我正在开发一个程序,我需要根据某些条件在数据集中显示特定的行。这些条件适用于我为机器学习模型创建的features
列。这个features
列是一个Vector列,当我尝试通过传递Vector值来过滤它时,我收到以下错误:
Exception in thread "main" java.lang.RuntimeException: Unsupported literal type class org.apache.spark.ml.linalg.DenseVector at org.apache.spark.sql.catalyst.expressions.Literal$.apply(literals.scala:75) at org.apache.spark.sql.functions$.lit(functions.scala:101)
这是给我错误的过滤部分:
dataset.where(dataset.col("features").notEqual(datapoint)); //datapoint is a Vector
有什么方法吗?
答案 0 :(得分:1)
您需要创建一个用于过滤Vector的udf。以下对我有用:
import org.apache.spark.ml.feature.VectorAssembler
import org.apache.spark.ml.linalg.Vectors
import org.apache.spark.sql.functions.udf
val df = sc.parallelize(Seq(
(1, 1, 1), (1, 2, 3), (1, 3, 5), (2, 4, 6),
(2, 5, 2), (2, 6, 1), (3, 7, 5), (3, 8, 16),
(1, 1, 1))).toDF("c1", "c2", "c3")
val dfVec = new VectorAssembler()
.setInputCols(Array("c1", "c2", "c3"))
.setOutputCol("features")
.transform(df)
def vectors_unequal(vec1: Vector) = udf((vec2: Vector) => !vec1.equals(vec2))
val vecToRemove = Vectors.dense(1,1,1)
val filtered = dfVec.where(vectors_unequal(vecToRemove)(dfVec.col("features")))
val filtered2 = dfVec.filter(vectors_unequal(vecToRemove)($"features")) // Also possible
dfVec show
收益:
+---+---+---+--------------+
| c1| c2| c3| features|
+---+---+---+--------------+
| 1| 1| 1| [1.0,1.0,1.0]|
| 1| 2| 3| [1.0,2.0,3.0]|
| 1| 3| 5| [1.0,3.0,5.0]|
| 2| 4| 6| [2.0,4.0,6.0]|
| 2| 5| 2| [2.0,5.0,2.0]|
| 2| 6| 1| [2.0,6.0,1.0]|
| 3| 7| 5| [3.0,7.0,5.0]|
| 3| 8| 16|[3.0,8.0,16.0]|
| 1| 1| 1| [1.0,1.0,1.0]|
+---+---+---+--------------+
filtered show
收益:
+---+---+---+--------------+
| c1| c2| c3| features|
+---+---+---+--------------+
| 1| 2| 3| [1.0,2.0,3.0]|
| 1| 3| 5| [1.0,3.0,5.0]|
| 2| 4| 6| [2.0,4.0,6.0]|
| 2| 5| 2| [2.0,5.0,2.0]|
| 2| 6| 1| [2.0,6.0,1.0]|
| 3| 7| 5| [3.0,7.0,5.0]|
| 3| 8| 16|[3.0,8.0,16.0]|
+---+---+---+--------------+
答案 1 :(得分:0)
我的情况: word2vec 后的原始数据:
result.show(10,false)
+-------------+-----------------------------------------------------------------------------------------------------------+
|ip |features |
+-------------+-----------------------------------------------------------------------------------------------------------+
|1.1.125.120 |[0.0,0.0,0.0,0.0,0.0] |
|1.11.114.150 |[0.0,0.0,0.0,0.0,0.0] |
|1.116.114.36 |[0.022845590487122536,-0.012075710110366344,-0.034423209726810455,-0.04642726108431816,0.09164007753133774]|
|1.117.21.102 |[0.0,0.0,0.0,0.0,0.0] |
|1.119.13.5 |[0.0,0.0,0.0,0.0,0.0] |
|1.119.130.2 |[0.0,0.0,0.0,0.0,0.0] |
|1.119.132.162|[0.0,0.0,0.0,0.0,0.0] |
|1.119.133.166|[0.0,0.0,0.0,0.0,0.0] |
|1.119.136.170|[0.0,0.0,0.0,0.0,0.0] |
|1.119.137.154|[0.0,0.0,0.0,0.0,0.0] |
+-------------+-----------------------------------------------------------------------------------------------------------+
我想删除嵌入零的ip:
import org.apache.spark.sql.functions.udf
import org.apache.spark.ml.linalg.Vector
val vecToSeq = udf((v: Vector) => v.toArray).asNondeterministic
val output = result.select($"ip",vecToSeq($"features").alias("features"))
val select_output = output.filter(output("features")!==Array(0,0,0,0,0))
select_output.show(5)
+-------------+--------------------+
| ip| features|
+-------------+--------------------+
| 1.116.114.36|[0.02284559048712...|
| 1.119.137.98|[-0.0244039318391...|
|1.119.177.102|[-0.0801128149032...|
|1.119.186.170|[0.01125990878790...|
|1.119.193.226|[0.04201301932334...|
+-------------+--------------------+