Merge lists of complex dicts with arbitrary keys - python

I have this code:
dotteds = ["apple.orange.banana", "a.b.c", "a.b.d"]
name = "name"
avtype = "type"
fields = "fields"
main_dictionary_list = []
for x in dotteds:
split_name = x.split('.')
if len(split_name) > 1:
value = {name: split_name[-1], avtype: 'string'}
dicts = []
for y in split_name:
dicts.append({name: y, avtype: {name: y, avtype: "record", fields: []}})
dicts[-1] = value
value = value['name']+split_name[-1]
for z in reversed(range(len(dicts))):
if z != 0:
dicts[z - 1]['type']['fields'].append(dicts[z])
main_dictionary_list.append(dicts[0])
else:
dicts = []
value = {name: split_name[-1], avtype: 'string'}
dicts.append(value)
main_dictionary_list.append(dicts[0])
print(main_dictionary_list)
Which gives me an output like this:
[{
'name': 'apple',
'type': {
'name': 'apple',
'type': 'record',
'fields': [{
'name': 'orange',
'type': {
'name': 'orange',
'type': 'record',
'fields': [{
'name': 'banana',
'type': 'string'
}
]
}
}
]
}
}, {
'name': 'a',
'type': {
'name': 'a',
'type': 'record',
'fields': [{
'name': 'b',
'type': {
'name': 'b',
'type': 'record',
'fields': [{
'name': 'c',
'type': 'string'
}
]
}
}
]
}
}, {
'name': 'a',
'type': {
'name': 'a',
'type': 'record',
'fields': [{
'name': 'b',
'type': {
'name': 'b',
'type': 'record',
'fields': [{
'name': 'd',
'type': 'string'
}
]
}
}
]
}
}
]
Ideally I need:
[{
'name': 'apple',
'type': {
'name': 'apple',
'type': 'record',
'fields': [{
'name': 'orange',
'type': {
'name': 'orange',
'type': 'record',
'fields': [{
'name': 'banana',
'type': 'string'
}
]
}
}
]
}
}, {
'name': 'a',
'type': {
'name': 'a',
'type': 'record',
'fields': [{
'name': 'b',
'type': {
'name': 'b',
'type': 'record',
'fields': [{
'name': 'c',
'type': 'string'
},
{
'name': 'd',
'type': 'string'
}
]
}
}
]
}
}
]
I need to be able to do this with any number of combinations:
apple.orange.banana, a.b.c, a.b.d, a.b.q.e.a.s.d, etc.
I cannot figure out how to combine the similar 'name: key' combinations. It's intended to be avro format.
I have also tried making the dotted values into a dictionary which is a bit of trouble on its own.

You can use recursion with collections.defaultdict:
from collections import defaultdict
def group(vals, last=None):
if any(len(i) == 1 for i in vals):
return [{'name':last, 'type':{'name':last, 'type':'record', 'fields':[{'name':i[0], 'type':'string'} if len(i) == 1 else group([i], i[0])[0] for i in vals]}}]
_d = defaultdict(list)
for i in vals:
_d[i[0]].append(i[1:])
return [{'name':a, 'type':group(b, last=a)} if last is None else
{'name':last, 'type':'record', 'fields':group(b, last=a)} for a, b in _d.items()]
import json
vals = ['apple.orange.banana', 'a.b.c', 'a.b.d']
print(json.dumps(group([i.split('.') for i in vals]), indent=4))
Output:
[
{
"name": "apple",
"type": [
{
"name": "apple",
"type": "record",
"fields": [
{
"name": "orange",
"type": {
"name": "orange",
"type": "record",
"fields": [
{
"name": "banana",
"type": "string"
}
]
}
}
]
}
]
},
{
"name": "a",
"type": [
{
"name": "a",
"type": "record",
"fields": [
{
"name": "b",
"type": {
"name": "b",
"type": "record",
"fields": [
{
"name": "c",
"type": "string"
},
{
"name": "d",
"type": "string"
}
]
}
}
]
}
]
}
]
vals = ['asd.2', 'asd.3', 'asd.5.3.4']
print(json.dumps(group([i.split('.') for i in vals]), indent=4))
