Question

我有两个文件。 functions.py有一个函数，并从该函数创建一个pyspark udf。 main.py尝试导入udf。但是，main.py似乎无法访问functions.py中的函数。

functions.py：

from pyspark.sql.functions import udf
from pyspark.sql.types import StringType

def do_something(x):
    return x + 'hello'

sample_udf = udf(lambda x: do_something(x), StringType())

main.py：

from functions import sample_udf, do_something
df = spark.read.load(file)
df.withColumn("sample",sample_udf(col("text")))

这会导致错误：

17/10/03 19:35:29 WARN TaskSetManager: Lost task 0.0 in stage 3.0 (TID 6, ip-10-223-181-5.ec2.internal, executor 3): org.apache.spark.api.python.PythonException: Traceback (most recent call last):
  File "/usr/lib/spark/python/pyspark/worker.py", line 164, in main
    func, profiler, deserializer, serializer = read_udfs(pickleSer, infile)
  File "/usr/lib/spark/python/pyspark/worker.py", line 93, in read_udfs
    arg_offsets, udf = read_single_udf(pickleSer, infile)
  File "/usr/lib/spark/python/pyspark/worker.py", line 79, in read_single_udf
    f, return_type = read_command(pickleSer, infile)
  File "/usr/lib/spark/python/pyspark/worker.py", line 55, in read_command
    command = serializer._read_with_length(file)
  File "/usr/lib/spark/python/pyspark/serializers.py", line 169, in _read_with_length
    return self.loads(obj)
  File "/usr/lib/spark/python/pyspark/serializers.py", line 454, in loads
    return pickle.loads(obj)
AttributeError: 'module' object has no attribute 'do_something'

如果我绕过do_something函数并将其放在udf中，例如：udf(lambda x: x + ' hello', StringType())，UDF导入正常 - 但我的函数稍微长一点，将它封装起来会很好在一个单独的功能。实现这一目标的正确方法是什么？

Answer 1

只需将其添加为答案： -

将py文件添加到sparkcontext中，以使其可供执行者使用。

sc.addPyFile("functions.py")
from functions import sample_udf

这是我的测试笔记本

https://databricks-prod-cloudfront.cloud.databricks.com/public/4027ec902e239c93eaaa8714f173bcfc/3669221609244155/3140647912908320/868274901052987/latest.html

谢谢，查尔斯。

Answer 2

我认为更干净的解决方案是使用udf装饰器定义udf函数：

from pyspark.sql.functions as F
from pyspark.sql.types import StringType

@F.udf
def sample_udf(x):
     return x + 'hello'

使用此解决方案，udf不会引用任何其他函数，并且您不需要在主代码中使用sc.addPyFile。

from functions import sample_udf, do_something
df = spark.read.load(file)
df.withColumn("sample",sample_udf(col("text")))
# It works :)

对于一些较旧的spark版本，装饰器不支持键入的udf，您可能需要按如下方式定义自定义装饰器：

from pyspark.sql.functions as F
from pyspark.sql.types as t

# Custom udf decorator which accept return type
def udf_typed(returntype=t.StringType()):
    def _typed_udf_wrapper(func):
        return F.udf(func, returntype)
    return _typed_udf_wrapper

@udf_typed(t.IntegerType())
def my_udf(x)
    return int(x)

如何将pyspark UDF导入主类

2 个答案: