在Java Spark Dataframe中比较日期。

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英文:

Comparing dates in Java Spark Dataframe

问题

以下是翻译好的内容:

我有下面的Spark DataFrame/数据集。Column_2列中的日期以字符串格式表示。

Column_1 Column_2
A        2020-08-05
B        2020-08-01
B        2020-09-20
B        2020-12-31
C        2020-05-10

我期望的输出DataFrame应该对于Column_1中的每个值只有一行,如果Column_1中的同一键对应多个日期,则应选择下一个可用日期。如果只有一行,则应保留日期。

期望的输出:

Column_1 Column_2
A        2020-08-05
B        2020-09-20
C        2020-05-10

是否有一种方法可以在Java Spark中实现这一目标?可能不需要使用UDF?

英文:

I have the below Spark dataframe/dataset. Column_2 has dates in string format.

Column_1 Column_2
A        2020-08-05
B        2020-08-01
B        2020-09-20
B        2020-12-31
C        2020-05-10

My expected output dataframe should have only one row per value in Column_1 and if there are multiple dates in column_2 for same key in column_1, then the next available date should be picked. if only one row is there, then the date should be retained

Expected Output:

Column_1 Column_2
A        2020-08-05
B        2020-09-20
C        2020-05-10

Is there a way to achieve this Java spark? possibly without using UDF?

答案1

得分: 1

也许这对你有帮助-

   dataset.show(false);
        dataset.printSchema();
        /**
         *+--------+----------+
         * |Column_1|Column_2  |
         * +--------+----------+
         * |A       |2020-08-05|
         * |D       |2020-08-01|
         * |D       |2020-08-02|
         * |B       |2020-08-01|
         * |B       |2020-09-20|
         * |B       |2020-12-31|
         * |C       |2020-05-10|
         * +--------+----------+
         *
         * root
         *  |-- Column_1: string (nullable = true)
         *  |-- Column_2: string (nullable = true)
         */

        dataset.withColumn("Column_2", to_date(col("Column_2")))
                .withColumn("count", count("Column_2").over(Window.partitionBy("Column_1")))
                .withColumn("positive", when(col("count").gt(1),
                        when(col("Column_2").gt(current_date()), col("Column_2"))
                ).otherwise(col("Column_2")))
                .withColumn("negative", when(col("count").gt(1),
                        when(col("Column_2").lt(current_date()), col("Column_2"))
                ).otherwise(col("Column_2")))
                .groupBy("Column_1")
                .agg(min("positive").as("positive"), max("negative").as("negative"))
                .selectExpr("Column_1", "coalesce(positive, negative) as Column_2")
                .show(false);
        /**
         * +--------+----------+
         * |Column_1|Column_2  |
         * +--------+----------+
         * |A       |2020-08-05|
         * |D       |2020-08-02|
         * |B       |2020-09-20|
         * |C       |2020-05-10|
         * +--------+----------+
         */
英文:

Perhaps this is helpful-

   dataset.show(false);
        dataset.printSchema();
        /**
         *+--------+----------+
         * |Column_1|Column_2  |
         * +--------+----------+
         * |A       |2020-08-05|
         * |D       |2020-08-01|
         * |D       |2020-08-02|
         * |B       |2020-08-01|
         * |B       |2020-09-20|
         * |B       |2020-12-31|
         * |C       |2020-05-10|
         * +--------+----------+
         *
         * root
         *  |-- Column_1: string (nullable = true)
         *  |-- Column_2: string (nullable = true)
         */

        dataset.withColumn("Column_2", to_date(col("Column_2")))
                .withColumn("count", count("Column_2").over(Window.partitionBy("Column_1")))
                .withColumn("positive", when(col("count").gt(1),
                        when(col("Column_2").gt(current_date()), col("Column_2"))
                ).otherwise(col("Column_2")))
                .withColumn("negative", when(col("count").gt(1),
                        when(col("Column_2").lt(current_date()), col("Column_2"))
                ).otherwise(col("Column_2")))
                .groupBy("Column_1")
                .agg(min("positive").as("positive"), max("negative").as("negative"))
                .selectExpr("Column_1", "coalesce(positive, negative) as Column_2")
                .show(false);
        /**
         * +--------+----------+
         * |Column_1|Column_2  |
         * +--------+----------+
         * |A       |2020-08-05|
         * |D       |2020-08-02|
         * |B       |2020-09-20|
         * |C       |2020-05-10|
         * +--------+----------+
         */

