我试图将一个较长的中空“数据”类转换为命名元组。我的类目前看起来是这样的:

class Node(object):
    def __init__(self, val, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

转换为namedtuple后,它看起来像:

from collections import namedtuple
Node = namedtuple('Node', 'val left right')

但这里有一个问题。我最初的类允许我只传入一个值,并通过为named/keyword参数使用默认值来处理默认值。喜欢的东西:

class BinaryTree(object):
    def __init__(self, val):
        self.root = Node(val)

但这在重构的命名tuple中不起作用,因为它期望我传递所有字段。我当然可以替换Node(val)到Node(val, None, None)的出现,但这不是我喜欢的。

那么,是否存在一个好技巧,可以让我的重写成功,而不增加大量的代码复杂性(元编程),或者我应该吞下药丸,继续“搜索和替换”?:)


当前回答

jterrace给出的使用recordtype的答案很好,但是库的作者建议使用他的namedlist项目,该项目同时提供了可变(namedlist)和不可变(namedtuple)实现。

from namedlist import namedtuple
>>> Node = namedtuple('Node', ['val', ('left', None), ('right', None)])
>>> Node(3)
Node(val=3, left=None, right=None)
>>> Node(3, 'L')
Node(val=3, left=L, right=None)

其他回答

在打字。在Python 3.6.1+中,您可以为NamedTuple字段提供默认值和类型注释。使用打字。如果你只需要前者:

from typing import Any, NamedTuple


class Node(NamedTuple):
    val: Any
    left: 'Node' = None
    right: 'Node' = None

用法:

>>> Node(1)
Node(val=1, left=None, right=None)
>>> n = Node(1)
>>> Node(2, left=n)
Node(val=2, left=Node(val=1, left=None, right=None), right=None)

Also, in case you need both default values and optional mutability, Python 3.7 is going to have data classes (PEP 557) that can in some (many?) cases replace namedtuples. Sidenote: one quirk of the current specification of annotations (expressions after : for parameters and variables and after -> for functions) in Python is that they are evaluated at definition time*. So, since "class names become defined once the entire body of the class has been executed", the annotations for 'Node' in the class fields above must be strings to avoid NameError.

这种类型提示被称为“前向引用”([1],[2]),在PEP 563中,Python 3.7+将有__future__导入(在4.0中默认启用),允许使用不带引号的前向引用,推迟它们的计算。

* AFAICT只有局部变量注释不计算在运行时。(来源:PEP 526)

一个稍微扩展的例子,用None初始化所有缺失的参数:

from collections import namedtuple

class Node(namedtuple('Node', ['value', 'left', 'right'])):
    __slots__ = ()
    def __new__(cls, *args, **kwargs):
        # initialize missing kwargs with None
        all_kwargs = {key: kwargs.get(key) for key in cls._fields}
        return super(Node, cls).__new__(cls, *args, **all_kwargs)

将其包装在函数中。

NodeT = namedtuple('Node', 'val left right')

def Node(val, left=None, right=None):
  return NodeT(val, left, right)

短的,简单的,不会导致人们不恰当地使用isinstance:

class Node(namedtuple('Node', ('val', 'left', 'right'))):
    @classmethod
    def make(cls, val, left=None, right=None):
        return cls(val, left, right)

# Example
x = Node.make(3)
x._replace(right=Node.make(4))

我不确定是否有一个简单的方法,只有内置的namedtuple。有一个很好的模块叫做recordtype,它有这个功能:

>>> from recordtype import recordtype
>>> Node = recordtype('Node', [('val', None), ('left', None), ('right', None)])
>>> Node(3)
Node(val=3, left=None, right=None)
>>> Node(3, 'L')
Node(val=3, left=L, right=None)