数据分组问题，但基于“窗口”

Question

全部，

我有一个数据集，其定义如下：

eno|date|attendance
1|01-Jan-2010|P
1|02-Jan-2010|P
1|03-Jan-2010|A
1|04-Jan-2010|P
1|05-Jan-2010|P
2|01-Jan-2010|P
2|02-Jan-2010|P
2|03-Jan-2010|P
2|04-Jan-2010|A
2|05-Jan-2010|P

对于每个员工，要求创建一个“间隔组”，该组基本上按时间顺序对出勤值进行分组。 组是将相似的出勤值组合在一起直到看到新的出勤值的东西。 所以预期的输出是：

eno|date|attendance|attendanceGroup
1|01-Jan-2010|P|1
1|02-Jan-2010|P|1
1|03-Jan-2010|A|2
1|04-Jan-2010|P|3
1|05-Jan-2010|P|3
2|01-Jan-2010|P|1
2|02-Jan-2010|P|1
2|03-Jan-2010|P|1
2|04-Jan-2010|A|2
2|05-Jan-2010|P|3

到目前为止，我所能做的就是获得前一行的出勤值，但完全不知道如何从这里开始......在此先感谢很多......

from datetime import datetime, timedelta
EmployeeAttendance = Row("eno", "date", "attendance")
EmpAttRowList = [EmployeeAttendance("1", datetime.now().date() - timedelta(days=100), "Y"),
                 EmployeeAttendance("1", datetime.now().date() - timedelta(days=99), "Y"),
                 EmployeeAttendance("1", datetime.now().date() - timedelta(days=98), "N"),
                 EmployeeAttendance("1", datetime.now().date() - timedelta(days=97), "Y"),
                 EmployeeAttendance("1", datetime.now().date() - timedelta(days=96), "Y"),
                 EmployeeAttendance("1", datetime.now().date() - timedelta(days=95), "N"),
                 EmployeeAttendance("1", datetime.now().date() - timedelta(days=94), "Y"),
                 EmployeeAttendance("1", datetime.now().date() - timedelta(days=93), "Y"),
                 EmployeeAttendance("2", datetime.now().date() - timedelta(days=100), "Y"),
                 EmployeeAttendance("2", datetime.now().date() - timedelta(days=99), "Y"),
                 EmployeeAttendance("2", datetime.now().date() - timedelta(days=98), "N"),
                 EmployeeAttendance("2", datetime.now().date() - timedelta(days=97), "Y"),
                 EmployeeAttendance("2", datetime.now().date() - timedelta(days=96), "Y"),
                 EmployeeAttendance("2", datetime.now().date() - timedelta(days=95), "N"),
                 EmployeeAttendance("2", datetime.now().date() - timedelta(days=94), "N"),
                 EmployeeAttendance("2", datetime.now().date() - timedelta(days=93), "N"),
                 EmployeeAttendance("2", datetime.now().date() - timedelta(days=92), "Y"),
                 EmployeeAttendance("2", datetime.now().date() - timedelta(days=91), "Y"),
                 EmployeeAttendance("2", datetime.now().date() - timedelta(days=90), "N"),
                 EmployeeAttendance("3", datetime.now().date() - timedelta(days=97), "Y"),
                 EmployeeAttendance("3", datetime.now().date() - timedelta(days=96), "Y"),
                 EmployeeAttendance("3", datetime.now().date() - timedelta(days=95), "Y"),
                 EmployeeAttendance("3", datetime.now().date() - timedelta(days=94), "N"),
                 EmployeeAttendance("3", datetime.now().date() - timedelta(days=93), "N"),
                 EmployeeAttendance("3", datetime.now().date() - timedelta(days=92), "Y"),
                 EmployeeAttendance("3", datetime.now().date() - timedelta(days=91), "Y"),
                 EmployeeAttendance("3", datetime.now().date() - timedelta(days=90), "Y"),
                 EmployeeAttendance("3", datetime.now().date() - timedelta(days=89), "Y")
                ]

df = spark.createDataFrame(EmpAttRowList, EmployeeAttendance)
window = Window.partitionBy(df['eno']).orderBy("date")
previousrowattendance = lag(df["attendance"]).over(window)

Answer 1

考虑到您已经使用上面提到的代码创建了数据框，您可以使用下面的代码来获取出席组。 让我知道它是否有效。

import pyspark.sql.functions as F
from pyspark.sql import Window

winSpec = Window.partitionBy('eno').orderBy('date')
df_unique = df.withColumn('prevAttendance', F.lag('attendance').over(winSpec))
df_unique = df_unique.filter((df_unique.attendance != df_unique.prevAttendance) | F.col('prevAttendance').isNull())
df_unique = df_unique.withColumn('attendanceGroup', F.row_number().over(winSpec))
df_unique = df_unique.withColumnRenamed('eno', 'eno_t').withColumnRenamed('date', 'date_t').drop('attendance').drop('prevAttendance')
df = df.join(df_unique, (df.eno == df_unique.eno_t) & (df.date == df_unique.date_t), 'left').drop('eno_t').drop('date_t')
df = df.withColumn('attendanceGroup', F.last('attendanceGroup', ignorenulls = True).over(winSpec))
df.orderBy('eno', 'date').show(10, False)

+---+----------+----------+---------------+
|eno|date      |attendance|attendanceGroup|
+---+----------+----------+---------------+
|1  |2019-08-16|Y         |1              |
|1  |2019-08-17|Y         |1              |
|1  |2019-08-18|N         |2              |
|1  |2019-08-19|Y         |3              |
|1  |2019-08-20|Y         |3              |
|1  |2019-08-21|N         |4              |
|1  |2019-08-22|Y         |5              |
|1  |2019-08-23|Y         |5              |
|2  |2019-08-16|Y         |1              |
|2  |2019-08-17|Y         |1              |
+---+----------+----------+---------------+
only showing top 10 rows

Answer 2

你可以试试这个：

1. 创建一个条件attendance != lag(attendance)的grp标志attendance != lag(attendance)以方便对标志求和

2. 新建一个由原来的 id eno和新创建的grp标志列分区的新窗口，并应用sum ，基本加 1 以从 1 开始计数器。

window = Window.partitionBy("eno").orderBy("date")

df = df.withColumn('grp', F.when(F.col("attendance") != F.lag(F.col("attendance")).over(window), 1).otherwise(0))
df = df.withColumn("group", 1 + F.sum(F.col("grp")).over(Window.partitionBy("eno", "grp").orderBy("date"))).drop("grp").orderBy("eno", "date")

输出

+---+----------+----------+-----+
|eno|      date|attendance|group|
+---+----------+----------+-----+
|  1|2019-08-17|         Y|    1|
|  1|2019-08-18|         Y|    1|
|  1|2019-08-19|         N|    2|
|  1|2019-08-20|         Y|    3|
|  1|2019-08-21|         Y|    1|
|  1|2019-08-22|         N|    4|
|  1|2019-08-23|         Y|    5|
|  1|2019-08-24|         Y|    1|
|  2|2019-08-17|         Y|    1|
|  2|2019-08-18|         Y|    1|
|  2|2019-08-19|         N|    2|
|  2|2019-08-20|         Y|    3|
|  2|2019-08-21|         Y|    1|
|  2|2019-08-22|         N|    4|
|  2|2019-08-23|         N|    1|
|  2|2019-08-24|         N|    1|
|  2|2019-08-25|         Y|    5|
|  2|2019-08-26|         Y|    1|
|  2|2019-08-27|         N|    6|
|  3|2019-08-20|         Y|    1|
+---+----------+----------+-----+
only showing top 20 rows