
On 2014/9/11 21:10, PO SU wrote:
Dear expeRts,
When i use the following codes:
pdf(file=1.pdf,width=15)
plot(1:3,main=你好)
dev.off()
#There were 12 warnings (use warnings() to see them)
I find that 你好 can't show correctly in pdf file, but i just
plot(1:3,main=你好) to R
¿En qué formato está: texto plano, separado por tabulaciones, de anchura
fija, EXCEL? ¿No puedes leerlo entero, filtrar por año y grabar cada parte?
Un saludo
Isidro Hidalgo Arellano
Observatorio Regional de Empleo
Consejería de Empleo y Economía
http://www.jccm.es
-Mensaje original-
Otra manera sería usar la función split. Que devuelve una lista de
data.frame.
Para expotar luego cada elemento puedes usar lapply, ejemplo:
data(iris)
lista-split(iris,iris$Species)
lapply(names(lista),function(x) write.table(x = lista[[x]],file =
paste0(x,.txt)) )
Saludos
El 11 de septiembre
Estimados, estoy creando un SpatialPolygonsDataFrame para crear el
polygons, debo crear de a uno los polygon, tengo 26 poligonos y quiero
saber si puedo escribir esto de una sola vez sin hacerlo uno para cada uno,
Ejemplo:
Sr1= Polygon(Calamuchita[,1:2])
Sr2= Polygon(Capital[,1:2])
Si bien prefiero algo como lo propuesto por Jorge, siguiendo tu planteo
creo que lo siguiente tiene que funcionar:
filename-test1.csv
DBa-read.table(filename,sep=;,header=TRUE) # en el archivo que
enviastes el separador es ; no ,
anios-unique(DBa$Year)
for(i in anios) {
Hola:
En casos como estos no hay que olvidar que el sistema operativo
un+x, olvidé casi todo los que sabía del /command/ de microsoft,
nos ofrece procedimienetos muy simples para realizar tareas
rutinarias. En este caso creo que lo más rápido es, dede una
terminal, en el directorio de datos:
Hola Marta,
Si el código exacto que utilizaste es el que has compartido, hay un error...
Te falta cerrar un corchete. Y he incluido un detalle para que los ficheros
tengan como nombre los años...
#--
filename-test1.csv
DBa-read.csv(filename,sep=*;* ,header=TRUE)
Muchas gracias a todos por su orientación. Me queda claro algunos puntos
adicionales que debo explorar para mejorar mis rutinas.
Saludos, buen día.
V
El 10 de septiembre de 2014, 18:00, Juan Diego Alcaraz-Hernández
jdalca...@gmail.com escribió:
Hola Victor
La pregunta es muy amplia, echale
51 matches
Mail list logo
