Question

我想我不会理解选择或放弃是如何工作的我正在爆炸数据集，我不希望将一些列复制到新生成的条目中。

val ds = spark.sparkContext.parallelize(Seq(
      ("2017-01-01 06:15:00", "ASC_a", "1"),
      ("2017-01-01 06:19:00", "start", "2"),
      ("2017-01-01 06:22:00", "ASC_b", "2"),
      ("2017-01-01 06:30:00", "end", "2"),
      ("2017-01-01 10:45:00", "ASC_a", "3"),
      ("2017-01-01 10:50:00", "start", "3"),
      ("2017-01-01 11:22:00", "ASC_c", "4"),
      ("2017-01-01 11:31:00", "end", "5" )
    )).toDF("timestamp", "status", "msg")
    ds.show()

    val foo = ds.select($"timestamp", $"msg")
    val bar = ds.drop($"status")
    foo.printSchema()
    bar.printSchema()
    println("foo " + foo.where($"status" === "end").count)
    println("bar" + bar.where($"status" === "end").count)

输出：

root
 |-- timestamp: string (nullable = true)
 |-- msg: string (nullable = true)

root
 |-- timestamp: string (nullable = true)
 |-- msg: string (nullable = true)

foo 2
吧2

为什么我仍然得到两个输出为2 a）没有选择状态
b）状态下降

编辑：
println("foo " + foo.where(foo.col("status") === "end").count)表示没有列状态。这应该与println("foo " + foo.where($"status" === "end").count)不一样吗？

Answer 1

为什么两个
的输出仍为2

因为优化器可以自由地重新组织执行计划。事实上，如果你检查它：

== Physical Plan ==
*Project [_1#4 AS timestamp#8, _3#6 AS msg#10]
+- *Filter (isnotnull(_2#5) && (_2#5 = end))
   +- *SerializeFromObject [staticinvoke(class org.apache.spark.unsafe.types.UTF8String, StringType, fromString, assertnotnull(input[0, scala.Tuple3, true])._1, true) AS _1#4, staticinvoke(class org.apache.spark.unsafe.types.UTF8String, StringType, fromString, assertnotnull(input[0, scala.Tuple3, true])._2, true) AS _2#5, staticinvoke(class org.apache.spark.unsafe.types.UTF8String, StringType, fromString, assertnotnull(input[0, scala.Tuple3, true])._3, true) AS _3#6]
      +- Scan ExternalRDDScan[obj#3]

您会看到过滤器尽早下推并在项目之前执行。所以它相当于：

SELECT _1 AS timetsatmp, _2 AS msg
FROM ds WHERE _2 IS NOT NULL AND _2 = 'end'

可以说这是一个小错误，代码应翻译为

SELECT * FROM (
  SELECT _1 AS timetsatmp, _2 AS msg FROM ds
) WHERE _2 IS NOT NULL AND _2 = 'end'

并抛出异常。

选择/删除不会真正删除列？

1 个答案: