在Spark中有一个dataframe df：

 |-- array_field: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- a: string (nullable = true)
 |    |    |-- b: long (nullable = true)
 |    |    |-- c: long (nullable = true)

如何将字段array_field.a重命名为array_field.a_renamed？

[更新]：

.withColumnRenamed()不适用于嵌套字段,所以我尝试了这个hacky和不安全的方法：

# First alter the schema:
schema = df.schema
schema['array_field'].dataType.elementType['a'].name = 'a_renamed'

ind = schema['array_field'].dataType.elementType.names.index('a')
schema['array_field'].dataType.elementType.names[ind] = 'a_renamed'

# Then set dataframe's schema with altered schema
df._schema = schema

我知道设置私有属性不是一个好习惯,但我不知道为df设置架构的其他方法

我认为我在正确的轨道上但是df.printSchema()仍然显示了array_field.a的旧名称,尽管df.schema == schema为True

解决方法:

Python

无法修改单个嵌套字段.您必须重新创建整个结构.在这种特殊情况下,最简单的解决方案是使用演员表.

首先是一堆进口：

from collections import namedtuple
from pyspark.sql.functions import col
from pyspark.sql.types import (
    ArrayType, LongType, StringType, StructField, StructType)

和示例数据：

Record = namedtuple("Record", ["a", "b", "c"])

df = sc.parallelize([([Record("foo", 1, 3)], )]).toDF(["array_field"])

让我们确认架构与您的情况相同：

df.printSchema()

root
 |-- array_field: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- a: string (nullable = true)
 |    |    |-- b: long (nullable = true)
 |    |    |-- c: long (nullable = true)

您可以将新模式定义为字符串：

str_schema = "array<struct<a_renamed:string,b:bigint,c:bigint>>"

df.select(col("array_field").cast(str_schema)).printSchema()

root
 |-- array_field: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- a_renamed: string (nullable = true)
 |    |    |-- b: long (nullable = true)
 |    |    |-- c: long (nullable = true)

或数据类型：

struct_schema = ArrayType(StructType([
    StructField("a_renamed", StringType()),
    StructField("b", LongType()),
    StructField("c", LongType())
]))

 df.select(col("array_field").cast(struct_schema)).printSchema()

root
 |-- array_field: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- a_renamed: string (nullable = true)
 |    |    |-- b: long (nullable = true)
 |    |    |-- c: long (nullable = true)

斯卡拉

Scala中可以使用相同的技术：

case class Record(a: String, b: Long, c: Long)

val df = Seq(Tuple1(Seq(Record("foo", 1, 3)))).toDF("array_field")

val strSchema = "array<struct<a_renamed:string,b:bigint,c:bigint>>"

df.select($"array_field".cast(strSchema))

要么

import org.apache.spark.sql.types._

val structSchema = ArrayType(StructType(Seq(
    StructField("a_renamed", StringType),
    StructField("b", LongType),
    StructField("c", LongType)
)))

df.select($"array_field".cast(structSchema))

可能的改进：

如果您使用富有表现力的数据操作或JSON处理库,则可以更容易地将数据类型转储到dict或JSON字符串,并从那里获取它(例如,Python / toolz)：

from toolz.curried import pipe, assoc_in, update_in, map
from operator import attrgetter

# Update name to "a_updated" if name is "a"
rename_field = update_in(
    keys=["name"], func=lambda x: "a_updated" if x == "a" else x)

updated_schema = pipe(
   #  Get schema of the field as a dict
   df.schema["array_field"].jsonValue(),
   # Update fields with rename
   update_in(
       keys=["type", "elementType", "fields"],
       func=lambda x: pipe(x, map(rename_field), list)),
   # Load schema from dict
   StructField.fromJson,
   # Get data type
   attrgetter("dataType"))

df.select(col("array_field").cast(updated_schema)).printSchema()