My question raised when I exploited this helpful answer provided by Trenton McKinney on the issue of flattening multiple nested JSON-files for handling in pandas.
Following his advice, I have used the flatten_json function described here to flatten a batch of nested json files. However, I have run into a problem with the uniformity of my JSON-files.
A single JSON-File looks roughly like this made-up example data:
{
"product": "example_productname",
"product_id": "example_productid",
"product_type": "example_producttype",
"producer": "example_producer",
"currency": "example_currency",
"client_id": "example_clientid",
"supplement": [
{
"supplementtype": "RTZ",
"price": 300000,
"rebate": "500",
},
{
"supplementtype": "CVB",
"price": 500000,
"rebate": "250",
},
{
"supplementtype": "JKL",
"price": 100000,
"rebate": "750",
},
],
}
Utilizing the referenced code, I will end up with data looking like this:
product
product_id
product_type
producer
currency
client_id
supplement_0_supplementtype
supplement_0_price
supplement_0_rebate
supplement_1_supplementtype
supplement_1_price
supplement_1_rebate
etc
example_productname
example_productid
example_type
example_producer
example_currency
example_clientid
RTZ
300000
500
CVB
500000
250
etc
example_productname2
example_productid2
example_type2
example_producer2
example_currency2
example_clientid2
CVB
500000
250
RTZ
300000
500
etc
There are multiple issues with this.
Firstly, in my data, there is a limited list of "supplements", however, they do not always appear, and if they do, they are not always in the same order. In the example table, you can see that the two "supplements" switched positions in the second row. I would prefer a fixed order of the "supplement columns".
Secondly, the best option would be a table like this:
product
product_id
product_type
producer
currency
client_id
supplement_RTZ_price
supplement_RTZ_rebate
supplement_CVB_price
supplement_CVB_rebate
etc
example_productname
example_productid
example_type
example_producer
example_currency
example_clientid
300000
500
500000
250
etc
I have tried editing the flatten_json function referenced, but I don't have an inkling of how to make this work.
The solution consists of simply editing the dictionary (thanks to Andrej Kesely). I just added a pass to exceptions in case some columns are inexistent.
d = {
"product": "example_productname",
"product_id": "example_productid",
"product_type": "example_producttype",
"producer": "example_producer",
"currency": "example_currency",
"client_id": "example_clientid",
"supplement": [
{
"supplementtype": "RTZ",
"price": 300000,
"rebate": "500",
},
{
"supplementtype": "CVB",
"price": 500000,
"rebate": "250",
},
{
"supplementtype": "JKL",
"price": 100000,
"rebate": "750",
},
],
}
for s in d["supplement"]:
try:
d["supplementtype_{}_price".format(s["supplementtype"])] = s["price"]
except:
pass
try:
d["supplementtype_{}_rebate".format(s["supplementtype"])] = s["rebate"]
except:
pass
del d["supplement"]
df = pd.DataFrame([d])
print(df)
product product_id product_type producer currency client_id supplementtype_RTZ_price supplementtype_RTZ_rebate supplementtype_CVB_price supplementtype_CVB_rebate supplementtype_JKL_price supplementtype_JKL_rebate
0 example_productname example_productid example_producttype example_producer example_currency example_clientid 300000 500 500000 250 100000 750
The used/referenced code:
def flatten_json(nested_json: dict, exclude: list=[''], sep: str='_') -> dict:
"""
Flatten a list of nested dicts.
