#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
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
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
I 05/09/2012 31/12/4712
>>
>>
>> df2[,c(1,2,7:8)]
>> MatNom DebutFin
>> 11 VERON 24/01/1995 31/12/1997
>> 26 BENARD 02/02/1995 12/03/1995
>> 36 BENARD 13/03/1995 30/06/1995
>> 46 BENARD 01/01/1996 31/01/199
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
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 04/09/2012
11 934 FORNI 05/09/2012 31/12/4712
Here, the dates look similar to the ones on df2 except for one row in df2.
A.K.
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) #
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
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