python - Haskell equivalent of data constructors in Python?

Question

In Haskell, I can define a binary tree as follows:

data Bint a = Leaf a | Branch a (Bint a) (Bint a)

then I can some operations on it as follows:

height (Leaf a) = 1
height (Branch a l r) = 1 + (max (height l) (height r))

count (Leaf a) = 1
count (Branch a l r) = 1 + (count l) + (count r)

I know Python doesn't has equivalent of data in Haskell. If it has, please tell.

So, how does one define a binary tree in Python and how to implement the above two functions in it?

score 4 · Accepted Answer

我将在这里寻找与 Haskell 非常相似的函数式编程。从某种意义上说，这不是很“pythonic”。特别是，它不是面向对象的。不过，它仍然有用且干净。

数据类型是一个类。具有多个数据构造函数的数据类型是一个包含有关其构造方式的额外信息的类。当然，它需要一些数据。使用构造函数确保所有树都是合法的：

class BinTree (object):
    def __init__(self, value=None, left=None, right=None):
        if left == None and right == None and value != None:                
            self.isLeaf = True
            self.value = value
        elif left != None and right != None and value == None:
            self.isLeaf = False
            self.value = (left, right)
        else:
            raise ArgumentError("some help message")

这个构造函数调用起来有点不方便，所以有一些易于使用的智能构造函数：

def leaf(value):
    return BinTree(value=value)

def branch(left, right):
    return BinTree(left=left, right=right)

我们如何把这些值拿出来？让我们也为此做一些助手：

def left(tree):
    if tree.isLeaf:
        raise ArgumentError ("tree is leaf")
    else:
        return tree.value[0]

def right(tree):
    if tree.isLeaf:
        raise ArgumentError ("tree is leaf")
    else:
        return tree.value[1]

def value(tree):
    if not tree.isLeaf:
        raise ArgumentError ("tree is branch")
    else:
        return tree.value

而已。你有一个纯“代数”数据类型，可以用函数访问：

def count(bin_tree):
    if bin_tree.isLeaf:
        return 1
    else:
        return count(left(bin_tree))+count(right(bin_tree))

score 3 · Accepted Answer

最接近的可能是带有方法的类：

class Leaf:
    def __init__(self, value):
        self.value = value

    def height(self):
        return 1

    def count(self):
        return 1


class Branch:
    def __init__(self, left, right):
        self.left = left
        self.right = right

    def height(self):
        return 1 + max(self.left.height(), self.right.height())

    def count(self):
        return 1 + self.left.count() + self.right.count()

虽然这有点惯用并且有其自身的优势，但它也缺乏 Haskell 版本的一些品质。最重要的是，函数定义必须与类一起在同一个模块中定义。您可以使用单调度泛型函数而不是方法来获取它。结果比方法和 Haskell 函数更开放，并且允许在有益的情况下将定义扩展到多个模块。

@singledispatch
def height(_):
    raise NotImplementedError()

@singledispatch
def count(_):
    raise NotImplementedError()

class Leaf:
    def __init__(self, value):
        self.value = value

@height.register(Leaf)
def height_leaf(leaf):
    return 1

@height.register(Leaf)
def count_leaf(leaf):
    return 1    

class Branch:
    def __init__(self, left, right):
        self.left = left
        self.right = right

@height.register(Branch)
def height_branch(b):
    return 1 + max(b.left.height(), b.right.height())

@count.register(Branch)
def count_branch(b):
    return 1 + b.left.count() + b.right.count()

score 1 · Accepted Answer

Python 没有 Haskell 的“数据构造函数”的概念。您可以像大多数其他 OOP 语言一样创建类。或者，您始终可以使用内置函数表示您的数据类型，并且只定义处理它们的函数（这是用于在heapq内置模块中实现堆的方法）。

python和haskell之间的差异很大，所以最好避免在haskell和python的语法/特性之间进行严格的比较，否则你最终会写出非pythonic和低效的python代码。