Output:
[
{
"name": "asd",
"type": [
{
"name": "asd",
"type": {
"name": "asd",
"type": "record",
"fields": [
{
"name": "2",
"type": "string"
},
{
"name": "3",
"type": "string"
},
{
"name": "5",
"type": "record",
"fields": [
{
"name": "5",
"type": "record",
"fields": [
{
"name": "3",
"type": {
"name": "3",
"type": "record",
"fields": [
{
"name": "4",
"type": "string"
}
]
}
}
]
}
]
}
]
}
}
]
}
]

Related

How to merge two DSL query for aggregation and filter

I need to search BusinessArea which is Research or Accounting this is array of fields(OR) statement
I need to search Role is Developer or Tester condition this is array of fields(OR) statement
I want to get the count of masterid of BusinessArea, designationNames, Role which is all the names
Name filter is "Group1"
Below is the dictionary
test= [ { 'masterid': '1', 'name': 'Group1', 'BusinessArea': [ 'Accounting','Research'], 'Designation': [ 'L1' 'L2' ] }, { 'masterid': '2', 'name': 'Group1', 'BusinessArea': ['Research','Accounting' ], 'Role': [ { 'id': '5032', 'name': 'Tester' }, { 'id': '5033', 'name': 'Developer' } ], 'Designation': [ 'L1' 'L2' ]}, { 'masterid': '3', 'name': 'Group1', 'BusinessArea': [ 'Engineering' ], 'Role': [ { 'id': '5032', 'name': 'Developer' }, { 'id': '5033', 'name': 'Developer', 'parentname': '' } ], 'Designation': [ 'L1' 'L2' ]}]
Below is the aggregation function
{
"size": 0,
"aggs": {
"countNames": {
"terms": {
"field": "BusinessArea.keyword"
}
},
"designationNames": {
"terms": {
"field": "Designation.keyword"
}
},
"Role": {
"terms": {
"field": "Role.name.keyword"
}
}
}
}
Below is the filtering query
{
"query": {
"bool": {
"must": [
{
"terms": {
"BusinessArea.keyword": [
"Research",
"Accounting"
]
}
},
{
"terms": {
"Role.name.keyword": [
"Developer",
"Tester"
]
}
}
]
}
}
}
"filter": [
"term": {
"name.keyword": "Group1"}]
I need to merge both query and output will be having from the both
Nice start !!! Now you can simply combine all those snippets like this:
{
"size": 0,
"query": {
"bool": {
"filter": [
{
"term": {
"name.keyword": "Group1"
}
},
{
"terms": {
"BusinessArea.keyword": [
"Research",
"Accounting"
]
}
},
{
"terms": {
"Role.name.keyword": [
"Developer",
"Tester"
]
}
}
]
}
},
"aggs": {
"countNames": {
"terms": {
"field": "BusinessArea.keyword"
}
},
"designationNames": {
"terms": {
"field": "Designation.keyword"
}
},
"Role": {
"terms": {
"field": "Role.name.keyword"
}
}
}
}

How to get the individual count of field from Elasticsearch

My content inside a dictionary is below
test=
[ { 'masterid': '1', 'name': 'Group1', 'BusinessArea': [ { 'id': '14', 'name': 'Accounting', 'parentname': 'Finance'}, { 'id': '3', 'name': 'Research', 'parentname': 'R & D' } ], 'Designation': [ { 'id': '16', 'name': 'L1' }, { 'id': '20', 'name': 'L2' }, { 'id': '25', 'name': 'L2' }] },
{ 'masterid': '2', 'name': 'Group1', 'BusinessArea': [ { 'id': '14', 'name': 'Research', 'parentname': '' }, { 'id': '3', 'name': 'Accounting', 'parentname': '' } ], 'Role': [ { 'id': '5032', 'name': 'Tester' }, { 'id': '5033', 'name': 'Developer' } ], 'Designation': [ { 'id': '16', 'name': 'L1' }, { 'id': '20', 'name': 'L2' }, { 'id': '25', 'name': 'L2' }]},
