你如何迭代两个字典并在同一条路径上获取值？

Question

我有两个字典,它们具有相似的结构......意味着它们(应该)具有相同的键结构,即使在嵌套键中也是如此.此外,这些dicts几乎可以有任何类型的嵌套结构... list,dict等...我希望能够遍历这些字典,并获取这两个值并从函数返回它们.

简单示例:

dict_a = {'a':1, 'b':2, 'c':{'d':3}}
dict_b = {'a':2, 'b':4, 'c':{'d':6}}
#Note the structure is the same for these dicts
#I want to be able to do something like:
>>get_values( dict_a, dict_b)
[(1,2),(2,4),(3,6)]

我通过遍历一个字典,然后将每个键(或索引,如果它遇到列表)附加到列表中......作为一种键路径来提出解决方案:

key_map = []#A list of all key-paths for a dictionary
generate_key_paths(dict_a, [], key_map)
def generate_key_paths(value, key_list,key_map ):

    new_list = [item for item in key_list]
    if isinstance( value, dict):
        #Handle list
        for key, val in value.iteritems():
            new_list.append( key)
            self._generate_key_paths( val, new_list, key_map )
            new_list = [item for item in key_list]

    elif isinstance( value, list ):
        #Handle list
        for idx,item in enumerate(value):
            new_list.append( idx )  
            self._generate_key_paths( item, new_list, key_map )
            new_list = [item for item in key_list]
    else:
        #Handle data--reached farthest point you can go
        #So just append (key-path, value) to key_map
        key_map.append((new_list, value ) )

然后,一旦你有一个关键路径,值元组的列表...采取路径,并尝试在第二个字典上达到它以获得其价值...

val_list = []
for item in key_map:
    value = get_value( item[0] )
    if value is not None:
        val_list.append( (item[1], value ) )
def get_value( key_list ):
    value = dict_b
    for item in key_list:
        try:
            value = value[item]
        except:
            value = None
            break
    return value

这对于字典可能具有的所有结构都很有效,但它似乎很多工作.有更多的pythonic方式实现这一目标吗？有更快,更有效的方式吗？

编辑:我正在寻找一个不是列表或字典的值,所以当达到这些值时,它应该在它们内部迭代,直到找到一个值.保证如果它是一个列表,它将是一个dicts列表,所以应该总是存在某种键:值关系.

例如,可能的dict可能如下所示:

dict_a = {'a':1, 'b':2, 'c':[{'d':5},{'e':6}]}

dict_b = {'a':2, 'b':4, 'c':[{'d':10},{'e':12}]}

回答: [(1,2), (2,4), (5,10), (6,12)]

Answer 1

你正在寻找一个等价的flatten(zipTree(...))(函数不存在,但其名称应该得到我的观点).

from collections import Mapping

def treezipFlat(t1,t2):
    if isinstance(t1,Mapping) and isinstance(t2,Mapping):
        assert set(t1)==set(t2)
        for k,v1 in t1.items():
            v2 = t2[k]
            for tuple in treezipFlat(v1,v2):
                yield tuple
    else:
        yield (t1,t2)

演示:

>>> dict_a = {'a':1, 'b':2, 'c':{'d':3}}
>>> dict_b = {'a':2, 'b':4, 'c':{'d':6}}
>>> list( treezipFlat(dict_a, dict_b) )
[(1, 2), (3, 6), (2, 4)]

您还可以通过增加函数来产生路径元组,如下所示:

from collections import Mapping

def treezipItems(t1,t2, path=[]):
    if isinstance(t1,Mapping) and isinstance(t2,Mapping):
        assert set(t1)==set(t2)
        for k,v1 in t1.items():
            v2 = t2[k]
            for tuple in treezipItems(v1,v2, path=path+[k]):
                yield tuple
    else:
        yield (path, (t1,t2))

>>> list( treezipItems(dict_a, dict_b) )
[(['a'], (1, 2)), (['c', 'd'], (3, 6)), (['b'], (2, 4))]

Imho我觉得这里的自然是一个叫做的函数treezip:

def treezip(t1,t2):
    if isinstance(t1,Mapping) and isinstance(t2,Mapping):
        assert set(t1)==set(t2)
        R = {}
        for k,v1 in t1.items():
            v2 = t2[k]
            R[k] = treezip(v1,v2)
        return R
    else:
        return (t1,t2)

>>> from pprint import pprint as pp
>>> treezip(dict_a, dict_b)
{'a': (1, 2), 'c': {'d': (3, 6)}, 'b': (2, 4)}

然后调用一个函数flattenValues(或者flattenItems如果你想保存键).