我有一个嵌套的源 json 文件,其中包含一个结构数组。结构的数量因行而异,我想使用 Spark (scala) 从结构的键/值动态创建新的数据帧列,其中键是列名,值是列值。< /p>
{"key1":{"key2":{"key3":"AK","key4":"EU","key5":{"key6":"001","key7":"N","values":[{"name":"valuesColumn1","value":"9.876"},{"name":"valuesColumn2","value":"1.2345"},{"name":"valuesColumn3","value":"8.675309"}]}}}}
scala> val df = spark.read.json("file:///tmp/nested_test.json")
root
|-- key1: struct (nullable = true)
| |-- key2: struct (nullable = true)
| | |-- key3: string (nullable = true)
| | |-- key4: string (nullable = true)
| | |-- key5: struct (nullable = true)
| | | |-- key6: string (nullable = true)
| | | |-- key7: string (nullable = true)
| | | |-- values: array (nullable = true)
| | | | |-- element: struct (containsNull = true)
| | | | | |-- name: string (nullable = true)
| | | | | |-- value: string (nullable = true)
df.select(
($"key1.key2.key3").as("key3"),
($"key1.key2.key4").as("key4"),
($"key1.key2.key5.key6").as("key6"),
($"key1.key2.key5.key7").as("key7"),
($"key1.key2.key5.values").as("values")).
show(truncate=false)
+----+----+----+----+----------------------------------------------------------------------------+
|key3|key4|key6|key7|values |
+----+----+----+----+----------------------------------------------------------------------------+
|AK |EU |001 |N |[[valuesColumn1, 9.876], [valuesColumn2, 1.2345], [valuesColumn3, 8.675309]]|
+----+----+----+----+----------------------------------------------------------------------------+
这里有一个包含 3 个结构体的数组,但是这 3 个结构体需要动态地溢出到 3 个单独的列中(3 个的数量可能相差很大),我不知道该怎么做。
请注意,为 values
数组中的每个数组元素生成了 3 个新列。
+----+----+----+----+-----------------------------------------+
|key3|key4|key6|key7|valuesColumn1|valuesColumn2|valuesColumn3|
+----+----+----+----+-----------------------------------------+
|AK |EU |001 |N |9.876 |1.2345 |8.675309 |
+----+----+----+----+-----------------------------------------+
我认为所需的解决方案是 something similar to what was discussed in this SO post,但有两个主要区别:
name
列驱动,列值需要由 value
驱动。...
| | | | |-- element: struct (containsNull = true)
| | | | | |-- name: string (nullable = true)
| | | | | |-- value: string (nullable = true)
答案 0 :(得分:1)
你可以这样做:
val sac = new SparkContext("local[*]", " first Program");
val sqlc = new SQLContext(sac);
import sqlc.implicits._;
import org.apache.spark.sql.functions.split
import scala.math._
import org.apache.spark.sql.types._
import org.apache.spark.sql.functions._
import org.apache.spark.sql.functions.{ min, max }
val json = """{"key1":{"key2":{"key3":"AK","key4":"EU","key5":{"key6":"001","key7":"N","values":[{"name":"valuesColumn1","value":"9.876"},{"name":"valuesColumn2","value":"1.2345"},{"name":"valuesColumn3","value":"8.675309"}]}}}}"""
val df1 = sqlc.read.json(Seq(json).toDS())
val df2 = df1.select(
($"key1.key2.key3").as("key3"),
($"key1.key2.key4").as("key4"),
($"key1.key2.key5.key6").as("key6"),
($"key1.key2.key5.key7").as("key7"),
($"key1.key2.key5.values").as("values")
)
val numColsVal = df2
.withColumn("values_size", size($"values"))
.agg(max($"values_size"))
.head()
.getInt(0)
val finalDFColumns = df2.select(explode($"values").as("values")).select("values.*").select("name").distinct.map(_.getAs[String](0)).orderBy($"value".asc).collect.foldLeft(df2.limit(0))((cdf, c) => cdf.withColumn(c, lit(null))).columns
val finalDF = df2.select($"*" +: (0 until numColsVal).map(i => $"values".getItem(i)("value").as($"values".getItem(i)("name").toString)): _*)
finalDF.columns.zip(finalDFColumns).foldLeft(finalDF)((fdf, column) => fdf.withColumnRenamed(column._1, column._2)).show(false)
finalDF.columns.zip(finalDFColumns).foldLeft(finalDF)((fdf, column) => fdf.withColumnRenamed(column._1, column._2)).drop($"values").show(false)
最终输出结果为:
+----+----+----+----+-------------+-------------+-------------+
|key3|key4|key6|key7|valuesColumn1|valuesColumn2|valuesColumn3|
+----+----+----+----+-------------+-------------+-------------+
|AK |EU |001 |N |9.876 |1.2345 |8.675309 |
+----+----+----+----+-------------+-------------+-------------+
希望我答对了你的问题!
