编写此UDF的目的是用变量替换列的值。 Python 2.7; Spark 2.2.0
import pyspark.sql.functions as func
def updateCol(col, st):
return func.expr(col).replace(func.expr(col), func.expr(st))
updateColUDF = func.udf(updateCol, StringType())
变量L_1至L_3的每一行都有更新的列。 这就是我的称呼方式:
updatedDF = orig_df.withColumn("L1", updateColUDF("L1", func.format_string(L_1))). \
withColumn("L2", updateColUDF("L2", func.format_string(L_2))). \
withColumn("L3", updateColUDF("L3",
withColumn("NAME", func.format_string(name)). \
withColumn("AGE", func.format_string(age)). \
select("id", "ts", "L1", "L2", "L3",
"NAME", "AGE")
错误是:
return Column(sc._jvm.functions.expr(str))
AttributeError: 'NoneType' object has no attribute '_jvm'
答案 0 :(得分:1)
答案 1 :(得分:1)
错误是因为您在udf中使用pyspark函数。了解L1,L2 ..变量的内容也将非常有帮助。
但是,如果我了解您要正确执行的操作,则不需要udf。我假设L1,L2等是常量,对吗?如果没有,请告知我相应地调整代码。这是一个示例:
from pyspark import SparkConf
from pyspark.sql import SparkSession, functions as F
conf = SparkConf()
spark_session = SparkSession.builder \
.config(conf=conf) \
.appName('test') \
.getOrCreate()
data = [{'L1': "test", 'L2': "data"}, {'L1': "other test", 'L2': "other data"}]
df = spark_session.createDataFrame(data)
df.show()
# +----------+----------+
# | L1| L2|
# +----------+----------+
# | test| data|
# |other test|other data|
# +----------+----------+
L1 = 'some other data'
updatedDF = df.withColumn(
"L1",
F.lit(L1)
)
updatedDF.show()
# +---------------+----------+
# | L1| L2|
# +---------------+----------+
# |some other data| data|
# |some other data|other data|
# +---------------+----------+
# or if you need to replace the value in a more complex way
pattern = '\w+'
updatedDF = updatedDF.withColumn(
"L1",
F.regexp_replace(F.col("L1"), pattern, "testing replace")
)
updatedDF.show()
# +--------------------+----------+
# | L1| L2|
# +--------------------+----------+
# |testing replace t...| data|
# |testing replace t...|other data|
# +--------------------+----------+
# or even something more complicated:
# set L1 value to L2 column when L2 column equals to data, otherwise, just leave L2 as it is
updatedDF = df.withColumn(
"L2",
F.when(F.col('L2') == 'data', L1).otherwise(F.col('L2'))
)
updatedDF.show()
# +----------+---------------+
# | L1| L2|
# +----------+---------------+
# | test|some other data|
# |other test| other data|
# +----------+---------------+
因此,您的示例将是:
DF = orig_df.withColumn("L1", pyspark_func.lit(L_1))
...
此外,请确保您在此点之前进行主动火花会话
我希望这会有所帮助。
编辑:如果L1,L2等是列表,则一个选择是用它们创建一个数据框并加入初始df。不幸的是,我们需要用于连接的索引,并且由于您的数据帧很大,因此我认为这不是一个非常有效的解决方案。我们还可以使用广播和udf或广播和加入。
这是一个如何进行联接的示例(我认为不太理想):
L1 = ['row 1 L1', 'row 2 L1']
L2 = ['row 1 L2', 'row 2 L2']
# create a df with indexes
to_update_df = spark_session.createDataFrame([{"row_index": i, "L1": row[0], "L2": row[1]} for i, row in enumerate(zip(L1, L2))])
# add indexes to the initial df
indexed_df = updatedDF.rdd.zipWithIndex().toDF()
indexed_df.show()
# +--------------------+---+
# | _1 | _2 |
# +--------------------+---+
# | [test, some other... | 0 |
# | [other test, othe... | 1 |
# +--------------------+---+
# bring the df back to its initial form
indexed_df = indexed_df.withColumn('row_number', F.col("_2"))\
.withColumn('L1', F.col("_1").getItem('L1'))\
.withColumn('L2', F.col("_1").getItem('L2')).\
select('row_number', 'L1', 'L2')
indexed_df.show()
# +----------+----------+---------------+
# |row_number| L1| L2|
# +----------+----------+---------------+
# | 0| test|some other data|
# | 1|other test| other data|
# +----------+----------+---------------+
# join with your results and keep the updated columns
final_df = indexed_df.alias('initial_data').join(to_update_df.alias('other_data'), F.col('row_index')==F.col('row_number'), how='left')
final_df = final_df.select('initial_data.row_number', 'other_data.L1', 'other_data.L2')
final_df.show()
# +----------+--------+--------+
# |row_number| L1| L2|
# +----------+--------+--------+
# | 0|row 1 L1|row 1 L2|
# | 1|row 2 L1|row 2 L2|
# +----------+--------+--------+
就性能而言,此^绝对可以更好。