我有一个看起来像这样的字典:
d = {'Food': {'Fruit' : {'Apples' : {'Golden Del.' : ['Yellow'],
'Granny Smith' : ['Green'],
'Fuji' : ['Red'],
},
'Cherries' : ['Red'],
'Bananas' : ['Yellow'],
'Grapes' : {'Red Grapes' : ['Red'],
'Green Grapes' : ['Green'],
},
},
'Dessert': {'Baked Ds' : {'Cakes' : {'Yellow Cake' : ['Yellow'],
'Red Velvet' : ['Red'],
},
'Cookies' : ['Yellow'],
},
},
'Steak' : ['Red'],
},
'Other': ['Blue'],
}
所以基本上是一个n-nested dict,其中每个值都是另一个dict或包含单个项目的列表。
假设我想通过单个列表项过滤此项,例如“红色”,结果将是:
d = {'Food': {'Fruit' : {'Apples' : {'Fuji' : ['Red'],
},
'Cherries' : ['Red'],
'Grapes' : {'Red Grapes' : ['Red'],
},
},
'Dessert': {'Baked Ds' : {'Cakes' : {'Red Velvet' : ['Red'],
},
},
},
'Steak' : ['Red'],
},
}
这样结构保持不变,但是除了列表项之外没有“红色”的所有内容都被删除,一直到层次结构。
有什么建议吗?我已经搞砸了一段时间并想出了这个,但它似乎不起作用:
def filterNestedDict(node, searchItem):
if isinstance(node,list):
return node
else:
for key, value in node.iteritems():
if isinstance(value,dict) and value is not {}:
return {key: filterNestedDict(value,searchItem)}
elif searchItem in value:
return {key: filterNestedDict(value,searchItem)}
return filterNestedDict(bigTree, searchItem)
我怀疑这只是一个递归问题,但任何建议都会非常感激。
谢谢!
答案 0 :(得分:4)
你非常接近,这应该适合你:
def filter_nested_dict(node, search_term):
if isinstance(node, list):
if node[0] == search_term:
return node
else:
return None
else:
dupe_node = {}
for key, val in node.iteritems():
cur_node = filter_nested_dict(val, search_term)
if cur_node:
dupe_node[key] = cur_node
return dupe_node or None
答案 1 :(得分:0)
from collections import Mapping
def yield_values_2(d):
for key in d.keys():
if isinstance(d[key], Mapping):
yield_values_2(d[key])
if not d[key]:
del d[key]
elif d[key] != ["Red"]:
del d[key]
d = {'Food': {'Fruit' : {'Apples' : {'Fuji' : ['Red'], 'Fuji2': ['Blue']
},
'Cherries' : ['Red'],
'Grapes' : {'Red Grapes' : ['Red'],
},
},
'Dessert': {'Baked Ds' : {'Cakes' : {'Red Velvet' : ['Red'],
},
},
},
'Steak' : ['Red'],
},
}
print d
yield_values_2(d)
print d