# 99 questions/Solutions/4

### From HaskellWiki

< 99 questions | Solutions(Difference between revisions)

(Add description for each type of solution. Change formatting, ensure that all code is enclosed in "haskell" braces.) |
m (Use headlines.) |
||

Line 1: | Line 1: | ||

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

− | The simple, recursive solution | + | == The simple, recursive solution == |

This is similar to the <hask>length</hask> from <hask>Prelude</hask>: | This is similar to the <hask>length</hask> from <hask>Prelude</hask>: | ||

<haskell> | <haskell> | ||

Line 10: | Line 10: | ||

The prelude for haskell 2010 can be found [http://www.haskell.org/onlinereport/haskell2010/haskellch9.html#x16-1710009 here.] | The prelude for haskell 2010 can be found [http://www.haskell.org/onlinereport/haskell2010/haskellch9.html#x16-1710009 here.] | ||

− | Same, but | + | == Same, but using an "accumulator" == |

<haskell> | <haskell> | ||

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

Line 19: | Line 19: | ||

</haskell> | </haskell> | ||

− | Using foldl/foldr | + | == Using foldl/foldr == |

<haskell> | <haskell> | ||

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

Line 30: | Line 30: | ||

</haskell> | </haskell> | ||

+ | == Zipping with an infinite list == | ||

We can also create an infinite list starting from 1. | 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: | Then we "zip" the two lists together and take the last element (which is a pair) from the result: | ||

Line 39: | Line 40: | ||

</haskell> | </haskell> | ||

− | We can also change each element into our list into a | + | == Mapping all elements to "1" == |

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

<haskell> | <haskell> | ||

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

## 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)