I have a flat text file that represents a hierarchy. It looks similar to this:
0 tom (1)
1 janet (8)
2 harry (1)
3 jules (1)
3 jacob (1)
1 mary (13)
2 jeff (1)
3 sam (2)
1 bob (28)
2 dick (1)
I want to read this in and build a nested dictionary (or some kind of data structure) to represent the hierarchy so it is easier to manage but I can't wrap my head around how to iterate and create a data structure. Maybe recursion?
The first number is the level of the hierarchy, the word is the name I want to store and the value in the parenthesis is the quantity that I also want to store.
I'd like to end up with something similar to this:
{
"tom": {
"quantity": 1,
"names": {
"janet": {
"quantity": 8,
"names": {
"harry": {
"quantity": 1,
"names": {
"jules": {
"quantity": 1
},
"jacob": {
"quantity": 1
}
}
}
}
},
"mary": {
"quantity": 13,
"names": {
"jeff": {
"quantity": 1,
"names": {
"sam": {
"quantity": 2
}
}
}
}
},
"bob": {
"quantity": 28,
"names": {
"dick": {
"quantity": 1
}
}
}
}
}
}
You can use recursion:
import re
with open('test_hierarchy.txt') as f:
d = [[int((k:=re.findall('\d+|\w+', i))[0]), k[1], int(k[-1])] for i in f]
def to_tree(data):
if not data:
return {}
r, _key, _val = {}, None, []
for a, b, c in data:
if not a:
if _key is not None:
r[_key[0]] = {'quantity':_key[-1], 'names':to_tree(_val)}
_key, _val = (b, c), []
else:
_val.append([a-1, b, c])
r = {**r, _key[0]:{'quantity':_key[-1], 'names':to_tree(_val)}}
return {a:{'quantity':b['quantity']} if not b['names'] else b for a, b in r.items()}
import json
print(json.dumps(to_tree(d), indent=4))
Output:
{
"tom": {
"quantity": 1,
"names": {
"janet": {
"quantity": 8,
"names": {
"harry": {
"quantity": 1,
"names": {
"jules": {
"quantity": 1
},
"jacob": {
"quantity": 1
}
}
}
}
},
"mary": {
"quantity": 13,
"names": {
"jeff": {
"quantity": 1,
"names": {
"sam": {
"quantity": 2
}
}
}
}
},
"bob": {
"quantity": 28,
"names": {
"dick": {
"quantity": 1
}
}
}
}
}
}
Related
I am filtering in ElasticSearch. I want doc_count to return 0 on non-data dates, but it doesn't print those dates at all, only dates with data are returned to me. do you know how i can do it? Here is the Python output:
0 NaN
1 NaN
2 NaN
3 NaN
4 NaN
...
33479 {'date': '2022-04-13T08:08:00.000Z', 'value': 7}
33480 {'date': '2022-04-13T08:08:00.000Z', 'value': 7}
33481 {'date': '2022-04-13T08:08:00.000Z', 'value': 7}
33482 {'date': '2022-04-13T08:08:00.000Z', 'value': 7}
33483 {'date': '2022-04-13T08:08:00.000Z', 'value': 7}
And here is my ElasticSearch filter:
"from": 0,
"size": 0,
"query": {
"bool": {
"must":
[
{
"range": {
"#timestamp": {
"gte": "now-1M",
"lt": "now"
}
}
}
]
}
},
"aggs": {
"continent": {
"terms": {
"field": "source.geo.continent_name.keyword"
},
"aggs": {
"_source": {
"date_histogram": {
"field": "#timestamp", "interval": "8m"
}}}}}}
You need to set min_doc_count value to 0 for aggregation where you want result with zero doc_count.
