Haskell: How to remove a List from a List in Haskell
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I have a first list and a second list, and I want to write a function that takes two lists and returns a list containing only those elements in the second list that do not appear in the first. And I don't want to use built-in functions. For example:
> removeAll [1..3] [0..10]
[0,4,5,6,7,8,9,10]
> removeAll "aeiou" "supercalifragilisticexpialidocious"
"sprclfrglstcxpldcs"
removeAll _ =
removeAll y = y
removeAll (x:xs) (y:ys)
| x==y = removeAll xs ys
| otherwise = y:removeAll xs ys
haskell
edited Nov 22 '18 at 5:39
m0nhawk
15.9k83263
asked Nov 22 '18 at 5:30
zheng zhongzheng zhong
94
add a comment |
I have a first list and a second list, and I want to write a function that takes two lists and returns a list containing only those elements in the second list that do not appear in the first. And I don't want to use built-in functions. For example:
> removeAll [1..3] [0..10]
[0,4,5,6,7,8,9,10]
> removeAll "aeiou" "supercalifragilisticexpialidocious"
"sprclfrglstcxpldcs"
removeAll _ =
removeAll y = y
removeAll (x:xs) (y:ys)
| x==y = removeAll xs ys
| otherwise = y:removeAll xs ys
haskell
edited Nov 22 '18 at 5:39
m0nhawk
15.9k83263
asked Nov 22 '18 at 5:30
zheng zhongzheng zhong
94
1
What is your question?
– talex
Nov 22 '18 at 5:43
3
You can always look at the source of the built-in functions, so you shouldn't count those out.
– Mateen Ulhaq
Nov 22 '18 at 5:44
add a comment |
I have a first list and a second list, and I want to write a function that takes two lists and returns a list containing only those elements in the second list that do not appear in the first. And I don't want to use built-in functions. For example:
> removeAll [1..3] [0..10]
[0,4,5,6,7,8,9,10]
> removeAll "aeiou" "supercalifragilisticexpialidocious"
"sprclfrglstcxpldcs"
removeAll _ =
removeAll y = y
removeAll (x:xs) (y:ys)
| x==y = removeAll xs ys
| otherwise = y:removeAll xs ys
haskell
edited Nov 22 '18 at 5:39
m0nhawk
15.9k83263
asked Nov 22 '18 at 5:30
zheng zhongzheng zhong
94
I have a first list and a second list, and I want to write a function that takes two lists and returns a list containing only those elements in the second list that do not appear in the first. And I don't want to use built-in functions. For example:
> removeAll [1..3] [0..10]
[0,4,5,6,7,8,9,10]
> removeAll "aeiou" "supercalifragilisticexpialidocious"
"sprclfrglstcxpldcs"
removeAll _ =
removeAll y = y
removeAll (x:xs) (y:ys)
| x==y = removeAll xs ys
| otherwise = y:removeAll xs ys
haskell
haskell
edited Nov 22 '18 at 5:39
m0nhawk
15.9k83263
asked Nov 22 '18 at 5:30
zheng zhongzheng zhong
94
edited Nov 22 '18 at 5:39
m0nhawk
15.9k83263
asked Nov 22 '18 at 5:30
zheng zhongzheng zhong
94
edited Nov 22 '18 at 5:39
m0nhawk
15.9k83263
edited Nov 22 '18 at 5:39
m0nhawk
15.9k83263
edited Nov 22 '18 at 5:39
m0nhawk
15.9k83263
15.9k83263
asked Nov 22 '18 at 5:30
zheng zhongzheng zhong
94
asked Nov 22 '18 at 5:30
zheng zhongzheng zhong
94
asked Nov 22 '18 at 5:30
zheng zhongzheng zhong
94
94
1
What is your question?
– talex
Nov 22 '18 at 5:43
3
You can always look at the source of the built-in functions, so you shouldn't count those out.
– Mateen Ulhaq
Nov 22 '18 at 5:44
add a comment |
1
What is your question?
– talex
Nov 22 '18 at 5:43
3
You can always look at the source of the built-in functions, so you shouldn't count those out.
– Mateen Ulhaq
Nov 22 '18 at 5:44
1
1
What is your question?