"""
out = dict()
def flatten(x: (list, dict, str), name: str='', exclude=exclude):
if type(x) is dict:
for a in x:
if a not in exclude:
flatten(x[a], f'{name}{a}{sep}')
elif type(x) is list:
i = 0
for a in x:
flatten(a, f'{name}{i}{sep}')
i += 1
else:
out[name[:-1]] = x
flatten(nested_json)
return out
# list of files
files = ['test1.json', 'test2.json']
# list to add dataframe from each file
df_list = list()
# iterate through files
for file in files:
with open(file, 'r') as f:
# read with json
data = json.loads(f.read())
# flatten_json into a dataframe and add to the dataframe list
df_list.append(pd.DataFrame.from_dict(flatten_json(data), orient='index').T)
# concat all dataframes together
df = pd.concat(df_list).reset_index(drop=True)
You can modify the dictionary before you create dataframe from it:
d = {
"product": "example_productname",
"product_id": "example_productid",
"product_type": "example_producttype",
"producer": "example_producer",
"currency": "example_currency",
"client_id": "example_clientid",
"supplement": [
{
"supplementtype": "RTZ",
"price": 300000,
"rebate": "500",
},
{
"supplementtype": "CVB",
"price": 500000,
"rebate": "250",
},
{
"supplementtype": "JKL",
"price": 100000,
"rebate": "750",
},
],
}
for s in d["supplement"]:
d["supplementtype_{}_price".format(s["supplementtype"])] = s["price"]
d["supplementtype_{}_rebate".format(s["supplementtype"])] = s["rebate"]
del d["supplement"]
df = pd.DataFrame([d])
print(df)
Prints:
product product_id product_type producer currency client_id supplementtype_RTZ_price supplementtype_RTZ_rebate supplementtype_CVB_price supplementtype_CVB_rebate supplementtype_JKL_price supplementtype_JKL_rebate
0 example_productname example_productid example_producttype example_producer example_currency example_clientid 300000 500 500000 250 100000 750
Related
I am using Github's GraphQL API to fetch some issue details.
I used Python Requests to fetch the data locally.
This is how the output.json looks like
{
"data": {
"viewer": {
"login": "some_user"
},
"repository": {
"issues": {
"edges": [
{
"node": {
"id": "I_kwDOHQ63-s5auKbD",
"title": "test issue 1",
"number": 146,
"createdAt": "2023-01-06T06:39:54Z",
"closedAt": null,
"state": "OPEN",
"updatedAt": "2023-01-06T06:42:00Z",
"comments": {
"edges": [
{
"node": {
"id": "IC_kwDOHQ63-s5R2XCV",
"body": "comment 01"
}
},
{
"node": {
"id": "IC_kwDOHQ63-s5R2XC9",
"body": "comment 02"
}
}
]
},
"labels": {
"edges": []
}
},
"cursor": "Y3Vyc29yOnYyOpHOWrimww=="
},
{
"node": {
"id": "I_kwDOHQ63-s5auKm8",
"title": "test issue 2",
"number": 147,
"createdAt": "2023-01-06T06:40:34Z",
"closedAt": null,
"state": "OPEN",
"updatedAt": "2023-01-06T06:40:34Z",
"comments": {
"edges": []
},
"labels": {
"edges": [
{
"node": {
"name": "food"
}
},
{
"node": {
"name": "healthy"
}
}
]
}
},
"cursor": "Y3Vyc29yOnYyOpHOWripvA=="
}
]
}
}
}
}
The json was put inside a list using
result = response.json()["data"]["repository"]["issues"]["edges"]
And then this list was put inside a DataFrame
import pandas as pd
df = pd.DataFrame (result, columns = ['node', 'cursor'])
df
These are the contents of the data frame
id
title
number
createdAt
closedAt
state
updatedAt
comments
labels
I_kwDOHQ63-s5auKbD
test issue 1
146
2023-01-06T06:39:54Z
None
OPEN
2023-01-06T06:42:00Z
{'edges': [{'node': {'id': 'IC_kwDOHQ63-s5R2XCV","body": "comment 01"}},{'node': {'id': 'IC_kwDOHQ63-s5R2XC9","body": "comment 02"}}]}
{'edges': []}
I_kwDOHQ63-s5auKm8
test issue 2
147
2023-01-06T06:40:34Z
None
OPEN
2023-01-06T06:40:34Z
{'edges': []}
{'edges': [{'node': {'name': 'food"}},{'node': {'name': 'healthy"}}]}
I would like to split/explode the comments and labels columns.
The values in these columns are nested dictionaries
I would like there to be as many rows for a single issue, as there are comments & labels.
I would like to flatten out the data frame.
So this involves split/explode and concat.
There are several stackoverflow answers that delve on this topic. And I have tried the code from several of them.
I can not paste the links to those questions, because stackoverflow marks my question as spam due to many links.
But these are the steps I have tried
df3 = df2['comments'].apply(pd.Series)
Drill down further
df4 = df3['edges'].apply(pd.Series)
df4
Drill down further
df5 = df4['node'].apply(pd.Series)
df5
The last statement above gives me the KeyError: 'node'
I understand, this is because node is not a key in the DataFrame.
But how else can i split this dictionary and concatenate all columns back to my issues row?
This is how I would like the output to look like
id
title
number
createdAt
closedAt
state
updatedAt
comments
labels
I_kwDOHQ63-s5auKbD
test issue 1
146
2023-01-06T06:39:54Z
None
OPEN
2023-01-06T06:42:00Z
comment 01
Null
I_kwDOHQ63-s5auKbD
test issue 1
146
2023-01-06T06:39:54Z
None
OPEN
2023-01-06T06:42:00Z
comment 02
Null
I_kwDOHQ63-s5auKm8
test issue 2
147
2023-01-06T06:40:34Z
None
OPEN
2023-01-06T06:40:34Z
Null
food
I_kwDOHQ63-s5auKm8
test issue 2
147
2023-01-06T06:40:34Z
None
OPEN
2023-01-06T06:40:34Z
Null
healthy
If dct is your dictionary from the question you can try:
df = pd.DataFrame(d['node'] for d in dct['data']['repository']['issues']['edges'])
df['comments'] = df['comments'].str['edges']
df = df.explode('comments')
df['comments'] = df['comments'].str['node'].str['body']
df['labels'] = df['labels'].str['edges']
df = df.explode('labels')
df['labels'] = df['labels'].str['node'].str['name']
print(df.to_markdown(index=False))
Prints:
id
title
number
createdAt
closedAt
state
updatedAt
comments
labels
I_kwDOHQ63-s5auKbD
test issue 1
146
2023-01-06T06:39:54Z
OPEN
2023-01-06T06:42:00Z
comment 01
nan
I_kwDOHQ63-s5auKbD
test issue 1
146
2023-01-06T06:39:54Z
OPEN
2023-01-06T06:42:00Z
comment 02
nan
I_kwDOHQ63-s5auKm8
test issue 2
147
2023-01-06T06:40:34Z
OPEN
2023-01-06T06:40:34Z
nan
food
I_kwDOHQ63-s5auKm8
test issue 2
147
2023-01-06T06:40:34Z
OPEN
2023-01-06T06:40:34Z
nan
healthy
#andrej-kesely has answered my question.
I have selected his response as the answer for this question.
I am now posting a consolidated script that includes my poor code and andrej's great code.
In this script i want to fetch details from Github's GraphQL API Server.
And put it inside pandas.
Primary source for this script is this gist.
And a major chunk of remaining code is an answer by #andrej-kesely.
Now onto the consolidated script.
First import the necessary packages and set headers
import requests
import json
import pandas as pd
headers = {"Authorization": "token <your_github_personal_access_token>"}
Now define the query that will fetch data from github.
In my particular case, I am fetching issue details form a particular repo
it can be something else for you.
query = """
{
viewer {
login
}
repository(name: "your_github_repo", owner: "your_github_user_name") {
issues(states: OPEN, last: 2) {
edges {
node {
id
title
number
createdAt
closedAt
state
updatedAt
comments(first: 10) {
edges {
node {
id
body
}
}
}
labels(orderBy: {field: NAME, direction: ASC}, first: 10) {
edges {
node {
name
}
}
}
comments(first: 10) {
edges {
node {
id
body
}
}
}
}
cursor
}
}
}
}
"""
Execute the query and save the response
def run_query(query):
request = requests.post('https://api.github.com/graphql', json={'query': query}, headers=headers)
if request.status_code == 200:
return request.json()
else:
raise Exception("Query failed to run by returning code of {}. {}".format(request.status_code, query))
result = run_query(query)
And now is the trickiest part.
