Spark Scala API有Dataset#transform
方法,可以轻松链接自定义DataFrame转换,如下所示:
val weirdDf = df
.transform(myFirstCustomTransformation)
.transform(anotherCustomTransformation)
我没有看到pyspark in the documentation的等效transform
方法。
是否有PySpark方式链接自定义转换?
如果没有,如何对pyspark.sql.DataFrame
类进行猴子修补以添加transform
方法?
答案 0 :(得分:1)
实现:
from pyspark.sql.dataframe import DataFrame
def transform(self, f):
return f(self)
DataFrame.transform = transform
用法:
spark.range(1).transform(lambda df: df.selectExpr("id * 2"))
答案 1 :(得分:0)
使用SQLTransformer对象(或任何其他Transformer)的Transformer管道是Spark解决方案,可简化链接转换。例如:
from pyspark.ml.feature import SQLTransformer
from pyspark.ml import Pipeline, PipelineModel
df = spark.createDataFrame([
(0, 1.0, 3.0),
(2, 2.0, 5.0)
], ["id", "v1", "v2"])
sqlTrans = SQLTransformer(
statement="SELECT *, (v1 + v2) AS v3, (v1 * v2) AS v4 FROM __THIS__")
sqlSelectExpr = SQLTransformer(statement="SELECT *, (id * 2) AS v5 FROM __THIS__")
pipeline = Pipeline(stages=[sqlTrans, sqlSelectExpr])
pipelineModel = pipeline.fit(df)
pipelineModel.transform(df).show()
当所有转换都是上述简单表达式时,另一种链接方法是使用单个SQLTransformer和字符串操作:
transforms = ['(v1 + v2) AS v3',
'(v1 * v2) AS v4',
'(id * 2) AS v5',
]
selectExpr = "SELECT *, {} FROM __THIS__".format(",".join(transforms))
sqlSelectExpr = SQLTransformer(statement=selectExpr)
sqlSelectExpr.transform(df).show()
请记住,Spark SQL转换可以优化,并且比定义为Python用户定义函数(UDF)的转换要快。