我是火花的新手,我正在尝试计算以0为底,以8为上限的窗口运行总和。
下面是一个玩具示例(请注意,实际数据更接近数百万行):
import pyspark.sql.functions as F
from pyspark.sql import Window
import pandas as pd
from pyspark.sql.functions import pandas_udf, PandasUDFType
pdf = pd.DataFrame({'ids': [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3],
'day': [1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4],
'counts': [-3, 3, -6, 3, 3, 6, -3, -6, 3, 3, 3, -3]})
sdf = spark.createDataFrame(pdf)
sdf = sdf.orderBy(sdf.ids,sdf.day)
这将创建表格
+----+---+-------+
|aIds|day|eCounts|
+----+---+-------+
| 1| 1| -3|
| 1| 2| 3|
| 1| 3| -6|
| 1| 4| 3|
| 2| 1| 3|
| 2| 2| 6|
| 2| 3| -3|
| 2| 4| -6|
| 3| 1| 3|
| 3| 2| 3|
| 3| 3| 3|
| 3| 4| -3|
+----+---+-------+
下面是一个求和结果的示例,以及预期输出runSumCap
+----+---+-------+------+---------+
|aIds|day|eCounts|runSum|runSumCap|
+----+---+-------+------+---------+
| 1| 1| -3| -3| 0| <-- reset to 0
| 1| 2| 3| 0| 3|
| 1| 3| -6| -6| 0| <-- reset to 0
| 1| 4| 3| -3| 3|
| 2| 1| 3| 3| 3|
| 2| 2| 6| 9| 8| <-- reset to 8
| 2| 3| -3| 6| 5|
| 2| 4| -6| 0| 0| <-- reset to 0
| 3| 1| 3| 3| 3|
| 3| 2| 3| 6| 6|
| 3| 3| 3| 9| 8| <-- reset to 8
| 3| 4| -3| 6| 5|
+----+---+-------+------+---------+
我知道我可以将运行总和计算为
partition = Window.partitionBy('aIds').orderBy('aIds','day').rowsBetween(Window.unboundedPreceding, Window.currentRow)`
sdf1 = sdf.withColumn('runSum',F.sum(sdf.eCounts).over(partition))
sdf1.orderBy('aIds','day').show()
为了达到预期效果,我尝试查看@pandas_udf来修改总和:
@pandas_udf('double', PandasUDFType.GROUPED_AGG)
def runSumCap(counts):
#counts columns is passed as a pandas series
floor = 0
cap = 8
runSum = 0
runSumList = []
for count in counts.tolist():
runSum = runSum + count
if(runSum > cap):
runSum = 8
elif(runSum < floor ):
runSum = 0
runSumList += [runSum]
return pd.Series(runSumList)
partition = Window.partitionBy('aIds').orderBy('aIds','day').rowsBetween(Window.unboundedPreceding, Window.currentRow)
sdf1 = sdf.withColumn('runSum',runSumCap(sdf['counts']).over(partition))
但是,这不起作用,而且这似乎不是最有效的方法。 我该如何工作?有没有办法让它保持平行,还是我必须去熊猫数据框
编辑: 对当前列进行了一些整理以对数据集进行排序,并对我要实现的目标有了更多见解
EDIT2: @DrChess提供的答案几乎可以得出正确的结果,但是由于某种原因,该系列与正确的日期不匹配:
+----+---+-------+------+
|aIds|day|eCounts|runSum|
+----+---+-------+------+
| 1| 1| -3| 0|
| 1| 2| 3| 0|
| 1| 3| -6| 3|
| 1| 4| 3| 3|
| 2| 1| 3| 3|
| 2| 2| 6| 8|
| 2| 3| -3| 0|
| 2| 4| -6| 5|
| 3| 1| 3| 6|
| 3| 2| 3| 3|
| 3| 3| 3| 8|
| 3| 4| -3| 5|
+----+---+-------+------+
答案 0 :(得分:1)
不幸的是,类型为pandas_udf
的{{1}}的窗口函数不能用于有界窗口函数(GROUPED_AGG
)。当前仅适用于无边界窗口,即.rowsBetween(Window.unboundedPreceding, Window.currentRow)
。另外,输入是.rowsBetween(Window.unboundedPreceding, Window.unboundedFollowing)
,但是输出应该是所提供类型的常量。因此,您将无法实现部分聚合。
您可以使用pandas.Series
的{{1}} GROUPED_MAP
。
这里是一些代码:
pandas_udf
答案 1 :(得分:1)
我找到了一种方法,首先在每行中创建一个数组(使用collect_list作为窗口函数),该数组包含用于使运行总和到那时为止的值。 然后,我定义了一个udf(无法使用pandas_udf进行此工作),并且此工作正常。 下面是完整的可复制示例:
import pyspark.sql.functions as F
from pyspark.sql import Window
import pandas as pd
from pyspark.sql.functions import pandas_udf, PandasUDFType
from pyspark.sql.types import *
import numpy as np
def accumalate(iterable):
total = 0
ceil = 8
floor = 0
for element in iterable:
total = total + element
if (total > ceil):
total = ceil
elif (total < floor):
total = floor
return total
pdf = pd.DataFrame({'aIds': [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3],
'day': [1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4],
'eCounts': [-3, 3, -6, 3, 3, 6, -3, -6, 3, 3, 3, -3]})
sdf = spark.createDataFrame(pdf)
sdf = sdf.orderBy(sdf.aIds,sdf.day)
runSumCap = F.udf(accumalate,LongType())
partition = Window.partitionBy('aIds').orderBy('aIds','day').rowsBetween(Window.unboundedPreceding, Window.currentRow)
sdf1 = sdf.withColumn('splitWindow',F.collect_list(sdf.eCounts).over(partition))
sdf2 = sdf1.withColumn('runSumCap',runSumCap(sdf1.splitWindow))
sdf2.orderBy('aIds','day').show()
这会产生预期的结果:
+----+---+-------+--------------+---------+
|aIds|day|eCounts| splitWindow|runSumCap|
+----+---+-------+--------------+---------+
| 1| 1| -3| [-3]| 0|
| 1| 2| 3| [-3, 3]| 3|
| 1| 3| -6| [-3, 3, -6]| 0|
| 1| 4| 3|[-3, 3, -6, 3]| 3|
| 2| 1| 3| [3]| 3|
| 2| 2| 6| [3, 6]| 8|
| 2| 3| -3| [3, 6, -3]| 5|
| 2| 4| -6|[3, 6, -3, -6]| 0|
| 3| 1| 3| [3]| 3|
| 3| 2| 3| [3, 3]| 6|
| 3| 3| 3| [3, 3, 3]| 8|
| 3| 4| -3| [3, 3, 3, -3]| 5|
+----+---+-------+--------------+---------+