How to parse a dataframe column having dynamic records and write it into different columns by maintaining the relation with the table

Question:

cur = snow_connection.cursor()
data = snow_connection.cursor().execute("SELECT x,y,z FROM abc.testing_view2").fetchall()
df = pd.DataFrame(data)
df.columns = ['x','y','z']
print(df)

temp_col=df.iloc[:,2]
print(temp_col)
for i in range (len(df.iloc[:,0])):
    for j in range(len(df.iloc[:,1])):
        for k in range(len(temp_col)):
            json_dict = str(json.loads(str(k)))
            print(json_dict)
    # accessing values in the dictionary
            portfolio_name = str(json_dict[0])
            portfolio_id = str(json_dict[1])
            for l in range(len(portfolio_name)):
                for m in range(len(portfolio_id)):
    # created a new dictionary with all three columns
                    new_dict = {'name': portfolio_name, 'portfolio_id': portfolio_id,'x': ['X'], 'y': ['Y']}
                    print(new_dict)
                    new_data = []
                    new_data.append(new_dict)
                    new_df = pd.DataFrame(new_data)

The problem is column z(which is having json record with name and id) is having dynamic records i.e could contain only one record with name and id and more than 1 too. but it will be mapped with only one x and y. So i have to write this data into multiple rows i.e if column z contains one record then it should write the data into 4 columns x,y ,name and id and if column z contains 2 records then it should write the data and create 2 rows.

I am iterating into this column and writing it into a dictionary to maintain the relation however it is not reading the json dict i created for name and id.

Sample Data:

{
  "document_portfolio": "[{"name": "MWP FT Income ICP", "portfolio_id": "31a01afd-1f69-e617-ade3-d7c1895c4461"}, {"name": "MWP Tactical ETP IMG", "portfolio_id": "13281ca6-a9a7-d361-1cf2-07e6fa3e283a"}]",
  "document_uuid": "28d7ccb1-3f9f-fdd0-c757-6b134a74fdd3",
  "user_id": "00u9vj92B0ZPUMU9b5d5"
}

{
"document_portfolio": "[{"name": "tesying", "portfolio_id": "59d26651-3484-e7ef-9ece-f7194d7639e0"}]","document_uuid": "1cf1ca8e-f0e9-844b-11d6-0d05302fb777",
"user_id": "00u5flkeths3G668k5d7"
}

Expected output :

portfolio_id    name    portfolio_uuid  user_id

31a01afd-1f69-e617-ade3-d7c1895c4461    MWP FT Income ICP   28d7ccb1-3f9f-fdd0-c757-6b134a74fdd3    00u9vj92B0ZPUMU9b5d5

13281ca6-a9a7-d361-1cf2-07e6fa3e283a    MWP Tactical ETP IMG    28d7ccb1-3f9f-fdd0-c757-6b134a74fdd3    00u9vj92B0ZPUMU9b5d5        

Sample Snowflake logic:

    CREATE OR REPLACE VIEW port_test_vw AS
    SELECT 
      LATERAL FLATTEN(PARSE_JSON(data:document_portfolio):portfolio) flattened,
flattened:name::string AS portfolio_name,
flattened:portfolio_id::string AS portfolio_id,
FROM XYZ;
Asked By: Beginner

||

Answers:

for i in range (len(df.iloc[:,0])):
    for j in range(len(df.iloc[:,1])):
        for s in range(len(df.iloc[:,2])):
            for index,row in df.iterrows():
                json_dict = row['DOCUMENT_PORTFOLIO']
                data = json.loads(json_dict)
                for i,row in enumerate(data):
                    #for j in range(len(df['DOCUMENT_UUID'])):
                    if len(data) == 1:
                        result_dic = {}
                        portfolio_name = row['name']
                        result_dic[i] = portfolio_name
                        print(portfolio_name)
                        portfolio_id = row['portfolio_id']
                        result_dic[i] = portfolio_id
                        print(portfolio_id)
                        DOCUMENT_UUID = df['DOCUMENT_UUID']
                        result_dic[j] = DOCUMENT_UUID
                        print(DOCUMENT_UUID)
                    else:
                        exit()

output:
This is the output when I am not putting condition on length
Ali Mahbod – $1M Valued Client-Copy
5efbd6c7-2abe-5e78-b035-1cfffc18b9cd
[{‘name’: ‘Falip Large Caps 1’, ‘portfolio_id’: ‘7c3b5cd5-b788-8667-23b2-d25fb110e525’}, {‘name’: ‘Falip Large Caps 2’, ‘portfolio_id’: ‘7c464968-91e8-a4d5-9756-07efb0d0c7b6’}]
Falip Large Caps 1
7c3b5cd5-b788-8667-23b2-d25fb110e525
Falip Large Caps 2
7c464968-91

However on adding condition on length I am getting this error:

Traceback (most recent call last):
File "Document_custom_field.py", line 102, in
for index,row in name.iterrows():
AttributeError: ‘str’ object has no attribute ‘iterrows’

I tried converting it into dataframe again and loading the json the again but it didnt worked.

For now I am exiting the loop if length is more than one however it is not writing the records into one single dataframe as per the length condition.

Answered By: Beginner

I have fixed the above issue. Hope if someone is also having you can reuse this logic:

df = pd.DataFrame(data)
df.columns = ['DOCUMENT_ID', 'DOCUMENT_UUID', 'USER_ID', 'DOCUMENT_PORTFOLIO'] 
for row in data:
    document_id = row[0] 
    document_uuid = row[1] 
    user_id = row[2] 
    json_dict = row[3]
    portfolio_data = json.loads(json_dict)
    for item in portfolio_data:
        portfolio_name = item['name'] 
        portfolio_id = item['portfolio_id'] 
        new_row = {'DOCUMENT_ID': document_id, 'DOCUMENT_UUID': document_uuid, 'USER_ID': user_id, 'PORTFOLIO_NAME': portfolio_name, 'PORTFOLIO_ID': portfolio_id}
        df = df.append(new_row, ignore_index=True) 
        #print(df)
        table_name = 'XYZ' 
        with snow_connection.cursor() as cursor:
            cursor.execute(f"CREATE OR REPLACE TABLE {table_name}(DOCUMENT_ID INT,DOCUMENT_UUID TEXT,USER_ID TEXT,PORTFOLIO_NAME TEXT,PORTFOLIO_ID TEXT)") 
df.to_sql(table_name, snow_connection, index=False, if_exists='append') 
Answered By: Beginner

Here is a Snowflake SQL syntax for you:

with x as (
select parse_json('{
  "document_portfolio": [{"name": "MWP FT Income ICP", "portfolio_id": "31a01afd-1f69-e617-ade3-d7c1895c4461"}, {"name": "MWP Tactical ETP IMG", "portfolio_id": "13281ca6-a9a7-d361-1cf2-07e6fa3e283a"}],
  "document_uuid": "28d7ccb1-3f9f-fdd0-c757-6b134a74fdd3",
  "user_id": "00u9vj92B0ZPUMU9b5d5"
}')::variant as var
)
SELECT y.value:name::string AS portfolio_name,
       y.value:portfolio_id::string AS portfolio_id
FROM x,
lateral flatten(input=>var:document_portfolio) y;
Answered By: Mike Walton
CREATE OR REPLACE VIEW document_testing_view222 
AS SELECT 
data:document_portfolio::STRING AS document_portfolio,
doc_port.value:name::string AS portfolio_name,
doc_port.value:portfolio_id::string AS portfolio_id 
FROM pcs.document, 
lateral flatten(input=>PARSE_JSON(data:document_portfolio):document_portfolio) doc_port;
Answered By: Beginner