Difference between revisions of "99 questions/Solutions/4"
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< 99 questions  Solutions
(Add description for each type of solution. Change formatting, ensure that all code is enclosed in "haskell" braces.) 

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(*) Find the number of elements of a list. 
(*) Find the number of elements of a list. 

+  The simple, recursive solution. 

⚫  
<haskell> 
<haskell> 

myLength :: [a] > Int 
myLength :: [a] > Int 

myLength [] = 0 
myLength [] = 0 

myLength (_:xs) = 1 + myLength xs 
myLength (_:xs) = 1 + myLength xs 

+  </haskell> 

⚫  
⚫  
+  Same, but now we use an "accumulator" argument. 

⚫  
+  <haskell> 

⚫  
⚫  
where 
where 

myLength_acc [] n = n 
myLength_acc [] n = n 

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</haskell> 
</haskell> 

+  Using foldl/foldr: 

<haskell> 
<haskell> 

−  myLength 
+  myLength :: [a] > Int 
−  +  myLength1 = foldl (\n _ > n + 1) 0 

−  +  myLength2 = foldr (\_ n > n + 1) 0 

−  +  myLength3 = foldr (\_ > (+1)) 0 

−  +  myLength4 = foldr ((+) . (const 1)) 0 

−  +  myLength5 = foldr (const (+1)) 0 

+  myLength6 = foldl (const . (+1)) 0 

</haskell> 
</haskell> 

+  We can also create an infinite list starting from 1. 

+  Then we "zip" the two lists together and take the last element (which is a pair) from the result: 

<haskell> 
<haskell> 

−  myLength 
+  myLength :: [a] > Int 
−  +  myLength1 xs = snd $ last $ zip xs [1..]  Just for fun 

−  +  myLength2 = snd . last . (flip zip [1..])  Because pointfree is also fun 

+  myLength3 = fst . last . zip [1..]  same, but easier 

</haskell> 
</haskell> 

+  We can also change each element into our list into a '1' and then add them all together. 

<haskell> 
<haskell> 

+  myLength :: [a] > Int 

myLength = sum . map (\_>1) 
myLength = sum . map (\_>1) 

</haskell> 
</haskell> 

−  
⚫  
−  
−   length returns the length of a finite list as an Int. 

−  length :: [a] > Int 

−  length [] = 0 

−  length (_:l) = 1 + length l 

−  
⚫  
Revision as of 13:04, 15 May 2014
(*) Find the number of elements of a list.
The simple, recursive solution.
This is similar to the length
from Prelude
:
myLength :: [a] > Int
myLength [] = 0
myLength (_:xs) = 1 + myLength xs
The prelude for haskell 2010 can be found here.
Same, but now we use an "accumulator" argument.
myLength :: [a] > Int
myLength list = myLength_acc list 0
where
myLength_acc [] n = n
myLength_acc (_:xs) n = myLength_acc xs (n + 1)
Using foldl/foldr:
myLength :: [a] > Int
myLength1 = foldl (\n _ > n + 1) 0
myLength2 = foldr (\_ n > n + 1) 0
myLength3 = foldr (\_ > (+1)) 0
myLength4 = foldr ((+) . (const 1)) 0
myLength5 = foldr (const (+1)) 0
myLength6 = foldl (const . (+1)) 0
We can also create an infinite list starting from 1. Then we "zip" the two lists together and take the last element (which is a pair) from the result:
myLength :: [a] > Int
myLength1 xs = snd $ last $ zip xs [1..]  Just for fun
myLength2 = snd . last . (flip zip [1..])  Because pointfree is also fun
myLength3 = fst . last . zip [1..]  same, but easier
We can also change each element into our list into a '1' and then add them all together.
myLength :: [a] > Int
myLength = sum . map (\_>1)