Save data in item wihitn Scrapy and Python

Question

I need to store data in Scrapy Item in parts. To update only the information that I found at the moment.

I did it, but the code looks too verbose. Three lines I repeat very often:

price['pricereg'] = pricereg
price['priceprolong'] = priceprolong 
price['pricechange'] = pricechange 

Is it possible to make the code shorter?

EMPTY_PRICE = {
        'pricereg': None,
        'priceprolong': None,
        'pricechange': None,
    }

item['name'] = "some name"
        price = item.get('price', EMPTY_PRICE)
        price['pricereg'] = pricereg
        price['priceprolong'] = priceprolong
        item['price'] = price

item['name'] = "some name"  
        price = item.get('price', EMPTY_PRICE)
        price['pricechange'] = pricechange 
        item['price'] = price

Answer 1

Have you tried using a collections.namedtuple or a dataclasses.dataclass rather than a dict?

from collections import namedtuple

Item = namedtuple('Item', ['name', 'pricereg', 'priceprolong', 'price'],
                  defaults=['', None, None, None])

EMPTY_ITEM = Item()
item['name'] = Item('name', pricereg, priceprolong, price)

print(item['name']._asdict())

or

from dataclasses import dataclass

@dataclass
class Item:
    name: str
    pricereg: int|None = None
    priceprolong: int|None = None
    price: int|None = None

EMPTY_ITEM = Item()
item['name'] = Item('name', pricereg, priceprolong, price)

print(item['name'].pricereg)

alternatively, if you're just setting the dict in one go:

item['name'] = {'pricereg': pricereg, 
                'priceprolong': priceprolong,
                'pricechange': pricechange if pricechange else None, # For defaults 
                'price': price}

If it's to do with setting it up in parts, you can build the dict with the parts you have and do a dict.update

item['name'] = {'pricereg': pricereg, 
                'priceprolong': priceprolong}

tmp = {
       'pricechange': pricechange if pricechange else None, # For defaults 
       'price': price}
item['name'].update(tmp)