Accessing, selecting and modifying elements in a tree

For convenience purpose, the result of the find method returns a namedtuple Element where

• Element[0] is the same as Element.path : the path to the value,
• Element[1] is the same as Element.value : the value

predicates on path

As path is of the Path class, you can use endswith, startswith and countains methods.

>>> from vector_dict.VectorDict import convert_tree, VectorDict
>>> a = convert_tree( { 'a' : { 'a' : { 'r' : "yop" , 'b' : { 'c' :  1 }, 'd' : True } } } )
>>> a.pprint()
a->a->r = 'yop'
a->a->b->c = 1
a->a->d = True
>>> a.tprint()
{
    a : {
        a : {
            r : 'yop',
            b : {
                c : 1,
            },
            d : True,
        },
    },
}
>>> print "\n".join([
       "%r => %r"  %( e.path, e.value ) for e in a.find(
           lambda k, v : k.endswith( 'a','r' ) )
   ])
 ('a', 'a', 'r') => 'yop'

If you ask a condition on path only, it will return all nodes verifying the condition

code continued from first example

>>> a.build_path( "a", "a", "g", "a", 2 )
>>> [ e.path for e in a.find( lambda k,v : k.contains( 'a' ) ) ]
[('a',), ('a', 'a'), ('a', 'a', 'r'), ('a', 'a', 'b'), ('a', 'a', 'b', 'c'), ('a', 'a', 'd'), ('a', 'a', 'g'), ('a', 'a', 'g', 'a')]

Limiting the matching elements to leaves

If you dont want any answers returned in the form of a tree add the is_leaf clause :

code continued from first example

>>> from vector_dict.Clause import is_leaf
>>> [ e.path for e in a.find( lambda k,v : k.contains( 'a' )  and is_leaf(v)) ]
[('a', 'a', 'r'), ('a', 'a', 'b', 'c'), ('a', 'a', 'd'), ('a', 'a', 'g', 'a')]
>>> [ e.value for e in a.find( lambda k,v : k.contains( 'a' )  and is_leaf(v)) ]
['yop', 1, True, 2]
>>> [ e for e in a.find( lambda k,v : k.contains( 'a' )  and is_leaf(v)) ]
[Element(path=('a', 'a', 'r'), value='yop'), Element(path=('a', 'a', 'b', 'c'), value=1), Element(path=('a', 'a', 'd'), value=True), Element(path=('a', 'a', 'g', 'a'), value=2)]
>>> a.tprint()
{
    a : {
        a : {
            r : 'yop',
            b : {
                c : 1,
            },
            d : True,
            g : {
                a : 2,
            },
        },
    },
}

Searching a specified location of a tree

You can also narrow your search on a subtree with the given path by combining with at() method. For instance searching in ‘a’ and ‘b’ subtree

>>> from vector_dict.VectorDict import convert_tree, VectorDict,Element,Path
>>> a = convert_tree(dict(a=dict(x=1, y=2, z=3), b=dict( x=-1, y=-1, z = -2)))
>>> [ e for e in a.at( 'b' ).find( lambda k, v : k.endswith('x')) ]
[Element(path=('x',), value=-1)]
>>> [ e for e in a.at( 'a' ).find( lambda k, v : k.endswith('x')) ]
[Element(path=('x',), value=1)]

Mainpulating values yielded by the find method

Also with at you can manipulate values :

code continued from the above example

>>> [ a.at(e.path , lambda v : v * -10 ) for e in a.find( lambda k, v : k.endswith('x')) ]
[-10, 10]
>>> a.tprint()
{
    a : {
        y : 2,
        x : -10,
        z : 3,
    },
    b : {
        y : -1,
        x : 10,
        z : -2,
    },
}

Warning

• Since it is pretty not a good idea to change a collection while it is being iterated any in situ search / replace at the same time is strongly discouraged;
• Always work with the path when manipulating