可选关键字参数的命名元组和默认值 [英] Named tuple and default values for optional keyword arguments

问题描述

我正在尝试将冗长的空心数据类转换为命名元组。我的班级目前看起来像这样：

I'm trying to convert a longish hollow "data" class into a named tuple. My class currently looks like this:

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：

After conversion to namedtuple it looks like:

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

但是这里有一个问题。我的原始类允许我只传入一个值，并通过对named / keyword参数使用默认值来处理默认值。像这样的东西：

But there is a problem here. My original class allowed me to pass in just a value and took care of the default by using default values for the named/keyword arguments. Something like:

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

但这在我重构的命名元组的情况下不起作用因为它希望我能通过所有领域。我当然可以将 Node（val）的出现替换为 Node（val，None，None）

But this doesn't work in the case of my refactored named tuple since it expects me to pass all the fields. I can of course replace the occurrences of Node(val) to Node(val, None, None) but it isn't to my liking.

那么是否存在一个不错的技巧，可以使我的重写成功而不增加很多代码复杂性（元编程），或者我应该吞下药丸并继续进行搜索并替换？ ：）

So does there exist a good trick which can make my re-write successful without adding a lot of code complexity (metaprogramming) or should I just swallow the pill and go ahead with the "search and replace"? :)

推荐答案

Python 3.7

使用默认值参数。

>>> from collections import namedtuple
>>> fields = ('val', 'left', 'right')
>>> Node = namedtuple('Node', fields, defaults=(None,) * len(fields))
>>> Node()
Node(val=None, left=None, right=None)

或者更好的方法是使用新的数据类库，它比namedtuple更好。

Or better yet, use the new dataclasses library, which is much nicer than namedtuple.

>>> from dataclasses import dataclass
>>> from typing import Any
>>> @dataclass
... class Node:
...     val: Any = None
...     left: 'Node' = None
...     right: 'Node' = None
>>> Node()
Node(val=None, left=None, right=None)

在Python 3.7之前

将 Node .__ new __.__ defaults __ 设置为默认值。

>>> from collections import namedtuple
>>> Node = namedtuple('Node', 'val left right')
>>> Node.__new__.__defaults__ = (None,) * len(Node._fields)
>>> Node()
Node(val=None, left=None, right=None)

在Python 2.6之前

将 Node .__ new __。func_defaults 设置为默认值。

>>> from collections import namedtuple
>>> Node = namedtuple('Node', 'val left right')
>>> Node.__new__.func_defaults = (None,) * len(Node._fields)
>>> Node()
Node(val=None, left=None, right=None)

顺序

在所有版本的Python中，如果设置的默认值少于namedtuple中的默认值，则默认值将应用于最右边的参数。

Order

In all versions of Python, if you set fewer default values than exist in the namedtuple, the defaults are applied to the rightmost parameters. This allows you to keep some arguments as required arguments.

>>> Node.__new__.__defaults__ = (1,2)
>>> Node()
Traceback (most recent call last):
  ...
TypeError: __new__() missing 1 required positional argument: 'val'
>>> Node(3)
Node(val=3, left=1, right=2)

适用于Python 2.6到3.6的包装器

这里是一个包装器，您甚至可以（可选）将默认值设置为 None以外的值。这不支持必需的参数。

Wrapper for Python 2.6 to 3.6

Here's a wrapper for you, which even lets you (optionally) set the default values to something other than None. This does not support required arguments.

import collections
def namedtuple_with_defaults(typename, field_names, default_values=()):
    T = collections.namedtuple(typename, field_names)
    T.__new__.__defaults__ = (None,) * len(T._fields)
    if isinstance(default_values, collections.Mapping):
        prototype = T(**default_values)
    else:
        prototype = T(*default_values)
    T.__new__.__defaults__ = tuple(prototype)
    return T

示例：

>>> Node = namedtuple_with_defaults('Node', 'val left right')
>>> Node()
Node(val=None, left=None, right=None)
>>> Node = namedtuple_with_defaults('Node', 'val left right', [1, 2, 3])
>>> Node()
Node(val=1, left=2, right=3)
>>> Node = namedtuple_with_defaults('Node', 'val left right', {'right':7})
>>> Node()
Node(val=None, left=None, right=7)
>>> Node(4)
Node(val=4, left=None, right=7)

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