如何做到这一点 - python字典遍历和搜索 [英] How to do this - python dictionary traverse and search
问题描述
我有嵌套字典:
{'key0':{'attrs':{'entity':'p' ,'hash':'34nj3h43b4n3','id':'4130'},
u'key1':{'attrs':{'entity':'r',
'hash'
'id':'4130-1'},
u'key2':{'attrs':{'entity':'c',
'hash' $$$$$$$$$$$哈希':'34njasasasd3h43b4n3',
'id':'4130-2'},
u'key4':{'attrs':{'entity':'c',
'哈希':'34njawersd3h43bdsfsd4n3',
'id':'4130-2-1'}},
u'key5':{'attrs':{'entity':'c',
'hash':'34njawersd3h43bdsfsd4n3',
'id':'4130-2-2'}}}},
'someohterthing':'someothervalue',
'something':'somevalue'}
给定一个 id
- 所有
如 4130
至 4130-2-2
。
whats最简单的方法是导航到正确的字典?
喜欢给定的 id
是 4130-2-1
然后应该使用 key = key5
编辑(1):嵌套位于 1
到 4
级别,但我知道嵌套之前我解析。
编辑(2) strong>:修正代码。
**编辑(3):**再次固定代码 ids
的字符串值。请原谅造成混乱。这是最后我希望:)
你的结构是不愉快的不规则。这是一个带有访问者函数的版本,该函数遍历 attrs
子字典。
def walkDict(aDict,visitor,path =()):
在aDict中的k:
如果k =='attrs':
visitor ,aDict [k])
elif type(aDict [k])!= dict:
pass
else:
walkDict(aDict [k],visitor,path +(k, )
def printMe(path,element):
打印路径,元素
def filterFor(path,element):
if element [ id'] =='4130-2-2':
打印路径,元素
你会用这样的。
walkDict(myDict,filterFor)
/ pre>
这可以变成一个生成器,而不是一个访问者;它将
yield path,aDict [k]
而不是调用访问者函数。
你会使用它一个for循环。
for path,attrDict in walkDictIter(aDict):
#process attrDict ...
I have nested dictionaries:
{'key0': {'attrs': {'entity': 'p', 'hash': '34nj3h43b4n3', 'id': '4130'}, u'key1': {'attrs': {'entity': 'r', 'hash': '34njasd3h43b4n3', 'id': '4130-1'}, u'key2': {'attrs': {'entity': 'c', 'hash': '34njasd3h43bdsfsd4n3', 'id': '4130-1-1'}}}, u'key3': {'attrs': {'entity': 'r', 'hash': '34njasasasd3h43b4n3', 'id': '4130-2'}, u'key4': {'attrs': {'entity': 'c', 'hash': '34njawersd3h43bdsfsd4n3', 'id': '4130-2-1'}}, u'key5': {'attrs': {'entity': 'c', 'hash': '34njawersd3h43bdsfsd4n3', 'id': '4130-2-2'}}}}, 'someohterthing': 'someothervalue', 'something': 'somevalue'}
given an
id
- one of all theids
like4130
to4130-2-2
.
whats the easiest way to navigate to the correct dictionary?Like if the given
id
is4130-2-1
then it should reach the dictionary withkey=key5
non xml approaches please.
Edit(1): The nesting is between
1
to4
levels, but I know the nesting before I parse.Edit(2): Fixed the code.
**Edit(3):**Fixed code again for string values of
ids
. Please excuse for the confusion created. This is final I hope :)解决方案Your structure is unpleasantly irregular. Here's a version with a Visitor function that traverses the
attrs
sub-dictionaries.def walkDict( aDict, visitor, path=() ): for k in aDict: if k == 'attrs': visitor( path, aDict[k] ) elif type(aDict[k]) != dict: pass else: walkDict( aDict[k], visitor, path+(k,) ) def printMe( path, element ): print path, element def filterFor( path, element ): if element['id'] == '4130-2-2': print path, element
You'd use it like this.
walkDict( myDict, filterFor )
This can be turned into a generator instead of a Visitor; it would
yield path, aDict[k]
instead of invoking the visitor function.You'd use it in a for loop.
for path, attrDict in walkDictIter( aDict ): # process attrDict...
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