Question

我有一个Age课，一个csv文件和一个pyspark运行时会话

ages.csv

Name;Age
alpha;noise20noise
beta;noi 3 sE 0
gamma;n 4 oi 0 se
phi;n50ise
detla;3no5ise
kappa;No 4 i 5 sE
omega;25noIsE

（在解析“年龄”列之后）的含义实际上是：

Name;Age
alpha;20
beta;30
gamma;40
phi;50
detla;35
kappa;45
omega;25

定义的班级：年龄 age.py

import re

class Age:
    # age is a number representing the age of a person
    def __init__(self, age):
        self.age = age

    def __eq__(self, other):
        return self.age == self.__parse(other)

    def __lt__(self, other):
        return self.age < self.__parse(other)

    def __gt__(self, other):
        return self.age > self.__parse(other)

    def __le__(self, other):
        return self.age <= self.__parse(other)

    def __ge__(self, other):
        return self.age >= self.__parse(other)

    def __parse(self, age):
        return int(''.join(re.findall(r'\d', age)))

# Let's test this class
if __name__ == '__main__':
    print(Age(18) == 'noise18noise')
    print(Age(18) <= 'aka 1 fakj 8 jal')
    print(Age(18) >= 'jaa 18 ka')
    print(Age(18) < '1 kda 9')
    print(Age(18) > 'akfa 1 na 7 noise')

Output:
True
True
True
True
True

该测试确实有效。我想在pyspark中使用它

运行pyspark，阅读ages.csv并导入Age

Using Python version 3.6.7 (default, Oct 23 2018 19:16:44)
SparkSession available as 'spark'.
>>> ages = spark.read.csv('ages.csv', sep=';', header=True)
19/01/28 14:44:18 WARN ObjectStore: Failed to get database global_temp, returning NoSuchObjectException
>>> ages.show()
+-----+------------+
| Name|         Age|
+-----+------------+
|alpha|noise20noise|
| beta|  noi 3 sE 0|
|gamma| n 4 oi 0 se|
|  phi|      n50ise|
|detla|     3no5ise|
|kappa| No 4 i 5 sE|
|omega|     25noIsE|
+-----+------------+

现在我想让所有年龄为20岁的人

>>> from age import Age
>>> ages.filter(ages.Age == Age(20)).show()

这是我得到的错误

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/opt/spark-2.3.1-bin-hadoop2.7/python/pyspark/sql/column.py", line 116, in _
    njc = getattr(self._jc, name)(jc)
File "/opt/spark-2.3.1-bin-hadoop2.7/python/lib/py4j-0.10.7-src.zip/py4j/java_gateway.py", line 1248, in __call__
File "/opt/spark-2.3.1-bin-hadoop2.7/python/lib/py4j-0.10.7-src.zip/py4j/java_gateway.py", line 1218, in _build_args
File "/opt/spark-2.3.1-bin-hadoop2.7/python/lib/py4j-0.10.7-src.zip/py4j/java_gateway.py", line 1218, in <listcomp>
File "/opt/spark-2.3.1-bin-hadoop2.7/python/lib/py4j-0.10.7-src.zip/py4j/protocol.py", line 298, in get_command_part
AttributeError: 'Age' object has no attribute '_get_object_id'

所以我的第一个问题是如何解决此错误

这是我首次尝试解决此问题：我将class Age的定义更改为扩展str，如下所示：

age.py

...
class Age(str):
    ....

第二次尝试：

>>> ages.filter(ages.Age == Age(20)).show()
+----+---+
|Name|Age|
+----+---+
+----+---+

尽管如此，我们仍然有：

>>> 'noise20noise' == Age(20)
True

您可以看到AttributeError: 'Age' object has no attribute '_get_object_id'消失了，但却没有计算出正确的答案，这就是我的第二个问题

再次是我的尝试：我使用pyspark用户定义的函数

>>> import pyspark.sql.functions as F
>>> import pyspark.sql.types as T
>>> eq20 = F.udf(lambda c: c == Age(20), T.BooleanType())
>>> ages.filter(eq20(ages.Age)).show()
+-----+------------+
| Name|         Age|
+-----+------------+
|alpha|noise20noise|
+-----+------------+

现在这可行。但是，这是事实：我最喜欢第一个习语

>>> ages.filter(ages.Age == Age(20)).show()

更简单，更具表现力。我不想每次都定义一个像eq20, eq21, less_than50, greater_than30, etc这样的函数

我可以在Age类别中进行定义，但是我不知道该怎么做。尽管如此，这是我到目前为止使用python decorator

尝试过的方法

age.py

# other imports here
...

import pyspark.sql.functions as F
import pyspark.sql.types as T

def connect_to_pyspark(function):
    return F.udf(function, T.BooleanType())

class Age(str):
    ...

    @connect_to_pyspark
    def __eq__(self, other):
        return self.age == self.__parse(other)

    ...
    # do the same decorator for the other comparative methods

再次测试：

>>> ages.filter(ages.Age == Age(20)).show()
+----+---+
|Name|Age|
+----+---+
+----+---+

它不起作用。还是我的装饰工写得不好？

如何解决所有这些问题？我对第一个问题的解决方案是否足够好？如果没有，该做什么呢？如果是，该如何解决第二个问题？

Answer 1

获取ages.Age == Age(20)将会非常困难，因为spark不遵守用于实现__eq__的python约定。稍后会详细介绍，但是如果您可以做Age(20) == ages.Age，那么您可以有一些选择。恕我直言，最简单的方法是仅将解析逻辑包装在udf中：

parse_udf = F.udf(..., T.IntegerType())
class Age:
    ...
    def __eq__(self, other: Column):
        return F.lit(self.age) == parse_udf(other)

请注意，Age不是str的子类，只会给世界带来伤害。如果要使用装饰器，则装饰器不应返回udf，而应返回应用udf的函数。像这样：

import re
import pyspark.sql.functions as F
import pyspark.sql.types as T

def connect_to_pyspark(function):
  def helper(age, other):
    myUdf = F.udf(lambda item_from_other: function(age, item_from_other), T.BooleanType())
    return myUdf(other)
  return helper

class Age:

    def __init__(self, age):
      self.age = 45

    def __parse(self, other):
      return int(''.join(re.findall(r'\d', other)))

    @connect_to_pyspark
    def __eq__(self, other):
        return self.age == self.__parse(other)

ages.withColumn("eq20", Age(20) == df.Age).show()

有关为什么需要使用Ages(20) == ages.Age的更多信息。在python中，如果您执行a == b并且a的类不知道如何与b比较，则它应该返回NotImplemented，然后python会尝试b.__eq__(a)，但是spark永远不会返回{{ 1}}，因此NotImplemented中的__eq__仅在表达式（（..

如何在pyspark操作中轻松使用自定义类方法？

1 个答案: