Question

我有区域明智的客户数据来源，如下所示

region,source,consumer_id
APAC,mail,1
APAC,referral,2
APAC,mail,3
APAC,referral,5
APAC,mail,6
APAC,referral,7
APAC,referral,8
US East,mail,9
US East,referral,10
US East,walkIn,11
AUS,walkIn,12
AUS,referral,13

有人可以使用pyspark数据帧帮助获得像下面这样的区域明智的源数。

region,mail_source_cnt, referral_source_cnt, walkIn_source_cnt
APAC,3,4,0
US EAST,1,1,1
AUS,0,1,1

感谢您的帮助

Answer 1

您可以聚合以获取计数并将列旋转为

>>> from pyspark.sql import functions as F
>>> df.show()
+-------+--------+-----------+
| region|  source|consumer_id|
+-------+--------+-----------+
|   APAC|    mail|          1|
|   APAC|referral|          2|
|   APAC|    mail|          3|
|   APAC|referral|          5|
|   APAC|    mail|          6|
|   APAC|referral|          7|
|   APAC|referral|          8|
|US East|    mail|          9|
|US East|referral|         10|
|US East|  walkIn|         11|
|    AUS|  walkIn|         12|
|    AUS|referral|         13|
+-------+--------+-----------+
>>> df1 = df.groupby('region','source').count()
>>> df1.show()
+-------+--------+-----+
| region|  source|count|
+-------+--------+-----+
|US East|  walkIn|    1|
|    AUS|  walkIn|    1|
|   APAC|    mail|    3|
|   APAC|referral|    4|
|US East|    mail|    1|
|US East|referral|    1|
|    AUS|referral|    1|
+-------+--------+-----+
>>> df2 = df1.withColumn('ccol',F.concat(df1['source'],F.lit('_cnt'))).groupby('region').pivot('ccol').agg(F.first('count')).fillna(0)
>>> df2.show()
+-------+--------+------------+----------+
| region|mail_cnt|referral_cnt|walkIn_cnt|
+-------+--------+------------+----------+
|    AUS|       0|           1|         1|
|   APAC|       3|           4|         0|
|US East|       1|           1|         1|
+-------+--------+------------+----------+

使用group来获取多个字段的pyspark数据帧

1 个答案: