Find all occurences of a key in nested python dictionaries and lists - http://stackoverflow.com/questions/9807634/find-all-occurences-of-a-key-in-nested-python-dictionaries-and-lists

gistfile1.py

# 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))

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

x[1]['GroupId'] On Tuesday, 5 May, 2020, 09:00:28 pm IST, deepthimm <[email protected]> wrote: @deepthimm commented on this gist. [{'Description': 'test', 'GroupName': 'test', 'IpPermissions': [{'FromPort': 0, 'IpProtocol': 'tcp', 'IpRanges': [{'CidrIp': '0.0.0.0/0'}], 'Ipv6Ranges': [], 'PrefixListIds': [], 'ToPort': 0, 'UserIdGroupPairs': []}], 'OwnerId': '514788592734', 'GroupId': 'sg-09d0256a21b88c2f7', 'IpPermissionsEgress': [{'IpProtocol': '-1', 'IpRanges': [{'CidrIp': '0.0.0.0/0'}], 'Ipv6Ranges': [], 'PrefixListIds': [], 'UserIdGroupPairs': []}], 'Tags': [{'Key': 'Environment', 'Value': 'prd'}], 'VpcId': 'vpc-7db77d17'}, {'Description': 'unused', 'GroupName': 'unused', 'IpPermissions': [{'FromPort': 0, 'IpProtocol': 'tcp', 'IpRanges': [{'CidrIp': '0.0.0.0/0'}], 'Ipv6Ranges': [], 'PrefixListIds': [], 'ToPort': 0, 'UserIdGroupPairs': []}], 'OwnerId': '514788592734', 'GroupId': 'sg-0dc6f80055683ca6c', 'IpPermissionsEgress': [{'IpProtocol': '-1', 'IpRanges': [{'CidrIp': '0.0.0.0/0'}], 'Ipv6Ranges': [], 'PrefixListIds': [], 'UserIdGroupPairs': []}], 'Tags': [{'Key': 'Environment', 'Value': 'prd'}], 'VpcId': 'vpc-7db77d17'}] need to extract only GroupId from here. Can someone please help? — You are receiving this because you commented. Reply to this email directly, view it on GitHub, or unsubscribe.

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.