There is what i have in source:
"VirtualizationType": "hvm",
"Tags": [
{
"Value": "Test",
"Key": "Name"
}
]
and here is a bit of my code:
for item in result:
temp = {"Tags": item["Value"], "VirtualizationType": item["VirtualizationType"]}
output.append(temp)
I can easily export VirtualizationType but couldn`t do it with Tags
An example of how to create a dictionary in Python, from docs:
tel = {'jack': 4098, 'sape': 4139}
In this example jack, snape are the keys and 4098, 4139 are the values.
Values can also be strings and a list of dictionary like in your case,
your code can modified into this:
dictionary = {"Virtualization Type" : "hvm", "Tags" : [{"Value" : "Test", "Key" : "Name"}] } #Runs in the python interpreter with no errors
Related
I have the below python dictionary stored as dictPython
{
"paging": {"count": 10, "start": 0, "links": []},
"elements": [
{
"organizationalTarget~": {
"vanityName": "vv",
"localizedName": "ViV",
"name": {
"localized": {"en_US": "ViV"},
"preferredLocale": {"country": "US", "language": "en"},
},
"primaryOrganizationType": "NONE",
"locations": [],
"id": 109,
},
"role": "ADMINISTRATOR",
},
],
}
I need to get the values of vanityName, localizedName and also the values from name->localized and name->preferredLocale.
I tried dictPython.keys() and it returned dict_keys(['paging', 'elements']).
Also I tried dictPython.values() and it returned me what is inside of the parenthesis({}).
I need to get [vv, ViV, ViV, US, en]
I am writing this in a form of answer, so I can get to explain it better without the comments characters limit
a dict in python is an efficient key/value structure or data type
for example dict_ = {'key1': 'val1', 'key2': 'val2'} to fetch key1 we can do it in 2 different ways
dict_.get(key1) this returns the value of the key in this case val1, this method has its advantage, that if the key1 is wrong or not found it returns None so no exceptions are raised. You can do dict_.get(key1, 'returning this string if the key is not found')
dict_['key1'] doing the same .get(...) but will raise a KeyError if the key is not found
So to answer your question after this introduction,
a dict can be thought of as nested dictionaries and/or objects inside of one another
to get your values you can do the following
# Fetch base dictionary to make code more readable
base_dict = dict_["elements"][0]["organizationalTarget~"]
# fetch name_dict following the same approach as above code
name_dict = base_dict["name"]
localized_dict = name_dict["localized"]
preferred_locale_dict = name_dict ["preferredLocale"]
so now we fetch all of the wanted data in their corresponding locations from your given dictionary, now to print the results, we can do the following
results_arr = []
for key1, key2 in zip(localized_dict, preferredLocale_dict):
results_arr.append(localized_dict.get(key1))
results_arr.append(preferred_locale_dict.get(key2))
print(results_arr)
What about:
dic = {
"paging": {"count": 10, "start": 0, "links": []},
"elements": [
{
"organizationalTarget~": {
"vanityName": "vv",
"localizedName": "ViV",
"name": {
"localized": {"en_US": "ViV"},
"preferredLocale": {"country": "US", "language": "en"},
},
"primaryOrganizationType": "NONE",
"locations": [],
"id": 109,
},
"role": "ADMINISTRATOR",
},
],
}
base = dic["elements"][0]["organizationalTarget~"]
c = base["name"]["localized"]
d = base["name"]["preferredLocale"]
output = [base["vanityName"], base["localizedName"]]
output.extend([c[key] for key in c])
output.extend([d[key] for key in d])
print(output)
outputs:
['vv', 'ViV', 'ViV', 'US', 'en']
So something like this?
[[x['organizationalTarget~']['vanityName'],
x['organizationalTarget~']['localizedName'],
x['organizationalTarget~']['name']['localized']['en_US'],
x['organizationalTarget~']['name']['preferredLocale']['country'],
x['organizationalTarget~']['name']['preferredLocale']['language'],
] for x in s['elements']]
I'm trying to save the reference to a value in a json file where item orders could not be guaranteed. So far, what I have for a dataset like this one:
"Values": [
{
"Object": "DFC_Asset_05",
"Properties": [
{
"Property": "WeightKilograms",
"Value Offset": 5
},
{
"Property": "WeightPounds",
"Value Offset": 10
}
]
},
{
"Object": "DFC_Asset_05",
"Properties": [
{
"Property": "Name",
"Value Offset": 25
},
{
"Property": "ShortName",
"Value Offset": 119
}
]
}
]
and retrieving this object:
{
"Property": "ShortName",
"Value Offset": 119
}
Is a string like this:
reference = "[Object=DFC_Asset_06][Properties][Property=Name]"
Which looks nice and understandable in a string, but it's very unclean to find the referenced value as I must first parse the reference it with a regex then loop in the data to retrieve the matching item.
Am I doing this wrong? Is there a better way to do this? I looked at the reduce() function however it seems like it's made for dictionaries with static data. For example, I could not save the direct keys:
reference = "[1][Properties][1]"
reference_using_reduce = [1, "Properties", 1]
As they might not always be in that order
You can run "queries" on JSONs without referencing specific indexes using the pyjq module:
query = (
'.Values[]' # On all the items in "Values"
'|select(."Object" == "DFC_Asset_06")' # Find key "Object" which holds this value
'|."Properties"[]' # And get all the items of "Properties"
'|select(."Property" == "Name")' # Where the key "Property" holds the value "Name"
)
pyjq.first(query, d)
Result:
{'Property': 'Name', 'Value Offset': 25}
You can read more about jq in the documentations.
I want to delete the following 'date' and 'last_modified' keys from the following nested dictionary. Kindly suggest any elegant way to do this dynamically with in Python.
{
"total_pages":1,
"datasets":[
{
"dataset_name":"enterpriseqa-landing-zone_census2017",
"database":"enterpriseqa-landing-zone",
"table":"census2017",
"owner":"qadataengineer",
"zone":"landing",
"date":"2020-06-09T07:11:25+00:00",
"location":"s3://enterpriseqa-landing-zone/static/census2017/",
"count":"5507",
"classification":"csv",
"last_modified":"2020-06-09T07:15:49+00:00",
"type":"Static"
}
]
}
If d is your dictionary from the question, you can use this example to delete the keys:
for dataset in d['datasets']:
del dataset['date']
del dataset['last_modified']
Produces this dictionary:
{
"total_pages": 1,
"datasets": [
{
"dataset_name": "enterpriseqa-landing-zone_census2017",
"database": "enterpriseqa-landing-zone",
"table": "census2017",
"owner": "qadataengineer",
"zone": "landing",
"location": "s3://enterpriseqa-landing-zone/static/census2017/",
"count": "5507",
"classification": "csv",
"type": "Static"
}
]
}
You can do it like this:
keys = ["date", "last_modified"]
[[d.pop(key) for key in keys] for d in dictionary["datasets"]]
Where dictionary is your dictionary.
I hope everyone is doing well.
I need a little help where I need to get all the strings from a variable and need to store into a single list in python.
For example -
I have json file from where I am getting ids and all the ids are getting stored into a variable called id as below when I run print(id)
17298626-991c-e490-bae6-47079c6e2202
17298496-19bd-2f89-7b5f-881921abc632
17298698-3e17-7a9b-b337-aacfd9483b1b
172986ac-d91d-c4ea-2e50-d53700480dd0
172986d0-18aa-6f51-9c62-6cb087ad31e5
172986f4-80f0-5c21-3aee-12f22a5f4322
17298712-a4ac-7b36-08e9-8512fa8322dd
17298747-8cc6-d9d0-8d05-50adf228c029
1729875c-050f-9a99-4850-bb0e6ad35fb0
1729875f-0d50-dc94-5515-b4891c40d81c
17298761-c26b-3ce5-e77e-db412c38a5b4
172987c8-2b5d-0d94-c365-e8407b0a8860
1729881a-e583-2b54-3a52-d092020d9c1d
1729881c-64a2-67cf-d561-6e5e38ed14cb
172987ec-7a20-7eb6-3ebe-a9fb621bb566
17298813-7ac4-258b-d6f9-aaf43f9147b1
17298813-f1ef-d28a-0817-5f3b86c3cf23
17298828-b62b-9ee6-248b-521b0663226e
17298825-7449-2fcb-378e-13671cb4688a
I want these all values to be stored into a single list.
Can some please help me out with this.
Below is the code I am using:
import json
with open('requests.json') as f:
data = json.load(f)
print(type(data))
for i in data:
if 'traceId' in i:
id = i['traceId']
newid = id.split()
#print(type(newid))
print(newid)
And below is my json file looks like:
[
{
"id": "376287298-hjd8-jfjb-khkf-6479280283e9",
"submittedTime": 1591692502558,
"traceId": "17298626-991c-e490-bae6-47079c6e2202",
"userName": "ABC",
"onlyChanged": true,
"description": "Not Required",
"startTime": 1591694487929,
"result": "NONE",
"state": "EXECUTING",
"paused": false,
"application": {
"id": "16b22a09-a840-f4d9-f42a-64fd73fece57",
"name": "XYZ"
},
"applicationProcess": {
"id": "dihihdosfj9279278yrie8ue",
"name": "Deploy",
"version": 12
},
"environment": {
"id": "fkjdshkjdshglkjdshgldshldsh03r937837",
"name": "DEV"
},
"snapshot": {
"id": "djnglkfdglki98478yhgjh48yr844h",
"name": "DEV_snapshot"
},
},
{
"id": "17298495-f060-3e9d-7097-1f86d5160789",
"submittedTime": 1591692844597,
"traceId": "17298496-19bd-2f89-7b5f-881921abc632",
"userName": "UYT,
"onlyChanged": true,
"startTime": 1591692845543,
"result": "NONE",
"state": "EXECUTING",
"paused": false,
"application": {
"id": "osfodsho883793hgjbv98r3098w",
"name": "QA"
},
"applicationProcess": {
"id": "owjfoew028r2uoieroiehojehfoef",
"name": "EDC",
"version": 5
},
"environment": {
"id": "16cf69c5-4194-e557-707d-0663afdbceba",
"name": "DTESTU"
},
}
]
From where I am trying to get the traceId.
you could use simple split method like the follwing:
ids = '''17298626-991c-e490-bae6-47079c6e2202 17298496-19bd-2f89-7b5f-881921abc632 17298698-3e17-7a9b-b337-aacfd9483b1b 172986ac-d91d-c4ea-2e50-d53700480dd0 172986d0-18aa-6f51-9c62-6cb087ad31e5 172986f4-80f0-5c21-3aee-12f22a5f4322 17298712-a4ac-7b36-08e9-8512fa8322dd 17298747-8cc6-d9d0-8d05-50adf228c029 1729875c-050f-9a99-4850-bb0e6ad35fb0 1729875f-0d50-dc94-5515-b4891c40d81c 17298761-c26b-3ce5-e77e-db412c38a5b4 172987c8-2b5d-0d94-c365-e8407b0a8860 1729881a-e583-2b54-3a52-d092020d9c1d 1729881c-64a2-67cf-d561-6e5e38ed14cb 172987ec-7a20-7eb6-3ebe-a9fb621bb566 17298813-7ac4-258b-d6f9-aaf43f9147b1 17298813-f1ef-d28a-0817-5f3b86c3cf23 17298828-b62b-9ee6-248b-521b0663226e 17298825-7449-2fcb-378e-13671cb4688a'''
l = ids.split(" ")
print(l)
This will give the following result, I assumed that the separator needed is simple space you can adjust properly:
['17298626-991c-e490-bae6-47079c6e2202', '17298496-19bd-2f89-7b5f-881921abc632', '17298698-3e17-7a9b-b337-aacfd9483b1b', '172986ac-d91d-c4ea-2e50-d53700480dd0', '172986d0-18aa-6f51-9c62-6cb087ad31e5', '172986f4-80f0-5c21-3aee-12f22a5f4322', '17298712-a4ac-7b36-08e9-8512fa8322dd', '17298747-8cc6-d9d0-8d05-50adf228c029', '1729875c-050f-9a99-4850-bb0e6ad35fb0', '1729875f-0d50-dc94-5515-b4891c40d81c', '17298761-c26b-3ce5-e77e-db412c38a5b4', '172987c8-2b5d-0d94-c365-e8407b0a8860', '1729881a-e583-2b54-3a52-d092020d9c1d', '1729881c-64a2-67cf-d561-6e5e38ed14cb', '172987ec-7a20-7eb6-3ebe-a9fb621bb566', '17298813-7ac4-258b-d6f9-aaf43f9147b1', '17298813-f1ef-d28a-0817-5f3b86c3cf23', '17298828-b62b-9ee6-248b-521b0663226e', '17298825-7449-2fcb-378e-13671cb4688a']
Edit
You get list of lists because each iteration you read only 1 id, so what you need to do is to initiate an empty list and append each id to it in the following way:
l = []
for i in data
if 'traceId' in i:
id = i['traceId']
l.append(id)
you can append the ids variable to the list such as,
#list declaration
l1=[]
#this must be in your loop
l1.append(ids)
I'm assuming you get the id as a str type value. Using id.split() will return a list of all ids in one single Python list, as each id is separated by space here in your example.
