Removing Whitespace From a Whole Data Frame in R

Question

I've been trying to remove the white space that I have in a data frame (using R). The data frame is large (>1gb) and has multiple columns that contains white space in every data entry.

Is there a quick way to remove the white space from the whole data frame? I've been trying to do this on a subset of the first 10 rows of data using:

gsub( " ", "", mydata) 

This didn't seem to work, although R returned an output which I have been unable to interpret.

str_replace( " ", "", mydata)

R returned 47 warnings and did not remove the white space.

erase_all(mydata, " ")

R returned an error saying 'Error: could not find function "erase_all"'

I would really appreciate some help with this as I've spent the last 24hrs trying to tackle this problem.

Thanks!

Answer 1

A lot of the answers are older, so here in 2019 is a simple dplyr answer that will operate only on the character columns to remove trailing and leading whitespace.

library(dplyr)
library(stringr)

data %>%
  mutate_if(is.character, str_trim)

You can switch out the str_trim() function for other ones if you want a different flavor of whitespace removal.

Answer 2

You could use trimws function in R 3.2 on all the columns.

myData[,c(1)]=trimws(myData[,c(1)])

You can loop this for all the columns in your dataset. It has good performance with large datasets as well.

Answer 3

If you want to maintain the variable classes in your data.frame - you should know that using apply will clobber them because it outputs a matrix where all variables are converted to either character or numeric. Building upon the code of Fremzy and Anthony Simon Mielniczuk you can loop through the columns of your data.frame and trim the white space off only columns of class factor or character (and maintain your data classes):

for (i in names(mydata)) {
  if(class(mydata[, i]) %in% c("factor", "character")){
    mydata[, i] <- trimws(mydata[, i])
  }
}

Answer 4

R is simply not the right tool for such file size. However have 2 options :

Use ffdply and ff base

Use ff and ffbase packages:

library(ff)
library(ffabse)
x <- read.csv.ffdf(file=your_file,header=TRUE, VERBOSE=TRUE,
                 first.rows=1e4, next.rows=5e4)
x$split = as.ff(rep(seq(splits),each=nrow(x)/splits))
ffdfdply( x, x$split , BATCHBYTES=0,function(myData)        
             apply(myData,2,function(x)gsub('\\s+', '',x))

Use sed (my preference)

sed -ir "s/(\S)\s+(/S)/\1\2/g;s/^\s+//;s/\s+$//" your_file 

Answer 5

I think that a simple approach with sapply, also works, given a df like:

dat<-data.frame(S=LETTERS[1:10],
            M=LETTERS[11:20],
            X=c(rep("A:A",3),"?","A:A ",rep("G:G",5)),
            Y=c(rep("T:T",4),"T:T ",rep("C:C",5)),
            Z=c(rep("T:T",4),"T:T ",rep("C:C",5)),
            N=c(1:3,'4 ','5 ',6:10),
            stringsAsFactors = FALSE)

You will notice that dat$N is going to become class character due to '4 ' & '5 ' (you can check with class(dat$N))

To get rid of the spaces on the numeic column simply convert to numeric with as.numeric or as.integer.

dat$N<-as.numeric(dat$N)

If you want to remove all the spaces, do:

dat.b<-as.data.frame(sapply(dat,trimws),stringsAsFactors = FALSE)

And again use as.numeric on col N (ause sapply will convert it to character)

dat.b$N<-as.numeric(dat.b$N)

Answer 6

If you're dealing with large data sets like this, you could really benefit form the speed of data.table.

library(data.table)

setDT(df)

for (j in names(df)) set(df, j = j, value = df[[trimws(j)]]) 

I would expect this to be the fastest solution. This line of code uses the set operator of data.table, which loops over columns really fast. There is a nice explanation here: Fast looping with set.