我正在尝试为dataframe. 我传入的数据是从json. 这是我的初始数据:
json2 = sc.parallelize(['{"name": "mission", "pandas": {"attributes": "[0.4, 0.5]", "pt": "giant", "id": "1", "zip": "94110", "happy": "True"}}'])
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然后这里是如何指定架构:
schema = StructType(fields=[
StructField(
name='name',
dataType=StringType(),
nullable=True
),
StructField(
name='pandas',
dataType=ArrayType(
StructType(
fields=[
StructField(
name='id',
dataType=StringType(),
nullable=False
),
StructField(
name='zip',
dataType=StringType(),
nullable=True
),
StructField(
name='pt',
dataType=StringType(),
nullable=True
),
StructField(
name='happy',
dataType=BooleanType(),
nullable=False
),
StructField(
name='attributes',
dataType=ArrayType(
elementType=DoubleType(),
containsNull=False
),
nullable=True
)
]
),
containsNull=True
),
nullable=True
)
])
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当我使用sqlContext.createDataFrame(json2, schema)然后尝试对结果执行操作show()时,dataframe我收到以下错误: …
我的团队的观点建立在建立在视图之上的视图上,所以经常DROP TABLE CASCADE是灾难和大量试验和错误的处方.
我想要的是一个返回所有依赖对象的查询,这些对象需要在给定的情况下以正确的顺序重新创建schema,table以便它们可以自动化并在脚本中运行.我正在使用Redshift DROP TABLE文档http://docs.aws.amazon.com/redshift/latest/dg/r_DROP_TABLE.html上的依赖查询的修改版本.
它似乎是返回视图及其依赖项,但不是常规表.我觉得我很亲密,我错过了什么?
WITH dependencies AS (
SELECT DISTINCT
cls1.oid AS tbloid,
nsp1.nspname AS schemaname,
cls1.relname AS name,
nsp2.nspname AS refbyschemaname,
cls2.relname AS refbyname,
cls2.oid AS viewoid
FROM pg_catalog.pg_class cls1
JOIN pg_catalog.pg_depend dep1
ON cls1.relfilenode = dep1.refobjid
JOIN pg_catalog.pg_depend dep2
ON dep1.objid = dep2.objid
JOIN pg_catalog.pg_class cls2
ON dep2.refobjid = cls2.relfilenode
LEFT OUTER JOIN pg_namespace nsp1
ON cls1.relnamespace = nsp1.oid
LEFT OUTER JOIN pg_namespace nsp2
ON cls2.relnamespace …Run Code Online (Sandbox Code Playgroud) 我正在学习Python的布尔逻辑以及如何缩短内容.标题中的两个表达式是否相同?如果没有,它们之间有什么区别?
我想知道spark-redshift如果列内容太长,如何截断列,而不是返回错误.
我的团队用于Sentry跟踪错误,因此我不希望使用 Luigi 的内置电子邮件功能将所有报告保存在一个地方。
这就是我目前的设置方式,它似乎完全跳过了 Sentry:
if __name__ == '__main__':
try:
luigi.run()
except Exception as e:
client = Client(
***
)
client.captureException(tags={
sys.argv[0]
})
logger.critical('Error occurred: {e}'.format(e=e))
raise
Run Code Online (Sandbox Code Playgroud) 我有一个docker-compose文件,并希望能够从本地缓存中的图像中旋转出一张图像,而不是从dockerhub中拉出图像。我正在使用sbt docker插件,因此我可以看到正在创建的图像,并且可以docker images在命令行中看到。但是,当我这样做时,docker-compose up -d myimage它始终默认为远程映像。如何强制它使用本地图像?
这是我的撰写文件的相关部分:
spark-master:
image: gettyimages/spark:2.2.0-hadoop-2.7
command: bin/spark-class org.apache.spark.deploy.master.Master -h spark-master
hostname: spark-master
environment:
MASTER: spark://spark-master:7077
SPARK_CONF_DIR: /conf
SPARK_PUBLIC_DNS: localhost
expose:
- 7001
- 7002
- 7003
- 7004
- 7005
- 7006
- 7077
- 6066
ports:
- 4040:4040
- 6066:6066
- 7077:7077
- 8080:8080
volumes:
- ./conf/master:/conf
- ./data:/tmp/data
hydra-streams:
image: ****/hydra-spark-core
command: bin/spark-class org.apache.spark.deploy.worker.Worker spark://spark-master:7077
hostname: worker
environment:
SPARK_CONF_DIR: /conf
SPARK_WORKER_CORES: 2
SPARK_WORKER_MEMORY: 1g
SPARK_WORKER_PORT: 8881
SPARK_WORKER_WEBUI_PORT: 8091
SPARK_PUBLIC_DNS: …Run Code Online (Sandbox Code Playgroud) 我正在使用该模块simple-salesforce,我在文档中没有看到有关进行批量API调用的任何内容.有人知道怎么做吗?
package impatient.mapsAndTups.objects
abstract class UnitConversion {
def convert[T](x: T): T
}
class Inches2Centimeters extends UnitConversion {
override def convert[Int](x: Int): Int = x * 100
}
object Conversions extends App {
val c = new Inches2Centimeters()
println(c.convert(15))
}
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我不明白为什么前面的代码无法编译。我收到错误:
Error:(9, 46) value * is not a member of type parameter Int
override def convert[Int](x: Int): Int = x * 100
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我可以做什么来解决这个问题?
我正在为 JSON 有效负载创建一个 avro 模式,该有效负载似乎具有多个对象的数组。我不确定如何在模式中表示这一点。有问题的关键是content:
{
"id": "channel-id",
"name": "My Channel with a New Title",
"description": "Herpy me derpy merpus herpsum ner berp berps derp ter tee",
"privacyLevel": "<private|org>",
"planId": "some-plan-id",
"owner": "a-user-handle",
"curators": [
"user-handle-1",
"user-handle-2"
],
"members": 5,
"content": [
{
"id": "docker",
"slug": "docker",
"index": 1,
"type": "path"
},
{
"id": "such-linkage",
"slug": "such-linkage",
"index": 2,
"type": "external-link",
"details": {
"url": "http://some-dank-link.com",
"title": "My Dank Link",
"contentType": "External Link",
"level": "Beginner",
"duration": "PT34293H33M9S"
}
}, …Run Code Online (Sandbox Code Playgroud) returnGreater :: (Ord a) => a -> a -> a
returnGreater a b
| (a > b) = a
| otherwise = b
returnGreatest2 :: (Ord a, Num a) => a -> a -> a -> (a, a)
returnGreatest2 a b c
| (a > b) = (a, returnGreater b c)
| otherwise = (b, returnGreater a c)
sumOfSquares :: (Num a) => (a, a) -> a
sumOfSquares (a, b) = a^2 + b^2
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鉴于上述功能,我很困惑为什么let x = sumOfSquares . …
python ×4
apache-spark ×2
avro ×1
dataframe ×1
docker ×1
haskell ×1
luigi ×1
pyspark ×1
python-3.x ×1
salesforce ×1
scala ×1
sentry ×1
sql ×1