xists("specdata")[1] TRUE
> Error:
> >> bad restore file magic number (file may be corrupted) -- no data
> >> loadedIn addition: Warning message:file ‘001.csv’ has magic number
> >> '"Date' Use of save versions prior to 2 is deprecated
> >>
versions prior to 2 is deprecated
> pollutantmean("C:\Users\rhmichel\Desktop\rprog-data-specdata\specdata",
> "sulfate", 1:10)Error: '\U' used without hex digits in character string
> starting ""C:\U">
Thank you for any help you can provide me,Heather M
command I should be using to create a directory named
'specdata' in R 3.2.3
Heather Michel
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help
Hello
I'm newbee with R and RInside
My question is about stderr in R. Is there a way to collect R stderr in C++
program embedding R Thanks in advance
Michel
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antivirus Avast
Hello
I would like to find an elegant way of calculating
c(rep(1, 43), rep(2,43),, rep(10,43))
Any idea ?
Thank you
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idea ?
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PLEASE
Thank you Jeff and Mark for your help
Michel
Le 04/12/2014 15:09, Jeff Newmiller a écrit :
This is a poor approach from a usability perspective... I suggest you create
two separate functions rather than one.
However, you seem to be missing a crucial point in the use of ggplot, which also
Hello
Can one calculate the month number between two dates
D1 - 01/01/2007 and D2 - 01/04/2009 ?
Thank you
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Cirad
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PLEASE do read the posting guide http
Thank you David
Now, the problem is to list all the combinations which verify the
condition III (ie every Rapporteur has to have more or less the same
number of demandeur)
Have you any idea ?
Michel
Le 14/10/2014 13:18, david.kaeth...@dlr.de a écrit :
Hello,
here's a draft of a solution
be balanced and not too different
(Accepted differences : 1)
table(dfnew$Rapporteur1)
Rapporteur01 Rapporteur02 Rapporteur03 Rapporteur04 Rapporteur05
4 4 4 4
4
Thanks for your help
Michel
Dem - structure(list(Nom = c(John
Hello
I have the two dataframes Df1 and Df2 which have the common variable
AgeSexeCadNCad
I would like to add the new variable Df2$Pourcent which correspond at
the value of Df1$AgeSexeCadNCad.
Thank you for your help.
Michel
Df1 - structure(list(AgeSexeCadNCad = structure(1:36, .Label =
c(60
Thank you to Marc Schwartz, Rui Barrada and Sarah Goslee
Michel
Le 19/09/2014 19:46, Marc Schwartz a écrit :
On Sep 19, 2014, at 12:15 PM, Arnaud Michel michel.arn...@cirad.fr wrote:
Hello
I have the two dataframes Df1 and Df2 which have the common variable
AgeSexeCadNCad
I would like to add
)
twoord.plot(
TT$date, TT$y1, TT$date, TT$y2,
lylim = MinMaxr, rylim = MinMaxl,
ylab=YLlab, rylab=YRlab, lcol=2,rcol=4,
do.first=plot_bg();grid(col=\white\,lty=1),
lpch=18, rpch=20, axislab.cex=0.8, cex.main=2,
xtickpos=TT$date, xticklab=TT$datepos)
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port: 06.47.43.55.31
Perfect Jim, It's fine !
Thank you
Michel
Le 26/07/2014 12:16, Jim Lemon a écrit :
On Sat, 26 Jul 2014 09:36:49 AM Arnaud Michel wrote:
Hello
With package plotrix and twoord.plot function, I would like to put the
labels of the ticks values of x-axe which are date (
c(2006 Jan, 2007 Jan, 2008
Perfect Jim, It's fine !
Michel
Le 26/07/2014 12:16, Jim Lemon a écrit :
On Sat, 26 Jul 2014 09:36:49 AM Arnaud Michel wrote:
Hello
With package plotrix and twoord.plot function, I would like to put the
labels of the ticks values of x-axe which are date (
c(2006 Jan, 2007 Jan, 2008 Jan, 2009
help
Michel
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, aes(long,lat,group=group)) +
geom_polygon(aes(fill = value)) +
geom_polygon(data = map.df.l, aes(long,lat),
fill=NA, color = white,
size=0.1) + # white borders
coord_map(project=orthographic, xlim=c(-22,34), ylim=c(35,70)) + # proj
labs(title = Cartographie) +
theme_minimal()
--
Michel ARNAUD
), class = data.frame)
The purpose is to transform df1 it df2 by giving for every group of lines A, B
and C the value 1 if there is at least a value equal to 1 or a value 0 if there
is no value equal to 1
Thanks for your helps
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Hello
I would like to replace the for loop this below
T - as.matrix(T)
for(i in 1: nrow(TEMP)){
for(j in 1: nrow(TEMP)){if (i = j) T[i, j] - 0 }}
I don't find the function in the doc.
Thanks in advance for your help.
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Av
Hi
From the vector
X - c(A, A, B, C, B, A, C)
I would like to build the Dataframe :
data.frame( A=c(1,1,0,0,0,1,0), B=c(0,0,1,0,1,0,0), C=c(0,0,0,1,0,0,1))
Any ideas ?
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tel
Dear R Users
I have the vector
X - c( 6 , 4 ,12 , 3)
I would like to build a new vector by to transform it into
Y - c(rep(X[1], X[1]), rep(X[2], X[2]), rep(X[3], X[3]), rep(X[4], X[4]))
Have you a more elegant answer ?
PS : Sorry for this basic question
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Chargé de mission
Thank you
Michel
Le 09/12/2013 08:14, Berend Hasselman a écrit :
On 09-12-2013, at 08:04, Arnaud Michel michel.arn...@cirad.fr wrote:
Dear R Users
I have the vector
X - c( 6 , 4 ,12 , 3)
I would like to build a new vector by to transform it into
Y - c(rep(X[1], X[1]), rep(X[2], X[2]), rep(X
, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1,
0, 0, 0, 1, 0, 0, 0, 0), .Dim = c(5L, 5L))
Result
1 5
2 3
2 4
4 1
4 3
Thank you for your help
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Av Agropolis 34398 Montpellier cedex 5
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port
Thank you Pascal
Its fine
Michel
Le 20/11/2013 11:55, Pascal Oettli a écrit :
Hello,
One approach is:
m - structure(c(0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0,
0, 0, 1, 0, 0, 0, 0), .Dim = c(5L, 5L))
out - which(m==1, arr.ind=TRUE)
out[order(out[,1]),]
Regards,
Pascal
On 20
Thank you also for your help
Michel
Le 20/11/2013 19:04, Dennis Murphy a écrit :
Hi:
which(m == 1L, arr.ind = TRUE)
Dennis
On Wed, Nov 20, 2013 at 2:28 AM, Arnaud Michel michel.arn...@cirad.fr wrote:
Hi
I have the following problem
I would like to build, from a matrix filled with 0
for your help
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Av Agropolis 34398 Montpellier cedex 5
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(fill = Recrutement)) +
labs(title = Age, x=catégorie, y=Age) +
theme(legend.position = c(0.1,0.9), legend.background =
element_rect(colour = black))
p
any idea ?
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Av Agropolis 34398 Montpellier cedex 5
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fax
OK It is right
Thank you Petr
Michel
Le 11/10/2013 14:58, PIKAL Petr a écrit :
Hi
I usually use scale
something like
scale_fill_discrete(name = Fancy Title)
shall do the trick
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r
M1%*%M2
scale(M1,TRUE,FALSE)
Sorry but I'm a newbe
Thanks in advance
Michel
But no body ... may be my question appear too simple for r-help people
I'll try alone using books
Thank
.
Is it possible to use with mpfrMatrix the same as operations
M1%*%M2
scale(M1,TRUE,FALSE)
Sorry but I'm a newbe
Thanks in advance
Michel
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Thanks I'm lookin for yur example
-Message d'origine-
De : r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] De
la part de arun
Envoyé : vendredi 20 septembre 2013 18:11
À : R help
Objet : Re: [R] Compare two subsequent rows based on specific values of a
string
Hi,
May
Merci Arun
Michel
Le 17/09/2013 22:41, arun a écrit :
Hi Arnaud,
You could also try:
indx- Df1$Mat[-1]==Df1$Mat[-nrow(Df1)]
indx1-c(indx,FALSE)
indx2-c(FALSE,indx)
Df1[indx1,]
Df1[indx2,]
A.K.
From: arun smartpink...@yahoo.com
To: Arnaud Michel michel.arn
Masculin 13/03/1949
Any ideas ?
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Av Agropolis 34398 Montpellier cedex 5
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Thank you Arun
but the values of other columns may be different !!!
