当尝试从JSON对象中删除值时,为什么我收到错误';Unicode';Object不支持项目删除&Quot;?
问题描述
我试图遍历对象列表,从每个对象中删除一个元素。每个对象都是新的一行。然后,我尝试按原样保存新文件,而不保存对象中包含的元素。
{
"business_id": "fNGIbpazjTRdXgwRY_NIXA",
"full_address": "1201 Washington Ave
Carnegie, 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,但是这些信息并不总是相同的。有些包含所有日期,有些仅包含一天或两天信息。
这是我尝试过的代码:
import json
with open('data.json') as data_file:
data = json.load(data_file)
for element in data:
del element['hours']
但是,运行代码时出现错误:
TypeError:‘Unicode’对象不支持项删除
解决方案
假设您要覆盖同一文件:
import json
with 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)
可能就是您要找的,如果我理解您的要求的话。因为它将删除hours
项(如果存在),如果不存在则不会失败。
不过,我不确定我是否理解为什么小时键是否包含几天对您有影响,因为您只是想去掉整个键/值对,对吗?
现在,如果您确实希望使用del
而不是pop
,下面是如何使您的代码工作的方法:
import json
with 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)
如果要将其写入其他文件,只需更改第二个open
语句中的文件名。
您可能已经注意到,我必须更改缩进,以便在数据清理阶段关闭该文件,并且可以在结束时覆盖该文件。
with
是所谓的上下文管理器,无论它提供什么(这里是data_file文件描述符),仅在该上下文中可用。这意味着with
挡路的缩进一结束,文件就会关闭,上下文以及变为无效/过时的文件描述符也会结束。
如果不这样做,您将无法在写入模式下打开文件并获取要写入的新文件描述符。
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