How to add or update list in python

Question

I have the following list:

DATA

data_set = [(2, 2021, '2300'),(1, 2021, '2500'),(12, 2020, '2400'),(11, 2020, '1500')]

I want to format this list into the following one,

EXPECTED OUTPUT

    [
        {
            "2021-01-01T00": [
                {
                    "2021-02-01T00:00:00": 2300
                },
                {
                    "2021-01-01T00:00:00": 2500
                }
            ]
        },
        {
            "2020-01-01T00": [
                {
                    "2020-12-01T00:00:00": 2400
                },
                {
                    "2020-11-01T00:00:00": 1500
                }
            ]
        }
    ]

I don't know how to format these two years in two objects. I tried hard but as I'm new to python so please help if you know the way to format this data.

EDITED

SAMPLE CODE

result = []
for item in data_set:
    result.append({
        str(datetime.datetime(item[1], item[0], 1)): float(item[2])
    })

With above code I got the following output,

    [
        {
            "2021-02-01 00:00:00": "2300"
        },
        {
            "2021-01-01 00:00:00": "2500"
        },
        {
            "2020-12-01 00:00:00": "2400"
        },
        {
            "2020-11-01 00:00:00": "1500"
        }
    ]

But I want the records of each year separate as I described above in the EXPECTED OUTPUT.

Please explain the logic of how the output is parsed. Contrary to popular belief, stack overflow users are not blessed with psychic powers :) — MusHusKat
– MusHusKat, Commented Feb 3, 2021 at 15:03
No, its not hard coded. Each tuple in the list data_set contains (month,year,value) like (2,2021,'2300'). We can get date value by datetime.datetime(2021, 2, 1) — UzairBinShaukat
– UzairBinShaukat, Commented Feb 3, 2021 at 15:29

UzairBinShaukat · Accepted Answer · 2021-02-05 02:32:42Z

1

This should work!

data_set = [(2, 2021, '2300'),(1, 2021, '2500'),(12, 2020, '2400'),(11, 2020, '1500')]
import datetime
final_dict = dict()
for item in data_set:

    year_date = str(datetime.datetime(item[1], 1, 1))
    full_date = str(datetime.datetime(item[1], item[0], 1))
    
    current_months_dict = final_dict.get(year_date, list())
    
    current_months_dict.append({full_date: item[2]})
    final_dict[year_date] = current_months_dict

final_list = []
for key, value in final_dict.items():
    final_list.append({key:value})
print(final_list)

edited Feb 5, 2021 at 2:32

UzairBinShaukat

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answered Feb 3, 2021 at 15:57

MusHusKat

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6 Comments

UzairBinShaukat Over a year ago

Thanks for your suggestion but in the EXPECTED OUTPUT section there is a list with two dict of each year. But here in your solution, final result is dict.

UzairBinShaukat Over a year ago

As in the line current_months_dict = final_dict.get(year_date, dict()). For instance, final_dict is a list then how you will cover this line? And this is my question but the way to quest is not so clear.

MusHusKat Over a year ago

Is this what you want?

MusHusKat Over a year ago

@UzairBinShaukat - I changed the code to make the structures list. I think it is what you want.

UzairBinShaukat Over a year ago

Thank you for the suggestion but in the last line final_list = [item for item in final_dict.items()] each of the item in dict is being appended in the list including year_date key. It is not according to the EXPECTED OUTPUT.

|

Ajay · Accepted Answer · 2021-02-03 16:59:56Z

Using groupby from itertools

from itertools import groupby
from datetime import datetime

m=[{" ".join(map(str,elem[0:2])):int(elem[2]) for elem in list(g)} for (k,g) in groupby(data_set,lambda x: x[1])]
#grouping is done based on the year 1.e x[1] first element
[{'2 2021': 2300, '1 2021': 2500}, {'12 2020': 2400, '11 2020': 1500}]

#using datetime module to parse date as required
n=[ {datetime.strptime(k,'%m %Y').strftime("%Y-%m-%dT%H:%M:%S"):v for k,v in item.items() } for item in m ]
[{'2021-02-01T00:00:00': 2300, '2021-01-01T00:00:00': 2500},
{'2020-12-01T00:00:00': 2400, '2020-11-01T00:00:00': 1500}]

#creating nested dict using minimum of existing keys
o=[{min(item.keys()):[{k:v} for k,v in item.items()]} for item in n]
[{'2021-01-01T00:00:00': [{'2021-02-01T00:00:00': 2300},
{'2021-01-01T00:00:00': 2500}]},
{'2020-11-01T00:00:00': [{'2020-12-01T00:00:00': 2400},
{'2020-11-01T00:00:00': 1500}]}]

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