#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
96
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(df1[,1:2],as.character)
df2New<- data.frame(Deb=unique(with(df1,ave(Debut,INDX,FU
Arnaud Michel
To: Rui Barradas ; R help ; arun
Cc:
Sent: Wednesday, July 17, 2013 4:03 PM
Subject: Re: [R] simplify a dataframe
Thank you for the question (1)
Sorry for the imprecision for the question (2) :
Suppose the date frame df
df1 <- data.frame(
Debut =c ( "24/01/1995", "
Thank you for the question (1)
Sorry for the imprecision for the question (2) :
Suppose the date frame df
df1 <- data.frame(
Debut =c ( "24/01/1995", "01/05/1997" ,"31/12/1997", "02/02/1995"
,"28/02/1995"
,"01/03/1995", "13/03/1995", "01/01/1996", "31/01/1996") ,
Fin = c ( "30/04/1997", "30/12/
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, Arnaud Michel escreveu:
Hi Arun
I have two questio
1997
4.0 13/03/1995 30/06/1995
4.1 01/01/1996 31/01/1996
10 02/02/1995 12/03/1995
6 24/01/1995 31/08/1995
9 01/09/1995 29/02/2000
1 26/01/1995 31/08/2001
2 05/09/2012 31/12/2013
3 01/09/2004 31/08/2007
7 01/09/2001 31/08/2004
8 01/09/2007 04/09/2012
A.K.
__
in,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 -
From: arun
To: Arnaud Michel
Cc: R help
Sent: Sunday, July 14, 2013 2:39 PM
Subject
01/09/1995 29/02/2000
1 26/01/1995 31/08/2001
2 05/09/2012 31/12/2013
3 01/09/2004 31/08/2007
7 01/09/2001 31/08/2004
8 01/09/2007 04/09/2012
A.K.
From: Arnaud Michel
To: arun
Cc: R help ; jholt...@gmail.com; Rui Barradas
Sent: Sunday, July 14
An other remark :
If I calculate
df1$D<- as.numeric(as.Date(paste0(substr(df1$Debut,7,10),"-",
substr(df1$Debut,4,5),"-",substr(df1$Debut,1,2
and
df1$F <- as.numeric(as.Date(paste0(substr(df1$Fin,7,10),"-",
substr(df1$Fin,4,5),"-",substr(df1$Fin,1,2
and if there is no interruption of time f
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 with the expected output
> (df2).
I want to group the lines which have the same caracteristics (Matric
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 with the expected output
(df2). Also, in your example dataset:
df1$contrat[grep("^CDD",df1$contrat)]
#[1] "CDD détaché ext. Cirad" "CDD détaché ext. Cirad" "CDD détaché ext. Cirad
Here is how you can do it with the 'data.table' package:
> require(data.table)
> df1 <- data.table(df1)
> result <- df1[
+ , list(Debut = Debut[1L] # first entry
+ , Fin = Fin[1L]
+ )
+ , keyby = c("Matricule", "Nom", "Sexe", "DateNaissance", "contrat",
"Pays")
+ ]
> result
MatriculeN
Hello,
My solution is missing a row, but maybe you can find some inspiration.
cols <- c("Matricule", "Nom", "Sexe", "DateNaissance", "contrat", "Pays")
irow1 <- duplicated(df1[, cols])
irow2 <- c(FALSE, df1$Debut[-1] == df1$Fin[-nrow(df1)])
df3 <- df1[!irow1 & !irow2, ]
dim(df2); dim(df3) #
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