{ 'masterid': '3', 'name': 'Group1', 'BusinessArea': [ { 'id': '14', 'name': 'Engineering' }, { 'id': '3', 'name': 'Engineering', 'parentname': '' } ], 'Role': [ { 'id': '5032', 'name': 'Developer' }, { 'id': '5033', 'name': 'Developer', 'parentname': '' } ], 'Designation': [ { 'id': '16', 'name': 'L1' }, { 'id': '20', 'name': 'L2' }, { 'id': '25', 'name': 'L2' }]}]
Code is below to put into elastic search index
from elasticsearch import Elasticsearch
es = Elasticsearch()
es.indices.create(index='new')
for e in test:
es.index(index="new", body=e, id=e['id'])
I want to get the count of masterid of BusinessArea which is all the names
Here it is Accounting, Research Engineering
[ {
"name": "BusinessArea",
"values": [
{
"name": "Accounting",
"count": "2"
},
{
"name": "Research",
"count": "2"
},
{
"name": "Engineering",
"count": "1"
}]
}]
or can i have answer like below
{
"A": {
"Designation": [{
"key": "L1",
"doc_count": 3
},
{
"key": "L2",
"doc_count": 3
}
]
},
{
"B": {
"BusinessArea": [{
"key": "Accounting",
"doc_count": 2
},
{
"key": "Research",
"doc_count": 2
},
{
"key": "Engineering",
"doc_count": 1
}
]
}
}
If you want to get the individual count of the field you can use the terms aggregation that is a multi-bucket value source-based aggregation where buckets are dynamically built - one per unique value.
Search Query:
{
"size":0,
"aggs": {
"countNames": {
"terms": {
"field": "BusinessArea.name.keyword"
}
}
}
}
Search Result:
"aggregations": {
"countNames": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "Accounting",
"doc_count": 2
},
{
"key": "Research",
"doc_count": 2
},
{
"key": "Engineering",
"doc_count": 1
}
]
}
Update 1:
If you want to have an individual count of the field for Designation as well as BusinessArea
Search Query:
{
"size": 0,
"aggs": {
"countNames": {
"terms": {
"field": "BusinessArea.name.keyword"
}
},
"designationNames": {
"terms": {
"field": "Designation.name.keyword"
}
}
}
}
Search Result:
"aggregations": {
"designationNames": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "L1",
"doc_count": 3
},
{
"key": "L2",
"doc_count": 3
}
]
},
"countNames": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "Accounting",
"doc_count": 2
},
{
"key": "Research",
"doc_count": 2
},
{
"key": "Engineering",
"doc_count": 1
}
]
}
You can simply use the count API of elasticsearch to get the count of All the documents in the elasticsearch index or based on a condition as shown in the same doc.
For your case, it should be like
GET /<your-index-name>/_count?q=name:BusinessArea
Or, if masterid is the Unique-id in your document, you can simply use
GET /<your-index-name>/_count

merge complex list of nested dicts

I'm trying to merge nested Dicts in a list based on "name" like the following:
[
{
"name": "abc",
"metadata": [
{
"name": "foo",
"data": [
{
"version": "1.0"
}
]
},
{
"name": "foo",
"data": [
{
"version": "2.0"
}
]
},
{
"name": "bar",
"data": [
{
"version": "1.0"
}
]
}
]
},
{
"name": "xyz",
"metadata": [
{
"name": "bob",
"data": [
{
"version": "3.2"
}
]
},
{
"name": "alice",
"data": [
{
"version": "2.2"
}
]
}
]
},
{
"name": "xyz",
"metadata": [
{
"name": "mike",
"data": [
{
"version": "3.2"
}
]
},
{
"name": "alice",
"data": [
{
"version": "2.2"
}
]
}
]
}
]
Considering that the merged items should not have duplicates in the metadata, how can I do that in Python? Metadata entries should be unique, if name+data+version exist in the metadata, then the item should not be merged.
my desired output should look like this
[
{
"name": "abc",
"metadata": [
{
"name": "foo",
"data": [
{
"version": "1.0"
}
]
},
{
"name": "foo",
"data": [
{
"version": "2.0"
}
]
},
{
"name": "bar",
"data": [
{
"version": "1.0"
}
]
}
]
},
{
"name": "xyz",
"metadata": [
{
"name": "bob",
"data": [
{
"version": "3.2"
}
]
},
{
"name": "mike",
"data": [
{
"version": "3.2"
}
]
},
{
"name": "alice",
"data": [
{
"version": "2.2"
}
]
}
]
}
]
You can use itertools.groubpy:
import itertools
d = [{'name': 'abc', 'metadata': [{'name': 'foo', 'data': [{'version': '1.0'}]}, {'name': 'foo', 'data': [{'version': '2.0'}]}, {'name': 'bar', 'data': [{'version': '1.0'}]}]}, {'name': 'xyz', 'metadata': [{'name': 'bob', 'data': [{'version': '3.2'}]}, {'name': 'alice', 'data': [{'version': '2.2'}]}]}, {'name': 'xyz', 'metadata': [{'name': 'mike', 'data': [{'version': '3.2'}]}, {'name': 'alice', 'data': [{'version': '2.2'}]}]}]
new_d = [[a, list(b)] for a, b in itertools.groupby(sorted(d, key=lambda x:x['name']), key=lambda x:x['name'])]
result = [{'name':a, 'metadata':[c for j in b for c in j['metadata']]} for a, b in new_d]
final_result = [{**i, 'metadata':[c for d, c in enumerate(i['metadata']) if all(a != c for a in i['metadata'][:d])]} for i in result]
import json
print(json.dumps(final_result, indent=4))
Output:
[
{
"name": "abc",
"metadata": [
{
"name": "foo",
"data": [
{
"version": "1.0"
}
]
},
{
"name": "foo",
"data": [
{
"version": "2.0"
}
]
},
{
"name": "bar",
"data": [
{
"version": "1.0"
}
]
}
]
},
{
"name": "xyz",
"metadata": [
{
"name": "bob",
"data": [
{
"version": "3.2"
}
]
},
{
"name": "alice",
"data": [
{
"version": "2.2"
}
]
},
{
"name": "mike",
"data": [
{
"version": "3.2"
}
]
}
]
}
]

Adding new pairs to a json file

I have a json file I need to add pairs to, I convert it into a dict, but now I need to put my new values in a specific place.
This is some of the json file I convert:
"rootObject": {
"id": "6ff0010c-00fe-485b-b695-4ddd6aca4dcd",
"type": "IDO_GEAR",
"children": [
{
"id": "1dd94d1a-e52d-40b3-a82b-6db02a8fbbab",
"type": "IDO_SYSTEM_LOADCASE",
"children": [],
"childList": "SYSTEMLOADCASE",
"properties": [
{
"name": "IDCO_IDENTIFICATION",
"value": "1dd94d1a-e52d-40b3-a82b-6db02a8fbbab"
},
{
"name": "IDCO_DESIGNATION",
"value": "Lastfall 1"
},
{
"name": "IDSLC_TIME_PORTION",
"value": 100
},
{
"name": "IDSLC_DISTANCE_PORTION",
"value": 100
},
{
"name": "IDSLC_OPERATING_TIME_IN_HOURS",
"value": 1
},
{
"name": "IDSLC_OPERATING_TIME_IN_SECONDS",
"value": 3600
},
{
"name": "IDSLC_OPERATING_REVOLUTIONS",
"value": 1
},
{
"name": "IDSLC_OPERATING_DISTANCE",
"value": 1
},
{
"name": "IDSLC_ACCELERATION",
"value": 9.81
},
{
"name": "IDSLC_EPSILON_X",
"value": 0
},
{
"name": "IDSLC_EPSILON_Y",
"value": 0
},
{
"name": "IDSLC_EPSILON_Z",
"value": 0
},
{
"name": "IDSLC_CALCULATION_WITH_OWN_WEIGHT",
"value": "CO_CALCULATION_WITHOUT_OWN_WEIGHT"
},
{
"name": "IDSLC_CALCULATION_WITH_TEMPERATURE",
"value": "CO_CALCULATION_WITH_TEMPERATURE"
},
{
"name": "IDSLC_FLAG_FOR_LOADCASE_CALCULATION",
"value": "LB_CALCULATE_LOADCASE"
},
{
"name": "IDSLC_STATUS_OF_LOADCASE_CALCULATION",
"value": false
}
I want to add somthing like ENTRY_ONE and ENTRY_TWO like this:
"rootObject": {
"id": "6ff0010c-00fe-485b-b695-4ddd6aca4dcd",
"type": "IDO_GEAR",
"children": [
{
"id": "1dd94d1a-e52d-40b3-a82b-6db02a8fbbab",
"type": "IDO_SYSTEM_LOADCASE",
"children": [],
"childList": "SYSTEMLOADCASE",
"properties": [
{
"name": "IDCO_IDENTIFICATION",
"value": "1dd94d1a-e52d-40b3-a82b-6db02a8fbbab"
},
{
"name": "IDCO_DESIGNATION",
"value": "Lastfall 1"
},
{
"name": "IDSLC_TIME_PORTION",
"value": 100
},
{
"name": "IDSLC_DISTANCE_PORTION",
"value": 100
},
{
"name": "ENTRY_ONE",
"value": 100
},
{
"name": "ENTRY_TWO",
"value": 55
},
{
"name": "IDSLC_OPERATING_TIME_IN_HOURS",
"value": 1
},
{
"name": "IDSLC_OPERATING_TIME_IN_SECONDS",
"value": 3600
},
{
"name": "IDSLC_OPERATING_REVOLUTIONS",
"value": 1
},
{
"name": "IDSLC_OPERATING_DISTANCE",
"value": 1
},
{
"name": "IDSLC_ACCELERATION",
"value": 9.81
},
{
"name": "IDSLC_EPSILON_X",
"value": 0
},
{
"name": "IDSLC_EPSILON_Y",
"value": 0
},
{
"name": "IDSLC_EPSILON_Z",
"value": 0
},
{
"name": "IDSLC_CALCULATION_WITH_OWN_WEIGHT",
"value": "CO_CALCULATION_WITHOUT_OWN_WEIGHT"
},
{
"name": "IDSLC_CALCULATION_WITH_TEMPERATURE",
"value": "CO_CALCULATION_WITH_TEMPERATURE"
},
{
"name": "IDSLC_FLAG_FOR_LOADCASE_CALCULATION",
"value": "LB_CALCULATE_LOADCASE"
},
{
"name": "IDSLC_STATUS_OF_LOADCASE_CALCULATION",
"value": false
}
How can I add the entries so that they are under the IDCO_IDENTIFICATION tag, and not under the rootObject?
The way I see your json file, it WOULD be under rootObject as EVERYTHING is under that key. There's quite a few closing brackets and braces missing.
