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(*) Find the number of elements of a list.
 
(*) Find the number of elements of a list.
   
  +
The simple, recursive solution.
  +
This is similar to the <hask>length</hask> from <hask>Prelude</hask>:
 
<haskell>
 
<haskell>
 
myLength :: [a] -> Int
 
myLength :: [a] -> Int
 
myLength [] = 0
 
myLength [] = 0
 
myLength (_:xs) = 1 + myLength xs
 
myLength (_:xs) = 1 + myLength xs
  +
</haskell>
  +
The prelude for haskell 2010 can be found [http://www.haskell.org/onlinereport/haskell2010/haskellch9.html#x16-1710009 here.]
   
myLength' :: [a] -> Int
+
Same, but now we use an "accumulator" argument.
myLength' list = myLength_acc list 0 -- same, with accumulator
+
<haskell>
  +
myLength :: [a] -> Int
  +
myLength list = myLength_acc list 0
 
where
 
where
 
myLength_acc [] n = n
 
myLength_acc [] n = n
Line 13: Line 17:
 
</haskell>
 
</haskell>
   
  +
Using foldl/foldr:
 
<haskell>
 
<haskell>
myLength' = foldl (\n _ -> n + 1) 0
+
myLength :: [a] -> Int
myLength'' = foldr (\_ n -> n + 1) 0
+
myLength1 = foldl (\n _ -> n + 1) 0
myLength''' = foldr (\_ -> (+1)) 0
+
myLength2 = foldr (\_ n -> n + 1) 0
myLength'''' = foldr ((+) . (const 1)) 0
+
myLength3 = foldr (\_ -> (+1)) 0
myLength''''' = foldr (const (+1)) 0
+
myLength4 = foldr ((+) . (const 1)) 0
myLength'''''' = foldl (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' xs = snd $ last $ zip xs [1..] -- Just for fun
+
myLength :: [a] -> Int
myLength'' = snd . last . (flip zip [1..]) -- Because point-free is also fun
+
myLength1 xs = snd $ last $ zip xs [1..] -- Just for fun
myLength''' = fst . last . zip [1..] -- same, but easier
+
myLength2 = snd . last . (flip zip [1..]) -- Because point-free 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>
 
This is <hask>length</hask> in <hask>Prelude</hask>.
 
 
-- length returns the length of a finite list as an Int.
 
length :: [a] -> Int
 
length [] = 0
 
length (_:l) = 1 + length l
 
 
The prelude for haskell 2010 can be found [http://www.haskell.org/onlinereport/haskell2010/haskellch9.html#x16-1710009 here.]
 
   
   

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 point-free 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)