Question

我从PySpark开始，我在使用嵌套对象创建DataFrame时遇到了麻烦。

这是我的榜样。

我有用户。

$ cat user.json
{"id":1,"name":"UserA"}
{"id":2,"name":"UserB"}

用户有订单。

$ cat order.json
{"id":1,"price":202.30,"userid":1}
{"id":2,"price":343.99,"userid":1}
{"id":3,"price":399.99,"userid":2}

我喜欢加入它以获得这样一个结构，其中订单是嵌套在用户中的数组。

$ cat join.json
{"id":1, "name":"UserA", "orders":[{"id":1,"price":202.30,"userid":1},{"id":2,"price":343.99,"userid":1}]}
{"id":2,"name":"UserB","orders":[{"id":3,"price":399.99,"userid":2}]}

我该怎么做？是否有任何类型的嵌套连接或类似的东西？

>>> user = sqlContext.read.json("user.json")
>>> user.printSchema();
root
 |-- id: long (nullable = true)
 |-- name: string (nullable = true)

>>> order =  sqlContext.read.json("order.json")
>>> order.printSchema();
root
 |-- id: long (nullable = true)
 |-- price: double (nullable = true)
 |-- userid: long (nullable = true)

>>> joined = sqlContext.read.json("join.json")
>>> joined.printSchema();
root
 |-- id: long (nullable = true)
 |-- name: string (nullable = true)
 |-- orders: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- id: long (nullable = true)
 |    |    |-- price: double (nullable = true)
 |    |    |-- userid: long (nullable = true)

编辑：我知道有可能使用join和foldByKey这样做，但有没有更简单的方法？

EDIT2：我正在使用@ zero323的解决方案

def joinTable(tableLeft, tableRight, columnLeft, columnRight, columnNested, joinType = "left_outer"):
    tmpTable = sqlCtx.createDataFrame(tableRight.rdd.groupBy(lambda r: r.asDict()[columnRight]))
    tmpTable = tmpTable.select(tmpTable._1.alias("joinColumn"), tmpTable._2.data.alias(columnNested))
    return tableLeft.join(tmpTable, tableLeft[columnLeft] == tmpTable["joinColumn"], joinType).drop("joinColumn")

我添加第二个嵌套结构'lines'

>>> lines =  sqlContext.read.json(path + "lines.json")
>>> lines.printSchema();
root
 |-- id: long (nullable = true)
 |-- orderid: long (nullable = true)
 |-- product: string (nullable = true)

orders = joinTable(order, lines, "id", "orderid", "lines")
joined = joinTable(user, orders, "id", "userid", "orders")
joined.printSchema()

root
 |-- id: long (nullable = true)
 |-- name: string (nullable = true)
 |-- orders: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- id: long (nullable = true)
 |    |    |-- price: double (nullable = true)
 |    |    |-- userid: long (nullable = true)
 |    |    |-- lines: array (nullable = true)
 |    |    |    |-- element: struct (containsNull = true)
 |    |    |    |    |-- _1: long (nullable = true)
 |    |    |    |    |-- _2: long (nullable = true)
 |    |    |    |    |-- _3: string (nullable = true)

此行之后的行名称将丢失。有什么想法吗？

编辑3：我试图手动指定架构。

from pyspark.sql.types import *
fields = []
fields.append(StructField("_1", LongType(), True))
inner = ArrayType(lines.schema)
fields.append(StructField("_2", inner))
new_schema = StructType(fields)
print new_schema

grouped =  lines.rdd.groupBy(lambda r: r.orderid)
grouped =  grouped.map(lambda x: (x[0], list(x[1])))
g = sqlCtx.createDataFrame(grouped, new_schema)

错误：

TypeError: StructType(List(StructField(id,LongType,true),StructField(orderid,LongType,true),StructField(product,StringType,true))) can not accept object in type <class 'pyspark.sql.types.Row'>

Answer 1

这仅适用于Spark 2.0或更高版本

首先，我们需要一些进口产品：

from pyspark.sql.functions import struct, collect_list

其余的是一个简单的聚合和连接：

orders = spark.read.json("/path/to/order.json")
users = spark.read.json("/path/to/user.json")

combined = users.join(
    orders
        .groupBy("userId")
        .agg(collect_list(struct(*orders.columns)).alias("orders"))
        .withColumnRenamed("userId", "id"), ["id"])

对于示例数据，结果为：

combined.show(2, False)

+---+-----+---------------------------+
|id |name |orders                     |
+---+-----+---------------------------+
|1  |UserA|[[1,202.3,1], [2,343.99,1]]|
|2  |UserB|[[3,399.99,2]]             |
+---+-----+---------------------------+

with schema：

combined.printSchema()

root
 |-- id: long (nullable = true)
 |-- name: string (nullable = true)
 |-- orders: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- id: long (nullable = true)
 |    |    |-- price: double (nullable = true)
 |    |    |-- userid: long (nullable = true)

和JSON表示：

for x in combined.toJSON().collect():
    print(x)

{"id":1,"name":"UserA","orders":[{"id":1,"price":202.3,"userid":1},{"id":2,"price":343.99,"userid":1}]}
{"id":2,"name":"UserB","orders":[{"id":3,"price":399.99,"userid":2}]}

Answer 2

用于将数据框从嵌套平铺到正常使用 dff= df.select("column with multiple columns.*").toPandas()

Answer 3

首先，您需要使用userid作为第二个DataFrame的连接键：

user.join(order, user.id == order.userid)

然后，您可以使用map步骤将生成的记录转换为所需的格式。

Spark - 创建嵌套DataFrame

3 个答案: