Rename columns in multiple dataframes, R

11
votes

I am trying to rename columns of multiple data.frames.

To give an example, let's say I've a list of data.frames dfA, dfB and dfC. I wrote a function changeNames to set names accordingly and then used lapply as follows:

dfs <- list(dfA, dfB, dfC)
ChangeNames <- function(x) {
    names(x) <- c("A", "B", "C" )  
}
lapply(dfs, ChangeNames)

However, this doesn't work as expected. It seems that I am not assigning the new names to the data.frame, rather only creating the new names. What am I doing wrong here?

Thank you in advance!

3
rdataframe
After the line names(x) <- in your function, add return(x) or simply x. Else, you're returning just names(x). - Arun
Thankyou for your reply Arun! If I ad return(x), I will get an outprint of dfA, dfB and dfC, with the new names. But if I view names(dfA), names(dfB) and names(dfC) afterwards, they still have the old column names. My data frames are also very large, so I am not interested in viewing them. Only changing there column names. - user2706593
lapply does not modify the input. There's no "change by reference" happening here. Everything is being done on a copy. You'll have to assign the result back. do: dfs <- lapply(dfs, ChangeNames) - Arun
Ok, dfs is now one big list containing dfA, dfB and dfC, with the new columnnames. I am still interested in working with dfA, dfb an dfC individually, and individually they still have the old columnnames? How do I assign the result back to the individual dataframes? - user2706593
well, you should assign them back. dfA <- dfs[[1]]... ? - Arun

3 Answers

15
votes

There are two things here:

  • 1) You should return the value you want from your function. Else, the last value will be returned. In your case, that's names(x). So, instead you should add as the final line, return(x) or simply x. So, your function would look like:

    ChangeNames <- function(x) {
    names(x) <- c("A", "B", "C" )
    return(x)
}

  • 2) lapply does not modify your input objects by reference. It works on a copy. So, you'll have to assign the results back. Or another alternative is to use for-loops instead of lapply:

    # option 1
dfs <- lapply(dfs, ChangeNames)

# option 2
for (i in seq_along(dfs)) {
    names(dfs[[i]]) <- c("A", "B", "C")
}

Even using the for-loop, you'll still make a copy (because names(.) <- . does). You can verify this by using tracemem.

df <- data.frame(x=1:5, y=6:10, z=11:15)
tracemem(df)
# [1] "<0x7f98ec24a480>"
names(df) <- c("A", "B", "C")
tracemem(df)
# [1] "<0x7f98e7f9e318>"

If you want to modify by reference, you can use data.table package's setnames function:

df <- data.frame(x=1:5, y=6:10, z=11:15)
require(data.table)
tracemem(df)
# [1] "<0x7f98ec76d7b0>"
setnames(df, c("A", "B", "C"))
tracemem(df)
# [1] "<0x7f98ec76d7b0>"

You see that the memory location df is mapped to hasn't changed. The names have been modified by reference.

11
votes

If the dataframes were not in a list but just in the global environment, you could refer to them using a vector of string names.

dfs <- c("dfA", "dfB", "dfC")

for(df in dfs) {
  df.tmp <- get(df)
  names(df.tmp) <- c("A", "B", "C" ) 
  assign(df, df.tmp)
}

EDIT

To simplify the above code you could use

for(df in dfs)
  assign(df, setNames(get(df),  c("A", "B", "C")))

or using data.table which doesn't require reassigning.

for(df in c("dfA", "dfB"))
  data.table::setnames(get(df),  c("G", "H"))
-1
votes

I had the problem of importing a public data set and having to rename each dataframe and rename each column in each dataframe to trim whitespaces, lowercase, and replace internal spaces with periods.

Combining the above methods got me:

for (eachdf in dfs)
  df.tmp <- get(eachdf) 
    for (eachcol in 1:length(df.tmp))
      colnames(df.tmp)[eachcol] <-
      str_trim(str_to_lower(str_replace_all(colnames(df.tmp)[eachcol], " ", ".")))
      }
  assign(eachdf, df.tmp) 
}