Namedtuple类

Question

当我需要一个快速的一次性类时，我发现自己经常用python代码编写这个类。

class Struct(object):
   def __init__( self, **kwargs ):
      for k in kwargs:
         setattr(self,k,kwargs[k])

基本的想法是这样我可以快速地做如下事情：

foo = Struct( bar='one', baz=1 )
print foo.bar
foo.baz += 1
foo.novo = 42 # I don't do this as often.

当然，这不能很好地扩展，添加方法是疯狂的，但即使如此，我也有足够的数据，只丢弃类，我一直在使用它。

这就是我想的namedtuple会是什么样子。但是namedtuple的语法很大，很难处理。

在标准库中有什么东西我还没有找到，它能做得更好吗？

这种款式不好吗？或者它有什么隐藏的缺陷？

更新

两个具体的例子来说明为什么我不使用dict，这两个例子都可以用dict完成，但显然不是惯用的。

#I know an order preserving dict would be better but they don't exist in 2.6.
closure = Struct(count=0)
def mk_Foo( name, path ):
   closure.count += 1
   return (name, Foo( name, path, closure.count ))

d = dict([
   mk_Foo( 'a', 'abc' ),
   mk_Foo( 'b', 'def' ),
   # 20 or so more
   ] )


@contextmanager
def deleter( path ):
   control = Struct(delete=True,path=path)
   try:      
      yield control
   finally:
      if control.delete:
         shutil.rmtree(path)

with deleter( tempfile.mkdtemp() ) as tmp:
   # do stuff with tmp.path

   # most contexts don't modify the delete member
   # but occasionally it's needed
   if keep_tmp_dir:
      tmp.delete = False

Answer 1

从Python 3.3及以后的版本中，您可以使用types.SimpleNamespace：

>>> import types
>>> foo = types.SimpleNamespace(bar='one', baz=1)
>>> print(foo.bar)
one
>>> foo.baz += 1
>>> foo.novo = 42

内置类型大致相当于以下代码：

class SimpleNamespace:

    def __init__(self, **kwargs):
        self.__dict__.update(kwargs)

    def __repr__(self):
        keys = sorted(self.__dict__)
        items = ("{}={!r}".format(k, self.__dict__[k]) for k in keys)
        return "{}({})".format(type(self).__name__, ", ".join(items))

    def __eq__(self, other):
        return self.__dict__ == other.__dict__

更新

从Python3.7开始，您可以使用数据类模块：

from dataclasses import dataclass, field

@dataclass
class Struct:
    bar: str = field(default='one')
    baz: int = field(default=1)

您可以如下使用：

foo = Struct( bar='one', baz=1 )
print(foo.bar)
foo.baz += 1
foo.novo = 42

默认情况下，它包含相等测试和一个漂亮的报表：

>>> foo == Struct(bar='one', baz=2)
True
>>> foo
Struct(bar='one', baz=2)

Answer 2

这有一个python方法（它只是更新实例的dict而不是调用setattr） Recipe 52308

class Bunch(object):
    def __init__(self, **kwds):
        self.__dict__.update(kwds)

Answer 3

class t(dict):

   def __init__(self, **kwargs):
      for key, value in kwargs.items():
         dict.__setitem__(self, key, value)
   def __getattr__(self, key):
      return dict.__getitem__(self, key)
   def __setattr__(self, key, value):
      raise StandardError("Cannot set attributes of tuple")      
   def __setitem__(self, key, value):
      raise StandardError("Cannot set attributes of tuple")      
   def __delitem__(self, key):
      raise StandardError("Cannot delete attributes of tuple")

point = t(x=10, y=500, z=-50)
print point.x        # 10
print point.y        # 500
print point['z']     # -50
print point          # {'z': -50, 'y': 500, 'x': 10}
point.x = 100        # StandardError: cannot set attributes of tuple
point.y += 5         # StandardError: cannot set attributes of tuple
point.z = -1         # StandardError: cannot set attributes of tuple

def hypo(x, y, z):
   return (x**2 + y**2 + z**2)**0.5

print hypo(point)    # TypeError: unsupported operand type(s)
print hypo(**point)  # 502.593274925   

for k in point.items():
   print k           # ('y', 500)
                     # ('x', 10)
                     # ('z', -50)

for k in point.keys():
   print k           # x
                     # y
                     # z

for k in point.values():
   print k           # 500
                     # 10
                     # -50

print len(point)     # 3

print dict(point)    # {'y': 500, 'x': 10, 'z': -50}

这是我解决这个问题的办法。优美的语法，不可更改（至少不需要借助于某个讨厌的对象。setattr）体操，轻巧且可打印。尽管这件事你做不到，但你也不能用一个口述词

point = t(x=10, y=20, z=30)
d = point.x ** 2 + point.y ** 2 + point.z ** 2

与…有很好的对称性

point = (10, 20, 30)
d = point[0] ** 2 + point[1] ** 2 + point[2] ** 2

总的来说

point = {'x': 10, 'y': 20, 'z': 30}
d = point['x'] ** 2 + point['y'] ** 2 + point['z'] ** 2