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Multiple identical keys from Python dict to JSON

I am trying to create JSON object in Python, and it works just fine despite the fact that I can't get multiple keys with the same name - but I need to do it.
Here's a function:
findings = AutoTree()
findings['report']['numberOfConditions'] = num_cond
if r == 'Mammography':
f_temp = df['Relevant findings'].values.tolist()[0:8]
f_list = [x for i, x in enumerate(f_temp) if i == f_temp.index(x)]
f_num_total = len(f_list)
f_rand = random.randrange(1, f_num_total + 1)
iter_params_mass = ['shape', 'margin', 'density']
for i in range(num_cond):
br = get_birad(row, 2, 7)
cond = camelCase(get_cond_name())
findings[cond]['biRad'] = br
for k in range(f_rand + 1):
f = camelCase(random.choice(f_list))
#f = 'mass'
if f == 'mass':
rep_temp = create_rep(iter_params_mass, row, f, r)
findings[cond][f] = rep_temp
"""I also have a lot elif conditions, and it just grabs parameters."""
report = json.dumps(findings)
print(report)
Output:
{
"report":{
"id":85,
"name":"Lydia",
"age":39,
"relevantModality":"Mammography",
"numberOfConditions":2
},
"ductEctasia":{
"biRad":"birad[1]",
"calcifications":[
{
"typicallyBenign":"Vascular",
"suspiciousMorphology":"Coarse heterogeneous",
"distribution":"Diffuse"
}
],
"lymphNodes":[
{
"lymphNodes":"Lymph nodes \u2013 axillary"
}
]
}
}
And I want to have multiple "lymphNodes" and "calcifications" objects. Is it possible? Maybe, you can suggest another way to create JSON object, not nested dictionaries? The problem is that I need to create object respectively to random parameter chosen from the database.

Aggregating values in one column by their corresponding value in another from two files

had a question regarding summing the multiple values of duplicate keys into one key with the aggregate total. For example:
1:5
2:4
3:2
1:4
Very basic but I'm looking for an output that looks like:
1:9
2:4
3:2
In the two files I am using, I am dealing with a list of 51 users(column 1 of user_artists.dat) who have the artistID(column 2) and how many times that user has listened to that particular artist given by the weight(column 3).
I am attempting to aggregate the total times that artist has been played, across all users and display it in a format such as:
Britney Spears (289) 2393140. Any help or input would be so appreciated.
import codecs
#from collections import defaultdict
with codecs.open("artists.dat", encoding = "utf-8") as f:
artists = f.readlines()
with codecs.open("user_artists.dat", encoding = "utf-8") as f:
users = f.readlines()
artist_list = [x.strip().split('\t') for x in artists][1:]
user_stats_list = [x.strip().split('\t') for x in users][1:]
artists = {}
for a in artist_list:
artistID, name = a[0], a[1]
artists[artistID] = name
grouped_user_stats = {}
for u in user_stats_list:
userID, artistID, weight = u
grouped_user_stats[artistID] = grouped_user_stats[artistID].astype(int)
grouped_user_stats[weight] = grouped_user_stats[weight].astype(int)
for artistID, weight in u:
grouped_user_stats.groupby('artistID')['weight'].sum()
print(grouped_user_stats.groupby('artistID')['weight'].sum())
#if userID not in grouped_user_stats:
#grouped_user_stats[userID] = { artistID: {'name': artists[artistID], 'plays': 1} }
#else:
#if artistID not in grouped_user_stats[userID]:
#grouped_user_stats[userID][artistID] = {'name': artists[artistID], 'plays': 1}
#else:
#grouped_user_stats[userID][artistID]['plays'] += 1
#print('this never happens')
#print(grouped_user_stats)

how about:
import codecs
from collections import defaultdict
# read stuff
with codecs.open("artists.dat", encoding = "utf-8") as f:
artists = f.readlines()
with codecs.open("user_artists.dat", encoding = "utf-8") as f:
users = f.readlines()
# transform artist data in a dict with "artist id" as key and "artist name" as value
artist_repo = dict(x.strip().split('\t')[:2] for x in artists[1:])
user_stats_list = [x.strip().split('\t') for x in users][1:]
grouped_user_stats = defaultdict(lambda:0)
for u in user_stats_list:
#userID, artistID, weight = u
grouped_user_stats[u[0]] += int(u[2]) # accumulate weights in a dict with artist id as key and sum of wights as values
# extra: "fancying" the data transforming the keys of the dict in "<artist name> (artist id)" format
grouped_user_stats = dict(("%s (%s)" % (artist_repo.get(k,"Unknown artist"), k), v) for k ,v in grouped_user_stats.iteritems() )
# lastly print it
for k, v in grouped_user_stats.iteritems():
print k,v

Scraping JSON arrays nested tags

I am trying to scrape data from a JSON file. I am able to scrape data from some of the tags but few nested tags are giving problem. Following is a sample from the file -
{"orders":[{
"order_id":9000,
"flight_start":"2017-06-15T05:00:00.000Z",
"flight_end":"2017-06-22T05:00:00.000Z",
"spots":[{
"spot_id":7354259,
"spot_length":15}],
"constraints":{
"forbid":[{
"network":"BRVO"},
{"network":"DSE"},
{"network":"ESPN"},
{"network":"DFC"},
{"hours":[2,6],
"days_of_week":["Monday","Tuesday","Thursday","Friday"]},
{"hours":[2,6],
"days_of_week":["Saturday","Sunday"]}],
"allocation":[{
"hours":[6,9],
"impressions":{
"min":0.05,
"max":0.05},
"days_of_week":["Monday","Tuesday","Wednesday","Thursday","Friday"]},{
"hours":[20,0],
"impressions":{"min":0.5,"max":0.5},
"days_of_week":["Monday","Tuesday","Wednesday","Thursday","Friday"]},{
"budget":{
"min":1,
"max":1},
"spot_length":15}]}}]}
I am not able to scrape all values from network tag, it is only returning top value from all the network tabs for each order.
I am using the following code -
import urllib
import json
url = 'http://vw-test.elasticbeanstalk.com/test'
json_obj = urllib.request.urlopen(url).read().decode('UTF-8')
data = json.loads(json_obj)
for i in data["orders"]:
k = i["order_id"]
j = i["flight_start"]
l = i["flight_end"]
m = i ['spots']
for value in m:
a = value["spot_length"]
b = value["spot_id"]
n = i["constraints"]
c = n["forbid"]
d = c[0]
e = d["network"]
print(e)
If any one could help me figure this out I'll be very grateful.