So I can only assume you are meaning you want it directly under IDCO_IDENTIFICATION (which is nested under rootObject). But that doesn't match what you have as your example output either. You have the new ENTRY_ONE and ENTRY_TWO within the properties, within the children, within the rootObject, not "under" IDCO_IDENTIFICATION. So I'm going to follow what you are asking for from your example output.
import json
with open('C:/test.json') as f:
data = json.load(f)
new_elements = [{"name":"ENTRY_ONE", "value":100},
{"name":"ENTRY_TWO", "value":55}]
for each in new_elements:
data['rootObject']['children'][0]['properties'].append(each)
with open('C:/test.json', 'w') as f:
json.dump(data, f)
Output:
import pprint
pprint.pprint(data)
{'rootObject': {'children': [{'childList': 'SYSTEMLOADCASE',
'children': [],
'id': '1dd94d1a-e52d-40b3-a82b-6db02a8fbbab',
'properties': [{'name': 'IDCO_IDENTIFICATION',
'value': '1dd94d1a-e52d-40b3-a82b-6db02a8fbbab'},
{'name': 'IDCO_DESIGNATION',
'value': 'Lastfall 1'},
{'name': 'IDSLC_TIME_PORTION',
'value': 100},
{'name': 'IDSLC_DISTANCE_PORTION',
'value': 100},
{'name': 'IDSLC_OPERATING_TIME_IN_HOURS',
'value': 1},
{'name': 'IDSLC_OPERATING_TIME_IN_SECONDS',
'value': 3600},
{'name': 'IDSLC_OPERATING_REVOLUTIONS',
'value': 1},
{'name': 'IDSLC_OPERATING_DISTANCE',
'value': 1},
{'name': 'IDSLC_ACCELERATION',
'value': 9.81},
{'name': 'IDSLC_EPSILON_X',
'value': 0},
{'name': 'IDSLC_EPSILON_Y',
'value': 0},
{'name': 'IDSLC_EPSILON_Z',
'value': 0},
{'name': 'IDSLC_CALCULATION_WITH_OWN_WEIGHT',
'value': 'CO_CALCULATION_WITHOUT_OWN_WEIGHT'},
{'name': 'IDSLC_CALCULATION_WITH_TEMPERATURE',
'value': 'CO_CALCULATION_WITH_TEMPERATURE'},
{'name': 'IDSLC_FLAG_FOR_LOADCASE_CALCULATION',
'value': 'LB_CALCULATE_LOADCASE'},
{'name': 'IDSLC_STATUS_OF_LOADCASE_CALCULATION',
'value': False},
{'name': 'ENTRY_ONE',
'value': 100},
{'name': 'ENTRY_TWO',
'value': 55}],
'type': 'IDO_SYSTEM_LOADCASE'}],
'id': '6ff0010c-00fe-485b-b695-4ddd6aca4dcd',
'type': 'IDO_GEAR'}}

Python - Adding fields and labels to nested json file

I have a dataframe as follows:
Name_ID | URL | Count | Rating
------------------------------------------------
ABC | www.example.com/ABC | 10 | 5
123 | www.example.com/123 | 9 | 4
XYZ | www.example.com/XYZ | 5 | 2
ABC111 | www.example.com/ABC111 | 5 | 2
ABC121 | www.example.com/ABC121 | 5 | 2
222 | www.example.com/222 | 5 | 3
abc222 | www.example.com/abc222 | 4 | 2
ABCaaa | www.example.com/ABCaaa | 4 | 2
I am trying to create a JSON as follows:
{
"name": "sampledata",
"children": [
{
"name": 9,
"children": [
{
"name": 4,
"children": [
{
"name": "123",
"size": 100
}
]
}
]
},
{
"name": 10,
"children": [
{
"name": 5,
"children": [
{
"name": "ABC",
"size": 100
}
]
}
]
},
{
"name": 4,
"children": [
{
"name": 2,
"children": [
{
"name": "abc222",
"size": 50
},
{
"name": "ABCaaa",
"size": 50
}
]
}
]
},
{
"name": 5,
"children": [
{
"name": 2,
"children": [
{
"name": "ABC",
"size": 16
},
{
"name": "ABC111",
"size": 16
},
{
"name": "ABC121",
"size": 16
}
]
},
{
"name": 3,
"children": [
{
"name": "222",
"size": 50
}
]
}
]
}
]
}
In order to do that:
I am trying to add labels such as "name" and "children" to the json while creating it.
I tried something like
results = [{"name": i, "children": j} for i,j in results.items()]
But it won't label it properly I believe.
Also, add another field with the label `"size"which I am planning to calculate based on the formula:
(Rating*Count*10000)/number_of_children_to_the_immediate_parent
Here is my dirty code:
import pandas as pd
from collections import defaultdict
import json
data =[('ABC', 'www.example.com/ABC', 10 , 5), ('123', 'www.example.com/123', 9, 4), ('XYZ', 'www.example.com/XYZ', 5, 2), ('ABC111', 'www.example.com/ABC111', 5, 2), ('ABC121', 'www.example.com/ABC121', 5, 2), ('222', 'www.example.com/222', 5, 3), ('abc222', 'www.example.com/abc222', 4, 2), ('ABCaaa', 'www.example.com/ABCaaa', 4, 2)]
df = pd.DataFrame(data, columns=['Name', 'URL', 'Count', 'Rating'])
gp = df.groupby(['Count'])
dict_json = {"name": "flare"}
children = []
for name, group in gp:
temp = {}
temp["name"] = name
temp["children"] = []
rgp = group.groupby(['Rating'])
for n, g in rgp:
temp2 = {}
temp2["name"] = n
temp2["children"] = g.reset_index().T.to_dict().values()
for t in temp2["children"]:
t["size"] = (t["Rating"] * t["Count"] * 10000) / len(temp2["children"])
t["name"] = t["Name"]
del t["Count"]
del t["Rating"]
del t["URL"]
del t["Name"]
del t["index"]
temp["children"].append(temp2)
children.append(temp)
dict_json["children"] = children
print json.dumps(dict_json, indent=4)
Though the above code does print what I need, I am looking for more efficient and cleaner way to do the same, mainly because the actual dataset might be even more nested and complicated. Any help/suggestion will be much appreciated.
Quite an interesting problem and a great question!
You can improve your approach by reorganizing the code inside the loops and using list comprehensions. No need to delete things and introduce temp variables inside loops:
dict_json = {"name": "flare"}
children = []
for name, group in gp:
temp = {"name": name, "children": []}
rgp = group.groupby(['Rating'])
for n, g in rgp:
temp["children"].append({
"name": n,
"children": [
{"name": row["Name"],
"size": row["Rating"] * row["Count"] * 10000 / len(g)}
for _, row in g.iterrows()
]
})
children.append(temp)
dict_json["children"] = children
Or, a "wrapped" version:
dict_json = {
"name": "flare",
"children": [
{
"name": name,
"children": [
{
"name": n,
"children": [
{
"name": row["Name"],
"size": row["Rating"] * row["Count"] * 10000 / len(g)
} for _, row in g.iterrows()
]
} for n, g in group.groupby(['Rating'])
]
} for name, group in gp
]
}
I'm getting the following dictionary printed for you sample input dataframe:
{
"name": "flare",
"children": [
{
"name": 4,
"children": [
{
"name": 2,
"children": [
{
"name": "abc222",
"size": 40000
},
{
"name": "ABCaaa",
"size": 40000
}
]
}
]
},
{
"name": 5,
"children": [
{
"name": 2,
"children": [
{
"name": "XYZ",
"size": 33333
},
{
"name": "ABC111",
"size": 33333
},
{
"name": "ABC121",
"size": 33333
}
]
},
{
"name": 3,
"children": [
{
"name": "222",
"size": 150000
}
]
}
]
},
{
"name": 9,
"children": [
{
"name": 4,
"children": [
{
"name": "123",
"size": 360000
}
]
}
]
},
{
"name": 10,
"children": [
{
"name": 5,
"children": [
{
"name": "ABC",
"size": 500000
}
]
}
]
}
]
}
If I understand correctly what you wan't to do is put a groupby into a nested json, if that is the case then you could use pandas groupby and cast it into a nested list of lists as so:
lol = pd.DataFrame(df.groupby(['Count','Rating'])\
.apply(lambda x: list(x['Name_ID']))).reset_index().values.tolist()
lol should look something like this:
[['10', '5', ['ABC']],
['4', '2', ['abc222', 'ABCaaa']],
['5', '2', ['XYZ ', 'ABC111', 'ABC121']],
['5', '3', ['222 ']],
['9', '4', ['123 ']]]
after that you could loop over lol to put it into a dict, but since you want to set nested items you'l have to use autovivification (check it out):
class autovividict(dict):
def __missing__(self, key):
value = self[key] = type(self)()
return value
d = autovividict()
for l in lol:
d[l[0]][l[1]] = l[2]
now you can use the json pack for printing and exporting:
print json.dumps(d,indent=2)
In case you need more than one groupby, you could concat your groups with pandas, cast to lol, remove any nans, and then loop, let me know if a full example can help.