– talex
Nov 22 '18 at 5:43
What is your question?
– talex
Nov 22 '18 at 5:43
3
3
You can always look at the source of the built-in functions, so you shouldn't count those out.
– Mateen Ulhaq
Nov 22 '18 at 5:44
You can always look at the source of the built-in functions, so you shouldn't count those out.
– Mateen Ulhaq
Nov 22 '18 at 5:44
add a comment |
3 Answers
3
active
oldest
votes
There is a pretty straightforward recursive definition
removeAll :: Eq a => [a] -> [a] -> [a]
removeAll _ =
removeAll toRemove (x:xs)
| x `belongs` toRemove = removeAll toRemove xs
| otherwise = x:removeAll toRemove xs
where belongs _ = False
belongs a (y:ys)
| a == y = True
| otherwise = belongs a ys
The belongs function is just the prelude defined elem. Since you don't want to use built-in functions you have to define it yourself.
answered Nov 22 '18 at 7:36
lsmorlsmor
1,004114
It works, thank you very much!
– zheng zhong
Nov 22 '18 at 14:25
add a comment |
This is one simple way:
removeAll ignores = filter (x -> notElem x ignores)
which makes use of:
filternotElem
If you wanted to write it in "pure-pure" Haskell, with no auxiliary functions... well, it sounds like it would be pretty ugly since we'd need to do some sort of nested for-loop. Here's a compromise:
myFilter _ =
myFilter pred (x:xs)
| pred x = x : myFilter pred xs
| otherwise = myFilter pred xs
myNotElem e = True
myNotElem e (x:xs)
| e == x = False
| otherwise = myNotElem e xs
removeAll ignores = myFilter (x -> myNotElem x ignores)
answered Nov 22 '18 at 6:09
Mateen UlhaqMateen Ulhaq
11.6k114994
Thank you for your answer!
– zheng zhong
Nov 22 '18 at 14:26
add a comment |
A problem of your current code is that you tried to recurse in two lists simultaneously in one function. You need to divide and conquer. If elem is allowed, you may write:
removeAll :: Eq a => [a] -> [a] -> [a]
removeAll _ =
removeAll x (y:ys)
| myElem y x = removeAll x ys
| otherwise = y:removeAll x ys
If you do not want to use the built-in elem as well, you can define it similarly:
myElem :: Eq a => a -> [a] -> Bool
myElem _ = False
myElem x (y:ys)
| x == y = True
| otherwise = myElem x ys
answered Nov 23 '18 at 8:22
Yongwei WuYongwei Wu
2,1761832
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is a pretty straightforward recursive definition
removeAll :: Eq a => [a] -> [a] -> [a]
removeAll _ =
removeAll toRemove (x:xs)
| x `belongs` toRemove = removeAll toRemove xs
| otherwise = x:removeAll toRemove xs
where belongs _ = False
belongs a (y:ys)
| a == y = True
| otherwise = belongs a ys
The belongs function is just the prelude defined elem. Since you don't want to use built-in functions you have to define it yourself.
answered Nov 22 '18 at 7:36
lsmorlsmor
1,004114
It works, thank you very much!
– zheng zhong
Nov 22 '18 at 14:25
add a comment |
There is a pretty straightforward recursive definition
removeAll :: Eq a => [a] -> [a] -> [a]
removeAll _ =
removeAll toRemove (x:xs)
| x `belongs` toRemove = removeAll toRemove xs
| otherwise = x:removeAll toRemove xs
where belongs _ = False
belongs a (y:ys)
| a == y = True
| otherwise = belongs a ys
The belongs function is just the prelude defined elem. Since you don't want to use built-in functions you have to define it yourself.
answered Nov 22 '18 at 7:36
lsmorlsmor
1,004114
It works, thank you very much!
