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| # This is a really old post, in the comments (and stackoverflow too) you'll find better solutions. | |
| def find(key, dictionary): | |
| for k, v in dictionary.iteritems(): | |
| if k == key: | |
| yield v | |
| elif isinstance(v, dict): | |
| for result in find(key, v): | |
| yield result | |
| elif isinstance(v, list): | |
| for d in v: | |
| for result in find(key, d): | |
| yield result | |
| example = {'app_url': '', 'models': [{'perms': {'add': True, 'change': True, 'delete': True}, 'add_url': '/admin/cms/news/add/', 'admin_url': '/admin/cms/news/', 'name': ''}], 'has_module_perms': True, 'name': u'CMS'} | |
| list(find('admin_url', example)) |
for item in iterator: yield item can be replaced with yield from iterator.
Many thanks 👍 !!!
Thanks a Lot !!
thanks you so much
Nice! if any of your nested lists contains primitives (e.g, strings) and not always dicts, it will error out. In that case add a check for it and skip the element (added line 2 before last):
def find(key, dictionary):
for k, v in dictionary.iteritems():
if k == key:
yield v
elif isinstance(v, dict):
for result in find(key, v):
yield result
elif isinstance(v, list):
for d in v:
if isinstance(d, dict):
for result in find(key, d):
yield result
Python3 users should replace dictionary.iteritems() with dictionary.items(). iteritems was removed.
Thanks for sharing!
Awesome thanks
Nice work-just what I needed!
Beautiful, thank you!
Super sweet and clean. Thank you!
For Python 3.x users, just replace " for k, v in dictionary.iteritems():" with " for k, v in dictionary.items():"
Thanks for your nice example. I changed it a bit for my use case, where I needed to get all contained keys of a nested iterable as a list (also for the values) as can be seen in https://gist.github.com/PatrikHlobil/9d045e43fe44df2d5fd8b570f9fd78cc :
def iterate_all(iterable, returned="key"):
"""Returns an iterator that returns all keys or values
of a (nested) iterable.
Arguments:
- iterable: <list> or <dictionary>
- returned: <string> "key" or "value"
Returns:
- <iterator>
"""
if isinstance(iterable, dict):
for key, value in iterable.items():
if returned == "key":
yield key
elif returned == "value":
if not (isinstance(value, dict) or isinstance(value, list)):
yield value
else:
raise ValueError("'returned' keyword only accepts 'key' or 'value'.")
for ret in iterate_all_keys(value, returned=returned):
yield ret
elif isinstance(iterable, list):
for el in iterable:
for ret in iterate_all_keys(el, returned=returned):
yield retFor the example:
example = {'app_url': '', 'models': [{'perms': {'add': True, 'change': True, 'delete': True},
'add_url': '/admin/cms/news/add/', 'admin_url': '/admin/cms/news/', 'name': ''}],
'has_module_perms': True, 'name': u'CMS'}
It gives the following output:
list(iterate_all(example, "key"))
>>> ['app_url', 'models', 'perms', 'add', 'change', 'delete', 'add_url', 'admin_url', 'name', 'has_module_perms', 'name']
list(iterate_all(example, "value"))
>>>['', True, True, True, '/admin/cms/news/add/', '/admin/cms/news/', '', True, 'CMS']
Is there a way to respect the hierarchy? So for example, if i have:
[
{
"id": 800,
"children": [
{
"id": 801,
"children": [
{
"id": 804,
},
{
"id": 805,
}
},
]
I get: 800/801/804 , 800/801/805 (and every combination possible)
@PatrikHlobil You can use the module nested_lookup for fetching the keys(https://pypi.python.org/pypi/nested-lookup).
Example:
>>> example = {'app_url': '', 'models': [{'perms': {'add': True, 'change': True, 'delete': True},
... 'add_url': '/admin/cms/news/add/', 'admin_url': '/admin/cms/news/', 'name': ''}],
... 'has_module_perms': True, 'name': u'CMS'}
>>> from nested_lookup import get_all_keys
>>> get_all_keys(example)
['app_url', 'models', 'perms', 'add', 'change', 'delete', 'add_url', 'admin_url', 'name', 'has_module_perms', 'name']
>>>
Can you pl give me the explanation of the code.
Hey, can you give the code for deleting all the entries in dict matching the key we send.
Thanks...
