Python 通过JSON进行迭代需要比I';I’我想要
我正在读取一个.json文件并解析一些数据以保存到一个对象中。JSON中只有2000个左右的项目需要迭代,但我当前运行的脚本比我希望的要长得多Python 通过JSON进行迭代需要比I';I’我想要,python,refactoring,Python,Refactoring,我正在读取一个.json文件并解析一些数据以保存到一个对象中。JSON中只有2000个左右的项目需要迭代,但我当前运行的脚本比我希望的要长得多 data_file = 'v1/data/data.json' user = User.objects.get(username='lsv') format = Format(format='Limited') format.save() lost_cards = [] lost_cards_file = 'v1/data/LostCards.txt'
data_file = 'v1/data/data.json'
user = User.objects.get(username='lsv')
format = Format(format='Limited')
format.save()
lost_cards = []
lost_cards_file = 'v1/data/LostCards.txt'
with open(data_file) as file:
data = json.load(file)
for item in data:
if item['model'] == 'cards.cardmodel':
if len(Card.objects.filter(name=item['fields']['name'])) == 0:
print(f"card not found: {item['fields']['name']}")
lost_cards.append(item['fields']['name'])
try:
Rating(
card=Card.objects.get(name=item['fields']['name'], set__code=item['fields']['set']),
rating=item['fields']['rating'],
reason=item['fields']['reason'],
format=format,
rator=user
).save()
except Exception as e:
print(e, item['fields']['name'], item['fields']['set'])
break
with open(lost_cards_file, 'w') as file:
file.write(str(lost_cards))
代码按预期工作,但所花的时间比我希望的要长得多。我希望有一个内置的JSON或迭代器函数可以加速这个过程。有。它被称为模块
打开(数据文件,'r')作为输入文件:
字典\u from\u json=json.load(输入\u文件)
我应该这样做