如何将下面的json转换为后面的关系行?我一直坚持的一点是 explode()
由于类型不匹配,函数引发异常。我还没有找到一种方法来强制将数据转换成合适的格式,以便可以从中的每个对象创建行 source
钥匙在 sample_json
对象。
json输入
sample_json = """
{
"dc_id": "dc-101",
"source": {
"sensor-igauge": {
"id": 10,
"ip": "68.28.91.22",
"description": "Sensor attached to the container ceilings",
"temp":35,
"c02_level": 1475,
"geo": {"lat":38.00, "long":97.00}
},
"sensor-ipad": {
"id": 13,
"ip": "67.185.72.1",
"description": "Sensor ipad attached to carbon cylinders",
"temp": 34,
"c02_level": 1370,
"geo": {"lat":47.41, "long":-122.00}
},
"sensor-inest": {
"id": 8,
"ip": "208.109.163.218",
"description": "Sensor attached to the factory ceilings",
"temp": 40,
"c02_level": 1346,
"geo": {"lat":33.61, "long":-111.89}
},
"sensor-istick": {
"id": 5,
"ip": "204.116.105.67",
"description": "Sensor embedded in exhaust pipes in the ceilings",
"temp": 40,
"c02_level": 1574,
"geo": {"lat":35.93, "long":-85.46}
}
}
}"""
期望输出
dc_id source_name id description
-------------------------------------------------------------------------------
dc-101 sensor-gauge 10 Sensor attached to the container ceilings
dc-101 sensor-ipad 13 Sensor ipad attached to carbon cylinders
dc-101 sensor-inest 8 Sensor attached to the factory ceilings
dc-101 sensor-istick 5 Sensor embedded in exhaust pipes in the ceilings
Pypark代码
from pyspark.sql.functions import *
df_sample_data = spark.read.json(sc.parallelize([sample_json]))
df_expanded = df_sample_data.withColumn("one_source",explode_outer(col("source")))
display(df_expanded)
错误
analysisexception:无法解析“explode”( source
)'由于数据类型不匹配:函数explode的输入应该是数组或Map类型,而不是struct。。。。
我将这个databricks笔记本放在一起,以进一步演示挑战并清楚地显示错误。我将能够使用这个笔记本来测试这里提供的任何建议。
1条答案
按热度按时间vwhgwdsa1#
你不能使用
explode
但是你可以在结构中得到列名source
(与df.select("source.*").columns
)使用列表理解,您可以从每个嵌套结构中创建一个字段数组,然后分解以获得所需的结果: