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1

I am asked to create a haskell function that will perform the following:

group [1,1,2,3,3] = [[1,1],[2],[3,3]]

group [] = []

Essentially it groups the consecutive terms that are the same

I should use take and drop while:

group [] = []
group (x:xs) = (takeWhile cond xs ++ [x]) : group (dropWhile cond xs)
 where
 cond = (x -> x == head xs)

Here is what I have.

The above works for cases where a single element is at the end: group [1,1,2] == [[1,1],2] but not for cases where it's in the middle: group [1,1,1,2,4] == [[1,1,1],[4,2]] (it should be [[1,1,1],[4],[2]]

What is my mistake in the function?

haskell

share|improve this question

asked Mar 9 at 2:17

K Split XK Split X

9603619

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Two different variables are named x.
  
  – Jason Orendorff
  Mar 9 at 3:38
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  group [1,1,2] == [[1,1],2] I know that can't be true without even looking at the source of group, because the RHS is ill-typed.
  
  – Joseph Sible
  Mar 9 at 4:27
  

  
  
  
  

add a comment | 

1

I am asked to create a haskell function that will perform the following:

group [1,1,2,3,3] = [[1,1],[2],[3,3]]

group [] = []

Essentially it groups the consecutive terms that are the same

I should use take and drop while:

group [] = []
group (x:xs) = (takeWhile cond xs ++ [x]) : group (dropWhile cond xs)
 where
 cond = (x -> x == head xs)

Here is what I have.

The above works for cases where a single element is at the end: group [1,1,2] == [[1,1],2] but not for cases where it's in the middle: group [1,1,1,2,4] == [[1,1,1],[4,2]] (it should be [[1,1,1],[4],[2]]

What is my mistake in the function?

haskell

share|improve this question

asked Mar 9 at 2:17

K Split XK Split X

9603619

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Two different variables are named x.
  
  – Jason Orendorff
  Mar 9 at 3:38
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  group [1,1,2] == [[1,1],2] I know that can't be true without even looking at the source of group, because the RHS is ill-typed.
  
  – Joseph Sible
  Mar 9 at 4:27
  

  
  
  
  

add a comment | 

I am asked to create a haskell function that will perform the following:

group [1,1,2,3,3] = [[1,1],[2],[3,3]]

group [] = []

Essentially it groups the consecutive terms that are the same

I should use take and drop while:

group [] = []
group (x:xs) = (takeWhile cond xs ++ [x]) : group (dropWhile cond xs)
 where
 cond = (x -> x == head xs)

Here is what I have.

The above works for cases where a single element is at the end: group [1,1,2] == [[1,1],2] but not for cases where it's in the middle: group [1,1,1,2,4] == [[1,1,1],[4,2]] (it should be [[1,1,1],[4],[2]]

What is my mistake in the function?

haskell

share|improve this question

asked Mar 9 at 2:17

K Split XK Split X

9603619

group [1,1,2,3,3] = [[1,1],[2],[3,3]]

group [] = []

group [] = []
group (x:xs) = (takeWhile cond xs ++ [x]) : group (dropWhile cond xs)
 where
 cond = (x -> x == head xs)

share|improve this question

asked Mar 9 at 2:17

K Split XK Split X

9603619

share|improve this question

asked Mar 9 at 2:17

K Split XK Split X

9603619

asked Mar 9 at 2:17

K Split XK Split X

9603619

asked Mar 9 at 2:17

K Split XK Split X

9603619

1 Answer
1

active

oldest

votes

5

Short answer: Your mistake is the usage of head in your condition. You want to compare with x, not head xs.

Stylistically, note also that

1. 
   something ++ [x] is less practical (and "efficient", although it doesn't really matter here) as x : something here, as order doesn't matter,


2. a function as f = x -> something should really be written as f x = something,


3. and you probably shouldn't use the same variable name to refer to different values, it's get confusing very easily.

Your function cond is easily rewritten using a partially applied ==:

group [] = []
group (x:xs) = (x : takeWhile cond xs) : group (dropWhile cond xs)
 where cond = (==) x

This yields the expected result for group [1,1,1,2,4].

share|improve this answer

edited Mar 9 at 3:45

answered Mar 9 at 3:31

tolUenetolUene

323114

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Did you mean to say "less practical" instead of "more practical"?
  
  – Fyodor Soikin
  Mar 9 at 3:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @FyodorSoikin Yes, thank you. Edited my post.
  
  – tolUene
  Mar 9 at 3:46
  

  
  
  
  

add a comment | 

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5

Short answer: Your mistake is the usage of head in your condition. You want to compare with x, not head xs.

Stylistically, note also that

1. 
   something ++ [x] is less practical (and "efficient", although it doesn't really matter here) as x : something here, as order doesn't matter,


2. a function as f = x -> something should really be written as f x = something,


3. and you probably shouldn't use the same variable name to refer to different values, it's get confusing very easily.

Your function cond is easily rewritten using a partially applied ==:

group [] = []
group (x:xs) = (x : takeWhile cond xs) : group (dropWhile cond xs)
 where cond = (==) x

This yields the expected result for group [1,1,1,2,4].

share|improve this answer

edited Mar 9 at 3:45

answered Mar 9 at 3:31

tolUenetolUene

323114

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Did you mean to say "less practical" instead of "more practical"?
  
  – Fyodor Soikin
  Mar 9 at 3:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @FyodorSoikin Yes, thank you. Edited my post.
  
  – tolUene
  Mar 9 at 3:46
  

  
  
  
  

add a comment | 

5

Short answer: Your mistake is the usage of head in your condition. You want to compare with x, not head xs.

Stylistically, note also that

1. 
   something ++ [x] is less practical (and "efficient", although it doesn't really matter here) as x : something here, as order doesn't matter,


2. a function as f = x -> something should really be written as f x = something,


3. and you probably shouldn't use the same variable name to refer to different values, it's get confusing very easily.

Your function cond is easily rewritten using a partially applied ==:

group [] = []
group (x:xs) = (x : takeWhile cond xs) : group (dropWhile cond xs)
 where cond = (==) x

This yields the expected result for group [1,1,1,2,4].

share|improve this answer

edited Mar 9 at 3:45

answered Mar 9 at 3:31

tolUenetolUene

323114

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Did you mean to say "less practical" instead of "more practical"?
  
  – Fyodor Soikin
  Mar 9 at 3:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @FyodorSoikin Yes, thank you. Edited my post.
  
  – tolUene
  Mar 9 at 3:46
  

  
  
  
  

add a comment | 

Short answer: Your mistake is the usage of head in your condition. You want to compare with x, not head xs.

Stylistically, note also that

1. 
   something ++ [x] is less practical (and "efficient", although it doesn't really matter here) as x : something here, as order doesn't matter,


2. a function as f = x -> something should really be written as f x = something,


3. and you probably shouldn't use the same variable name to refer to different values, it's get confusing very easily.

Your function cond is easily rewritten using a partially applied ==:

group [] = []
group (x:xs) = (x : takeWhile cond xs) : group (dropWhile cond xs)
 where cond = (==) x

This yields the expected result for group [1,1,1,2,4].

share|improve this answer

edited Mar 9 at 3:45

answered Mar 9 at 3:31

tolUenetolUene

323114

group [] = []
group (x:xs) = (x : takeWhile cond xs) : group (dropWhile cond xs)
 where cond = (==) x

share|improve this answer

edited Mar 9 at 3:45

answered Mar 9 at 3:31

tolUenetolUene

323114

answered Mar 9 at 3:31

tolUenetolUene

323114

answered Mar 9 at 3:31

tolUenetolUene

323114