Michel
Le 17/09/2013 20:56, arun a écrit :
Hi,
Try:
Df1[duplicated(Df1),]
Df1[duplicated(Df1,fromLast=TRUE),]
A.K.
- Original Message -
From: Arnaud Michel michel.arn...@cirad.fr
To: R help r-help@r-project.org
Cc
you for your help
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Av Agropolis 34398 Montpellier cedex 5
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Hi Tsjerk
Thank you but the color always remains black !
I would want that the color changes on the same graph (color = 3 on the
4 first steps, col = 4 on 5 following steps
Michel
Le 16/09/2013 09:01, Tsjerk Wassenaar a écrit :
Hi Michel,
lines(xx,yy,col=zz-2,type=s)
If you use
Thanks Pascal and Tsjerk
Michel
Le 16/09/2013 09:42, Pascal Oettli a écrit :
Hi,
Maybe the following might help you:
s - seq(length(xx)-1)
plot(xx, yy, type=n)
segments(xx[s], yy[s], xx[s+1], yy[s], col=zz, lwd=2)
segments(xx[s+1], yy[s], xx[s+1], yy[s+1], col='grey')
Regards,
Pascal
, 75,
76, 77, 78, 79, 80, 81, 82, 83, 84, 86, 87,
88, 89, 90, 91, 92, 93, 94, 95, 96, 98, 99,
100, 101, 102, 103, 105, 106, 107, 108, 109,
110), class = data.frame)
Any idea ?
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Av Agropolis 34398 Montpellier cedex 5
tel
Thanks to all three for your fast answer
Michel
Le 08/09/2013 18:41, Renaud Lancelot a écrit :
paste(df1$Cat,
formatC(df1$Ech, flag = 0, width = max(nchar(df1$Ech))),
sep = .)
2013/9/8 Arnaud Michel michel.arn...@cirad.fr
mailto:michel.arn...@cirad.fr
Hello
I have
= .) )
# user system elapsed
# 0.170.000.17
# R Lancelot
system.time(df1$CatEch3 -
paste(df1$Cat, formatC(df1$Ech, flag = 0,
width = max(nchar(df1$Ech))), sep = .))
# user system elapsed
# 0.340.000.35
Thanks
Michel
Le 08/09/2013 19:15, Arnaud Michel a écrit :
Thanks
Thank you Arun
Michel
Le 01/09/2013 07:36, arun a écrit :
Hi Arnaud,
No problem.
Try,
x- 1:10
set.seed(28)
y1- rnorm(10)
set.seed(485)
y2- rnorm(10,25)
library(plotrix)
twoord.plot(x,y1,y2,lylim=c(-2,2),rylim=c(20,30),ylab=y1,rylab=y2,lcol=2,rcol=4,main=y1,
y2 vs. x)
A.K
Thank you Arun
But have you a solution if y1 and y2 have not the same unit (ex : the
unit of y1 is meter and the unit of y2 is Kg) and if I want the axe of
y1 at the left of the plot and the axe of y2 at the rigth of the plot
Michel
Le 31/08/2013 21:27, arun a écrit :
Hi,
May
Hello,
I have 3 vectors x, y1 and y2
I would like to represent on the same plot the two graph (y1, x) and
(y2, x).
Is it possible with ggplot ? other package ?
Thanks for your help
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Av Agropolis 34398 Montpellier cedex 5
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elapsed
14.030.00 14.04
system.time(droplevels(PaysContrat1[with(PaysContrat1,ave(seq_along(Matricule),Matricule,FUN=min))
,] ))
user system elapsed
0.2 0.0 0.2
Michel
Le 24/07/2013 15:29, arun a écrit :
Hi Michel,
You could try:
df1New-droplevels(TEST[with(TEST,ave
But I just noticed that the two solutions are not comparable :
the change concern only Nom and Prenom (solution Berend) and not also
Sexe or Date.de.naissance orother variables (solution Arun) that can
changed. But my question was badly put.
Michel
Le 25/07/2013 08:06, Arnaud Michel a écrit
Le 25/07/2013 08:50, Berend Hasselman a écrit :
On 25-07-2013, at 08:35, Arnaud Michel michel.arn...@cirad.fr wrote:
But I just noticed that the two solutions are not comparable :
the change concern only Nom and Prenom (solution Berend) and not also Sexe or
Date.de.naissance orother variables
, Jeanine,
Jeannine, Michel, Michele, Michèle, Michelle, Victor
), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L,
1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class =
factor),
Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L,
1L, 3L, 3L, 3L), .Label = c(03/09
Thank you Berend
It is exactly what I wanted.
Michel
Le 24/07/2013 09:48, Berend Hasselman a écrit :
On 24-07-2013, at 08:39, Arnaud Michel michel.arn...@cirad.fr wrote:
Hello
I have the following problem :
The dataframe TEST has multiple lines for a same person because :
there are differents
Hi Arun,
Merci à toi
Bien amicalement
Michel
Le 24/07/2013 15:29, arun a écrit :
Hi Michel,
You could try:
df1New-droplevels(TEST[with(TEST,ave(seq_along(Matricule),Matricule,FUN=min)),])
row.names(df1New)-1:nrow(df1New)
df2New-droplevels(TEST[with(TEST,ave(seq_along(Matricule),Matricule,FUN
=sample.data) than to put sample.data$ in
front of every variable name; and easier to read as well.
Terry Therneau (author of coxph function)
Thank you,
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Ph.D. Student
Department of Wildland Resources
Utah State University
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to me that yes.
Thanks for your help
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is that interaction gives you directly factor, paste gives you
character vector, but it may be convenient too for your purpose.
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Arnaud Michel
Sent: Monday, July 22
Hi
You can do a rotation and use gvisColumnChart instead gvisBarChart
plot(gvisColumnChart(MyData, xvar=Names1, yvar=c(Values1,
Values2),options=list(width=2500,height=1000)))
Michel
Le 17/07/2013 15:57, Christofer Bogaso a écrit :
Hello Arnaud,
Thank you for your pointer. However I need
28/02/1995 11
01/03/1995 *12/03/1995* 11
*13/03/1995* *30/06/1995*4
*01/01/1996* 30/01/19965
31/01/1996 *31/01/1996*5
Thanks for your help
--
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Av Agropolis 34398 Montpellier cedex 5
tel
helps
Michel
Le 17/07/2013 19:57, Rui Barradas a écrit :
Hello,
As for question (1), try the following.
y2 - cumsum(c(TRUE, diff(x1) 0))
identical(as.integer(y1), y2) # y1 is of class numeric
As for question (2) I'm not understanding it.
Hope this helps,
Rui Barradas
Em 17-07-2013 18:21
Thanks Arun and Rui for your helps
Michel
Le 17/07/2013 22:20, arun a écrit :
#or
library(plyr)
res-ddply(df1,.(INDX),summarize,Debut=head(Debut,1),Fin=tail(Fin,1))
res$INDX-factor(res$INDX,levels=unique(df1$INDX))
res[order(res$INDX),-1]
# DebutFin
#3 24/01/1995 31/12/1997
#4 02
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Michel
Also, in your example dataset:
df1$contrat[grep(^CDD,df1$contrat)]
#[1] CDD détaché ext. Cirad CDD détaché ext. Cirad CDD détaché ext.
Cirad
#[4] CDD détaché ext. Cirad CDD détaché ext.Cirad CDD détaché ext.
Cirad
#[7] CDD détaché ext. Cirad CDD détaché ext.Cirad CDD détaché
for the lines i and i+1
then df1$F[i] + 1 == df1$D[i+1]
Michel
Le 14/07/2013 18:17, Arnaud Michel a écrit :
Hi,
Excuse me for the indistinctness
Le 13/07/2013 17:18, arun a écrit :
Hi,
when the value of Debut of lines i = value Fin of lines i-1
That part is not clear esp. when it is looked upon
Super !!!
Thank you very much Arun
Michel
Le 15/07/2013 03:47, arun a écrit :
HI Michel,
This gives the same order as that of df2.
df1$contrat[grep(^CDD,df1$contrat)]- CDD détaché ext. Cirad
df1[48,8]- 31/12/2013
indx-as.numeric(interaction(df1[,1:6],drop=TRUE))
lst1-split(df1,indx)
lst2-lst1
, contrat, Pays, Debut, Fin), class =
data.frame, row.names = c(NA,
-11L))
Thank you for your help
--
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Av Agropolis 34398 Montpellier cedex 5
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, Début5, Fin5,
Cat6, Début6, Fin6, Cat7, Début7, Fin7, Cat8, Début8,
Fin8, Cat9, Début9, Fin9), class = data.frame, row.names = c(NA,
-4L))
Any idea ?
