我有一个代表文件夹树的字典：
folders = [{
"NAME": " Folder 1",
"ID": "869276"
}, {
"ID": "869277",
"NAME": "- Sub-folder 1"
}, {
"ID": "869279",
"NAME": "-- Sub-sub-folder 1"
}, {
"NAME": "--- Sub-sub-folder 1 2",
"ID": "869285"
}, {
"NAME": "--- Sub-sub-folder 1 3",
"ID": "869286"
}, {
"NAME": "-- Sub-sub-folder 2",
"ID": "869280"
}, {
"ID": "869281",
"NAME": " Folder 2"
}, {
"ID": "869282",
"NAME": "- Sub-folder 2"
}, {
"NAME": "- Sub-folder 2 1",
"ID": "869283"
}, {
"NAME": "-- Sub-Sub-folder 2 1",
"ID": "869284"
}]
更明确的表示：
Folder 1
- Sub-folder 1
-- Sub-sub-folder 1
--- Sub-sub-folder 1 2
--- Sub-sub-folder 1 3
-- Sub-sub-folder 2
Folder 2
- Sub-folder 2
- Sub-folder 2 1
-- Sub-Sub-folder 2 1
我需要将这个字典组织成一个新的字典，其中每个文件夹都有父文件夹的值，比如
[{
"NAME": " Folder 1",
"ID": "869276",
"PARENT": "0"
}, {
"ID": "869277",
"NAME": "- Sub-folder 1",
"PARENT": "869276"
},
...
]
所以我想的是计算＆＃39; - ＆＃39;在文件夹名称之前跟踪文件夹深度：
for folder in folders:
# Folders in root have a whitespace before the name
depth = folder['NAME'].split(' ')[0].count('-')
if depth == 0:
parent = '0'
else:
#for each previous_folder:
previous_depth = previous_folder['NAME'].split(' ')[0].count('-')
if previous_depth < depth:
parent = prvious_folder['ID']
else:
#keep looking...
问题是用实际工作代码填充注释行。如何从当前文件夹开始与列表中的每个先前文件夹进行交互？我如何继续循环？
答案 0 :(得分：1)
我认为诀窍是跟踪当前父母，祖父母等的列表中的祖先。您可以将它们从列表中删除以返回到基因库中。我有一些你可以删除的调试打印但它帮助我看看算法是如何进展的。我创建了一个名为“”的虚拟根来处理顶级文件夹。您可以将其重命名为任何内容，如果您不希望它显示，也可以重命名。
+--------------------+--------------+--------------------+-------------+--------------+---------------------+-------------------+-----------------------+-------------+--------------------+
| created_at| screen_name| text|retweet_count|favorite_count|in_reply_to_status_id|in_reply_to_user_id|in_reply_to_screen_name|user_mentions| hashtags|
+--------------------+--------------+--------------------+-------------+--------------+---------------------+-------------------+-----------------------+-------------+--------------------+
|2017-03-13 23:00:...| danielmellen|#DevOps understan...| 0| 0| null| null| null| []| [devops]|
|2017-03-13 23:00:...| RebacaInc|Automation of ent...| 0| 0| null| null| null| []|[googlecloud, orc...|
|2017-03-13 23:00:...| CMMIAppraiser|Get your Professi...| 0| 0| null| null| null| []| [broadsword]|
|2017-03-13 23:00:...| usxtron|and when the syst...| 0| 0| null| null| null| []| [cloud]|
|2017-03-13 23:00:...| SearchCRM|.#Automation and ...| 0| 0| null| null| null| []|[automation, chat...|
|2017-03-13 23:00:...| careers_tech|SummitSync - Juni...| 0| 0| null| null| null| []|[junior, cloud, e...|
|2017-03-13 23:00:...| roy_lauzon|Both the #DevOps ...| 0| 0| null| null| null| []|[devops, cybersec...|
|2017-03-13 23:00:...| nosqlgal|Introducing #Couc...| 0| 0| null| null| null| []| [couchbase, nosql]|
|2017-03-13 23:00:...| jordanfarrer|Ran into a weird ...| 0| 0| null| null| null| []| [docker]|
|2017-03-13 23:00:...| BGrieveSTL|#purestorage + #a...| 0| 0| null| null| null| []|[purestorage, azure]|
