PySpark.将数据框传递给pandas_udf并返回一系列 [英] PySpark. Passing a Dataframe to a pandas_udf and returning a series

问题描述

我正在使用PySpark的新pandas_udf装饰器，并且试图将其接受多列作为输入并返回一系列作为输入，但是，我得到了TypeError: Invalid argument

I'm using PySpark's new pandas_udf decorator and I'm trying to get it to take multiple columns as an input and return a series as an input, however, I get a TypeError: Invalid argument

示例代码

@pandas_udf(df.schema, PandasUDFType.SCALAR)
def fun_function(df_in):
    df_in.loc[df_in['a'] < 0] = 0.0
    return (df_in['a'] - df_in['b']) / df_in['c']

推荐答案

A SCALAR udf expects pandas series as input instead of a data frame. For your case, there's no need to use a udf. Direct calculation from columns a, b, c after clipping should work:

import pyspark.sql.functions as f

df = spark.createDataFrame([[1,2,4],[-1,2,2]], ['a', 'b', 'c'])

clip = lambda x: f.when(df.a < 0, 0).otherwise(x)
df.withColumn('d', (clip(df.a) - clip(df.b)) / clip(df.c)).show()

#+---+---+---+-----+
#|  a|  b|  c|    d|
#+---+---+---+-----+
#|  1|  2|  4|-0.25|
#| -1|  2|  2| null|
#+---+---+---+-----+

如果必须使用pandas_udf，则返回类型必须为double，而不是df.schema，因为您只返回熊猫系列而不是熊猫数据框；而且您还需要将列作为Series传递给函数，而不是整个数据帧:

And if you have to use a pandas_udf, your return type needs to be double, not df.schema because you only return a pandas series not a pandas data frame; And also you need to pass columns as Series into the function not the whole data frame:

@pandas_udf('double', PandasUDFType.SCALAR)
def fun_function(a, b, c):
    clip = lambda x: x.where(a >= 0, 0)
    return (clip(a) - clip(b)) / clip(c)

df.withColumn('d', fun_function(df.a, df.b, df.c)).show()
#+---+---+---+-----+                                                             
#|  a|  b|  c|    d|
#+---+---+---+-----+
#|  1|  2|  4|-0.25|
#| -1|  2|  2| null|
#+---+---+---+-----+

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