PySpark

Question

我需要将列表传递给 UDF，列表将确定距离的 score/category。现在，我将所有距离硬编码为第 4 个分数。

a= spark.createDataFrame([("A", 20), ("B", 30), ("D", 80)],["Letter", "distances"])

from pyspark.sql.functions import udf
def cate(label, feature_list):
    if feature_list == 0:
        return label[4]
label_list = ["Great", "Good", "OK", "Please Move", "Dead"]
udf_score=udf(cate, StringType())
a.withColumn("category", udf_score(label_list,a["distances"])).show(10)

当我尝试类似的操作时，出现此错误。

Py4JError: An error occurred while calling z:org.apache.spark.sql.functions.col. Trace:
py4j.Py4JException: Method col([class java.util.ArrayList]) does not exist
    at py4j.reflection.ReflectionEngine.getMethod(ReflectionEngine.java:318)
    at py4j.reflection.ReflectionEngine.getMethod(ReflectionEngine.java:339)
    at py4j.Gateway.invoke(Gateway.java:274)
    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)

Answer 1

尝试柯里化该函数，以便 DataFrame 调用中的唯一参数是您希望该函数作用的列的名称：

udf_score=udf(lambda x: cate(label_list,x), StringType())
a.withColumn("category", udf_score("distances")).show(10)

Answer 2

from pyspark.sql.functions import udf, col

#sample data
a= sqlContext.createDataFrame([("A", 20), ("B", 30), ("D", 80)],["Letter", "distances"])
label_list = ["Great", "Good", "OK", "Please Move", "Dead"]

def cate(label, feature_list):
    if feature_list == 0:
        return label[4]
    else:  #you may need to add 'else' condition as well otherwise 'null' will be added in this case
        return 'I am not sure!'

def udf_score(label_list):
    return udf(lambda l: cate(l, label_list))
a.withColumn("category", udf_score(label_list)(col("distances"))).show()

输出为：

+------+---------+--------------+
|Letter|distances|      category|
+------+---------+--------------+
|     A|       20|I am not sure!|
|     B|       30|I am not sure!|
|     D|       80|I am not sure!|
+------+---------+--------------+

Answer 3

我认为将列表作为变量的默认值传递可能会有所帮助

from pyspark.sql.functions import udf, col

#sample data
a= sqlContext.createDataFrame([("A", 20), ("B", 30), ("D", 80),("E",0)],["Letter", "distances"])
label_list = ["Great", "Good", "OK", "Please Move", "Dead"]

#Passing List as Default value to a variable
def cate( feature_list,label=label_list):
    if feature_list == 0:
        return label[4]
    else:  #you may need to add 'else' condition as well otherwise 'null' will be added in this case
        return 'I am not sure!'

udfcate = udf(cate, StringType())

a.withColumn("category", udfcate("distances")).show()

输出：

+------+---------+--------------+
|Letter|distances|      category|
+------+---------+--------------+
|     A|       20|I am not sure!|
|     B|       30|I am not sure!|
|     D|       80|I am not sure!|
|     E|        0|          Dead|
+------+---------+--------------+

PySpark - 将列表作为参数传递给 UDF

PySpark - Pass list as parameter to UDF

python

user-defined-functions

spark-dataframe