in R, use gsub to remove all punctuation except period - r

Question

I am new to R so I hope you can help me.
I want to use gsub to remove all punctuation except for periods and minus signs so I can keep decimal points and negative symbols in my data.
Example
My data frame z has the following data:
[,1] [,2]
[1,] "1" "6"
[2,] "2#" "7.235"
[3,] "3" "8"
[4,] "4" "$9"
[5,] "£5" "-10"
I want to use gsub("[[:punct:]]", "", z) to remove the punctuation.
Current output
> gsub("[[:punct:]]", "", z)
[,1] [,2]
[1,] "1" "6"
[2,] "2" "7235"
[3,] "3" "8"
[4,] "4" "9"
[5,] "5" "10"
I would like, however, to keep the "-" sign and the "." sign.
Desired output
PSEUDO CODE:
> gsub("[[:punct:]]", "", z, except(".", "-") )
[,1] [,2]
[1,] "1" "6"
[2,] "2" "7.235"
[3,] "3" "8"
[4,] "4" "9"
[5,] "5" "-10"
Any ideas how I can make some characters exempt from the gsub() function?

Answer 1

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

Answer 2

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.

Answer 3

Here are some options to restrict a generic character class in R using both base R (g)sub and the stringr remove/replace functions:
(g)sub with perl=TRUE
You may use the [[:punct:]] bracket expression with the [:punct:] POSIX character class and restrict it with the (?!\.) negative lookahead that will require that the immediately following char on the right is not equal to .:
(?!\.)[[:punct:]] # Excluding a dot only
(?![.-])[[:punct:]] # Excluding a dot and hyphen
To match one or more occurrences, wrap it with a non-capturing group and then set the + quantifier to the group:
(?:(?!\.)[[:punct:]])+ # Excluding a dot only
(?:(?![.-])[[:punct:]])+ # Excluding a dot and hyphen
Note that when you remove found matches, both expressions will yield the same results, however, when you need to replace with some other string/char, the quantification will allow changing whole consecutive character chunks with a single occurrence of the replacement pattern.
With stringr replace/remove functions
Before going into details, mind that the PCRE [[:punct:]] used with (g)sub will not match the same chars in the stringr regex functions that are powered by the ICU regex library. You need to use [\p{P}\p{S}] instead, see R/regex with stringi/ICU: why is a '+' considered a non-[:punct:] character?
The ICU regex library has a nice feature that can be used with character classes, called character class subtraction.
So, you write your character class, say, all punctuation matching class like [\p{P}\p{S}], and then you want to "exclude" (=subtract) a char or two or three, or a whole subclass of chars. You may use two notations:
[\p{P}\p{S}&&[^.]] # Excluding a dot
[\p{P}\p{S}--[.]] # Excluding a dot
[\p{P}\p{S}&&[^.-]] # Excluding a dot and hyphen
[\p{P}\p{S}--[.-]] # Excluding a dot and hyphen
To match 1+ consecutive occurrences with this approach, you do not need any wrapping groups, simply use +:
[\p{P}\p{S}&&[^.]]+ # Excluding a dot
[\p{P}\p{S}--[.]]+ # Excluding a dot
[\p{P}\p{S}&&[^.-]]+ # Excluding a dot and hyphen
[\p{P}\p{S}--[.-]]+ # Excluding a dot and hyphen
See R demo tests with outputs:
x <- "Abc.123#&*xxx(x-y-z)???? some#other!chars."
gsub("(?!\\.)[[:punct:]]", "", x, perl=TRUE)
## => [1] "Abc.123xxxxyz someotherchars."
gsub("(?!\\.)[[:punct:]]", "~", x, perl=TRUE)
## => [1] "Abc.123~~~xxx~x~y~z~~~~~ some~other~chars."
gsub("(?:(?!\\.)[[:punct:]])+", "~", x, perl=TRUE)
## => [1] "Abc.123~xxx~x~y~z~ some~other~chars."
library(stringr)
stringr::str_remove_all(x, "[\\p{P}\\p{S}&&[^.]]") # Same as "[\\p{P}\\p{S}--[.]]"
## => [1] "Abc.123xxxxyz someotherchars."
stringr::str_replace_all(x, "[\\p{P}\\p{S}&&[^.]]", "~")
## => [1] "Abc.123~~~xxx~x~y~z~~~~~ some~other~chars."
stringr::str_replace_all(x, "[\\p{P}\\p{S}&&[^.]]+", "~") # Same as "[\\p{P}\\p{S}--[.]]+"
## => [1] "Abc.123~xxx~x~y~z~ some~other~chars."

Answer 4

Another way to think about it is what do you want to keep? You can use regular expressions to both keep information as well as omit it. I have a lot of data frames that I need to clean units out of and convert from multiple rows in one pass and I find it easiest to use something from the apply family in these instances.
Recreating the example:
a <- c('1', '2#', '3', '4', '£5')
b <- c('6', '7.235', '8', '$9', '-10')
z <- matrix(data = c(a, b), nrow = length(a), ncol=2)
Then use apply in conjunction with gsub.
apply(z, 2, function(x) as.numeric(gsub('[^0-9\\.\\-]', '', x)))
[,1] [,2]
[1,] 1 6.000
[2,] 2 7.235
[3,] 3 8.000
[4,] 4 9.000
[5,] 5 -10.000
This instructs R to match everything except digits, periods, and hyphens/dashes. Personally, I find it much cleaner and easier to use in these situations and gives the same output.
Also, the documentation has a good explanation of these powerful but confusing regular expressions.
https://stat.ethz.ch/R-manual/R-devel/library/base/html/regex.html
Or ?regex