Ufdjrw

Can I ask an author to send me his ebook?

Can a Wizard take the Magic Initiate feat and select spells from the Wizard list?

“Since the train was delayed for more than an hour, passengers were given a full refund.” – Why is there no article before “passengers”?

Short story about an alien named Ushtu(?) coming from a future Earth, when ours was destroyed by a nuclear explosion

Why not use the yoke to control yaw, as well as pitch and roll?

How do I deal with an erroneously large refund?

Reflections in a Square

How to leave only the following strings?

Why are two-digit numbers in Jonathan Swift's "Gulliver's Travels" (1726) written in "German style"?

Is there a way to convert Wolfram Language expression to string?

Meaning of "Not holding on that level of emuna/bitachon"

What is the evidence that custom checks in Northern Ireland are going to result in violence?

Why does my GNOME settings mention "Moto C Plus"?

Grounding PCB Within Aluminum Enclosure

Converting a text document with special format to Pandas DataFrame

Does Prince Arnaud cause someone holding the Princess to lose?

Can the van der Waals coefficients be negative in the van der Waals equation for real gases?

Who can become a wight?

Is my guitar’s action too high?

Why did Bronn offer to be Tyrion Lannister's champion in trial by combat?

When speaking, how do you change your mind mid-sentence?

How to ask rejected full-time candidates to apply to teach individual courses?

Assertions In A Mock Callout Test

Why isn't everyone flabbergasted about Bran's "gift"?

How to fix this haskell function using take and drop while?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;

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

9753619

• 
  
  
  

  
  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

9753619

• 
  
  
  

  
  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

9753619

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

9753619

share|improve this question

asked Mar 9 at 2:17

K Split XK Split X

9753619

asked Mar 9 at 2:17

K Split XK Split X

9753619

asked Mar 9 at 2:17

K Split XK Split X

9753619

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 | 

Your Answer

StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55073376%2fhow-to-fix-this-haskell-function-using-take-and-drop-while%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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