即使python确实有一些函数特性，它也不是函数式语言，所以你必须彻底改变程序的范式，才能获得可读、pythonic和高效的程序。

使用类的可能实现可能是：

class Bint(object):
    def __init__(self, value, left=None, right=None):
        self.a = value
        self.left = left
        self.right = right

    def height(self):
        left_height = self.left.height() if self.left else 0
        right_height = self.right.height() if self.right else 0
        return 1 + max(left_height, right_height)

    def count(self):
        left_count = self.left.count() if self.left else 0
        right_height = self.right.count() if self.right else 0
        return 1 + left_count + right_count

代码可以简化一点，为leftand提供一个“更智能”的默认值right：

class Nil(object):
    def height(self):
        return 0
    count = height

nil = Nil()

class Bint(object):
    def __init__(self, value, left=nil, right=nil):
        self.value = value
        self.left = left
        self.right = right
    def height(self):
        return 1 + max(self.left.height(), self.right.height())
    def count(self):
        return 1 + self.left.count() + self.right.count()

请注意，这些实现允许节点只有一个孩子。

但是，您不必使用类来定义数据类型。例如，您可以说 aBint可以是单个元素的列表、根的值或具有三个元素的列表：值、左孩子和右孩子。

在这种情况下，您可以将函数定义为：

def height(bint):
    if len(bint) == 1:
        return 1
    return 1 + max(height(bint[1]), height(bint[2]))

def count(bint):
    if len(bint) == 1:
    return 1 + count(bint[1]) + count(bint[2])

另一种方法是使用namedtuples：

from collections import namedtuple

Leaf = namedtuple('Leaf', 'value')
Branch = namedtuple('Branch', 'value left right')

def height(bint):
    if len(bint) == 1: # or isinstance(bint, Leaf)
        return 1
    return 1 + max(height(bint.left), height(bint.right))

def count(bint):
    if len(bint) == 1:  # or isinstance(bint, Leaf)
    return 1 + count(bint.left) + count(bint.right)

score 0 · Accepted Answer

自上次更新以来五年，但这是我的答案。

使data Tree a = Leaf a | Branch (Tree a) (Tree a) deriving (Show)python...

没有类型检查（更像 python）

class Tree:
    def __init__(self):
        pass

    def height(self):
        pass

    def count(self):
        pass

class Leaf(Tree):
    def __init__(self, value):
        self.value = value

    def __str__(self):
        return("Leaf " + str(self.value))

    def height(self):
        return 1

    def count(self):
        return 1

class Branch(Tree):
    def __init__(self, left, right):
        if isinstance(left, Tree) and isinstance(right, Tree):
            self.left = left
            self.right = right
        else:
            raise ValueError

    def __str__(self):
        return("Branch (" + str(self.left) + " " +
               str(self.right) + ")")

    def height(self):
        return 1 + max(self.left.height(), self.right.height())

    def count(self):
        return 1 + self.left.count() + self.right.count()


#usage : Branch(Leaf(5), Branch(Leaf(3), Leaf(2)))

使用类型检查（更像是 haskell）

`height`, `count`Tree 类的方法

class Tree:
    def __init__(self, tree, type_):
        def typecheck(subtree):
            if isinstance(subtree, Leaf):
                if not isinstance(subtree.value, type_):
                    print(subtree.value)
                    raise ValueError
            elif isinstance(subtree, Branch):
                typecheck(subtree.left)
                typecheck(subtree.right)
            else:
                raise ValueError
        typecheck(tree)
        self.tree = tree
        self.type_ = type_

    def __str__(self):
        return ("Tree " + self.type_.__name__ + "\n" + str(self.tree))

    def height(self):
        if isinstance(self, Leaf):
            return 1
        elif isinstance(self, Branch):
            return 1 + max(self.left.height(), self.right.height())
        else:
            return self.tree.height()

    def count(self):
        if isinstance(self, Leaf):
            return 1
        elif isinstance(self, Branch):
            return 1 + self.left.count() + self.right.count()
        else:
            return self.tree.count()


class Leaf(Tree):
    def __init__(self, value):
            self.value = value

    def __str__(self):
            return("Leaf " + str(self.value))


class Branch(Tree):
    def __init__(self, left, right):
        if isinstance(left, Tree) and isinstance(right, Tree):
            self.left = left
            self.right = right
        else:
            raise ValueError

    def __str__(self):
        return("Branch (" + str(self.left) + " " +
               str(self.right) + ")")


