in R, use gsub to remove all punctuation except period

You may try this code. I found it quite handy.

x <- c('6,345', '7.235', '8', '$9', '-10')
gsub("[^[:alnum:]\\-\\.\\s]", "", x)

[1] "6345"  "7.235" "8"     "9"     "-10"

x <- c('1', '2@', '3', '4', '£5')
gsub("[^[:alnum:]\\-\\.\\s]", "", x)

[1] "1" "2" "3" "4" "5"

This code{gsub("[^[:alnum:]]", "", x))} removes everything that does not include alphanumeric terms. Then we add to the exception list. Here we add hyphen(\-), full-stop(\.) and space(\s) to get gsub("[^[:alnum:]\-\.\s]", "", x). Now it removes everything that is not alphanumeric, hyphen, full stop and space.

You can put back some matches like this:

 sub("([.-])|[[:punct:]]", "\\1", as.matrix(z))
     X..1. X..2.  
[1,] "1"   "6"    
[2,] "2"   "7.235"
[3,] "3"   "8"    
[4,] "4"   "9"    
[5,] "5"   "-10"  

Here I am keeping the . and -.

And I guess , the next step is to coerce you result to a numeric matrix, SO here I combine the 2 steps like this:

matrix(as.numeric(sub("([.-])|[[:punct:]]", "\\1", as.matrix(z))),ncol=2)
   [,1]    [,2]
[1,]    1   6.000
[2,]    2   7.235
[3,]    3   8.000
[4,]    4   9.000
[5,]    5 -10.000