Removing commented lines of text in a string

Question

Suppose that I have a string str. str contains headings -- which are written in commented lines -- and data. For example:

str = "# Heading 1
# Heading 2
@ Heading 3
@ Heading 4
Data 1
Data 2
Data 3";

where # and @ denote commented lines.

Now, I would like to write a function removeComments to remove all lines in str that are commented. That is, I would like to remove all lines in str that start with the character # or the character @.

I have come up with one way to do this, I think. My procedure is to first split str into lines using StringSplit with the "\n" delimiter; then use DeleteCases to delete all lines in that list which begin with either # or @; and, finally, convert the list back to a string using Riffle and StringJoin. So, I have the following:

removeComments[str_String] := Module[{result},
  result = StringSplit[str, "\n"];
  result = 
   DeleteCases[
    result, _?(StringMatchQ[#, ("#" ~~ ___) | ("@" ~~ ___)] &)];
  result = StringJoin[Riffle[result, "\n"]];
  result
  ]
removeComments[str]

which gives the following correct output (a string):

Data 1
Data 2
Data 3

However, now suppose that one of my data items (for example, a single space in front of Data 2) has one or more leading spaces:

str = "# Heading 1
# Heading 2
@ Heading 3
@ Heading 4
Data 1
 Data 2
Data 3";

removeComments[str]

In this case, I get the incorrect output:

Data 1
Data 3

That is, Data 2 is omitted altogether. Something is clearly wrong with my approach, but I haven't been able to find my mistake (although it is probably obvious). Could you please help me find what is wrong? Thanks for your time.

Answer 1

Corrected answer

My original answer (see edit history) was not correct, unless all of your data lines are contiguous.

str = "# Heading 1\n# Heading 2\n@ Heading 3\n@ Heading 4\nData 1\n Data 2\nData 3\n@ \
Heading 5\nData 4";

StringReplace[str, Shortest[StartOfLine ~~ "@" | "#" ~~ ___ ~~ "\n"] :> ""]

"Data 1\n Data 2\nData 3\nData 4"

Or:

"" <> StringSplit[str, Shortest[StartOfLine ~~ "@" | "#" ~~ ___ ~~ "\n"]]

"Data 1\n Data 2\nData 3\nData 4"

And the RE equivalent:

First @ StringPattern`PatternConvert[Shortest[StartOfLine ~~ "@" | "#" ~~ ___ ~~ "\n"]]

"(?ms)^[@#].*?\n"

"" <> StringSplit[str, RegularExpression["(?ms)^[@#].*?\n"]]

"Data 1\n Data 2\nData 3\nData 4"

Answer 2

This answer only discusses the implied question of why the OP's code fails. I think this should be made clear because it puzzled me until Mr.Wizard and rm -rf enlightened me. I would like keep others from falling into the same trap.

I was studying the code posted in this question, trying to learn why it failed, when I ran into the following.

StringMatchQ[" ", "@"]

True

I found this very surprising and posted a question on it for a very short period. Mr.Wizard and rm -rf quickly pointed out that @ was a wild card character and that my code should be

StringMatchQ[" ", "\\@"]

False

This same mistake also occurs in the code posted in the question above. With the following minor change, the OP's code works.

removeComments[str_String] := 
  Module[{result},
    result = StringSplit[str, "\n"];
    result = 
      DeleteCases[result, _?(StringMatchQ[#, ("#" ~~ ___) | ("\\@" ~~ ___)] &)];
    StringJoin[Riffle[result, "\n"]]]

removeComments[
  "# Heading 1\n# Heading 2\n@ Heading 3\n@ Heading 4\nData 1\n Data 2\nData 3"]

Data 1
$\ $Data 2
Data 3