删除JSON对象中的元素
我试图遍历对象列表,从每个对象中删除一个元素。每个对象都是换行符。我试图然后按原样保存新文件,而对象中不包含任何元素。我知道这可能是一个简单的任务,但似乎无法完成这项工作。如果有人可以伸出援手,将不胜感激。谢谢。
{"business_id": "fNGIbpazjTRdXgwRY_NIXA",
"full_address": "1201 Washington Ave\nCarnegie, PA 15106",
"hours": {
"Monday": {
"close": "23:00",
"open": "11:00"
},
"Tuesday": {
"close": "23:00",
"open": "11:00"
},
"Friday": {
"close": "23:00",
"open": "11:00"
},
"Wednesday": {
"close": "23:00",
"open": "11:00"
},
"Thursday": {
"close": "23:00",
"open": "11:00"
},
"Saturday": {
"close": "23:00",
"open": "11:00"
}
},
"open": true,
"categories": ["Bars", "American (Traditional)", "Nightlife", "Lounges", "Restaurants"],
"city": "Carnegie",
"review_count": 7,
"name": "Rocky's Lounge",
"neighborhoods": [],
"longitude": -80.0849416,
"state": "PA",
"stars": 4.0,
"latitude": 40.3964688,
"attributes": {
"Alcohol": "full_bar",
"Noise Level": "average",
"Music": {
"dj": false
},
"Attire": "casual",
"Ambience": {
"romantic": false,
"intimate": false,
"touristy": false,
"hipster": false,
"divey": false,
"classy": false,
"trendy": false,
"upscale": false,
"casual": false
},
"Good for Kids": true,
"Wheelchair Accessible": true,
"Good For Dancing": false,
"Delivery": false,
"Dogs Allowed": false,
"Coat Check": false,
"Smoking": "no",
"Accepts Credit Cards": true,
"Take-out": true,
"Price Range": 1,
"Outdoor Seating": false,
"Takes Reservations": false,
"Waiter Service": true,
"Wi-Fi": "free",
"Caters": false,
"Good For": {
"dessert": false,
"latenight": false,
"lunch": false,
"dinner": false,
"brunch": false,
"breakfast": false
},
"Parking": {
"garage": false,
"street": false,
"validated": false,
"lot": true,
"valet": false
},
"Has TV": true,
"Good For Groups": true
},
"type": "business"
}
我需要删除小时元素中包含的信息,但是信息并不总是相同的。有些包含全天,有些仅包含一两天的信息。我尝试使用的代码是Pyton,我整天都在搜索该代码以解决问题。我对Python不太熟练。任何帮助,将不胜感激。
import jsonwith open('data.json') as data_file:
data = json.load(data_file)
for element in data:
del element['hours']
对不起,只是添加我在运行代码时遇到的错误是TypeError:“ unicode”对象不支持项目删除
回答:
假设您要覆盖相同的文件:
import jsonwith open('data.json', 'r') as data_file:
data = json.load(data_file)
for element in data:
element.pop('hours', None)
with open('data.json', 'w') as data_file:
data = json.dump(data, data_file)
dict.pop(<key>,
not_found=None)如果我了解您的要求,那么您可能正在寻找。因为hours
如果存在,它将删除密钥,如果不存在,它将不会失败。
但是我不确定我是否理解小时键是否包含几天对您有什么影响,因为您只想摆脱整个键/值对,对吗?
现在,如果您确实要使用del
而不是pop
,请按以下步骤使代码正常工作:
import jsonwith open('data.json') as data_file:
data = json.load(data_file)
for element in data:
if 'hours' in element:
del element['hours']
with open('data.json', 'w') as data_file:
data = json.dump(data, data_file)
因此,如您所见,我添加了代码以将数据写回到文件中。如果要将其写入另一个文件,只需在第二个open语句中更改文件名。
您可能已经注意到,我不得不更改了缩进,以便在数据清理阶段关闭文件,并在最后将其覆盖。
with
是所谓的上下文管理器,它提供的任何内容(此处为data_file文件描述符)
在该上下文中可用。这意味着,一旦with
块的缩进结束,文件就会关闭并且上下文也将结束,文件描述符也变得无效/过时。
如果不这样做,您将无法以写入模式打开文件并获得要写入的新文件描述符。
我希望已经足够清楚了…
这次,您似乎显然需要执行以下操作:
with open('dest_file.json', 'w') as dest_file: with open('source_file.json', 'r') as source_file:
for line in source_file:
element = json.loads(line.strip())
if 'hours' in element:
del element['hours']
dest_file.write(json.dumps(element))
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