Change values in a dateframe

On 24-07-2013, at 08:39, Arnaud Michel <michel.arnaud at cirad.fr> 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 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, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 8L,
5L, 6L, 9L, 3L, 3L, 7L), .Label = c("CHICHE", "GEOF", "GUTIER",
"JACQUE", "LANGUE", "LANGUE-LOPEZ", "RIVIER", "TRU", "VINCENT"
), class = "factor"), Prenom = structure(c(8L, 3L, 4L, 5L, 1L,
2L, 2L, 9L, 6L, 7L, 7L), .Label = c("Edgar", "Elodie", "Jeanine",
"Jeannine", "Michel", "Michele", "Mich?le", "Michelle", "Victor"
), class = "factor"), Sexe = structure(c(1L, 1L, 1L, 2L, 2L,
1L, 1L, 2L, 1L, 1L, 1L), .Label = c("F?minin", "Masculin"), class = "factor"),
    Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L,
    1L, 3L, 3L, 3L), .Label = c("03/09/1940", "04/03/1946", "07/12/1947",
    "18/11/1945", "27/09/1947", "29/12/1936", "30/03/1935"), class = "factor")), .Names = c("Matricule",
"Nom", "Prenom", "Sexe", "Date.de.naissance"), class = "data.frame", row.names = c(NA,
-11L))


I would want to make homogeneous the information and would like built 2 dataframes :
df1 wich has the value of Nom and Prenom of the first lines of TEST when there are different values. The other values (Matricule or Sexe or Date.de.naissance) are unchanged

df1 <- structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 6L,
5L, 5L, 7L, 3L, 3L, 3L), .Label = c("CHICHE", "GEOF", "GUTIER",
"JACQUE", "LANGUE", "TRU", "VINCENT"), class = "factor"), Prenom = structure(c(6L,
3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L, 5L, 5L), .Label = c("Edgar",
"Elodie", "Jeanine", "Michel", "Michele", "Michelle", "Victor"
), class = "factor"), Sexe = structure(c(1L, 1L, 1L, 2L, 2L,
1L, 1L, 2L, 1L, 1L, 1L), .Label = c("F?minin", "Masculin"), class = "factor"),
    Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L,
    1L, 3L, 3L, 3L), .Label = c("03/09/1940", "04/03/1946", "07/12/1947",
    "18/11/1945", "27/09/1947", "29/12/1936", "30/03/1935"), class = "factor")), .Names = c("Matricule",
"Nom", "Prenom", "Sexe", "Date.de.naissance"), class = "data.frame", row.names = c(NA,
-11L))

df2 wich has the value of Nom and Prenom of the last lines of TEST when there are different values. The other values (Matricule or Sexe or Date.de.naissance) are unchanged.

df2 <- structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 3L, 6L,
4L, 4L, 7L, 5L, 5L, 5L), .Label = c("CHICHE", "GEOF", "JACQUE",
"LANGUE-LOPEZ", "RIVIER", "TRU", "VINCENT"), class = "factor"),
    Prenom = structure(c(6L, 3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L,
    5L, 5L), .Label = c("Edgar", "Elodie", "Jeannine", "Michel",
    "Mich?le", "Michelle", "Victor"), class = "factor"), Sexe = structure(c(1L,
    1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L), .Label = c("F?minin",
    "Masculin"), class = "factor"), Date.de.naissance = structure(c(4L,
    2L, 2L, 7L, 6L, 5L, 5L, 1L, 3L, 3L, 3L), .Label = c("03/09/1940",
    "04/03/1946", "07/12/1947", "18/11/1945", "27/09/1947", "29/12/1936",
    "30/03/1935"), class = "factor")), .Names = c("Matricule",
"Nom", "Prenom", "Sexe", "Date.de.naissance"), class = "data.frame", row.names = c(NA,
-11L))

Something like this

r1 <- droplevels(do.call(rbind,lapply(split(TEST,TEST$Matricule),
                     FUN=function(x) {x[,c("Nom","Prenom")] <- x[1,c("Nom","Prenom"),drop=TRUE];x})))
rownames(r1) <- NULL
r1

r2 <- droplevels(do.call(rbind,lapply(split(TEST,TEST$Matricule),
                     FUN=function(x) {x[,c("Nom","Prenom")] <- x[nrow(x),c("Nom","Prenom"),drop=TRUE];x})))
rownames(r2) <- NULL
r2

#> identical(r1,df1)
#[1] TRUE
#> identical(r2,df2)
#[1] TRUE

Note: I had to change the Prenom and Sexe columns because of encoding issues. but that shouldn't have any influence on the above.

Berend