我想将下面的JSON数据转换为avro格式,我使用下面的代码片段以avro格式编写JSON数据,但收到一个错误。如果有人能帮上忙,那就太好了
from fastavro import writer, reader, schema
from rec_avro import to_rec_avro_destructive, from_rec_avro_destructive, rec_avro_schema
def getweatherdata():
url = 'https://api.openweathermap.org/data/2.5/onecall?lat=33.441792&lon=-94.037689&exclude=hourly,daily&appid=' + apikey
response = requests.get(url)
data = response.text
return data
def turntoavro():
avro_objects = (to_rec_avro_destructive(rec) for rec in getweatherdata())
with open('json_in_avro.avro', 'wb') as f_out:
writer(f_out, schema.parse_schema(rec_avro_schema()), avro_objects)
turntoavro()
Error details:
File "fastavro/_write.pyx", line 269, in fastavro._write.write_record
TypeError: Expected dict, got str
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "datalake.py", line 30, in <module>
turntoavro()
File "datalake.py", line 26, in turntoavro
writer(f_out, schema.parse_schema(rec_avro_schema()), avro_objects)
File "fastavro/_write.pyx", line 652, in fastavro._write.writer
File "fastavro/_write.pyx", line 605, in fastavro._write.Writer.write
File "fastavro/_write.pyx", line 341, in fastavro._write.write_data
File "fastavro/_write.pyx", line 278, in fastavro._write.write_record
AttributeError: 'str' object has no attribute 'get'
样本数据:
{
"lat": 33.44,
"lon": -94.04,
"timezone": "America/Chicago",
"timezone_offset": -18000
}
为了检索对请求的响应,您使用了
response.text
,它以字符串而不是JSON格式返回响应。您必须使用response.json()
将其转换为JSON格式:正如其中一个答案中提到的,您可能希望使用
response.json()
而不是response.text
,这样您就可以得到一个实际的JSON字典但是,另一个问题是
getweatherdata()
返回单个字典,因此当您执行avro_objects = (to_rec_avro_destructive(rec) for rec in getweatherdata())
操作时,您正在迭代该字典中的键。相反,您应该执行avro_objects = [to_rec_avro_destructive(getweatherdata())]
我相信这个代码应该适用于您:
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