Thank you
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Project
Zimbabwe ZW Abaco
South Africa ZA Abaco
Madagascar MG Abaco
Tanzania TZ Adaptclone
Mali ML Abaco
Mali ML Adaptclone
..
Any idea ?
Michel
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Thank you arun !
I don't know the library reshape2
Michel
Le 20/06/2013 19:55, arun a écrit :
Hi,
Not sure if you wanted the entries with 0.
library(reshape2)
dfMelt-melt(df,id.var=c(Country,Iso))
#subset those with 1
dfNew- subset(dfMelt,value==1,select=-4) row.names(dfNew)- 1:nrow(dfNew
correspond with the country : Ex ZW
is ISO of Zimbabwe)
PaysPaysIso NbrProj
Zimbabwe ZW 1
Burkina FasoBF2
I know associate Country and Number of projets but how associate Iso
Any idea ?
Thank you for your help
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Chargé de mission auprès du
idea ?
Thank you for your help
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Dear R users, I'm trying to get the 'Digitize' package but it has been removed
from the CRAN repository. I tried to download the .tar.gz file from the archive
but I haven't been able to install it. Anyone has a clue on how I could proceed
to access this package? Thanks Jean-Michel Fortin
*1.5, width= 556*1.5
))
plot(G1)
Thank you
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, have a nice week-end.
Michel
On Sat, Jul 30, 2011 at 1:28 AM, Achim Zeileis achim.zeil...@uibk.ac.atwrote:
On Fri, 29 Jul 2011, Michel Lutz wrote:
Achim,
Thank you so much for this prompt answer. Really appreciated !
Anyway, I am still a bit lost... don't you mind if I ask you somme
.
Even is it's painful for me, because I don't know what happened in the
system...
Time for me to investigate.
Thank you very very much, Achim. Enjoy your week-end.
Michel
On Sat, Jul 30, 2011 at 11:34 AM, Achim Zeileis achim.zeil...@uibk.ac.atwrote:
On Sat, 30 Jul 2011, Michel Lutz wrote
. What 'tol' stands for? Seems
it is not a 'breackpoints' attributes.
Any help would greatly appreciated.
Many thanks in advance,
Regards,
Michel
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don't know
what to do.
* *But the tests need to be adjusted*
Are such adjustements implement in breakpoints? (no mention in the durab
example, basic function settings are used).
In advance, thank you very much, and sorry for the disturbance.
Michel
On Fri, Jul 29, 2011 at 10:58 AM, Achim Zeileis
, ...) vcovHC(x, type = HC, ...))
plot(stab.model)
bp.mes - breakpoints(model.mes, data = D)
Fstats works, by breakpoints tells me:
Erreur dans chol2inv(qr.R(fm$qr)) :
l'élément (5, 5) est nul, donc l'inverse ne peut être calculé
I tried and tried again, no clue
Thanks
Michel
On Fri, Jul 29
I
should build my input csv file differently, but I am not sure how to do...
Is someone able to help me ?
Thank you very much !
REgards,
Michel
-- Forwarded message --
From: Red Roo redr...@yahoo.com
Date: Tue, Aug 24, 2010 at 3:46 PM
Subject: Re: [R-sig-hpc] Holtman's
Dear Rexperts,
I am using R to query google.
I am getting different results (in size) for manual queries and queries sent
through getForm of RCurl.
It seems that RCurl limits the size of the text retrieved (the maximum I
could get is around 32 k bits).
Any idea how to get around this ?
)
without any change. The means are the same for the two packages.
I saw a few posts with discrepancies but not with the same database
Thanks in advance
Michel Boutsen
Brussel's University
Department of Biostatistics 1 1 6 1 2.31 0
2 1 6 1 4.06 1
3 1 6 1 3.28 0
4 1 6 0 3.2 0
5
but with no explanations
Thanks in advance
Michel Boutsen
Brussel's University
Department of Biostatistics
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Hello
Is there exists a package for multivariate random forest, namely for
multivariate response data ? It seems to be impossible with the
randomForest function and I did not find any information about this
in the help pages ...
Thank you for your help
Bertrand
Hello
Is there exists a package for multivariate random forest, namely for
multivariate response data ? It seems to be impossible with the
randomForest function and I did not find any information about this
in the help pages ...
Thank you for your help
Bertrand
Hello
Is there exists a package for multivariate random forest, namely for
multivariate response data ? It seems to be impossible with the
randomForest function and I did not find any information about this
in the help pages ...
Thank you for your help
Bertrand
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