# 99 questions/Solutions/4

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< 99 questions | Solutions(Difference between revisions)

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(*) Find the number of elements of a list. |
(*) Find the number of elements of a list. |
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+ | == The simple, recursive solution == |
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+ | This is similar to the <hask>length</hask> from <hask>Prelude</hask>: |
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<haskell> |
<haskell> |
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myLength :: [a] -> Int |
myLength :: [a] -> Int |
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myLength [] = 0 |
myLength [] = 0 |
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myLength (_:xs) = 1 + myLength xs |
myLength (_:xs) = 1 + myLength xs |
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+ | </haskell> |
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+ | The prelude for haskell 2010 can be found [http://www.haskell.org/onlinereport/haskell2010/haskellch9.html#x16-1710009 here.] |
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− | myLength' :: [a] -> Int |
+ | == Same, but using an "accumulator" == |

− | myLength' list = myLength_acc list 0 -- same, with accumulator |
+ | <haskell> |

+ | myLength :: [a] -> Int |
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+ | myLength list = myLength_acc list 0 |
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where |
where |
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myLength_acc [] n = n |
myLength_acc [] n = n |
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</haskell> |
</haskell> |
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+ | == Using foldl/foldr == |
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<haskell> |
<haskell> |
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− | 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 |

+ | myLength5 = foldr (const (+1)) 0 |
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+ | myLength6 = foldl (const . (+1)) 0 |
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</haskell> |
</haskell> |
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+ | == Zipping with an infinite list == |
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+ | We can also create an infinite list starting from 1. |
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+ | Then we "zip" the two lists together and take the last element (which is a pair) from the result: |
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<haskell> |
<haskell> |
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− | 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 |
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</haskell> |
</haskell> |
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+ | == Mapping all elements to "1" == |
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+ | We can also change each element into our list into a "1" and then add them all together. |
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<haskell> |
<haskell> |
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+ | myLength :: [a] -> Int |
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myLength = sum . map (\_->1) |
myLength = sum . map (\_->1) |
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</haskell> |
</haskell> |
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− | This is <hask>length</hask> in <hask>Prelude</hask>. |
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− | A fancier one! :-) |
+ | [[Category:Programming exercise spoilers]] |

− | <haskell> |
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− | myLength = foldl (const . (+1)) 0 |
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− | </haskell> |

## Latest revision as of 13:21, 15 May 2014

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

## Contents |

## [edit] 1 The simple, recursive solution

This is similar to thelength

Prelude

myLength :: [a] -> Int myLength [] = 0 myLength (_:xs) = 1 + myLength xs

The prelude for haskell 2010 can be found here.

## [edit] 2 Same, but using an "accumulator"

myLength :: [a] -> Int myLength list = myLength_acc list 0 where myLength_acc [] n = n myLength_acc (_:xs) n = myLength_acc xs (n + 1)

## [edit] 3 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

## [edit] 4 Zipping with an infinite list

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

## [edit] 5 Mapping all elements to "1"

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)