|2017-03-13 23:00:...| Hotelbeds_API|"How to Quickly O...| 0| 0| null| null| null| []| [api, feedly]|
|2017-03-13 23:00:...| ScalaWilliam|Principles behind...| 0| 0| null| null| null| []| [agile]|
|2017-03-13 23:00:...| PRFT_Oracle|[On-Demand Webina...| 0| 0| null| null| null| []| [cloud]|
|2017-03-13 23:00:...| PDF_filler|Now you can #secu...| 0| 0| null| null| null| []|[secure, data, ap...|
|2017-03-13 23:00:...|lgoncalves1979|10 Mistakes We Ma...| 0| 0| null| null| null| []|[coaching, scrumm...|
|2017-03-13 23:00:...| Jelecos|Vanguard CIO: Why...| 0| 0| null| null| null| []|[microservices, cio]|
|2017-03-13 23:00:...| DJGaryBaldy|Why bother with W...| 0| 0| null| null| null| []| [automation]|
|2017-03-13 23:00:...| 1codeblog|Apigee Edge Produ...| 0| 0| null| null| null| []|[cloud, next17, g...|
|2017-03-13 23:00:...| CloudRank|Why and when shou...| 0| 0| null| null| null| []|[machinelearning,...|
|2017-03-13 23:00:...| forgeaheadio|5 essentials for ...| 0| 0| null| null| null| []|[hybrid, cloud, h...|
+--------------------+--------------+--------------------+-------------+--------------+---------------------+-------------------+-----------------------+-------------+--------------------+
only showing top 20 rows
输出：
folders = [{
"NAME": " Folder 1",
"ID": "869276"
}, {
"ID": "869277",
"NAME": "- Sub-folder 1"
}, {
"ID": "869279",
"NAME": "-- Sub-sub-folder 1"
}, {
"NAME": "--- Sub-sub-folder 1 2",
"ID": "869285"
}, {
"NAME": "--- Sub-sub-folder 1 3",
"ID": "869286"
}, {
"NAME": "-- Sub-sub-folder 2",
"ID": "869280"
}, {
"ID": "869281",
"NAME": " Folder 2"
}, {
"ID": "869282",
"NAME": "- Sub-folder 2"
}, {
"NAME": "- Sub-folder 2 1",
"ID": "869283"
}, {
"NAME": "-- Sub-Sub-folder 2 1",
"ID": "869284"
}]
# id to folder index (with virtual root) for printing
folders_by_id = {folder['ID']:folder for folder in folders}
folders_by_id['<root>'] = {'NAME':'<root>', 'ID':-1}
# current ancestors stack
parents = ['<root>']
for folder in folders:
depth = folder['NAME'].split(' ')[0].count('-') + 1 # w/ virtual root
print('state', 'parents', [folders_by_id[_id] for _id in parents], 'name', folder['NAME'], 'depth', depth)
while depth < len(parents):
old = parents.pop()
print('removing', old)
folder['PARENT'] = parents[-1]
parents.append(folder['ID'])
print()
print('++++++++++++++++++++++++++++++ showing parents +++++++++++++++++++++++++++++++')
for folder in folders:
parent = folders_by_id[folder['PARENT']]
print('{padding}{parent} ({p_id}) --> {child} ({c_id})'.format(
padding=' ' * parent['NAME'].count('-'), parent=parent['NAME'],
p_id= parent['ID'], child=folder['NAME'], c_id=folder['ID']))
答案 1 :(得分：0)
如果您将每个深度级别的父ID存储在一个dict中，只需几行代码就可以完成：
new_a = list()
names_list = list()
for el in a:
if el['name'] not in names_list:
new_a.append({'name':el['name'],'id':el['id']})
names_list.append(el['name'])
else:
for new_el in new_a:
if el['name'] == new_el['name']:
new_el['id'] += el['id']