{
"from": 0,
"size": 0,
"query": {
"bool": {
"must": [
{
"range": {
"#timestamp": {
"gte": "now-1M",
"lt": "now"
}
}
}
]
}
},
"aggs": {
"continent": {
"terms": {
"field": "source.geo.continent_name.keyword",
"min_doc_count": 0
},
"aggs": {
"_source": {
"date_histogram": {
"field": "#timestamp",
"interval": "8m",
"min_doc_count": 0
}
}
}
}
}
}
So say I have a json with the following structure:
{
"json_1": {
"1": {
"banana": 0,
"corn": 5,
"apple": 5
},
"2": {
"melon": 10
},
"3": {
"onion": 9,
"garlic": 4
}
}
}
but I also have another json with the same structure but a little different data:
{
"json_2": {
"1": {
"banana": 2,
"corn": 3
},
"2": {
"melon": 1,
"watermelon": 5
},
"3": {
"onion": 4,
"garlic": 1
}
}
}
whats a fast algorithm to combine these two jsons into one so that for each number i would have the json_1 amount and the json_2 amount for every fruit and if for example one json doesn't have a fruit that the other one has it will not combine them:
{
"combined": {
"1": {
"banana": {
"json_1": 0,
"json_2": 2
},
"corn": {
"json_1": 5,
"json_2": 3
}
},
"2": {
"melon": {
"json_1": 10,
"json_2": 1
}
},
"3": {
"onion": {
"json_1": 9,
"json_2": 4
},
"garlic": {
"json_1": 4,
"json_2": 1
}
}
}
}
I have this json file loaded in Python with json.loads('myfile.json'):
[
{
"cart": {
"items": {
"3154ba405e5c5a22bbdf9bf1": {
"item": {
"_id": "3154ba405e5c5a22bbdf9bf1",
"title": "Drink alla cannella",
"price": 5.65,
"__v": 0
},
"qty": 1,
"price": 5.65
}
},
"totalQty": 1,
"totalPrice": 5.65
}
},
{
"cart": {
"items": {
"6214ba405e4c5a31bbdf9ad7": {
"item": {
"_id": "6214ba405e4c5a31bbdf9ad7",
"title": "Drink alla menta",
"price": 5.65,
"__v": 0
},
"qty": 2,
"price": 11.3
}
},
"totalQty": 2,
"totalPrice": 11.3
}
}
]
How I can access to both totalQty and totalPrice fields at same time and sum them?
How I can access to both Title fields to print it?
Let's assume that you have the JSON data available as a string then:
jdata = '''
[
{
"cart": {
"items": {
"3154ba405e5c5a22bbdf9bf1": {
"item": {
"_id": "3154ba405e5c5a22bbdf9bf1",
"title": "Drink alla cannella",
"price": 5.65,
"__v": 0
},
"qty": 1,
"price": 5.65
}
},
"totalQty": 1,
"totalPrice": 5.65
}
},
{
"cart": {
"items": {
"6214ba405e4c5a31bbdf9ad7": {
"item": {
"_id": "6214ba405e4c5a31bbdf9ad7",
"title": "Drink alla menta",
"price": 5.65,
"__v": 0
},
"qty": 2,
"price": 11.3
}
},
"totalQty": 2,
"totalPrice": 11.3
}
}
]
'''
totalQty = 0
totalPrice = 0
for d in json.loads(jdata):
c = d['cart']
totalQty += c['totalQty']
totalPrice += c['totalPrice']
for sd in c['items'].values():
print(sd['item']['title'])
print(f'{totalQty:d}', f'{totalPrice:.2f}')
Output:
3 16.95
Note:
I suspect that what you really want to do is multiply those two values
Let's say I have a collection like the following. For every document that contains animals.horse, I want to set animals.goat equal to animals.horse (so the horses don't get lonely or outnumbered).
[
{
"_id": 1,
"animals": {
"goat": 1
}
},
{
"_id": 2,
"animals": {
"cow": 1,
"horse": 2,
"goat": 1
}
},
{
"_id": 3,
"animals": {
"horse": 5
}
},
{
"_id": 4,
"animals": {
"cow": 1
}
}
]
In Mongo shell, this works as desired:
db.collection.update(
{"animals.horse": { "$gt": 0 }},
[ { "$set": { "animals.goat": "$animals.horse" } } ],
{ "multi": true }
)
which achieves the desired result:
[
{
"_id": 1,
"animals": {
"goat": 1
}
},
{
"_id": 2,
"animals": {
"cow": 1,
"goat": 2,
"horse": 2
}
},
{
"_id": 3,
"animals": {
"goat": 5,
"horse": 5
}
},
{
"_id": 4,
"animals": {
"cow": 1
}
}
]
However, this doesn't work in pymongo -- the collection is unaltered.
db.collection.update_many( filter = {'animals.horse': {'$gt':0} },
update = [ {'$set': {'animals.goat': '$animals.horse' } } ],
upsert = True
)
What am I doing wrong?