This is broken, except for the limited example you give where you have a dict at the root of your data-structure and restrictions on lists.
If you just add another key to your example:
example['xyz'] = [[dict(admin_url="broken")]]
You don't get two results, as one would expect, you get an attributeerror.
That StackOverflow post has much better answers, you should not indiscriminately copy the one with the unrealistic assumptions. You should also indicate this is Python2 only because of your use of .iteritems()
Thank you very much, this saved my day.
Thanks! 🙏
i have this produced this raw data and now i only want latitude and longitude pairs, so can anybody help me...........
[11:40 PM, 1/12/2020] Vaibhav Kaushik: {"Response": {"MetaInfo": {"Timestamp": "2020-01-12T18:02:12.167+0000"}, "View": [{"Result": [{"Location": {"Address": {"AdditionalData": [{"key": "CountryName", "value": "India"}, {"key": "StateName", "value": "Uttar Pradesh"}, {"key": "CountyName", "value": "Gautam Buddha Nagar"}], "City": "Noida", "Country": "IND", "County": "Gautam Buddha Nagar", "District": "Sector 125", "Label": "Amity Road, Sector 125, Noida 201303, India", "PostalCode": "201303", "State": "UP", "Street": "Amity Road"}, "DisplayPosition": {"Latitude": 28.54712, "Longitude": 77.33482}, "LocationId": "NT_8RywhVafbhYYOKQDacPkWA", "LocationType": "point", "MapView": {"BottomRight": {"Latitude": 28.54405, "Longitude": 77.33731}, "TopLeft": {"Latitude": 28.5493, "Longitude": 77.33237}}, "NavigationPosition": [{"Latitude": 28.54712, "Longitude": 77.33482}]}, "MatchLevel": "street", "MatchQuality": {"City": 1.0, "Country": 1.0, "District": 1.0, "State": 1.0, "Street": [1.0]}, "Relevance": 1.0}], "ViewId": 0, "_type": "SearchResultsViewType"}]}}
Hey Vaibhav, you can use the following -
def find(key, dictionary):
for k, v in dictionary.items():
if k == key:
yield dictionary
elif isinstance(v, dict):
for result in find(key, v):
yield result
elif isinstance(v, list):
for d in v:
if isinstance(d, dict):
for result in find(key, d):
yield result
print(list(find("Latitude", data)))
I used it on your data and it is showing the following result, hope this helps you -
[{'Latitude': 28.54712, 'Longitude': 77.33482}, {'Latitude': 28.54405, 'Longitude': 77.33731},
{'Latitude': 28.5493, 'Longitude': 77.33237}, {'Latitude': 28.54712, 'Longitude': 77.33482}]
{'destination_addresses': ['Chennai, Tamil Nadu 600009, India'], 'origin_addresses': ['Chennai, Tamil Nadu 600101, India'], 'rows': [{'elements': [{'distance': {'text': '7.7 mi', 'value': 12327}, 'duration': {'text': '38 mins', 'value': 2285}, 'status': 'OK'}]}], 'status': 'OK'}
Hi Friends, from this JSON i need to extract the value '7.7' which is under the key 'text' under 'distance'. Can any one help in this
Hello all.. Could anyone help me Python parse and print < "name" , "raw" > (values only), from the JSON below?
{"count":57,"next":null,"previous":null,"results":[{"id":229,"state":"complete","substate":null,"exceptions":[],"name":"Sender Account Number","output_name":null,"field_definition_attributes":{"required":false,"data_type":"Account Number","multiline":false,"routing":false,"supervision_override":null},"transcription":{"raw":"1957-3549-2","normalized":"195735492","source":"machine_transcription","data_deleted":false,"user_transcribed":null},"field_image_url":"/api/v5/image/613cf762-b4bc-46f0-a511-3dc8bb37eae3?start_x=0.3110429607297866&start_y=0.1052441592299208&end_x=0.5696909842243418&end_y=0.16043316955780607","page_id":5}
So for the 'Item' included above, starting at "results": and nesting node to "transcription", the parsed values need to be:
Sender Account Number, 1957-3549-2
(repeating items) 'for'
Much thanks for any assistance!!
Thanks for the help.
Python3 users should replace dictionary.iteritems() with dictionary.items(). iteritems was removed.
This is exactly what I needed to deal with nested dict's from an API request; thank you!!