In my query response, there are several nested dictionaries.
I would like to split them - more details in my question above.
This magic code from #andrej-kesely does that for you.
df = pd.DataFrame(d['node'] for d in result['data']['repository']['issues']['edges'])
df['comments'] = df['comments'].str['edges']
df = df.explode('comments')
df['comments'] = df['comments'].str['node'].str['body']
df['labels'] = df['labels'].str['edges']
df = df.explode('labels')
df['labels'] = df['labels'].str['node'].str['name']
print(df)
I have a nested JSON data. I want to get the value of key "name" inside the dictionary "value" based on the key "id" in "key" dictionary (let the user enter the id). I don't want to use indexing which, because places are changing on every url differently. Also data is large, so I need one row solution (without for loop).
Code
import requests, re, json
r = requests.get('https://www.trendyol.com/apple/macbook-air-13-m1-8gb-256gb-ssd-altin-p-67940132').text
json_data1 = json.loads(re.search(r"window.__PRODUCT_DETAIL_APP_INITIAL_STATE__=({.*}});window", r).group(1))
print(json_data1)
print('json_data1:',json_data1['product']['attributes'][0]['value']['name'])
Output
{'product': {'attributes': [{'key': {'name': 'İşlemci Tipi', 'id': 168}, 'value': {'name': 'Apple M1', 'id': 243383}, 'starred': True, 'description': '', 'mediaUrls': []}, {'key': {'name': 'SSD Kapasitesi', 'id': 249}..........
json_data1: Apple M1
JSON Data
{
"product": {
"attributes": [
{
"key": { "name": "İşlemci Tipi", "id": 168 },
"value": { "name": "Apple M1", "id": 243383 },
"starred": true,
"description": "",
"mediaUrls": []
},
{
"key": { "name": "SSD Kapasitesi", "id": 249 },
"value": { "name": "256 GB", "id": 3376 },
"starred": true,
"description": "",
"mediaUrls": []
},
.
.
.
]
}
}
Expected Output is getting value by key id: (type must be str)
input >> id: 168
output >> name: Apple M1
Since you originally didn't want a for loop, but now it's a matter of speed,
Here's a solution with for loop, you can test it and see if it's faster than the one you already had
import json
with open("file.json") as f:
data = json.load(f)
search_key = int(input("Enter id: "))
for i in range(0, len(data['product']['attributes'])):
if search_key == data['product']['attributes'][i]['key']['id']:
print(data['product']['attributes'][i]['value']['name'])
Input >> Enter id: 168
Output >> Apple M1
I found the solution with for loop. It works fast so I preferred it.
for i in json_data1['product']['attributes']:
cpu = list(list(i.values())[0].values())[1]
if cpu == 168:
print(list(list(i.values())[1].values())[0])
Iteration is unavoidable if the index is unknown, but the cost can be reduced substantially by using a generator expression and Python's built-in next function:
next((x["value"]["name"] for x in data["product"]["attributes"] if x["key"]["id"] == 168), None)
To verify that a generator expression is in fact faster than a for loop, here is a comparison of the running time of xFranko's solution and the above:
import time
def time_func(func):
def timer(*args):
time1 = time.perf_counter()
func(*args)
time2 = time.perf_counter()
return (time2 - time1) * 1000
return timer
number_of_attributes = 100000
data = {
"product": {
"attributes": [
{
"key": { "name": "İşlemci Tipi", "id": i },
"value": { "name": "name" + str(i), "id": 243383 },
"starred": True,
"description": "",
"mediaUrls": []
} for i in range(number_of_attributes)
]
}
}
def getName_generator(id):
return next((x["value"]["name"] for x in data["product"]["attributes"] if x["key"]["id"] == id), None)
def getName_for_loop(id):
return_value = None
for i in range(0, len(data['product']['attributes'])):
if id == data['product']['attributes'][i]['key']['id']:
return_value = data['product']['attributes'][i]['value']['name']
return return_value
print("Generator:", time_func(getName_generator)(0))
print("For loop:", time_func(getName_for_loop)(0))
print()
print("Generator:", time_func(getName_generator)(number_of_attributes - 1))
print("For loop:", time_func(getName_for_loop)(number_of_attributes - 1))
My results:
Generator: 0.0075999999999964984
For loop: 43.73920000000003
Generator: 23.633300000000023
For loop: 49.839699999999986
Conclusion:
For large data sets, a generator expression is indeed faster, even if it has to traverse the entire set.
I have a generator being returned from:
data = public_client.get_product_trades(product_id='BTC-USD', limit=10)
How do i turn the data in to a pandas dataframe?
the method DOCSTRING reads:
"""{"Returns": [{
"time": "2014-11-07T22:19:28.578544Z",
"trade_id": 74,
"price": "10.00000000",
"size": "0.01000000",
"side": "buy"
}, {
"time": "2014-11-07T01:08:43.642366Z",
"trade_id": 73,
"price": "100.00000000",
"size": "0.01000000",
"side": "sell"
}]}"""
I have tried:
df = [x for x in data]
df = pd.DataFrame.from_records(df)
but it does not work as i get the error:
AttributeError: 'str' object has no attribute 'keys'
When i print the above "x for x in data" i see the list of dicts but the end looks strange, could this be why?
print(list(data))
[{'time': '2020-12-30T13:04:14.385Z', 'trade_id': 116918468, 'price': '27853.82000000', 'size': '0.00171515', 'side': 'sell'},{'time': '2020-12-30T12:31:24.185Z', 'trade_id': 116915675, 'price': '27683.70000000', 'size': '0.01683711', 'side': 'sell'}, 'message']
It looks to be a list of dicts but the end value is a single string 'message'.
Based on the updated question:
df = pd.DataFrame(list(data)[:-1])
Or, more cleanly:
df = pd.DataFrame([x for x in data if isinstance(x, dict)])
print(df)
time trade_id price size side
0 2020-12-30T13:04:14.385Z 116918468 27853.82000000 0.00171515 sell
1 2020-12-30T12:31:24.185Z 116915675 27683.70000000 0.01683711 sell
Oh, and BTW, you'll still need to change those strings into something usable...
So e.g.:
df['time'] = pd.to_datetime(df['time'])
for k in ['price', 'size']:
df[k] = pd.to_numeric(df[k])
You could access the values in the dictionary and build a dataframe from it (although not particularly clean):
dict_of_data = [{
"time": "2014-11-07T22:19:28.578544Z",
"trade_id": 74,
"price": "10.00000000",
"size": "0.01000000",
"side": "buy"
}, {
"time": "2014-11-07T01:08:43.642366Z",
"trade_id": 73,
"price": "100.00000000",
"size": "0.01000000",
"side": "sell"
}]
import pandas as pd
list_of_data = [list(dict_of_data[0].values()),list(dict_of_data[1].values())]
pd.DataFrame(list_of_data, columns=list(dict_of_data[0].keys())).set_index('time')
its straightforward just use the pd.DataFrame constructor:
#list_of_dicts = [{
# "time": "2014-11-07T22:19:28.578544Z",
# "trade_id": 74,
# "price": "10.00000000",
# "size": "0.01000000",
# "side": "buy"
# }, {
# "time": "2014-11-07T01:08:43.642366Z",
# "trade_id": 73,
# "price": "100.00000000",
# "size": "0.01000000",
# "side": "sell"
#}]
# or if you take it from 'data'
list_of_dicts = data[:-1]
df = pd.DataFrame(list_of_dicts)
df
Out[4]:
time trade_id price size side
0 2014-11-07T22:19:28.578544Z 74 10.00000000 0.01000000 buy
1 2014-11-07T01:08:43.642366Z 73 100.00000000 0.01000000 sell
UPDATE
according to the question update, it seems you have json data that is still string...
import json
data = json.loads(data)
data = data['Returns']
pd.DataFrame(data)
time trade_id price size side
0 2014-11-07T22:19:28.578544Z 74 10.00000000 0.01000000 buy
1 2014-11-07T01:08:43.642366Z 73 100.00000000 0.01000000 sell
I have a JSON which is in nested form. I would like to extract specific data from json and put into csv using pandas python.
data = {
"class":"hudson.model.Hudson",
"jobs":[
{
"_class":"hudson.model.FreeStyleProject",
"name":"git_checkout",
"url":"http://localhost:8080/job/git_checkout/",
"builds":[
{
"_class":"hudson.model.FreeStyleBuild",
"duration":1201,
"number":6,
"result":"FAILURE",
"url":"http://localhost:8080/job/git_checkout/6/"
}
]
},
{
"_class":"hudson.model.FreeStyleProject",
"name":"output",
"url":"http://localhost:8080/job/output/",
"builds":[
]
},
{
"_class":"org.jenkinsci.plugins.workflow.job.WorkflowJob",
"name":"pipeline_test",
"url":"http://localhost:8080/job/pipeline_test/",
"builds":[
{
"_class":"org.jenkinsci.plugins.workflow.job.WorkflowRun",
"duration":9274,
"number":85,
"result":"SUCCESS",
"url":"http://localhost:8080/job/pipeline_test/85/"
},
{
"_class":"org.jenkinsci.plugins.workflow.job.WorkflowRun",
"duration":4251,
"number":84,
"result":"SUCCESS",
"url":"http://localhost:8080/job/pipeline_test/84/"
}
]
}
]
}
From the above JSON i want to fetch jobs name value and builds result value . I am new to python any help will be appreciated .
Till now i have tried
main_data = data['jobs]
json_normalize(main_data,['builds'],
record_prefix='jobs_', errors='ignore')
which gives information only build key values and not the name of job .
Can anyone help ?
Expected Output:
Considering only first build result value you can need to be in csv column you can achieve this using pandas.
data = {
"class": "hudson.model.Hudson",
"jobs": [
{
"_class": "hudson.model.FreeStyleProject",
"name": "git_checkout",
"url": "http://localhost:8080/job/git_checkout/",
"builds": [
{
"_class": "hudson.model.FreeStyleBuild",
"duration": 1201,
"number": 6,
"result": "FAILURE",
"url": "http://localhost:8080/job/git_checkout/6/"
}
]
},
{
"_class": "hudson.model.FreeStyleProject",
"name": "output",
"url": "http://localhost:8080/job/output/",
"builds": []
},
{
"_class": "org.jenkinsci.plugins.workflow.job.WorkflowJob",
"name": "pipeline_test",
"url": "http://localhost:8080/job/pipeline_test/",
"builds": [
{
"_class": "org.jenkinsci.plugins.workflow.job.WorkflowRun",
"duration": 9274,
"number": 85,
"result": "SUCCESS",
"url": "http://localhost:8080/job/pipeline_test/85/"
},
{
"_class": "org.jenkinsci.plugins.workflow.job.WorkflowRun",
"duration": 4251,
"number": 84,
"result": "SUCCESS",
"url": "http://localhost:8080/job/pipeline_test/84/"
}
]
}
]
}
main_data = data.get('jobs')
res = {'name':[], 'result':[]}
for name_dict in main_data:
res['name'].append(name_dict.get('name','NA'))
resultval = name_dict['builds'][0].get('result') if len(name_dict['builds'])>0 else 'NA'
res['result'].append(resultval)
print(res)
import pandas as pd
df = pd.DataFrame(res)
df.to_csv("/home/file_timer/jobs.csv", index=False)
Check the csv file output
name,result
git_checkout,FAILURE
output,NA
pipeline_test,SUCCESS
If 'NA' result want to skip then
main_data = data.get('jobs')
res = {'name':[], 'result':[]}
for name_dict in main_data:
if len(name_dict['builds'])==0:
continue
res['name'].append(name_dict.get('name', 'NA'))
resultval = name_dict['builds'][0].get('result')
res['result'].append(resultval)
print(res)
import pandas as pd
df = pd.DataFrame(res)
df.to_csv("/home/akash.pagar/shell_learning/file_timer/jobs.csv", index=False)
Output will bw like
name,result
git_checkout,FAILURE
pipeline_test,SUCCESS
Simply with build number,
for job in data.get('jobs'):
for build in job.get('builds'):
print(job.get('name'), build.get('number'), build.get('result'))
gives the result
git_checkout 6 FAILURE
pipeline_test 85 SUCCESS
pipeline_test 84 SUCCESS
If you want to get the result of latest build, and pretty sure about the build number always in decending order,
for job in data.get('jobs'):
if job.get('builds'):
print(job.get('name'), job.get('builds')[0].get('result'))
and if you are not sure the order,
for job in data.get('jobs'):
if job.get('builds'):
print(job.get('name'), sorted(job.get('builds'), key=lambda k: k.get('number'))[-1].get('result'))
then the result will be:
git_checkout FAILURE
pipeline_test SUCCESS
Assuming last build is the last element of its list and you don't care about jobs with no builds, this does:
import pandas as pd
#data = ... #same format as in the question
z = [(job["name"], job["builds"][-1]["result"]) for job in data["jobs"] if len(job["builds"])]
df = pd.DataFrame(data=z, columns=["name", "result"])
#df.to_csv #TODO
Also we don't necessarily need pandas to create the csv file.
You could do:
import csv
#z = ... #see previous code block
with open("f.csv", 'w') as fp:
csv.writer(fp).writerows([("name", "result")] + z)
I'm comparing json files between two different API endpoints to see which json records need an update, which need a create and what needs a delete. So, by comparing the two json files, I want to end up with three json files, one for each operation.
The json at both endpoints is structured like this (but they use different keys for same sets of values; different problem):
{
"records": [{
"id": "id-value-here",
"c": {
"d": "eee"
},
"f": {
"l": "last",
"f": "first"
},
"g": ["100", "89", "9831", "09112", "800"]
}, {
…
}]
}
So the json is represented as a list of dictionaries (with further nested lists and dictionaries).
If a given json endpoint (j1) id value ("id":) exists in the other endpoint json (j2), then that record should be added to j_update.
So far I have something like this, but I can see that .values() doesn't work because it's trying to operate on the list instead of on all the listed dictionaries(?):
j_update = {r for r in j1['records'] if r['id'] in
j2.values()}
This doesn't return an error, but it creates an empty set using test json files.
Seems like this should be simple, but tripping over the nesting I think of dictionaries in a list representing the json. Do I need to flatten j2, or is there a simpler dictionary method python has to achieve this?
