Difference between revisions of "関数"

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[[Category:Glossary]] [[Category:Language]]
 
[[Category:Glossary]] [[Category:Language]]
Mathematically speaking, a [http://mathworld.wolfram.com/Function.html function] relates all values in a set <math>A</math> to values in a set <math>B</math>. The function <math>square\ x = x^2</math>, given that <math>x</math> is an integer, will map all elements of the set of integers into another set -- in this case the set of square integers. In Haskell functions can be specified as below in the examples, with an optional [[type specification]] that gives the compiler (and other programmers) a hint as to the use of the function.
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数学的に言えば、[http://mathworld.wolfram.com/Function.html 関数]というのは<math>A</math>という集合のすべての値を<math>B</math>という値の集合に関連付けるものです。関数<math>square\ x = x^2</math><math>x</math>が整数なら、整数の集合の中にある全ての要素を他の集合--この場合整数の2乗の集合に対応させます。Haskellの関数は以下の例で表されます。付属的に [[type specification]] がコンパイラ(や利用しようとするプログラマ)に関数の使い方のヒントを与えます。
  
 
==Examples==
 
==Examples==
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</haskell>
 
</haskell>
  
In other words, a function has ''input'' and ''output'', and it describes how to produce the output from its input. Functions can be ''applied'', which just means that you give an input value as argument to the function and can then expect to receive the corresponding output value.
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言い換えれば、関数というのは "入力" と "出力" を持っていて、入力からどのように出力を生成するかを記述したものといえます。関数は "適用する" ことができます。つまり単純に入力値を関数の引数として与えて、対応する出力値として受け取るということです。
  
Haskell functions are ''first class'' entities, which means that they
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Haskellの関数は "ファーストクラス" エンティティです。つまり
* can be given names
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* 名前が与えられる
* can be the value of some expression
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* ある式の値である
* can be members of a list
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* リストのメンバになれる
* can be elements of a tuple
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* タプルの要素になれる
* can be passed as parameters to a function
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* 関数の引数として与えられる
* can be returned from a function as a result
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* 関数の結果として返される
(quoted from Davie's ''Introduction to Functional Programming Systems using Haskell.'')
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(Davieの "Introduction to Functional Programming Systems using Haskell."より引用)
===<hask>map</hask> example===
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As an example of the power of first-class functions, consider the function ''map'':  
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===<hask>map</hask> の例===
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ファーストクラス関数の威力を見られる例として、"map"関数を考えてみましょう。
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''map'':  
 
<haskell>
 
<haskell>
 
map :: (a -> b) -> [a] -> [b]
 
map :: (a -> b) -> [a] -> [b]
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</haskell>
 
</haskell>
  
(Note this is a  [[Higher order function]].)
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(これは[[Higher order function]]であることを覚えていてください)
 
This function takes two arguments: a function ''f'' which maps ''a''s to ''b''s, and a list ''xs'' of ''a''s. It returns a list of ''b''s which are the results of applying ''f'' to every member of ''xs''. So <hask>map square [1,1,2,3,5,8]</hask> would yield the list <hask>[1,1,4,9,25,64]</hask>. When you realize that the list of ''b''s that ''map'' returns can itself be a list of functions, things start to get interesting.
 
This function takes two arguments: a function ''f'' which maps ''a''s to ''b''s, and a list ''xs'' of ''a''s. It returns a list of ''b''s which are the results of applying ''f'' to every member of ''xs''. So <hask>map square [1,1,2,3,5,8]</hask> would yield the list <hask>[1,1,4,9,25,64]</hask>. When you realize that the list of ''b''s that ''map'' returns can itself be a list of functions, things start to get interesting.
  

Revision as of 15:16, 5 December 2009

数学的に言えば、関数というのはAという集合のすべての値をBという値の集合に関連付けるものです。関数square\ x = x^2xが整数なら、整数の集合の中にある全ての要素を他の集合--この場合整数の2乗の集合に対応させます。Haskellの関数は以下の例で表されます。付属的に type specification がコンパイラ(や利用しようとするプログラマ)に関数の使い方のヒントを与えます。

Examples

square :: Int -> Int
square x = x * x

言い換えれば、関数というのは "入力" と "出力" を持っていて、入力からどのように出力を生成するかを記述したものといえます。関数は "適用する" ことができます。つまり単純に入力値を関数の引数として与えて、対応する出力値として受け取るということです。

Haskellの関数は "ファーストクラス" エンティティです。つまり

  • 名前が与えられる
  • ある式の値である
  • リストのメンバになれる
  • タプルの要素になれる
  • 関数の引数として与えられる
  • 関数の結果として返される

(Davieの "Introduction to Functional Programming Systems using Haskell."より引用)

map の例

ファーストクラス関数の威力を見られる例として、"map"関数を考えてみましょう。 map:

map :: (a -> b) -> [a] -> [b]
map f xs = [f x | x <- xs]

(これはHigher order functionであることを覚えていてください) This function takes two arguments: a function f which maps as to bs, and a list xs of as. It returns a list of bs which are the results of applying f to every member of xs. So map square [1,1,2,3,5,8] would yield the list [1,1,4,9,25,64]. When you realize that the list of bs that map returns can itself be a list of functions, things start to get interesting.

Suppose you have some data structure (e.g. Set) that has the function insert :: Int -> Set -> Set, which takes an integer and a set, and returns the set created by inserting the given integer into the given set. And suppose you have mySet and myList, a set and a list of values to be added to the set, respectively. One could write a function to recurse over the list of integers, each time inserting a single member of myList, but with first-class functions this is not necessary. Look at the expression map insert myList -- what is the type of the list which it produces? Since insert takes an Int and a Set, but only Ints were given, the resulting list will be of functions that take a set and return a set. Conceptually, the code map insert [1,2,3] will return the list [(insert 1) (insert 2) (insert 3)].

Composition / folding example

Haskell supports a Function composition operator:

(.) :: (b -> c) -> (a ->b) -> (a->c)
(f . g) x = f (g x)

So, for example, ((insert 1) . (insert 2) . (insert 3)) mySet is the same as insert 1 (insert 2 (insert 3 mySet)). We're almost there -- what we need now is a function that can automatically put the composition operator between every element of map insert myList. Such code is included in Haskell, and it's known as folding.

Several variants of the fold function are defined, but the basic concept is the same: given a function and a list, "collapse" the list by applying the function "between" its elements. This is easiest to see with simple binary operators, but it is not limited to them. For example, foldr1 (+) [1,1,2,3,5] eventually creates the expression 1+1+2+3+5, and thus returns 12. In the set example, foldr1 (.) (map insert myList) gives us what we want, the successive insertion of every element of myList. What is the type of this expression? It is Set -> Set, meaning it will take a set and return a set -- in this case, the set it returns will have every element of myList inserted. To complete the example,

newSet = (foldr1 (.) (map insert myList)) mySet

will define newSet as mySet with the elements of myList inserted.

See also

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