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