I would like to create a dictionary containing a nested structure of dictionaries, like bellow :
{
"Jaque": {
"ES": {
"Madrid": [
{
"experience": 9
}
]
},
"FR": {
"Lyon": [
{
"experience": 11.4
}
],
"Paris": [
{
"experience": 20
}
]
}
},
"James": {
"UK": {
"London": [
{
"experience": 10.9
}
]
}
},
"Henry": {
"UK": {
"London": [
{
"experience": 15
}
]
}
},
"Joe": {
"US": {
"Boston": [
{
"experience": 100
}
]
}
}
}
}
My input is a list of dictionaries of this format:
c = [{
"country": "US",
"city": "Boston",
"name": "Joe",
"experience": 100
},
{
"country": "FR",
"city": "Paris",
"name": "Jaque",
"experience": 20
},
{
"country": "FR",
"city": "Lyon",
"name": "Jaque",
"experience": 11.4
},
{
"country": "ES",
"city": "Madrid",
"name": "Jaque",
"experience": 9
},
{
"country": "UK",
"city": "London",
"name": "Henry",
"experience": 15
},
{
"country": "UK",
"city": "London",
"name": "James",
"experience": 10.9
}
]
My first approach was to create the nested dict, step by step:
dd = dict.fromkeys([i.get("name") for i in c],defaultdict(dict))
#will create
# dd = {'Joe': defaultdict(<class 'dict'>, {}), 'Jaque': defaultdict(<class 'dict'>, {}), 'James': defaultdict(<class 'dict'>, {}), 'Henry': defaultdict(<class 'dict'>, {})}
for i in dd:
for j in c:
#verify if name from d is in dict j
if i in j.values():
dd[i]=dict(zip([a.get("country") for a in c if i in a.values() ],[b.get("city") for b in c if i in b.values() ]))
# dd will become
#{'Joe': {'US': 'Boston'}, 'Jaque': {'FR': 'Lyon', 'ES': 'Madrid'}, 'Henry': {'UK': 'London'}, 'James': {'UK': 'London'}}
Now I can't figure a way to create/update the nested structure of dict dd. Is there a more dynamic way to create dict? Thx
You could use itertools.groupby to organize the list similarly to your expected output and then loop to convert to a dict.
from itertools import groupby
from operator import itemgetter
data = [{"country": "US", "city": "Boston", "name": "Joe", "experience": 100 }, {"country": "FR", "city": "Paris", "name": "Jaque", "experience": 20 }, {"country": "FR", "city": "Lyon", "name": "Jaque", "experience": 11.4 }, {"country": "ES", "city": "Madrid", "name": "Jaque", "experience": 9 }, {"country": "UK", "city": "London", "name": "Henry", "experience": 15 }, {"country": "UK", "city": "London", "name": "James", "experience": 10.9 } ]
result = {}
for key, values in groupby(sorted(data, key=itemgetter('name')), key=itemgetter('name')):
result[key] = {
v['country']: {v['city']: [{'experience': v['experience']}]} for v in values
}
print(result)
# {'Henry': {'UK': {'London': [{'experience': 15}]}}, 'James': {'UK': {'London': [{'experience': 10.9}]}}, 'Jaque': {'FR': {'Lyon': [{'experience': 11.4}]}, 'ES': {'Madrid': [{'experience': 9}]}}, 'Joe': {'US': {'Boston': [{'experience': 100}]}}}
You can use recursion with itertools.groupby:
from itertools import groupby
def group(d, keys = None):
key, *keys = keys
new_d = {a:list(b) for a, b in groupby(sorted(d, key=lambda x:x[key]), key=lambda x:x[key])}
t = {a:[{c:d for c, d in k.items() if c != key} for k in b] for a, b in new_d.items()}
return {a:group(b, keys) if not all(len(i) == 1 for i in b) else b for a, b in t.items()}
result = group(data, keys = ['name', 'country', 'city', 'experience'])
import json
print(json.dumps(result, indent=4)))
Output:
{
"Henry": {
"UK": {
"London": [
{
"experience": 15
}
]
}
},
"James": {
"UK": {
"London": [
{
"experience": 10.9
}
]
}
},
"Jaque": {
"ES": {
"Madrid": [
{
"experience": 9
}
]
},
"FR": {
"Lyon": [
{
"experience": 11.4
}
],
"Paris": [
{
"experience": 20
}
]
}
},
"Joe": {
"US": {
"Boston": [
{
"experience": 100
}
]
}
}
}