# 99 questions/Solutions/4

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< 99 questions | Solutions(Difference between revisions)

(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. | ||

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

− | myLength | + | <haskell> |

+ | myLength :: [a] -> Int | ||

+ | myLength list = myLength_acc list 0 | ||

where | where | ||

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

Line 13: | Line 19: | ||

</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 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> | ||

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[[Category:Programming exercise spoilers]] | [[Category:Programming exercise spoilers]] |

## Revision as of 13:04, 15 May 2014

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

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.

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)