The operation
(filter (`notElem` "'\"").[(1,'a','%',"yes")])
gives an error. How can apply this filter on that list properly?
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Don Stewart
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thetux4
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4You're having enough issues with beginner material in Haskell, that I'd suggest joining the #haskell irc channel for help, rather than asking lots of easy questions on SO. – Don Stewart Apr 15 '11 at 23:58
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Note, for those answering, this is a follow-on from http://stackoverflow.com/questions/5683210/showing-a-haskell-list-of-tuples-with-custom-syntax – Don Stewart Apr 16 '11 at 00:00
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1Id rather read some list tutorials actually. – thetux4 Apr 16 '11 at 00:01
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2"Learn You a Haskell" is a good one: http://learnyouahaskell.com/ – Don Stewart Apr 16 '11 at 00:05
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2You might benefit from Real World Haskell as well. I did. http://book.realworldhaskell.org/read/getting-started.html – Tim Perry Apr 16 '11 at 01:08
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See also http://stackoverflow.com/questions/3030675/haskell-function-composition-and-function-application-idioms-correct-use – Don Stewart Apr 16 '11 at 20:21
2 Answers
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You've got a couple of serious problems. First, your syntax is wacky (. definitely shouldn't be there). But the bigger problem is that what you're trying to filter is of the type [(Int,Char,Char,[Char])] (that is, a list containing a 4-tuple).
And your list has only one element, which is (1,'a','%',"yes"). So filtering that is useless anyway. When function you provide for filtering must be of type a -> Boolean, where a is the type of all the elements of the list.
Seems like you wanted some sort of wonky heterogenous list or something.
Jonathan Sterling
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The . operator in Haskell is function composition -- it composes two functions together.
So your code,
(`notElem` "'\"") . [(1,'a','%',"yes")]
looks like the composition of the notElem function and some list. That's just wrong.
Remove the ., and make sure to show the list first:
> filter (`notElem` "'\"") (show [(1,'a','%',"yes")])
"[(1,a,%,yes)]"
Don Stewart
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Ahh allright. Isn't it possible doing this without using show and assigning the filtered list into a new variable? – thetux4 Apr 15 '11 at 23:59
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@thetux4 What result is it that you're trying to achieve from the filtering? It's really not clear. – Jonathan Sterling Apr 16 '11 at 00:02
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@Jonathan I actually want to get [(1,a,%,yes)] after filtering but without using show. – thetux4 Apr 16 '11 at 00:04
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@thetux4: Why *don't* you want to use `show`? The questions you've been asking are all aimed at very small goals; what's the overarching problem you're trying to solve? – Antal Spector-Zabusky Apr 16 '11 at 02:42
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