----------- EDIT with Explanation----------
此块获取要为数组结构创建的列数。
val numColsVal = df2
.withColumn("values_size", size($"values"))
.agg(max($"values_size"))
.head()
.getInt(0)
finalDFColumns
是使用所有预期的列作为空值输出创建的 DF。
下面的块返回需要从数组结构中创建的不同列。
df2.select(explode($"values").as("values")).select("values.*").select("name").distinct.map(_.getAs[String](0)).orderBy($"value".asc).collect
下面的块将上述新列与 df2
中的其他列组合在一起,这些列用空/空值初始化。
foldLeft(df2.limit(0))((cdf, c) => cdf.withColumn(c, lit(null)))
如果您打印输出,则将这两个块组合起来:
+----+----+----+----+------+-------------+-------------+-------------+
|key3|key4|key6|key7|values|valuesColumn1|valuesColumn2|valuesColumn3|
+----+----+----+----+------+-------------+-------------+-------------+
+----+----+----+----+------+-------------+-------------+-------------+
现在我们已经准备好了结构。我们需要这里相应列的值。下面的块为我们提供了值:
df2.select($"*" +: (0 until numColsVal).map(i => $"values".getItem(i)("value").as($"values".getItem(i)("name").toString)): _*)
结果如下:
+----+----+----+----+--------------------+---------------+---------------+---------------+
|key3|key4|key6|key7| values|values[0][name]|values[1][name]|values[2][name]|
+----+----+----+----+--------------------+---------------+---------------+---------------+
| AK| EU| 001| N|[[valuesColumn1, ...| 9.876| 1.2345| 8.675309|
+----+----+----+----+--------------------+---------------+---------------+---------------+
现在我们需要像上面第一个块中那样重命名列。因此,我们将使用 zip
函数合并列,然后使用 foldLeft 方法重命名输出列,如下所示:
finalDF.columns.zip(finalDFColumns).foldLeft(finalDF)((fdf, column) => fdf.withColumnRenamed(column._1, column._2)).show(false)
这导致以下结构:
+----+----+----+----+--------------------+-------------+-------------+-------------+
|key3|key4|key6|key7| values|valuesColumn1|valuesColumn2|valuesColumn3|
+----+----+----+----+--------------------+-------------+-------------+-------------+
| AK| EU| 001| N|[[valuesColumn1, ...| 9.876| 1.2345| 8.675309|
+----+----+----+----+--------------------+-------------+-------------+-------------+
我们快到了。我们现在只需要像这样删除不需要的 values
列:
finalDF.columns.zip(finalDFColumns).foldLeft(finalDF)((fdf, column) => fdf.withColumnRenamed(column._1, column._2)).drop($"values").show(false)
从而导致预期的输出如下 -
+----+----+----+----+-------------+-------------+-------------+
|key3|key4|key6|key7|valuesColumn1|valuesColumn2|valuesColumn3|
+----+----+----+----+-------------+-------------+-------------+
|AK |EU |001 |N |9.876 |1.2345 |8.675309 |
+----+----+----+----+-------------+-------------+-------------+
我不确定我是否能够清楚地解释它。但是,如果您尝试打破上述语句/代码并尝试打印它,您将了解我们是如何到达输出的。您可以在互联网上找到此逻辑中使用的不同函数的示例说明。
答案 1 :(得分:0)
我发现这种方法效果更好,并且使用爆炸和枢轴更容易理解:
val json = """{"key1":{"key2":{"key3":"AK","key4":"EU","key5":{"key6":"001","key7":"N","values":[{"name":"valuesColumn1","value":"9.876"},{"name":"valuesColumn2","value":"1.2345"},{"name":"valuesColumn3","value":"8.675309"}]}}}}"""
val df = spark.read.json(Seq(json).toDS())
// schema
df.printSchema
root
|-- key1: struct (nullable = true)
| |-- key2: struct (nullable = true)
| | |-- key3: string (nullable = true)
| | |-- key4: string (nullable = true)
| | |-- key5: struct (nullable = true)
| | | |-- key6: string (nullable = true)
| | | |-- key7: string (nullable = true)
| | | |-- values: array (nullable = true)
| | | | |-- element: struct (containsNull = true)
| | | | | |-- name: string (nullable = true)
| | | | | |-- value: string (nullable = true)
// create final df
val finalDf = df.
select(
$"key1.key2.key3".as("key3"),
$"key1.key2.key4".as("key4"),
$"key1.key2.key5.key6".as("key6"),
$"key1.key2.key5.key7".as("key7"),
explode($"key1.key2.key5.values").as("values")
).
groupBy(
$"key3", $"key4", $"key6", $"key7"
).
pivot("values.name").
agg(min("values.value")).alias("values.name")
// result
finalDf.show
+----+----+----+----+-------------+-------------+-------------+
|key3|key4|key6|key7|valuesColumn1|valuesColumn2|valuesColumn3|
+----+----+----+----+-------------+-------------+-------------+
| AK| EU| 001| N| 9.876| 1.2345| 8.675309|
+----+----+----+----+-------------+-------------+-------------+