setup
from io import StringIO
import pandas as pd
txt = """Name_ID,URL,Count,Rating
ABC,www.example.com/ABC,10,5
123,www.example.com/123,9,4
XYZ,www.example.com/XYZ,5,2
ABC111,www.example.com/ABC111,5,2
ABC121,www.example.com/ABC121,5,2
222,www.example.com/222,5,3
abc222,www.example.com/abc222,4,2
ABCaaa,www.example.com/ABCaaa,4,2"""
df = pd.read_csv(StringIO(txt))
size
pre-calculate it
df['size'] = df.Count.mul(df.Rating) \
.mul(10000) \
.div(df.groupby(
['Count', 'Rating']).Name_ID.transform('count')
).astype(int)
solution
create recursive function
def h(d):
if isinstance(d, pd.Series): d = d.to_frame().T
rec_cond = d.index.nlevels > 1 or d.index.nunique() > 1
return {'name': str(d.index[0]), 'size': str(d['size'].iloc[0])} if not rec_cond else \
[dict(name=str(n), children=h(g.xs(n))) for n, g in d.groupby(level=0)]
demo
import json
my_dict = dict(name='flare', children=h(df.set_index(['Count', 'Rating', 'Name_ID'])))
json.dumps(my_dict)
'{"name": "flare", "children": [{"name": "4", "children": [{"name": "2", "children": [{"name": "ABCaaa", "children": {"name": "ABCaaa", "size": "40000"}}, {"name": "abc222", "children": {"name": "abc222", "size": "40000"}}]}]}, {"name": "5", "children": [{"name": "2", "children": [{"name": "ABC111", "children": {"name": "ABC111", "size": "33333"}}, {"name": "ABC121", "children": {"name": "ABC121", "size": "33333"}}, {"name": "XYZ", "children": {"name": "XYZ", "size": "33333"}}]}, {"name": "3", "children": {"name": "222", "size": "150000"}}]}, {"name": "9", "children": [{"name": "4", "children": {"name": "123", "size": "360000"}}]}, {"name": "10", "children": [{"name": "5", "children": {"name": "ABC", "size": "500000"}}]}]}'
my_dict
{'children': [{'children': [{'children': [{'children': {'name': 'ABCaaa',
'size': '40000'},
'name': 'ABCaaa'},
{'children': {'name': 'abc222', 'size': '40000'}, 'name': 'abc222'}],
'name': '2'}],
'name': '4'},
{'children': [{'children': [{'children': {'name': 'ABC111', 'size': '33333'},
'name': 'ABC111'},
{'children': {'name': 'ABC121', 'size': '33333'}, 'name': 'ABC121'},
{'children': {'name': 'XYZ', 'size': '33333'}, 'name': 'XYZ'}],
'name': '2'},
{'children': {'name': '222', 'size': '150000'}, 'name': '3'}],
'name': '5'},
{'children': [{'children': {'name': '123', 'size': '360000'}, 'name': '4'}],
'name': '9'},
{'children': [{'children': {'name': 'ABC', 'size': '500000'}, 'name': '5'}],
'name': '10'}],
'name': 'flare'}

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