Elasticsearch Aggregation to pandas Dataframe

Question

I am working with some ElasticSearch data and i would like to generate the tables from the aggregations like in Kibana. A sample output of the aggregation is below, based on the following code :

    s.aggs.bucket("name1", "terms", field="field1").bucket(
        "name2", "terms", field="innerField1"
    ).bucket("name3", "terms", field="InnerAgg1")
     response = s.execute()
   resp_dict = response.aggregations.name.buckets




{
    "key": "Locationx",
    "doc_count": 12,
    "name2": {
        "doc_count_error_upper_bound": 0,
        "sum_other_doc_count": 0,
        "buckets": [{
            "key": "Sub-Loc1",
            "doc_count": 1,
            "name3": {
                "doc_count_error_upper_bound": 0,
                "sum_other_doc_count": 0,
                "buckets": [{
                    "key": "super-Loc1",
                    "doc_count": 1
                }]
            }
        }, {
            "key": "Sub-Loc2",
            "doc_count": 1,
            "name3": {
                "doc_count_error_upper_bound": 0,
                "sum_other_doc_count": 0,
                "buckets": [{
                    "key": "super-Loc1",
                    "doc_count": 1
                }]
            }
        }]
    }
}

In this case, the expected output would be:

Now, I have tried a variety of methods, with a short description of what went wrong :

Pandasticsearch = completely failed even with just 1 dictionary. The dictionary was not created, as it was struggling with keys, even with each dictionary being dealt with separately:

for d in resp_dict :
    x= d.to_dict()
    pandas_df = Select.from_dict(x).to_pandas()
    print(pandas_df)

In particular, the error that was recieved related to the the fact that the dictionary was not made and thus ['took'] was not a key.

Pandas (pd.Dataframe.from_records()) = only gave me the first aggregation, with a column containing the inner dictionary, and using pd.apply(pd.Series) on it gave another table of resulting dictionaries.

StackOverflow posts recursive function = the dictionary looks completely different than the example used,and tinkering led me nowhere unless i drastically change the input.

TheDonk · Accepted Answer · 2020-01-24 08:28:10Z

Struggling with the same problem, I've come to believe the reason for this being that the response_dict are not normal dicts, but an elasticsearch_dsl.utils.AttrList of elasticsearch_dsl.utils.AttrDict.

If you have an AttrList of AttrDicts, it's possible to do:

resp_dict = response.aggregations.name.buckets
new_response = [i._d_ for i in resp_dict]

To get a list of normal dicts instead. This will probably play nicer with other libraries.

Edit:

I wrote a recursive function which at least handles some cases, not extensively tested yet though and not wrapped in a nice module or anything. It's just a script. The one_lvl function keeps track of all the siblings and siblings of parents in the tree in a dictionary called tmp, and recurses when it finds a new named aggregation. It assumes a lot about the structure of the data, which I'm not sure is warranted in the general case.

The lvl stuff is necessary I think because you might have duplicate names, so key exists at several aggregation-levels for instance.

#!/usr/bin/env python3

from elasticsearch_dsl.query import QueryString
from elasticsearch_dsl import Search, A
from elasticsearch import Elasticsearch
import pandas as pd

PORT = 9250
TIMEOUT = 10000
USR = "someusr"
PW = "somepw"
HOST = "test.com"
INDEX = "my_index"
QUERY = "foobar"

client = Elasticsearch([HOST], port = PORT, http_auth=(USR, PW), timeout = TIMEOUT)

qs = QueryString(query = QUERY)
s = Search(using=client, index=INDEX).query(qs)

s = s.params(size = 0)

agg= {
    "dates" : A("date_histogram", field="date", interval="1M", time_zone="Europe/Berlin"),
    "region" : A("terms", field="region", size=10),
    "county" : A("terms", field="county", size = 10)
}

s.aggs.bucket("dates", agg["dates"]). \
       bucket("region", agg["region"]). \
       bucket("county", agg["county"])

resp = s.execute()

data = {"buckets" : [i._d_ for i in resp.aggregations.dates]}
rec_list = ["buckets"] + [*agg.keys()]

def get_fields(i, lvl):
    return {(k + f"{lvl}"):v for k, v in i.items() if k not in rec_list}

def one_lvl(data, tmp, lvl, rows, maxlvl):
    tmp = {**tmp, **get_fields(data, lvl)}

    if "buckets" not in data:
        rows.append(tmp)

    for d in data:
        if d in ["buckets"]:
            for v, b in enumerate(data[d]):
                tmp = {**tmp, **get_fields(data[d][v], lvl)}
                for k in b:
                    if k in agg.keys():
                        one_lvl(data[d][v][k], tmp, lvl+1, rows, maxlvl)
                    else:
                        if lvl == maxlvl:
                            tmp = {**tmp, (k + f"{lvl}") : data[d][v][k]}
                            rows.append(tmp)

    return rows


rows = one_lvl(data, {}, 1, [], len(agg))
df = pd.DataFrame(rows)

Honestly the best thing that can be done at this time. I would have loved to find a better way, but turning it into a bunch of dicts and using those is the way I did it
I made an attempt at finding a recursive solution, edited the post.

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