– zheng zhong
Nov 22 '18 at 14:25
add a comment |
There is a pretty straightforward recursive definition
removeAll :: Eq a => [a] -> [a] -> [a]
removeAll _ =
removeAll toRemove (x:xs)
| x `belongs` toRemove = removeAll toRemove xs
| otherwise = x:removeAll toRemove xs
where belongs _ = False
belongs a (y:ys)
| a == y = True
| otherwise = belongs a ys
The belongs function is just the prelude defined elem. Since you don't want to use built-in functions you have to define it yourself.
answered Nov 22 '18 at 7:36
lsmorlsmor
1,004114
There is a pretty straightforward recursive definition
removeAll :: Eq a => [a] -> [a] -> [a]
removeAll _ =
removeAll toRemove (x:xs)
| x `belongs` toRemove = removeAll toRemove xs
| otherwise = x:removeAll toRemove xs
where belongs _ = False
belongs a (y:ys)
| a == y = True
| otherwise = belongs a ys
The belongs function is just the prelude defined elem. Since you don't want to use built-in functions you have to define it yourself.
answered Nov 22 '18 at 7:36
lsmorlsmor
1,004114
answered Nov 22 '18 at 7:36
lsmorlsmor
1,004114
answered Nov 22 '18 at 7:36
lsmorlsmor
1,004114
answered Nov 22 '18 at 7:36
lsmorlsmor
1,004114
1,004114
It works, thank you very much!
– zheng zhong
Nov 22 '18 at 14:25
add a comment |
It works, thank you very much!
– zheng zhong
Nov 22 '18 at 14:25
It works, thank you very much!
– zheng zhong
Nov 22 '18 at 14:25
It works, thank you very much!
– zheng zhong
Nov 22 '18 at 14:25
add a comment |
This is one simple way:
removeAll ignores = filter (x -> notElem x ignores)
which makes use of:
filternotElem
If you wanted to write it in "pure-pure" Haskell, with no auxiliary functions... well, it sounds like it would be pretty ugly since we'd need to do some sort of nested for-loop. Here's a compromise:
myFilter _ =
myFilter pred (x:xs)
| pred x = x : myFilter pred xs
| otherwise = myFilter pred xs
myNotElem e = True
myNotElem e (x:xs)
| e == x = False
| otherwise = myNotElem e xs
removeAll ignores = myFilter (x -> myNotElem x ignores)
answered Nov 22 '18 at 6:09
Mateen UlhaqMateen Ulhaq
11.6k114994
Thank you for your answer!
– zheng zhong
Nov 22 '18 at 14:26
add a comment |
This is one simple way:
removeAll ignores = filter (x -> notElem x ignores)
which makes use of:
filternotElem
If you wanted to write it in "pure-pure" Haskell, with no auxiliary functions... well, it sounds like it would be pretty ugly since we'd need to do some sort of nested for-loop. Here's a compromise:
myFilter _ =
myFilter pred (x:xs)
| pred x = x : myFilter pred xs
| otherwise = myFilter pred xs
myNotElem e = True
myNotElem e (x:xs)
| e == x = False
| otherwise = myNotElem e xs
removeAll ignores = myFilter (x -> myNotElem x ignores)
answered Nov 22 '18 at 6:09
Mateen UlhaqMateen Ulhaq
11.6k114994
Thank you for your answer!
– zheng zhong
Nov 22 '18 at 14:26
add a comment |
This is one simple way:
removeAll ignores = filter (x -> notElem x ignores)
which makes use of:
filternotElem
If you wanted to write it in "pure-pure" Haskell, with no auxiliary functions... well, it sounds like it would be pretty ugly since we'd need to do some sort of nested for-loop. Here's a compromise:
myFilter _ =
myFilter pred (x:xs)
| pred x = x : myFilter pred xs
| otherwise = myFilter pred xs
myNotElem e = True
myNotElem e (x:xs)
| e == x = False
| otherwise = myNotElem e xs
removeAll ignores = myFilter (x -> myNotElem x ignores)
answered Nov 22 '18 at 6:09
Mateen UlhaqMateen Ulhaq
11.6k114994
This is one simple way:
removeAll ignores = filter (x -> notElem x ignores)
which makes use of:
filternotElem
If you wanted to write it in "pure-pure" Haskell, with no auxiliary functions... well, it sounds like it would be pretty ugly since we'd need to do some sort of nested for-loop. Here's a compromise:
myFilter _ =
myFilter pred (x:xs)
| pred x = x : myFilter pred xs
| otherwise = myFilter pred xs
myNotElem e = True
myNotElem e (x:xs)
| e == x = False
| otherwise = myNotElem e xs
removeAll ignores = myFilter (x -> myNotElem x ignores)
answered Nov 22 '18 at 6:09
Mateen UlhaqMateen Ulhaq
11.6k114994
edited Nov 22 '18 at 6:16
answered Nov 22 '18 at 6:09
Mateen UlhaqMateen Ulhaq
11.6k114994
answered Nov 22 '18 at 6:09
Mateen UlhaqMateen Ulhaq
11.6k114994
answered Nov 22 '18 at 6:09
Mateen UlhaqMateen Ulhaq
11.6k114994
11.6k114994
Thank you for your answer!