答案2

得分: 0

创建DataFrame首先

df_b = spark.createDataFrame([("A","2020-08-05"),("B","2020-08-01"),("B","2020-09-20"),("B","2020-12-31"),("C","2020-05-10")],["col1","col2"])
_w = W.partitionBy("col1").orderBy("col1")
df_b = df_b.withColumn("rn", F.row_number().over(_w))

在这里的逻辑是选择每个组的第二个元素,如果任何组有多行。为了做到这一点,我们可以首先为每个组分配一个行号,然后在每个组中的行数大于1的情况下选择每个组的前两行。

case = F.expr("""
            CASE WHEN rn =1 THEN 1
                    WHEN rn =2 THEN 1
              END""")
df_b = df_b.withColumn('case_condition', case)
df_b = df_b.filter(F.col("case_condition") == F.lit("1"))

中间输出

+----+----------+---+--------------+
|col1|      col2| rn|case_condition|
+----+----------+---+--------------+
|   B|2020-08-01|  1|             1|
|   B|2020-09-20|  2|             1|
|   C|2020-05-10|  1|             1|
|   A|2020-08-05|  1|             1|
+----+----------+---+--------------+

现在,最后只需取每个组的最后一个元素--

df = df_b.groupBy("col1").agg(F.last("col2").alias("col2")).orderBy("col1")
df.show()
+----+----------+
|col1|      col2|
+----+----------+
|   A|2020-08-05|
|   B|2020-09-20|
|   C|2020-05-10|
+----+----------+
英文:

Create the DataFrame First

df_b = spark.createDataFrame([("A","2020-08-05"),("B","2020-08-01"),("B","2020-09-20"),("B","2020-12-31"),("C","2020-05-10")],[ "col1","col2"])
_w = W.partitionBy("col1").orderBy("col1")
df_b = df_b.withColumn("rn", F.row_number().over(_w))

The logic here to pick the second element of each group if any group has a more than one row. In order to do that we can first assign a row number to every group and we will pick first element of every group where row count is 1 and , first 2 row of every group where row count is more than 1 in every group.

case = F.expr("""
CASE WHEN rn =1 THEN 1
WHEN rn =2 THEN 1
END""")
df_b = df_b.withColumn('case_condition', case)
df_b = df_b.filter(F.col("case_condition") == F.lit("1")) 

Intermediate Output

+----+----------+---+--------------+
|col1|      col2| rn|case_condition|
+----+----------+---+--------------+
|   B|2020-08-01|  1|             1|
|   B|2020-09-20|  2|             1|
|   C|2020-05-10|  1|             1|
|   A|2020-08-05|  1|             1|
+----+----------+---+--------------+

Now, finally just take the last element of every group --

df = df_b.groupBy("col1").agg(F.last("col2").alias("col2")).orderBy("col1")
df.show()
+----+----------+
|col1|      col2|
+----+----------+
|   A|2020-08-05|
|   B|2020-09-20|
|   C|2020-05-10|
+----+----------+

答案3

得分: 0

SCALA: 这将提供结果。

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

val w = Window.partitionBy("Column_1")

df.withColumn("count", count("Column_2").over(w))
  .withColumn("later", expr("IF(Column_2 > date(current_timestamp), True, False)"))
  .filter("count = 1 or (count != 1 and later = True)")
  .groupBy("Column_1")
  .agg(min("Column_2").alias("Column_2"))
  .orderBy("Column_1")
  .show(false)

+--------+----------+
|Column_1|Column_2  |
+--------+----------+
|A       |2020-08-05|
|B       |2020-09-20|
|C       |2020-05-10|
+--------+----------+

有一个例外情况,如果Column_1的日期计数大于1并且没有日期晚于current_timestamp,则不会为Column_1的值提供结果。

英文:

SCALA: This will give the result.

import org.apache.spark.sql.expressions.Window
val w = Window.partitionBy("Column_1")
df.withColumn("count", count("Column_2").over(w))
.withColumn("later", expr("IF(Column_2 > date(current_timestamp), True, False)"))
.filter("count = 1 or (count != 1 and later = True)")
.groupBy("Column_1")
.agg(min("Column_2").alias("Column_2"))
.orderBy("Column_1")
.show(false)
+--------+----------+
|Column_1|Column_2  |
+--------+----------+
|A       |2020-08-05|
|B       |2020-09-20|
|C       |2020-05-10|
+--------+----------+

It has an exception that if the count of the dates for the Column_1 is larger than 1 and there is no date after the current_timestamp, it will not give the result for the value of Column_1.

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  • 本文由 发表于 2020年8月4日 15:17:00
  • 转载请务必保留本文链接:https://go.coder-hub.com/63241916.html
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