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
--
Michel ARNAUD
DGDRD-Drh - TA 174/04
tel : 04.67.61.75.38
port: 06.47.43.55.31
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R-help@r-project.org mailing list -- To U
. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
On December 4, 2014 2:09:12 AM PST, Arnaud Michel
wrote:
Hello
I have a function named FctModele13 in which
1) I calculate a dataframe named Total and
2) I used ggplot2.
I have the following
Hello
I have a function named FctModele13 in which
1) I calculate a dataframe named Total and
2) I used ggplot2.
I have the following problem. I cannot produce simultaneously
* the graphic by ggplot2
* the dataframe
My simplified code is the following one :
TT <- FctModele13(ListePlusde50a
Hello
Can one calculate the month number between two dates
D1 <- "01/01/2007" and D2 <- "01/04/2009" ?
Thank you
--
Michel ARNAUD
Cirad
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
"C", "D", "B", "D", "B", "C", "D", "B", "B", "B", "A", "C", "D", "B",
"A", "D", "D"),
Unite = c("Unite8", "Un
Hello
I have 2 df Dem and Rap.
I would want to build all the df (dfnew) by associating these two df
(Dem and Rap) in the following way :
For each value of Dem$Nom (dfnew$Demandeur), I associate 2 different
values of Rap$Nom (dfnew$Rapporteur1 and dfnew$Rapporteur2) in such a way
* for each d
Thank you to Marc Schwartz, Rui Barrada and Sarah Goslee
Michel
Le 19/09/2014 19:46, Marc Schwartz a écrit :
On Sep 19, 2014, at 12:15 PM, Arnaud Michel wrote:
Hello
I have the two dataframes Df1 and Df2 which have the common variable
AgeSexeCadNCad
I would like to add the new variable Df2
Hello
I have the two dataframes Df1 and Df2 which have the common variable
AgeSexeCadNCad
I would like to add the new variable Df2$Pourcent which correspond at
the value of Df1$AgeSexeCadNCad.
Thank you for your help.
Michel
Df1 <- structure(list(AgeSexeCadNCad = structure(1:36, .Label =
c("6
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&q
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", "
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 Jan", "2010 Jan", "2011
Jan", "2012 Jan") with if possible, the angle of labels and x-axe = 40
Any idea ?
Thank you fo
Hello
I would like to map the population of the European countries in 2011.
I am using the spatial shapefiles of Europe published by EUROSTAT.
I applyed the script below of Markus Kainubut but I had a problem with
the map.
Any ideas ?
Thanks for your help
##
Dear All,
From the dataframe df1
df1 <-
structure(list(Nom = structure(1:9, .Label = c("A1", "A2", "A3",
"B1", "B2", "C1", "C2", "C3", "C4"), class = "factor"), Pays1 = c(1,
1, 0, 0, 1, 0, 0, 0, 0), Pays2 = c(0, 0, 0, 1, 1, 0, 1, 0, 1),
Pays3 = c(0, 0, 0, 0, 1, 0, 0, 0, 0), Pays4 = c(1, 0, 0
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.
--
Michel ARNAUD
Chargé de mission auprès du DRH
DGDRD-Drh - TA 174/04
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 ?
--
Michel ARNAUD
Chargé de mission auprès du DRH
DGDRD-Drh - TA 174/04
Av Agropolis 34398 Montpellier cedex 5
tel : 04.67.61
Thank you
Michel
Le 09/12/2013 08:14, Berend Hasselman a écrit :
On 09-12-2013, at 08:04, Arnaud Michel 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[3], X[3]), r
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
--
Michel ARNAUD
Chargé de mission au
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 wrote:
Hi
I have the following problem
I would like to build, from a matrix filled with 0 and with 1, a matrix
or a
On 20 November 2013 19:28, Arnaud Michel wrote:
Hi
I have the following problem
I would like to build, from a matrix filled with 0 and with 1, a matrix
or a data.frame which contains, in every line, the number of the line
and the number of the column of the matrix for which the value is equal
to 1.
E
Hi
I have the following problem
I would like to build, from a matrix filled with 0 and with 1, a matrix
or a data.frame which contains, in every line, the number of the line
and the number of the column of the matrix for which the value is equal
to 1.
Exemple :
dput(m)
structure(c(0, 0, 0, 1, 0
Hello
I had draw the results of PCA (Principal Components Analysis) (1)
Is it possible to put on this graphic the 75% ellipse confidence of each
sex calculated by dataEllipse (library(car)) ?