– zheng zhong
Nov 22 '18 at 14:26
add a comment |
Thank you for your answer!
– zheng zhong
Nov 22 '18 at 14:26
Thank you for your answer!
– zheng zhong
Nov 22 '18 at 14:26
Thank you for your answer!
– zheng zhong
Nov 22 '18 at 14:26
add a comment |
A problem of your current code is that you tried to recurse in two lists simultaneously in one function. You need to divide and conquer. If elem is allowed, you may write:
removeAll :: Eq a => [a] -> [a] -> [a]
removeAll _ =
removeAll x (y:ys)
| myElem y x = removeAll x ys
| otherwise = y:removeAll x ys
If you do not want to use the built-in elem as well, you can define it similarly:
myElem :: Eq a => a -> [a] -> Bool
myElem _ = False
myElem x (y:ys)
| x == y = True
| otherwise = myElem x ys
answered Nov 23 '18 at 8:22
Yongwei WuYongwei Wu
2,1761832
add a comment |
A problem of your current code is that you tried to recurse in two lists simultaneously in one function. You need to divide and conquer. If elem is allowed, you may write:
removeAll :: Eq a => [a] -> [a] -> [a]
removeAll _ =
removeAll x (y:ys)
| myElem y x = removeAll x ys
| otherwise = y:removeAll x ys
If you do not want to use the built-in elem as well, you can define it similarly:
myElem :: Eq a => a -> [a] -> Bool
myElem _ = False
myElem x (y:ys)
| x == y = True
| otherwise = myElem x ys
answered Nov 23 '18 at 8:22
Yongwei WuYongwei Wu
2,1761832
add a comment |
A problem of your current code is that you tried to recurse in two lists simultaneously in one function. You need to divide and conquer. If elem is allowed, you may write:
removeAll :: Eq a => [a] -> [a] -> [a]
removeAll _ =
removeAll x (y:ys)
| myElem y x = removeAll x ys
| otherwise = y:removeAll x ys
If you do not want to use the built-in elem as well, you can define it similarly:
myElem :: Eq a => a -> [a] -> Bool
myElem _ = False
myElem x (y:ys)
| x == y = True
| otherwise = myElem x ys
answered Nov 23 '18 at 8:22
Yongwei WuYongwei Wu
2,1761832
A problem of your current code is that you tried to recurse in two lists simultaneously in one function. You need to divide and conquer. If elem is allowed, you may write:
removeAll :: Eq a => [a] -> [a] -> [a]
removeAll _ =
removeAll x (y:ys)
| myElem y x = removeAll x ys
| otherwise = y:removeAll x ys
If you do not want to use the built-in elem as well, you can define it similarly:
myElem :: Eq a => a -> [a] -> Bool
myElem _ = False
myElem x (y:ys)
| x == y = True
| otherwise = myElem x ys
answered Nov 23 '18 at 8:22
Yongwei WuYongwei Wu
2,1761832
answered Nov 23 '18 at 8:22
Yongwei WuYongwei Wu
2,1761832
answered Nov 23 '18 at 8:22
Yongwei WuYongwei Wu
2,1761832
answered Nov 23 '18 at 8:22
Yongwei WuYongwei Wu
2,1761832
2,1761832
add a comment |
add a comment |
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1
What is your question?
– talex
Nov 22 '18 at 5:43
3
You can always look at the source of the built-in functions, so you shouldn't count those out.
– Mateen Ulhaq
Nov 22 '18 at 5:44