CODEWITHSUNDEEP
r
Rene Martinez
Rene Martinez

Reputation: 3

rename specific columns in dataframes in a list

I have a list with several dataframes, each dataframe correspond to an election, say 2010, 2013, ..., 2021.

For each dataframe, I want to add the year to every column name except the first three columns. So I'm trying to add the year to [,4:end_column] using a for loop:

for (i in names(SECCIONES_year)) {
  year <- (substr(i, nchar(i)-4, nchar(i)))
  end_column <- ncol(SECCIONES_year[[i]])

  past_names <- names(SECCIONES_year[[i]][ ,4:end_column])

  colnames(SECCIONES_year[[i]][,4:end_column]) <- paste(past_names, year, sep="")

}

Unfortunately, it doesn't work but it does if I work with all column names as:

colnames(SECCIONES_year[[i]]) <- paste(past_names, year, sep=)

I tried with colnames, names and setNames but failed every try.

Upvotes: 0

Views: 73

Answers (2)

Ren&#233; Mart&#237;nez
Ren&#233; Mart&#237;nez

Reputation: 309

Theres no need for comma in the second bracket, before "4":

colnames(SECCIONES_year[[i]])[4:ncol(SECCIONES_year[[i]])] <- paste(past_names, year, sep="")

Upvotes: 0

akrun
akrun

Reputation: 886938

With base R, an opiton is to loop over the list as well as the names of the list with Map, extract the substring from the names (could be done outside the loop as well as the substr is vectorized), then assign the subset of column names of the data.frame elements by pasteing the substring 'year' with the corresponding column names, and return the data.frame

SECCIONES_year <- Map(function(x, nm)
         {
        year <- substr(nm, nchar(nm)-4, nchar(nm))
        names(x)[4:ncol(x)] <- paste0(names(x)[4:ncol(x)], year)
        x  
    }, 

      SECCIONES_year, names(SECCIONES_year))

In the OP's code, it could be assignment error i.e. instead of colnames(SECCIONES_year[[i]][,4:end_column]) it would be colnames(SECCIONES_year[[i]])[,4:end_column] i.e. as an example

lst1 <- list(data.frame(col1 = 1:5, col2 = 6:10),
    data.frame(col3 = 3:5, col4 = 4:6))
names(lst1) <- c("a", "b")

Checking the output with both the methods

# OP's method
> for(nm in names(lst1)) colnames(lst1[[nm]][2]) <- "hello"
> lst1
$a
  col1 col2
1    1    6
2    2    7
3    3    8
4    4    9
5    5   10

$b
  col3 col4
1    3    4
2    4    5
3    5    6
# modified
> for(nm in names(lst1)) colnames(lst1[[nm]])[2] <- "hello"
> lst1
$a
  col1 hello
1    1     6
2    2     7
3    3     8
4    4     9
5    5    10

$b
  col3 hello
1    3     4
2    4     5
3    5     6

Upvotes: 3

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