My code is
1) PCA data frame with 169 rows and 2 columns
library(ggplot2)
p <- ggplot(PCA, aes(x=F1, y=
-bounces@r-
project.org] On Behalf Of Arnaud Michel
Sent: Friday, October 11, 2013 11:02 AM
To: R help
Subject: [R] ggplot2 : to change the title of legend
Hello
I don't arrive to change the title of the legend My code is :
library(ggplot2)
ma <- max(General$AgeChangCat) ; mi <- min(Gene
Hello
I don't arrive to change the title of the legend
My code is :
library(ggplot2)
ma <- max(General$AgeChangCat) ; mi <- min(General$AgeChangCat)
Test$Recrutement <- factor(Test$CadNonCadRecrut)
p <-
ggplot(Test, aes(x=factor(Cat1), y=AgeChangCat )) +
ylim(mi,ma) +
geom_point() +
geom_boxplot(a
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
To: Arnaud Michel
Cc: R help
Sent
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
To: R help
Cc:
Sent: Tuesday, September 17, 2013 2:14 PM
Hi
I have a dataframe Df1
dput(Df1)
structure(list(Mat = c(141, 141, 157, 157, 188, 188, 232, 232,
253, 253, 253, 254, 254, 254, 254, 256, 256, 264, 264), Prenom =
c("Pierre",
"Pierre", "Jean-Claude", "Jean-Claude", "Jean-Louis", "Jean-Louis",
"Philippe", "Philippe", "Christophe", "Christophe",
+1], col='grey')
Regards,
Pascal
On 16/09/2013 15:42, Arnaud Michel wrote:
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, 29
ot;)
>
> 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 <mailto:michel.arn...@cirad.fr>> wrote:
>
>
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,
Hi
I would like to eliminate a large number of lines of the dataframe df1
The lines to delete are given here by the values of Mat (ex : 2,4,7,10).
but I have a large number (300) values of Mat
dput(df1)
structure(list(Mat = c(1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3,
3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5
t, sprintf("%02d", df1$Ech), sep = ".") )
# user system elapsed
# 0.170.000.17
# R Lancelot
system.time(df1$CatEch3 <-
paste(df1$Cat, formatC(df1$Ech, flag = "0",
width = max(nchar(df1$Ech))), sep = "."))
# user system elapsed
# 0.340
Thanks to all three for your fast answer
Michel
Le 08/09/2013 18:41, Renaud Lancelot a écrit :
> paste(df1$Cat,
> formatC(df1$Ech, flag = "0", width = max(nchar(df1$Ech))),
> sep = ".")
>
>
>
> 2013/9/8 Arnaud Michel <mailto:michel.arn...@c
Hello
I have a large dataframe (nrow=55000).
This below df1 an extract of the original dataframe
dput(df1)
structure(list(Cat = c(6, 6, 6, 6, 6, 6, 6, 6, 4, 4, 4, 4, 4,
4, 8, 8, 9, 9, 9, 9, 9, 9, 9, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7,
7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8,
8
col=c("red","blue"))
#or
library(ggplot2)
dat1<- data.frame(x,y1,y2)
ggplot(dat1,aes(x))+geom_line(aes(y=y1,colour="y1"))+
geom_line(aes(y=y2,colour="y2")) +ylab("y1:y2")
A.K.
- Original Message -
From: Arnaud Michel
To: R help
Cc:
Sent
=4,main="y1,
y2 vs. x")
A.K.
----- Original Message -
From: Arnaud Michel
To: arun
Cc: R help
Sent: Sunday, September 1, 2013 1:18 AM
Subject: Re: [R] To represent on the same plot the relation (y1, x) and (y2, x)
Thank you Arun
But have you a solution if y1 and y2 have not the
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
--
Michel ARNAUD
Chargé de mission auprès du DRH
DGDRD-Drh - TA 174/04
Av Agropolis 34398 Montpellier cedex 5
tel
Le 25/07/2013 08:50, Berend Hasselman a écrit :
On 25-07-2013, at 08:35, Arnaud Michel 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 (solution Arun) that
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
identical(df2New,df2)
#[1] TRUE
A.K.
- Original Message -
From: Arnaud Michel
To: R help
Cc:
Sent: Wednesday, July 24, 2013 2:39 AM
Subject: [R] Change values in a dateframe
Hello
I have the following problem :
The dataframe TEST has multiple lines for a same person because :
there are
Matricule,FUN=max)),])
row.names(df2New)<-1:nrow(df2New)
identical(df1New,df1)
#[1] TRUE
identical(df2New,df2)
#[1] TRUE
A.K.
- Original Message -
From: Arnaud Michel
To: R help
Cc:
Sent: Wednesday, July 24, 2013 2:39 AM
Subject: [R] Change values in a dateframe
Hello
I have t
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 wrote:
Hello
I have the following problem :
The dataframe TEST has multiple lines for a same person because :
there are differents values of Nom or
Hello
I have the following problem :
The dataframe TEST has multiple lines for a same person because :
there are differents values of Nom or differents values of Prenom
but the values of Matricule or Sexe or Date.de.naissance are the same.
TEST <- structure(list(Matricule = c(66L, 67L, 67L, 68L,
s 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, 2013 10:57 AM
To: R help
Subjec
Hi
To calculate the value of the interaction between factors of a dataframe
df, does exist any function which could replace the function when the
dataframe df has the numbers of rows of df is large (~55000) and also
the numbers of combinaison of the three factors is large. The calcul abort.
T
#4 02/02/1995 12/03/1995
#1 13/03/1995 30/06/1995
#2 01/01/1996 31/01/1996
A.K.
- Original Message -
From: arun
To: Arnaud Michel
Cc: R help ; Rui Barradas
Sent: Wednesday, July 17, 2013 4:14 PM
Subject: Re: [R] simplify a dataframe
Hi,
You could try:
df1[,1:2]<-lapply
1/1996 31/01/1996
Thank you for your 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 i
Hi Arun
I have two questions always about the question of symplify a dataframe
I would like
1) to transform the vector x1 into the vector y1
x1 <- c(1,1,1,-1000, 1,-1000, 1,1,1,1,1,1,-1000)
y1 <- c(1,1,1,1,2,2, 3,3,3,3,3,3,3)
2) to transform the vectors Deb
1:150)
library(googleVis)
plot(gvisBarChart(MyData, xvar="Names1", yvar=c("Values1",
"Values2"),options=list(width=1200,height=1500,hAxis.gridlines =
"{count: 10}",hAxis.minValue = 0, hAxis.maxValue = 100)))
However this is not clearly working. Can someone point
You can try with list options :
plot(gvisBarChart(MyData, xvar="Names1", yvar=c("Values1", "Values2"),
options=list(width=1200,height=1500)))
Le 15/07/2013 20:00, Christofer Bogaso a écrit :
Hello again,
Let say I have following data-frame:
MyData <- data.frame(Names1 = paste("XXX", 1:150),
nrow(x)])!=1))),function(x)
data.frame(x[1,1:6],Debut=head(x$Debut,1),Fin=tail(x$Fin,1),stringsAsFactors=FALSE)))}))
row.names(res)<- 1:nrow(res)
df2[11,8]<- "31/12/2013"
names(res)[1]<- "Mat"
identical(res,df2)
#[1] TRUE
A.K.
- Original Message -
Fro
,2
and if there is no interruption of time 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
6 BENARD 01/01/1996 31/01/1996 #missing this row
> 58 DALNIC 24/01/1995 31/08/1995
> 68 DALNIC 01/09/1995 29/02/2000
> 7 934 FORNI 26/01/1995 31/08/2001
> 8 934 FORNI 01/09/2001 31/08/2004
> 9 934 FORNI 01/09/2004 31/08/2007
> 10 934 FORNI 01/09/2007 04/09/201
Hello
I have the following problem : group the lines of a dataframe when no
information change (Matricule, Nom, Sexe, DateNaissance, Contrat, Pays)
and when the value of Debut of lines i = value Fin of lines i-1
I can obtain it with a do loop. Is it possible to avoid the loop ?
The dataframe
Hello
I would like to transform the dataframe df1 into df2 (ie copy the data
from several lines for a men/women to only one line by individu men/women)
dput(df1)
structure(list(Mat = c(934L, 934L, 934L, 935L, 935L, 936L, 936L,
936L, 936L, 937L, 937L, 937L, 937L), Nom = structure(c(2L, 2L,
2L,
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:
Hello
I have the following dataframe :
df <- data.frame(
Country=c("Zimbabwe","Burkina Faso","South Africa","Madagascar","Tanzania",
"Mali","Mozambique","Madagascar","Ghana","Nigeria","Kenya","Burkina Faso",
"South Africa","Tanzania","Kenya","Ethiopia" ) ,
Iso=c("ZW","BF","ZA","MG","TZ","ML","M
Hello
I have the following dataframe
df <- data.frame(
Project=c("Abaco","Abaco","Abac","Abaco","Abaco","Abaco",
"Abaco","Adaptclone","Adaptclone","Adaptclone","Adaptclone","Adaptclone",
"Adaptclone","Adopt","Adopt","Adopt"),
Country=c("Zimbabwe","Burkina Faso","South Africa","Madagascar","Tanz
Dear all,
Without a loop, I would like transform 3 numeric vectors empty of 0/1 of
same length
Vec1 : transform 1 to A and 0 to ""
Vec2 : transform 1 to B and 0 to ""
Vec3 : transform 1 to C and 0 to ""
to obtain only 1 vector Vec who is the paste of the 3 vectors (Ex : ABC,
BC, AC, AB,...)
An
Hi
I am using the package googleVis and the function gvisGeoChart
Is it possible to put a title on the map ?
Here is the call of the function :
library(googleVis)
G1 <- gvisGeoChart(PaysProjets, locationvar='Pays', colorvar='NbProj',
options=list(
region= "world",
displayMode="regions",
height=3
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