在 PySpark 中查找从中提取最小值超过 window.partitionBy 的行值

Question

我有一个像这样的 PySpark 数据框：

+--------+-------------+--------------+-----------------------+
|material|purchase_date|mkt_prc_usd_lb|min_mkt_prc_over_1month|
+--------+-------------+--------------+-----------------------+
|  Copper|   2019-01-09|        2.6945|                 2.6838|
|  Copper|   2019-01-23|        2.6838|                 2.6838|
|    Zinc|   2019-01-23|        1.1829|                 1.1829|
|    Zinc|   2019-06-26|        1.1918|                 1.1918|
|Aluminum|   2019-01-02|        0.8363|                 0.8342|
|Aluminum|   2019-01-09|        0.8342|                 0.8342|
|Aluminum|   2019-01-23|        0.8555|                 0.8342|
|Aluminum|   2019-04-03|        0.8461|                 0.8461|
+--------+-------------+--------------+-----------------------+

最后一列 'min_mkt_prc_over_1month' 计算为 material 一个月内的最小值 'mkt_prc_usd_lb'（第 3 列），即（-15 天，至 +15 天）material, purchase_date window:

密码是：


w2 = (Window()
           .partitionBy("material")
           .orderBy(col("purchase_date").cast("timestamp").cast("long"))
           .rangeBetween(-days(15), days(15)))

现在，我想看看当金额 was/will 最小时 'purchase_date' 是多少？

预期输出：（从前两行开始）

+--------+-------------+--------------+-----------------------+------------------+
|material|purchase_date|mkt_prc_usd_lb|min_mkt_prc_over_1month|date_of_min_price |
+--------+-------------+--------------+-----------------------+------------------+
|  Copper|   2019-01-09|        2.6945|                 2.6838|        2019-01-23|
|  Copper|   2019-01-23|        2.6838|                 2.6838|        2019-01-23|
+--------+-------------+--------------+-----------------------+------------------+

Answer 1

试试这个。我们可以在 two prc are the same to populate it with purchase date、otherwise to put Null 的位置创建一个列，然后我们可以使用 First with ignoreNulls=True，在我们的 newly created column using our window w2..

from pyspark.sql.functions import *
from pyspark.sql.window import Window

days= lambda i: i * 86400
w2 = (Window()
           .partitionBy("material")
           .orderBy(col("purchase_date").cast("timestamp").cast("long"))
           .rangeBetween(-days(15), days(15)))


df.withColumn("first",\
              expr("""IF(mkt_prc_usd_lb=min_mkt_prc_over_1month,purchase_date,null)"""))\
  .withColumn("date_of_min_price", first("first", True).over(w2)).drop("first")\
  .show()

#+--------+-------------+--------------+-----------------------+-----------------+
#|material|purchase_date|mkt_prc_usd_lb|min_mkt_prc_over_1month|date_of_min_price|
#+--------+-------------+--------------+-----------------------+-----------------+
#|  Copper|   2019-01-09|        2.6945|                 2.6838|       2019-01-23|
#|  Copper|   2019-01-23|        2.6838|                 2.6838|       2019-01-23|
#|    Zinc|   2019-01-23|        1.1829|                 1.1829|       2019-01-23|
#|    Zinc|   2019-06-26|        1.1918|                 1.1918|       2019-06-26|
#|Aluminum|   2019-01-02|        0.8363|                 0.8342|       2019-01-09|
#|Aluminum|   2019-01-09|        0.8342|                 0.8342|       2019-01-09|
#|Aluminum|   2019-01-23|        0.8555|                 0.8342|       2019-01-09|
#|Aluminum|   2019-04-03|        0.8461|                 0.8461|       2019-04-03|
#+--------+-------------+--------------+-----------------------+-----------------+

在 PySpark 中查找从中提取最小值超过 window.partitionBy 的行值

Find the row value from which minimum value was extracted over window.partitionBy in PySpark

pyspark

apache-spark-sql

window

pyspark-dataframes

partition-by