如何分组并合并spark dataframe组的行

Question

假设我有一个这样的table，

A  | B  |    C     | D  |  E  | F
x1 | 5  | 20200115 | 15 | 4.5 | 1
x1 | 10 | 20200825 | 15 | 5.6 | 19
x2 | 10 | 20200115 | 15 | 4.1 | 1
x2 | 10 | 20200430 | 15 | 9.1 | 1

我希望在 col A 上合并这些行并生成这样的数据框

A  | B  |    C     | D  |  E  | F
x1 | 15 | 20200825 | 15 | 5.6 | 19
x2 | 10 | 20200115 | 15 | 4.1 | 1
x2 | 10 | 20200430 | 15 | 9.1 | 1

基本上，如果A列组的B列总和等于D列的值，那么，

B 列的新值将是 B 列的总和
C、E、F 列将根据 C 列的最新日期（YYYYmmDD 中的日期）选取

由于对于X2组，上述条件不成立（即B列的总和为20大于D列的15），我想保留目标中的两条记录

假设：在我的数据中，给定组的 D 列将相同（在本例中为 15）

我看了一堆分组和开窗（分区）的例子，但在我看来这是不同的，我无法缩小路线范围。

我可以将分组数据通过管道传输到 UDF 并执行某些操作吗？

PS: 在 pyspark 中构建这个，如果你的示例可以在 pyspark 中，那就太好了

Answer 1

试试这个-

使用sum + max加窗函数

df.show(false)
    df.printSchema()
    /**
      * +---+---+--------+---+---+---+
      * |A  |B  |C       |D  |E  |F  |
      * +---+---+--------+---+---+---+
      * |x1 |5  |20200115|15 |4.5|1  |
      * |x1 |10 |20200825|15 |5.6|19 |
      * |x2 |10 |20200115|15 |4.1|1  |
      * |x2 |10 |20200430|15 |9.1|1  |
      * +---+---+--------+---+---+---+
      *
      * root
      * |-- A: string (nullable = true)
      * |-- B: integer (nullable = true)
      * |-- C: integer (nullable = true)
      * |-- D: integer (nullable = true)
      * |-- E: double (nullable = true)
      * |-- F: integer (nullable = true)
      */

    val w = Window.partitionBy("A")
    df.withColumn("sum", sum("B").over(w))
      .withColumn("latestC", max("C").over(w))
      .withColumn("retain",
        when($"sum" === $"D", when($"latestC" === $"C", true).otherwise(false) )
          .otherwise(true) )
      .where($"retain" === true)
      .withColumn("B", when($"sum" === $"D", when($"latestC" === $"C", $"sum").otherwise($"B") )
        .otherwise($"B"))
      .show(false)

    /**
      * +---+---+--------+---+---+---+---+--------+------+
      * |A  |B  |C       |D  |E  |F  |sum|latestC |retain|
      * +---+---+--------+---+---+---+---+--------+------+
      * |x1 |15 |20200825|15 |5.6|19 |15 |20200825|true  |
      * |x2 |10 |20200115|15 |4.1|1  |20 |20200430|true  |
      * |x2 |10 |20200430|15 |9.1|1  |20 |20200430|true  |
      * +---+---+--------+---+---+---+---+--------+------+
      */

Answer 2

在 pyspark 中，我会这样做：

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

b = ["A", "B", "C", "D", "E", "F"]
a = [
    ("x1", 5, "20200115", 15, 4.5, 1),
    ("x1", 10, "20200825", 15, 5.6, 19),
    ("x2", 10, "20200115", 15, 4.1, 1),
    ("x2", 10, "20200430", 15, 9.1, 1),
]

df = spark.createDataFrame(a, b)


df = df.withColumn("B_sum", F.sum("B").over(W.partitionBy("A")))

process_df = df.where("D >= B_Sum")
no_process_df = df.where("D < B_sum").drop("B_sum")


process_df = (
    process_df.withColumn(
        "rng", F.row_number().over(W.partitionBy("A").orderBy(F.col("C").desc()))
    )
    .where("rng=1")
    .select("A", F.col("B_sum").alias("B"), "C", "D", "E", "F",)
)

final_output = process_df.unionByName(no_process_df)
+---+---+--------+---+---+---+
|  A|  B|       C|  D|  E|  F|
+---+---+--------+---+---+---+
| x1| 15|20200825| 15|5.6| 19|
| x2| 10|20200115| 15|4.1|  1|
| x2| 10|20200430| 15|9.1|  1|
+---+---+--------+---+---+---+

如何分组并合并spark dataframe组的行

How to group by and merge these rows of spark dataframe's group

grouping

pandas

apache-spark

apache-spark-sql

pyspark