Haskell - Removing non-letter characters but ignoring white spaces?
I am very new to Haskell. I am trying to return a list of strings from a given string (which could contain non-letter characters) but I get a single string in the list.
The below code shows What I have tried so far:
toLowerStr xs = map toLower xs
--drop non-letters characters
dropNonLetters xs = words $ (filter (x -> x `elem` ['a'..'z'])) $ toLowerStr xs
- lowercase all the characters by using
toLowerfunction - remove non-letter characters by using
filterfunction - return a list of strings by using
wordsfunction
I think the filter function is removing the white spaces and therefore it becomes a single string. I tried using isSpace function but I don't know exactly how to implement it in this case.
What is it that I am doing wrong? I get this output:
λ> dropNonLetters "ORANGE, apple! APPLE!!"
["orangeappleapple"]
But I want to achieve the below output:
λ> dropNonLetters "ORANGE, apple! APPLE!!"
["orange","apple","apple"]
haskell
edited Dec 21 '17 at 14:10
AJFarmar
8,87622752
asked Dec 21 '17 at 13:57
Big SmileBig Smile
643514
add a comment |
I am very new to Haskell. I am trying to return a list of strings from a given string (which could contain non-letter characters) but I get a single string in the list.
The below code shows What I have tried so far:
toLowerStr xs = map toLower xs
--drop non-letters characters
dropNonLetters xs = words $ (filter (x -> x `elem` ['a'..'z'])) $ toLowerStr xs
- lowercase all the characters by using
toLowerfunction - remove non-letter characters by using
filterfunction - return a list of strings by using
wordsfunction
I think the filter function is removing the white spaces and therefore it becomes a single string. I tried using isSpace function but I don't know exactly how to implement it in this case.
What is it that I am doing wrong? I get this output:
λ> dropNonLetters "ORANGE, apple! APPLE!!"
["orangeappleapple"]
But I want to achieve the below output:
λ> dropNonLetters "ORANGE, apple! APPLE!!"
["orange","apple","apple"]
haskell
edited Dec 21 '17 at 14:10
AJFarmar
8,87622752
asked Dec 21 '17 at 13:57
Big SmileBig Smile
643514
What with tabs, ..
– Willem Van Onsem
Dec 21 '17 at 13:58
@WillemVanOnsem spaces between words
– Big Smile
Dec 21 '17 at 13:58
add a comment |
I am very new to Haskell. I am trying to return a list of strings from a given string (which could contain non-letter characters) but I get a single string in the list.
The below code shows What I have tried so far:
toLowerStr xs = map toLower xs
--drop non-letters characters
dropNonLetters xs = words $ (filter (x -> x `elem` ['a'..'z'])) $ toLowerStr xs
- lowercase all the characters by using
toLowerfunction - remove non-letter characters by using
filterfunction - return a list of strings by using
wordsfunction
I think the filter function is removing the white spaces and therefore it becomes a single string. I tried using isSpace function but I don't know exactly how to implement it in this case.
What is it that I am doing wrong? I get this output:
λ> dropNonLetters "ORANGE, apple! APPLE!!"
["orangeappleapple"]
But I want to achieve the below output:
λ> dropNonLetters "ORANGE, apple! APPLE!!"
["orange","apple","apple"]
haskell
edited Dec 21 '17 at 14:10
AJFarmar
8,87622752
asked Dec 21 '17 at 13:57
Big SmileBig Smile
643514
I am very new to Haskell. I am trying to return a list of strings from a given string (which could contain non-letter characters) but I get a single string in the list.
The below code shows What I have tried so far:
toLowerStr xs = map toLower xs
--drop non-letters characters
dropNonLetters xs = words $ (filter (x -> x `elem` ['a'..'z'])) $ toLowerStr xs
- lowercase all the characters by using
toLowerfunction - remove non-letter characters by using
filterfunction - return a list of strings by using
wordsfunction
I think the filter function is removing the white spaces and therefore it becomes a single string. I tried using isSpace function but I don't know exactly how to implement it in this case.
What is it that I am doing wrong? I get this output:
λ> dropNonLetters "ORANGE, apple! APPLE!!"
["orangeappleapple"]
But I want to achieve the below output:
λ> dropNonLetters "ORANGE, apple! APPLE!!"
["orange","apple","apple"]
haskell
haskell
edited Dec 21 '17 at 14:10
AJFarmar
8,87622752
asked Dec 21 '17 at 13:57
Big SmileBig Smile
643514
edited Dec 21 '17 at 14:10
AJFarmar
8,87622752
asked Dec 21 '17 at 13:57
Big SmileBig Smile
643514
edited Dec 21 '17 at 14:10
AJFarmar
8,87622752
edited Dec 21 '17 at 14:10
AJFarmar
8,87622752
edited Dec 21 '17 at 14:10
AJFarmar
8,87622752
8,87622752
asked Dec 21 '17 at 13:57
Big SmileBig Smile
643514
asked Dec 21 '17 at 13:57
Big SmileBig Smile
643514
asked Dec 21 '17 at 13:57
Big SmileBig Smile
643514
643514
What with tabs, ..
– Willem Van Onsem
Dec 21 '17 at 13:58
@WillemVanOnsem spaces between words
– Big Smile
Dec 21 '17 at 13:58
add a comment |
What with tabs, ..
– Willem Van Onsem
Dec 21 '17 at 13:58
@WillemVanOnsem spaces between words
– Big Smile
Dec 21 '17 at 13:58
What with tabs, ..
– Willem Van Onsem
Dec 21 '17 at 13:58
What with tabs, ..
– Willem Van Onsem
Dec 21 '17 at 13:58
@WillemVanOnsem spaces between words
– Big Smile
Dec 21 '17 at 13:58
@WillemVanOnsem spaces between words
– Big Smile
Dec 21 '17 at 13:58
add a comment |
1 Answer
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I think the filter function is removing the white spaces and therefore it becomes a single string.
That is correct. As filter predicate you write x -> x `elem` ['a'..'z']. ['a'..'z'] is a list that contains lowercase letters, so for whitespace, the predicate will fail, and thus you should allow spaces as well.
We can for instance add the space character to the list:
dropNonLetters xs = words $ (filter (x -> x `elem` (' ':['a'..'z'])))) $ toLowerStr xsBut this is inelegant and does not really explain itself. The Data.Char module however ships with two functions that are interesting here: isLower :: Char -> Bool, and isSpace :: Char -> Bool. We can use this like:
dropNonLetters xs = words $ (filter (x -> isLower x || isSpace x)) $ toLowerStr xsisLower and isSpace are not only more "descriptive" and elegant. Usually these functions will be faster than a membership check (which will usually be done in O(n)), and furthermore it will also take into account tabs, new lines, etc.
We can also perform an eta-reduction on the function:
dropNonLetters = words . (filter (x -> isLower x || isSpace x)) . toLowerStrThis then produces:
Prelude Data.Char> dropNonLetters "ORANGE, apple! APPLE!!"
["orange","apple","apple"]
I advise you to rename the function dropNonLetters, since now it does not fully explain that it will generate a list of words. Based on the name, I would think that it only drops non-letters, not that it converts the string to lowercase nor that it constructs words.
edited Nov 14 '18 at 2:01
Jon Purdy
38.1k768125
answered Dec 21 '17 at 14:03
Willem Van OnsemWillem Van Onsem
147k16141231
Thank you very much this worked and I will rename the function. You are missing->in the second block of the code.
– Big Smile
Dec 21 '17 at 14:11
@BigSmile: thanks, it is fixed now :)
– Willem Van Onsem
Dec 21 '17 at 14:12
add a comment |
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votes
I think the filter function is removing the white spaces and therefore it becomes a single string.
That is correct. As filter predicate you write x -> x `elem` ['a'..'z']. ['a'..'z'] is a list that contains lowercase letters, so for whitespace, the predicate will fail, and thus you should allow spaces as well.
We can for instance add the space character to the list:
dropNonLetters xs = words $ (filter (x -> x `elem` (' ':['a'..'z'])))) $ toLowerStr xsBut this is inelegant and does not really explain itself. The Data.Char module however ships with two functions that are interesting here: isLower :: Char -> Bool, and isSpace :: Char -> Bool. We can use this like:
dropNonLetters xs = words $ (filter (x -> isLower x || isSpace x)) $ toLowerStr xsisLower and isSpace are not only more "descriptive" and elegant. Usually these functions will be faster than a membership check (which will usually be done in O(n)), and furthermore it will also take into account tabs, new lines, etc.
We can also perform an eta-reduction on the function:
dropNonLetters = words . (filter (x -> isLower x || isSpace x)) . toLowerStrThis then produces:
Prelude Data.Char> dropNonLetters "ORANGE, apple! APPLE!!"
["orange","apple","apple"]
I advise you to rename the function dropNonLetters, since now it does not fully explain that it will generate a list of words. Based on the name, I would think that it only drops non-letters, not that it converts the string to lowercase nor that it constructs words.
edited Nov 14 '18 at 2:01
Jon Purdy
38.1k768125
answered Dec 21 '17 at 14:03
Willem Van OnsemWillem Van Onsem
147k16141231
Thank you very much this worked and I will rename the function. You are missing->in the second block of the code.
– Big Smile
Dec 21 '17 at 14:11
@BigSmile: thanks, it is fixed now :)
– Willem Van Onsem
Dec 21 '17 at 14:12
add a comment |
I think the filter function is removing the white spaces and therefore it becomes a single string.
That is correct. As filter predicate you write x -> x `elem` ['a'..'z']. ['a'..'z'] is a list that contains lowercase letters, so for whitespace, the predicate will fail, and thus you should allow spaces as well.
We can for instance add the space character to the list:
dropNonLetters xs = words $ (filter (x -> x `elem` (' ':['a'..'z'])))) $ toLowerStr xsBut this is inelegant and does not really explain itself. The Data.Char module however ships with two functions that are interesting here: isLower :: Char -> Bool, and isSpace :: Char -> Bool. We can use this like:
dropNonLetters xs = words $ (filter (x -> isLower x || isSpace x)) $ toLowerStr xsisLower and isSpace are not only more "descriptive" and elegant. Usually these functions will be faster than a membership check (which will usually be done in O(n)), and furthermore it will also take into account tabs, new lines, etc.
We can also perform an eta-reduction on the function:
dropNonLetters = words . (filter (x -> isLower x || isSpace x)) . toLowerStrThis then produces:
Prelude Data.Char> dropNonLetters "ORANGE, apple! APPLE!!"
["orange","apple","apple"]
I advise you to rename the function dropNonLetters, since now it does not fully explain that it will generate a list of words. Based on the name, I would think that it only drops non-letters, not that it converts the string to lowercase nor that it constructs words.
edited Nov 14 '18 at 2:01
Jon Purdy
38.1k768125
answered Dec 21 '17 at 14:03
Willem Van OnsemWillem Van Onsem
147k16141231
Thank you very much this worked and I will rename the function. You are missing->in the second block of the code.
– Big Smile
Dec 21 '17 at 14:11
@BigSmile: thanks, it is fixed now :)
– Willem Van Onsem
Dec 21 '17 at 14:12
add a comment |
I think the filter function is removing the white spaces and therefore it becomes a single string.
That is correct. As filter predicate you write x -> x `elem` ['a'..'z']. ['a'..'z'] is a list that contains lowercase letters, so for whitespace, the predicate will fail, and thus you should allow spaces as well.
We can for instance add the space character to the list:
dropNonLetters xs = words $ (filter (x -> x `elem` (' ':['a'..'z'])))) $ toLowerStr xsBut this is inelegant and does not really explain itself. The Data.Char module however ships with two functions that are interesting here: isLower :: Char -> Bool, and isSpace :: Char -> Bool. We can use this like:
dropNonLetters xs = words $ (filter (x -> isLower x || isSpace x)) $ toLowerStr xsisLower and isSpace are not only more "descriptive" and elegant. Usually these functions will be faster than a membership check (which will usually be done in O(n)), and furthermore it will also take into account tabs, new lines, etc.
We can also perform an eta-reduction on the function:
dropNonLetters = words . (filter (x -> isLower x || isSpace x)) . toLowerStrThis then produces:
Prelude Data.Char> dropNonLetters "ORANGE, apple! APPLE!!"
["orange","apple","apple"]
I advise you to rename the function dropNonLetters, since now it does not fully explain that it will generate a list of words. Based on the name, I would think that it only drops non-letters, not that it converts the string to lowercase nor that it constructs words.
edited Nov 14 '18 at 2:01
Jon Purdy
38.1k768125
answered Dec 21 '17 at 14:03
Willem Van OnsemWillem Van Onsem
147k16141231
I think the filter function is removing the white spaces and therefore it becomes a single string.
That is correct. As filter predicate you write x -> x `elem` ['a'..'z']. ['a'..'z'] is a list that contains lowercase letters, so for whitespace, the predicate will fail, and thus you should allow spaces as well.
We can for instance add the space character to the list:
dropNonLetters xs = words $ (filter (x -> x `elem` (' ':['a'..'z'])))) $ toLowerStr xsBut this is inelegant and does not really explain itself. The Data.Char module however ships with two functions that are interesting here: isLower :: Char -> Bool, and isSpace :: Char -> Bool. We can use this like:
dropNonLetters xs = words $ (filter (x -> isLower x || isSpace x)) $ toLowerStr xsisLower and isSpace are not only more "descriptive" and elegant. Usually these functions will be faster than a membership check (which will usually be done in O(n)), and furthermore it will also take into account tabs, new lines, etc.
We can also perform an eta-reduction on the function:
dropNonLetters = words . (filter (x -> isLower x || isSpace x)) . toLowerStrThis then produces:
Prelude Data.Char> dropNonLetters "ORANGE, apple! APPLE!!"
["orange","apple","apple"]
I advise you to rename the function dropNonLetters, since now it does not fully explain that it will generate a list of words. Based on the name, I would think that it only drops non-letters, not that it converts the string to lowercase nor that it constructs words.
edited Nov 14 '18 at 2:01
Jon Purdy
38.1k768125
answered Dec 21 '17 at 14:03
Willem Van OnsemWillem Van Onsem
147k16141231
edited Nov 14 '18 at 2:01
Jon Purdy
38.1k768125
edited Nov 14 '18 at 2:01
Jon Purdy
38.1k768125
edited Nov 14 '18 at 2:01
Jon Purdy
38.1k768125
38.1k768125
answered Dec 21 '17 at 14:03
Willem Van OnsemWillem Van Onsem
147k16141231
answered Dec 21 '17 at 14:03
Willem Van OnsemWillem Van Onsem
147k16141231
answered Dec 21 '17 at 14:03
Willem Van OnsemWillem Van Onsem
147k16141231
147k16141231
Thank you very much this worked and I will rename the function. You are missing->in the second block of the code.
– Big Smile
Dec 21 '17 at 14:11
@BigSmile: thanks, it is fixed now :)
– Willem Van Onsem
Dec 21 '17 at 14:12
add a comment |
Thank you very much this worked and I will rename the function. You are missing->in the second block of the code.
– Big Smile
Dec 21 '17 at 14:11
@BigSmile: thanks, it is fixed now :)
– Willem Van Onsem
Dec 21 '17 at 14:12
Thank you very much this worked and I will rename the function. You are missing
-> in the second block of the code.– Big Smile
Dec 21 '17 at 14:11
Thank you very much this worked and I will rename the function. You are missing
-> in the second block of the code.– Big Smile
Dec 21 '17 at 14:11
@BigSmile: thanks, it is fixed now :)
– Willem Van Onsem
Dec 21 '17 at 14:12
@BigSmile: thanks, it is fixed now :)
– Willem Van Onsem
Dec 21 '17 at 14:12
add a comment |
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What with tabs, ..
– Willem Van Onsem
Dec 21 '17 at 13:58
@WillemVanOnsem spaces between words
– Big Smile
Dec 21 '17 at 13:58