id = """17298626-991c-e490-bae6-47079c6e2202 17298496-19bd-2f89-7b5f-881921abc632
17298698-3e17-7a9b-b337-aacfd9483b1b 172986ac-d91d-c4ea-2e50-d53700480dd0
172986d0-18aa-6f51-9c62-6cb087ad31e5 172986f4-80f0-5c21-3aee-12f22a5f4322
17298712-a4ac-7b36-08e9-8512fa8322dd 17298747-8cc6-d9d0-8d05-50adf228c029
1729875c-050f-9a99-4850-bb0e6ad35fb0 1729875f-0d50-dc94-5515-b4891c40d81c
17298761-c26b-3ce5-e77e-db412c38a5b4 172987c8-2b5d-0d94-c365-e8407b0a8860
1729881a-e583-2b54-3a52-d092020d9c1d 1729881c-64a2-67cf-d561-6e5e38ed14cb
172987ec-7a20-7eb6-3ebe-a9fb621bb566 17298813-7ac4-258b-d6f9-aaf43f9147b1
17298813-f1ef-d28a-0817-5f3b86c3cf23 17298828-b62b-9ee6-248b-521b0663226e
17298825-7449-2fcb-378e-13671cb4688a"""
id_list = id.split()
print(id_list)
Output:
['17298626-991c-e490-bae6-47079c6e2202', '17298496-19bd-2f89-7b5f-881921abc632',
'17298698-3e17-7a9b-b337-aacfd9483b1b', '172986ac-d91d-c4ea-2e50-d53700480dd0',
'172986d0-18aa-6f51-9c62-6cb087ad31e5', '172986f4-80f0-5c21-3aee-12f22a5f4322',
'17298712-a4ac-7b36-08e9-8512fa8322dd', '17298747-8cc6-d9d0-8d05-50adf228c029',
'1729875c-050f-9a99-4850-bb0e6ad35fb0', '1729875f-0d50-dc94-5515-b4891c40d81c',
'17298761-c26b-3ce5-e77e-db412c38a5b4', '172987c8-2b5d-0d94-c365-e8407b0a8860',
'1729881a-e583-2b54-3a52-d092020d9c1d', '1729881c-64a2-67cf-d561-6e5e38ed14cb',
'172987ec-7a20-7eb6-3ebe-a9fb621bb566', '17298813-7ac4-258b-d6f9-aaf43f9147b1',
'17298813-f1ef-d28a-0817-5f3b86c3cf23', '17298828-b62b-9ee6-248b-521b0663226e',
'17298825-7449-2fcb-378e-13671cb4688a']
split() splits by default with space as a separator. You can use the sep argument to use any other separator if needed.
I have below sample data in JSON format :
project_cost_details is my database result set after querying.
{
"1": {
"amount": 0,
"breakdown": [
{
"amount": 169857,
"id": 4,
"name": "SampleData",
"parent_id": "1"
}
],
"id": 1,
"name": "ABC PR"
}
}
Here is full json : https://jsoneditoronline.org/?id=2ce7ab19af6f420397b07b939674f49c
Expected output :https://jsoneditoronline.org/?id=56a47e6f8e424fe8ac58c5e0732168d7
I have this sample JSON which i created using loops in code. But i am stuck at how to convert this to expected JSON format. I am getting sequential changes, need to convert to tree like or nested JSON format.
Trying in Python :
project_cost = {}
for cost in project_cost_details:
if cost.get('Parent_Cost_Type_ID'):
project_id = str(cost.get('Project_ID'))
parent_cost_type_id = str(cost.get('Parent_Cost_Type_ID'))
if project_id not in project_cost:
project_cost[project_id] = {}
if "breakdown" not in project_cost[project_id]:
project_cost[project_id]["breakdown"] = []
if 'amount' not in project_cost[project_id]:
project_cost[project_id]['amount'] = 0
project_cost[project_id]['name'] = cost.get('Title')
project_cost[project_id]['id'] = cost.get('Project_ID')
if parent_cost_type_id == cost.get('Cost_Type_ID'):
project_cost[project_id]['amount'] += int(cost.get('Amount'))
#if parent_cost_type_id is None:
project_cost[project_id]["breakdown"].append(
{
'amount': int(cost.get('Amount')),
'name': cost.get('Name'),
'parent_id': parent_cost_type_id,
'id' : cost.get('Cost_Type_ID')
}
)
from this i am getting sample JSON. It will be good if get in this code only desired format.
Also tried this solution mention here : https://adiyatmubarak.wordpress.com/2015/10/05/group-list-of-dictionary-data-by-particular-key-in-python/
I got approach to convert sample JSON to expected JSON :
data = [
{ "name" : "ABC", "parent":"DEF", },
{ "name" : "DEF", "parent":"null" },
{ "name" : "new_name", "parent":"ABC" },
{ "name" : "new_name2", "parent":"ABC" },
{ "name" : "Foo", "parent":"DEF"},
{ "name" : "Bar", "parent":"null"},
{ "name" : "Chandani", "parent":"new_name", "relation": "rel", "depth": 3 },
{ "name" : "Chandani333", "parent":"new_name", "relation": "rel", "depth": 3 }
]
result = {x.get("name"):x for x in data}
#print(result)
tree = [];
for a in data:
#print(a)
if a.get("parent") in result:
parent = result[a.get("parent")]
else:
parent = ""
if parent:
if "children" not in parent:
parent["children"] = []
parent["children"].append(a)
else:
tree.append(a)
Reference help : http://jsfiddle.net/9FqKS/ this is a JavaScript solution i converted to Python
It seems that you want to get a list of values from a dictionary.
result = [value for key, value in project_cost_details.items()]