archery 1.2.2

mdct

dict，始终支持所有线性代数属性

基本上是：任意基向量的dict（递归）。

要了解其使用和实现的更多信息：

• Video presentation in FOSDEM 2017
• or look at the presentation

示例：

>>> from archery import mdict
>>> point = mdict(x=1, y=1, z=1)
>>> point2 = mdict(x=1, y=-1)
>>> print( (2 * point + point2)/4)
>>> # OUT : {'y': 0.25, 'x': 0.75, 'z': 0.5}
>>> print(point - point2)
>>> # OUT : {'y': 2, 'x': 0, 'z': 1}
>>> b=mdict(x=2, z=-1)
>>> a=mdict(x=1, y=2.0)
>>> a+b
>>> # OUT: {'y': 2.0, 'x': 3, 'z': -1}
>>> b-a
>>> # OUT: {'y': -2.0, 'x': 1, 'z': -1}
>>> -(a-b)
>>> # OUT: {'y': -2.0, 'x': 1, 'z': -1}
>>> a+1
>>> # OUT: {'y': 3.0, 'x': 2}
>>> -1-a
>>> # OUT: {'y': -3.0, 'x': -2}
>>> a*b
>>> # OUT: {'x': 2}
>>> a/b
>>> # OUT: {'x': 0}
>>> 1.0*a/b
>>> # OUT: {'x': 0.5}

vdict

在欧几里得意义中定义abs（），dot（），cos（）的dict

示例：

>>> from archery import vdict as Point
>>>
>>> u = Point(x=1, y=1)
>>> v = Point(x=1, y=0)
>>> u.cos(v)
>>> 0.7071067811865475
>>> u.dot(v)
>>> # OUT: 1
>>> u.cos(2*v)
>>> # OUT: 0.7071067811865475
>>> u.dot(2*v)
>>> #OUT: 2
>>> abs(u)
>>> #OUT: 1.4142135623730951
>>> u3 = Point(x=1, y=1, z=2)
>>> u4 = Point(x=1, y=3, z=4)
>>> u3 + u4
>>> #OUT: dict(x=2, y=4, z=6)
>>> assert u4 + u4 == 2*u4
>>> from archery import vdict
>>> from math import acos, pi
>>> point = vdict(x=1, y=1, z=1)
>>> point2 = vdict(x=1, y=-1)
>>> point2 = mdict(x=1, y=-1)
>>> print( (2 * point + point2)/4)
>>> # OUT : {'y': 0.25, 'x': 0.75, 'z': 0.5}
>>> print(acos(vdict(x=1,y=0).cos(vdict(x=1, y=1)))*360/2/pi)
>>> # OUT : 45.0
>>> print(abs(vdict(x=1, y=1)))
>>> # OUT : 1.41421356237
>>> print(vdict(x=1,y=0,z=3).dot(vdict(x=1, y=1, z=-1)))
>>> #OUT -2

SDICT

dict用于搜索值/键/path并具有特殊兴趣。

基本上，它以元组的形式返回一个迭代器，它包含所有键和值。 这是一个巧妙的技巧，如果您将它与make_from_path结合使用，它将帮助您在dict中准确地选择所需内容：

>>> from archery import sdict, make_from_path
>>> tree = sdict(
...      a = 1,
...      b = dict(
...          c = 3.0,
...          d = dict(e=True)
...      ),
...      point = dict( x=1, y=1, z=0),
... )
>>> list(tree.leaf_search(lambda x: type(x) is float ))
>>> #Out: [3.0]
>>> res = list(tree.search(lambda x: ("point") in x ))
>>> ## equivalent to list(tree.search(lambda x: Path(x).contains("point")))
>>> print(res)
>>> #Out: [('point', 'y', 1), ('point', 'x', 1), ('point', 'z', 0)]
>>> make_from_path(dict(), res)
>>> # {('point', 'y', 1): {('point', 'x', 1): ('point', 'z', 0)}}

资源

票务：https://github.com/jul/archery/issues?state=open 来源：https://github.com/jul/archery 最新文档：http://archery.readthedocs.org/en/latest/index.html