The json data in your question isn't complete. Making some assumptions, this could work:
for i in data["orders"]:
k = i["order_id"]
j = i["flight_start"]
l = i["flight_end"]
m = i ['spots']
for value in m:
a = value["spot_length"]
b = value["spot_id"]
n = i["constraints"]
c = n["forbid"]
d = c[0]
networks = [d["network"] for d in c if "network" in d]
print(networks)

Adding to JSON in Python and converting to an object

I have a JSON array shown below.
[
"3D3iAR9M4HDETajfD79gs9BM8qhMSq5izX",
"35xfg4UnpEJeHDo55HNwJbr1V3G1ddCuVA"
]
I would like to add a value in the form of the string (self.tx_amount_5) so I get a JSON OBJECT something like this:
{
"3D3iAR9M4HDETajfD79gs9BM8qhMSq5izX" : 100000
"35xfg4UnpEJeHDo55HNwJbr1V3G1ddCuVA" : 100000
}
The part of code that has generated the first JSON array is:
r = requests.get('http://api.blockcypher.com/v1/btc/main/addrs/A/balance')
balance = r.json()['balance']
with open("Entries#x1.csv") as f,open("winningnumbers.csv") as nums:
nums = set(imap(str.rstrip, nums))
r = csv.reader(f)
results = defaultdict(list)
for row in r:
results[sum(n in nums for n in islice(row, 1, None))].append(row[0])
self.number_matched_0 = results[0]
self.number_matched_1 = results[1]
self.number_matched_2 = results[2]
self.number_matched_3 = results[3]
self.number_matched_4 = results[4]
self.number_matched_5 = results[5]
self.number_matched_5_json = json.dumps(self.number_matched_5, sort_keys = True, indent = 4)
print(self.number_matched_5_json)
if len(self.number_matched_3) == 0:
print('Nobody matched 3 numbers')
else:
self.tx_amount_3 = int((balance*0.001)/ len(self.number_matched_3))
if len(self.number_matched_4) == 0:
print('Nobody matched 4 numbers')
else:
self.tx_amount_4 = int((balance*0.1)/ len(self.number_matched_4))
if len(self.number_matched_5) == 0:
print('Nobody matched 3 numbers')
else:
self.tx_amount_5 = int((balance*0.4)/ len(self.number_matched_5))

If I understand correctly, you can create the dictionary like this:
import json
s="""[
"3D3iAR9M4HDETajfD79gs9BM8qhMSq5izX",
"35xfg4UnpEJeHDo55HNwJbr1V3G1ddCuVA"
]"""
d = {el: self.tx_amount_5 for el in json.loads(s)}
print(d)
which produces
{'3D3iAR9M4HDETajfD79gs9BM8qhMSq5izX': 100000,
'35xfg4UnpEJeHDo55HNwJbr1V3G1ddCuVA': 100000}

python generating nested dictionary key error

I am trying to create a nested dictionary from a mysql query but I am getting a key error
result = {}
for i, q in enumerate(query):
result['data'][i]['firstName'] = q.first_name
result['data'][i]['lastName'] = q.last_name
result['data'][i]['email'] = q.email
error
KeyError: 'data'
desired result
result = {
'data': {
0: {'firstName': ''...}
1: {'firstName': ''...}
2: {'firstName': ''...}
}
}

You wanted to create a nested dictionary
result = {} will create an assignment for a flat dictionary, whose items can have any values like "string", "int", "list" or "dict"
For this flat assignment
python knows what to do for result["first"]
If you want "first" also to be another dictionary you need to tell Python by an assingment
result['first'] = {}.
otherwise, Python raises "KeyError"
I think you are looking for this :)
>>> from collections import defaultdict
>>> mydict = lambda: defaultdict(mydict)
>>> result = mydict()
>>> result['Python']['rules']['the world'] = "Yes I Agree"
>>> result['Python']['rules']['the world']
'Yes I Agree'

result = {}
result['data'] = {}
for i, q in enumerate(query):
result['data']['i'] = {}
result['data'][i]['firstName'] = q.first_name
result['data'][i]['lastName'] = q.last_name
result['data'][i]['email'] = q.email
Alternatively, you can use you own class which adds the extra dicts automatically
class AutoDict(dict):
def __missing__(self, k):
self[k] = AutoDict()
return self[k]
result = AutoDict()
for i, q in enumerate(query):
result['data'][i]['firstName'] = q.first_name
result['data'][i]['lastName'] = q.last_name
result['data'][i]['email'] = q.email

result['data'] does exist. So you cannot add data to it.
Try this out at the start:
result = {'data': []};

You have to create the key data first:
result = {}
result['data'] = {}
for i, q in enumerate(query):
result['data'][i] = {}
result['data'][i]['firstName'] = q.first_name
result['data'][i]['lastName'] = q.last_name
result['data'][i]['email'] = q.email