====edit j1 and j2====
have same structure, use different keys; toy data
j1
{
"records": [{
"field_5": 2329309841,
"field_12": {
"email": "cmix#etest.com"
},
"field_20": {
"last": "Mixalona",
"first": "Clara"
},
"field_28": ["9002329309999", "9002329309112"],
"field_44": ["1002329309832"]
}, {
"field_5": 2329309831,
"field_12": {
"email": "mherbitz345#test.com"
},
"field_20": {
"last": "Herbitz",
"first": "Michael"
},
"field_28": ["9002329309831", "9002329309112", "8002329309999"],
"field_44": ["1002329309832"]
}, {
"field_5": 2329309855,
"field_12": {
"email": "nkatamaran#test.com"
},
"field_20": {
"first": "Noriss",
"last": "Katamaran"
},
"field_28": ["9002329309111", "8002329309112"],
"field_44": ["1002329309877"]
}]
}
j2
{
"records": [{
"id": 2329309831,
"email": {
"email": "mherbitz345#test.com"
},
"name_primary": {
"last": "Herbitz",
"first": "Michael"
},
"assign": ["8003329309831", "8007329309789"],
"hr_id": ["1002329309877"]
}, {
"id": 2329309884,
"email": {
"email": "yinleeshu#test.com"
},
"name_primary": {
"last": "Lee Shu",
"first": "Yin"
},
"assign": ["8002329309111", "9003329309831", "9002329309111", "8002329309999", "8002329309112"],
"hr_id": ["1002329309832"]
}, {
"id": 23293098338,
"email": {
"email": "amlouis#test.com"
},
"name_primary": {
"last": "Maxwell Louis",
"first": "Albert"
},
"assign": ["8002329309111", "8007329309789", "9003329309831", "8002329309999", "8002329309112"],
"hr_id": ["1002329309877"]
}]
}
If you read the json it will output a dict. You are looking for a particular key in the list of the values.
if 'records' in j2:
r = j2['records'][0].get('id', []) # defaults if id does not exist
It it prettier to do a recursive search but i dunno how you data is organized to quickly come up with a solution.
To give an idea for recursive search consider this example
def resursiveSearch(dictionary, target):
if target in dictionary:
return dictionary[target]
for key, value in dictionary.items():
if isinstance(value, dict):
target = resursiveSearch(value, target)
if target:
return target
a = {'test' : 'b', 'test1' : dict(x = dict(z = 3), y = 2)}
print(resursiveSearch(a, 'z'))
You tried:
j_update = {r for r in j1['records'] if r['id'] in j2.values()}
Aside from the r['id'/'field_5] problem, you have:
>>> list(j2.values())
[[{'id': 2329309831, ...}, ...]]
The id are buried inside a list and a dict, thus the test r['id'] in j2.values() always return False.
The basic solution is fairly simple.
First, create a set of j2 ids:
>>> present_in_j2 = set(record["id"] for record in j2["records"])
Then, rebuild the json structure of j1 but without the j1 field_5 that are not present in j2:
>>> {"records":[record for record in j1["records"] if record["field_5"] in present_in_j2]}
{'records': [{'field_5': 2329309831, 'field_12': {'email': 'mherbitz345#test.com'}, 'field_20': {'last': 'Herbitz', 'first': 'Michael'}, 'field_28': ['9002329309831', '9002329309112', '8002329309999'], 'field_44': ['1002329309832']}]}
It works, but it's not totally satisfying because of the weird keys of j1. Let's try to convert j1 to a more friendly format:
def map_keys(json_value, conversion_table):
"""Map the keys of a json value
This is a recursive DFS"""
def map_keys_aux(json_value):
"""Capture the conversion table"""
if isinstance(json_value, list):
return [map_keys_aux(v) for v in json_value]
elif isinstance(json_value, dict):
return {conversion_table.get(k, k):map_keys_aux(v) for k,v in json_value.items()}
else:
return json_value
return map_keys_aux(json_value)
The function focuses on dictionary keys: conversion_table.get(k, k) is conversion_table[k] if the key is present in the conversion table, or the key itself otherwise.
>>> j1toj2 = {"field_5":"id", "field_12":"email", "field_20":"name_primary", "field_28":"assign", "field_44":"hr_id"}
>>> mapped_j1 = map_keys(j1, j1toj2)
Now, the code is cleaner and the output may be more useful for a PUT:
>>> d1 = {record["id"]:record for record in mapped_j1["records"]}
>>> present_in_j2 = set(record["id"] for record in j2["records"])
>>> {"records":[record for record in mapped_j1["records"] if record["id"] in present_in_j2]}
{'records': [{'id': 2329309831, 'email': {'email': 'mherbitz345#test.com'}, 'name_primary': {'last': 'Herbitz', 'first': 'Michael'}, 'assign': ['9002329309831', '9002329309112', '8002329309999'], 'hr_id': ['1002329309832']}]}