我必须在pyspark数据帧中找到特定数据点的邻居。
a= spark.createDataFrame([("A", [0,1]), ("B", [5,9]), ("D", [13,5])],["Letter", "distances"])
我创建了这个函数,它将接收数据帧(DB),然后使用欧氏距离检查最接近固定点(Q)的数据点。它将根据某些epsilon值(eps)过滤掉相关数据点并返回子集。
def rangequery(DB, Q, eps):
distance_udf = F.udf(lambda x: float(distance.euclidean(x, Q)), FloatType())
df_neigh =DB.withColumn('euclid_distances', distance_udf(F.col('distances')))
return df_neigh.filter(df_neigh['euclid_distances'] <= eps)
但现在我需要为数据框中的每一个点运行此函数
所以我做了以下事情。
def check_neighbours(distance):
df = rangequery(a,distances, 9)
if df.count()>=1:
return "Has Neighbours"
else:
return "No Neighbours"
udf_neigh=udf(check_neighbours, StringType())
a.withColumn("label", udf_neigh( a["distances"])).show()
当我尝试运行此代码时出现以下错误。
PicklingError: Could not serialize object: Py4JError: An error occurred while calling o380.__getnewargs__. Trace:
py4j.Py4JException: Method __getnewargs__([]) does not exist
at py4j.reflection.ReflectionEngine.getMethod(ReflectionEngine.java:318)
at py4j.reflection.ReflectionEngine.getMethod(ReflectionEngine.java:326)
at py4j.Gateway.invoke(Gateway.java:272)
at py4j.commands.AbstractCommand.invokeMethod(AbstractCommand.java:132)
at py4j.commands.CallCommand.execute(CallCommand.java:79)
at py4j.GatewayConnection.run(GatewayConnection.java:214)
at java.lang.Thread.run(Thread.java:745)
答案 0 :(得分:0)
从this answer借用 ,这是一种方法。请考虑以下示例:
from pyspark.sql.functions import col, udf
# create dummy dataset
DB = sqlCtx.createDataFrame(
[("A", [0,1]), ("B", [5,9]), ("D", [13,5])],
["Letter", "distances"]
)
# Define your distance metric as a udf
from scipy.spatial import distance
distance_udf = udf(lambda x, y: float(distance.euclidean(x, y)), FloatType())
# Use crossJoin() to compute distances.
eps = 9 # minimum distance
DB.alias("l")\
.crossJoin(DB.alias("r"))\
.where(distance_udf(col("l.distances"), col("r.distances")) < eps)\
.groupBy("l.letter", "l.distances")\
.count()\
.withColumn("count", col("count") - 1)\
.withColumn("label", udf(lambda x: "Has Neighbours" if x >= 1 else "No Neighbours")(col("count")))\
.sort('letter')\
.show()
输出:
+------+---------+-----+--------------+
|letter|distances|count| label|
+------+---------+-----+--------------+
| A| [0, 1]| 0| No Neighbours|
| B| [5, 9]| 1|Has Neighbours|
| D| [13, 5]| 1|Has Neighbours|
+------+---------+-----+--------------+
.withColumn("count", col("count") - 1)
完成的地方,因为我们知道每列都将自己作为一个平凡的邻居。 (您可以根据需要删除此行。)
您编写的代码不起作用,因为正如linked post中的@ user8371915所述:
您无法在
中引用分布式DataFrame
udf