6

I want to remove all characters in a string except:

  • - or _ or .
  • A thru Z
  • a thru z
  • 0 to 9
  • space

On linux command line, using sed I would do this:

$ echo "testing-#$% yes.no" | sed 's/[^-_.a-zA-Z0-9 ]//g'

Output: 

testing- yes.no

How can I achieve the same effect in Red language with PARSE? I looked at:

However, I could not codify it. I tried: 

>> parse "mystring%^&" [#a - #z #A - #Z #0 - #9]
== false
>> parse "mystring%^&" [#a-#z#A-#Z#0-#9]        
== false
rnso
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  • Remember to use `trim` if you want to remove some chars. `trim/with "testing-#$% yes.no" "-#$%." == "testing yesno"` – endo64 Sep 22 '17 at 13:56

3 Answers3

6

First note the difference between ISSUE! and CHAR!

#a #b #c  ; issues
#"a" #"b" #"c"   ; chars

You can then establish a character set (BITSET! type) either for the characters you want to keep or those you wish to discard. We'll do the former here:

good-chars: charset [#"a" - #"z" #"A" - #"Z" #"0" - #"9"]

Now that we have that, we can approach this in some different ways:

Parse

A fairly basic parse loop—skips any good-chars and removes anything else.

parse "mystring%^&" [any [some good-chars | remove skip]]

Remove-Each

Hopefully self-explanatory:

remove-each char "mystring%^&" [not find good-chars char]
rgchris
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  • On Red command line, your first answer (parse...) `produces ==true` as output. The second answer (remove-each...) produces no output. Why is this happening? – rnso Sep 21 '17 at 02:15
  • @rnso Parse returns **true** if the rule matches. In Red, you can wrap the rule thus: `take parse "mystring%^&" [collect [keep any [some good-chars | remove skip]]]` but that copies the string. Apparently **remove-each** returns the product of the last iteration—in Rebol 2 it would return the modified series. Note that both my answers as-is modify the existing string, they do not create new strings. – rgchris Sep 21 '17 at 16:31
  • @rnso If your string is set to a word, you can use ALSO: **also mystring parse mystring [any [...]]** or: **also mystring remove-each char mystring [...]** – rgchris Sep 21 '17 at 16:33
  • Good information. You may add these comments to your answer above since many readers just read answers and not comments. – rnso Sep 21 '17 at 16:37
  • @rnso It's implied in that I answered your question as asked. Your chosen answer actually creates several new strings in the process. – rgchris Sep 21 '17 at 19:49
4

First, characters must be in quotes, #a is issue!, char! is #"a". You've got the specification right, but you must pass it to charset function, to make a bitset! form it.

Then you can parse your string, keeping valid characters and skiping invalid:

>> chars: charset [#"a" - #"z" #"A" - #"Z" #"0" - #"9"]
== make bitset! #{000000000000FFC07FFFFFE07FFFFFE0}
>> rejoin parse "mystring%^&asdf" [collect some [keep chars | skip]]
== "mystringasdf"
rebolek
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2

An alternative solution to PARSE would be to use REPLACE here with a COMPLEMENT CHARSET:

replace/all "mystring%^&" complement charset [{-_. } #"a" - #"z" #"0" - #"9"] {}

NB. Above works in Rebol (2 & 3). Unfortunately it currently hangs in Red (tested on 0.63 on MacOS).

draegtun
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