Revisions to Removing commented lines of text in a string

Commonmark migration

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edited Jun 16, 2020 at 9:23

1

###Corrected answer

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"

###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"

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"

added 3 characters in body

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edited Sep 1, 2013 at 11:40

Mr.Wizard

273.1k
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595
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###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"

###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"

###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"

added 126 characters in body

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edited Aug 30, 2013 at 22:35

Mr.Wizard

273.1k
34
595
1.4k

###Corrected answer

My original answer (preserved belowsee 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"

Old answer

Please tell me if this works as desiredOr:

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

StringCases[strStringSplit[str, StartOfLineShortest[StartOfLine ~~ Except["@""@" | "#"]"#" ~~ __ ~~ EndOfLine]"\n"]]

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

The regular expression for this pattern is more compact but IMO harder to readAnd the RE equivalent:

First @ StringPattern`PatternConvert[StartOfLineStringPattern`PatternConvert[Shortest[StartOfLine ~~ Except["@""@" | "#"]"#" ~~ __ ~~ EndOfLine]"\n"]]

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

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

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

###Corrected answer

My original answer (preserved below) 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"

Old answer

Please tell me if this works as desired:

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

StringCases[str, StartOfLine ~~ Except["@" | "#"] ~~ __ ~~ EndOfLine]

{"Data 1\n Data 2\nData 3"}

The regular expression for this pattern is more compact but IMO harder to read:

First @ StringPattern`PatternConvert[StartOfLine ~~ Except["@" | "#"] ~~ __ ~~ EndOfLine]

"(?ms)^[^@#].+$"

StringCases[str, RegularExpression["(?ms)^[^@#].+$"]]

{"Data 1\n Data 2\nData 3"}

###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"

deleted 416 characters in body

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edited Aug 30, 2013 at 22:30

Mr.Wizard

273.1k
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595
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added 442 characters in body

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edited Aug 30, 2013 at 22:21

Mr.Wizard

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595
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added 281 characters in body

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edited Aug 30, 2013 at 21:55

Mr.Wizard

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595
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created Aug 30, 2013 at 21:49

Mr.Wizard

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595
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