Question

这可能是最容易通过示例解释的。假设我有一个用户登录网站的DataFrame，例如：

scala> df.show(5)
+----------------+----------+
|       user_name|login_date|
+----------------+----------+
|SirChillingtonIV|2012-01-04|
|Booooooo99900098|2012-01-04|
|Booooooo99900098|2012-01-06|
|  OprahWinfreyJr|2012-01-10|
|SirChillingtonIV|2012-01-11|
+----------------+----------+
only showing top 5 rows

我想在此添加一个列，指示他们何时成为网站上的活跃用户。但有一点需要注意：有一段时间用户被认为是活动的，在此期间之后，如果他们再次登录，他们的became_active日期会重置。假设这段时间 5天。然后从上表得到的所需表格将是这样的：

+----------------+----------+-------------+
|       user_name|login_date|became_active|
+----------------+----------+-------------+
|SirChillingtonIV|2012-01-04|   2012-01-04|
|Booooooo99900098|2012-01-04|   2012-01-04|
|Booooooo99900098|2012-01-06|   2012-01-04|
|  OprahWinfreyJr|2012-01-10|   2012-01-10|
|SirChillingtonIV|2012-01-11|   2012-01-11|
+----------------+----------+-------------+

因此，特别是，SirChillingtonIV的became_active日期已重置，因为他们的第二次登录是在活动期过后，但Booooooo99900098的became_active日期未重置第二次他/她登录，因为它属于活跃期。

我最初的想法是使用lag的窗口函数，然后使用lag ged值填充became_active列;例如，大致类似的东西：

import org.apache.spark.sql.expressions.Window
import org.apache.spark.sql.functions._

val window = Window.partitionBy("user_name").orderBy("login_date")
val df2 = df.withColumn("tmp", lag("login_date", 1).over(window))

然后，如果became_active是tmp（即，如果它是第一次登录）或null，则填写login_date - tmp >= 5日期的规则将是became_active = login_date 1}}然后tmp;否则，转到tmp中的下一个最新值并应用相同的规则。这表明了一种递归方法，我很难想象实现的方法。

我的问题：这是一种可行的方法吗？如果是这样，我怎样才能回过头来＃34;并查看Column的早期值，直到找到我停止的位置？据我所知，我无法迭代Spark SQL x ^ x == 0 y ^ y ^ y == y z ^ 0 == z的值。还有另一种方法来实现这个结果吗？

Answer 1

这是诀窍。导入一堆函数：

import org.apache.spark.sql.expressions.Window
import org.apache.spark.sql.functions.{coalesce, datediff, lag, lit, min, sum}

定义窗口：

val userWindow = Window.partitionBy("user_name").orderBy("login_date")
val userSessionWindow = Window.partitionBy("user_name", "session")

找到新会话开始的点数：

val newSession =  (coalesce(
  datediff($"login_date", lag($"login_date", 1).over(userWindow)),
  lit(0)
) > 5).cast("bigint")

val sessionized = df.withColumn("session", sum(newSession).over(userWindow))

查找每个会话的最早日期：

val result = sessionized
  .withColumn("became_active", min($"login_date").over(userSessionWindow))
  .drop("session")

将数据集定义为：

val df = Seq(
  ("SirChillingtonIV", "2012-01-04"), ("Booooooo99900098", "2012-01-04"),
  ("Booooooo99900098", "2012-01-06"), ("OprahWinfreyJr", "2012-01-10"), 
  ("SirChillingtonIV", "2012-01-11"), ("SirChillingtonIV", "2012-01-14"),
  ("SirChillingtonIV", "2012-08-11")
).toDF("user_name", "login_date")

结果是：

+----------------+----------+-------------+
|       user_name|login_date|became_active|
+----------------+----------+-------------+
|  OprahWinfreyJr|2012-01-10|   2012-01-10|
|SirChillingtonIV|2012-01-04|   2012-01-04| <- The first session for user
|SirChillingtonIV|2012-01-11|   2012-01-11| <- The second session for user
|SirChillingtonIV|2012-01-14|   2012-01-11| 
|SirChillingtonIV|2012-08-11|   2012-08-11| <- The third session for user
|Booooooo99900098|2012-01-04|   2012-01-04|
|Booooooo99900098|2012-01-06|   2012-01-04|
+----------------+----------+-------------+

Answer 2

重构以上答案以使用Pyspark

在Pyspark中，您可以执行以下操作。

create data frame

df = sqlContext.createDataFrame(
[
("SirChillingtonIV", "2012-01-04"), 
("Booooooo99900098", "2012-01-04"), 
("Booooooo99900098", "2012-01-06"), 
("OprahWinfreyJr", "2012-01-10"), 
("SirChillingtonIV", "2012-01-11"), 
("SirChillingtonIV", "2012-01-14"), 
("SirChillingtonIV", "2012-08-11")
], 
("user_name", "login_date"))

上面的代码创建了如下所示的数据框

+----------------+----------+
|       user_name|login_date|
+----------------+----------+
|SirChillingtonIV|2012-01-04|
|Booooooo99900098|2012-01-04|
|Booooooo99900098|2012-01-06|
|  OprahWinfreyJr|2012-01-10|
|SirChillingtonIV|2012-01-11|
|SirChillingtonIV|2012-01-14|
|SirChillingtonIV|2012-08-11|
+----------------+----------+

现在，我们首先要找出login_date天之间的差异超过5天。

为此，请执行以下操作。

必要的进口

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


# defining window partitions  
login_window = Window.partitionBy("user_name").orderBy("login_date")
session_window = Window.partitionBy("user_name", "session")

session_df = df.withColumn("session", f.sum((f.coalesce(f.datediff("login_date", f.lag("login_date", 1).over(login_window)), f.lit(0)) > 5).cast("int")).over(login_window))

如果date_diff是NULL，则运行上面的代码行时，coalesce函数会将NULL替换为0。

+----------------+----------+-------+
|       user_name|login_date|session|
+----------------+----------+-------+
|  OprahWinfreyJr|2012-01-10|      0|
|SirChillingtonIV|2012-01-04|      0|
|SirChillingtonIV|2012-01-11|      1|
|SirChillingtonIV|2012-01-14|      1|
|SirChillingtonIV|2012-08-11|      2|
|Booooooo99900098|2012-01-04|      0|
|Booooooo99900098|2012-01-06|      0|
+----------------+----------+-------+


# add became_active column by finding the `min login_date` for each window partitionBy `user_name` and `session` created in above step
final_df = session_df.withColumn("became_active", f.min("login_date").over(session_window)).drop("session")

+----------------+----------+-------------+
|       user_name|login_date|became_active|
+----------------+----------+-------------+
|  OprahWinfreyJr|2012-01-10|   2012-01-10|
|SirChillingtonIV|2012-01-04|   2012-01-04|
|SirChillingtonIV|2012-01-11|   2012-01-11|
|SirChillingtonIV|2012-01-14|   2012-01-11|
|SirChillingtonIV|2012-08-11|   2012-08-11|
|Booooooo99900098|2012-01-04|   2012-01-04|
|Booooooo99900098|2012-01-06|   2012-01-04|
+----------------+----------+-------------+

具有复杂条件的Spark SQL窗口函数

2 个答案: