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Magician Joined: 24 Feb 2001 Posts: 342 Location: USA

I'd like to write a function that takes a string as an argument, and returns that string with all instances of items from a certain string list removed. for example:




mob = "The monkey"


word_list = {The |A |An }


mob_to_hit = @strip_article(@mob)

and the value of @mob_to_hit would end up being "monkey"
i cant seem to figure out what the correct value for @strip_article should be. any advice is appreciated!

Emit

Posted: Fri Nov 16, 2001 9:55 pm

This might be a little more complicated than you had in mind, but it's all I could think of atm.

#ALIAS stripmob {
#VARIABLE Mob_To_Hit %null
#VARIABLE Mob "%-1"
#VARIABLE Mob %replace( @Mob, " ", "| |")
#FORALL {@Mob} {#IF (!%ismember( %i, @Word_List)) {#VARIABLE Mob_To_Hit %additem( %i, @Mob_To_Hit)}}
#VARIABLE Mob_To_Hit %replace( %replace( @Mob_To_Hit, "|", ""), " ", " ")
}

What it does is save whatever you input after the alias into the variable @Mob. Then it checks that variable against the words in the @Word_List, and copies only those words into @Mob_To_Hit that aren't in @Word_List.

Acaila

Magician Joined: 24 Feb 2001 Posts: 342 Location: USA

i wrote an alias that does the job like this:




#alias stripper {


	#forall @words_to_strip {


		mob_to_hit = %replace(@mob,%i,"")


		}


	}

but i was looking to make it a function so i could call it in the middle of another function call

i cant see how though, because i can't find a looping function, and i don't think you can put commands into custom functions, and i can't see how to do it without looping

Emit

Newbie Joined: 10 Oct 2000 Posts: 9 Location: China

Maybe Zugg need to add a command
"#return" for Usr-defind func.
Then we can use #func more freely.

GURU Joined: 14 Oct 2000 Posts: 556 Location: USA

A clumsy way of using a function to perform this would be the following:




#FUNCTION strip_article {%delitem("A", %delitem("An", %delitem("The", %replace(%1, " ","|"))))}

This converts the phrase into a string list then uses nested %delitem functions to seek out and remove the three articles. Obviously, as the number of articles grows, this function will become rather unwieldy.

Can you say with certainty that the last word in the phrase will always be the keyword? If so, this function will serve you:




#FUNCTION last_word {%word(%1, %numwords(%1))}

Troubadour

Magician Joined: 24 Feb 2001 Posts: 342 Location: USA

unfortunately, the key word is guaranteed not to be the last word

and the real list of key words has about 8 words/phrases. i'll just use the alias, but thanks for the help anyway
Emit