问题描述
说我有一份杂志订阅清单，像这样:
Say I have a list of magazine subscriptions, like so:
subscription_id user_id created_at
12384 1 2018-08-10
83294 1 2018-06-03
98234 1 2018-04-08
24903 2 2018-05-08
32843 2 2018-03-06
09283 2 2018-04-07
现在，我想添加一列，指出用户在此当前订阅之前有多少个先前的订阅.例如，如果这是用户的第一个预订，则新列的值应为0.如果他们在此预订之前开始有一个预订，则新列的值应为1.这是完整的期望输出:
Now I want to add a column that states how many previous subscriptions a user had, before this current subscription. For example, if this is the user's first subscription, the new column's value should be 0. If they had one subscription starting before this subscription, the new column's value should be 1. Here is the full desired output:
subscription_id user_id created_at users_previous_subs
12384 1 2018-08-10 2
83294 1 2018-06-03 1
98234 1 2018-04-08 0
24903 2 2018-05-08 2
32843 2 2018-04-06 1
09283 2 2018-03-07 0
我该如何做到这一点，最好是在PySpark中，因此不使用 shift
How can I accomplish this, preferably in PySpark, therefore not using shift
让我知道是否不清楚.谢谢！
Let me know if this is not clear. Thanks!!
推荐答案
这可以归结为行编号的计算.
from pyspark.sql import Window
from pyspark.sql import functions as func
#Define a window
w = Window.partitionBy(df.user_id).orderBy(df.created_at)
#Add an extra column with rownumber
df.withColumn('prev_subs',func.row_number().over(w)-1)
df.show()
如果可以建立联系(即，给定日期的用户超过1行)，请使用 dense_rank .
If there can be ties, (i.e. more than 1 row with a given date for a user), use dense_rank.
df.withColumn('prev_subs',func.dense_rank().over(w)-1)
这篇关于Pyspark-GroupBy和Count与WHERE结合的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！