#usage tree1 = Tree(Branch(Leaf(5), Branch(Leaf(3), Leaf(2))), int)
#usage tree1.height() -> 3
#usage tree1.count() -> 5

`height`, `count`Leaf 和 Branch 类的方法

class Tree:
    def __init__(self, tree, type_):
        def typecheck(subtree):
            if isinstance(subtree, Leaf):
                if not isinstance(subtree.value, type_):
                    print(subtree.value)
                    raise ValueError
            elif isinstance(subtree, Branch):
                typecheck(subtree.left)
                typecheck(subtree.right)
            else:
                raise ValueError
        typecheck(tree)
        self.tree = tree
        self.type_ = type_

    def __str__(self):
        return ("Tree " + self.type_.__name__ + "\n" + str(self.tree))

    def height(self):
        return self.tree.height()

    def count(self):
        return self.tree.count()


class Leaf(Tree):
    def __init__(self, value):
            self.value = value

    def __str__(self):
            return("Leaf " + str(self.value))

    def height(self):
        return 1

    def count(self):
        return 1


class Branch(Tree):
    def __init__(self, left, right):
        if isinstance(left, Tree) and isinstance(right, Tree):
            self.left = left
            self.right = right
        else:
            raise ValueError

    def __str__(self):
        return("Branch (" + str(self.left) + " " +
               str(self.right) + ")")

    def height(self):
        return 1 + max(self.left.height(), self.right.height())

    def count(self):
        return 1 + self.left.count() + self.right.count()

#usage Tree(Branch(Leaf(5), Branch(Leaf(3), Leaf(2))), int)
#usage tree1.height() -> 3
#usage tree1.count() -> 5

`height`,`count`类外的方法（最像haskell）

class Tree:
    def __init__(self, tree, type_):
        def typecheck(subtree):
            if isinstance(subtree, Leaf):
                if not isinstance(subtree.value, type_):
                    print(subtree.value)
                    raise ValueError
            elif isinstance(subtree, Branch):
                typecheck(subtree.left)
                typecheck(subtree.right)
            else:
                raise ValueError
        typecheck(tree)
        self.tree = tree
        self.type_ = type_

    def __str__(self):
        return ("Tree " + self.type_.__name__ + "\n" + str(self.tree))


class Leaf(Tree):
    def __init__(self, value):
            self.value = value

    def __str__(self):
            return("Leaf " + str(self.value))


class Branch(Tree):
    def __init__(self, left, right):
        if isinstance(left, Tree) and isinstance(right, Tree):
            self.left = left
            self.right = right
        else:
            raise ValueError

    def __str__(self):
        return("Branch (" + str(self.left) + " " +
               str(self.right) + ")")

def height(tree):
    if not isinstance(tree, Tree):
        raise ValueError
    if isinstance(tree, Leaf):
        return 1
    elif isinstance(tree, Branch):
        return 1 + max(height(tree.left), height(tree.right))
    else:
        return height(tree.tree)

def count(tree):
    if not isinstance(tree, Tree):
        raise ValueError
    if isinstance(tree, Leaf):
        return 1
    elif isinstance(tree, Branch):
        return 1 + count(tree.left) + count(tree.right)
    else:
        return count(tree.tree)

#usage tree1 = Tree(Branch(Leaf(5), Branch(Leaf(3), Leaf(2))), int)
#usage height(tree1) -> 3
#usage count(tree1) -> 5

python - Haskell equivalent of data constructors in Python?

4 回答 4

没有类型检查（更像 python）

使用类型检查（更像是 haskell）

height, countTree 类的方法

height, countLeaf 和 Branch 类的方法

height,count类外的方法（最像haskell）

Related

Reference

`height`, `count`Tree 类的方法

`height`, `count`Leaf 和 Branch 类的方法

`height`,`count`类外的方法（最像haskell）