Question

我的数据框类似于：

id  | date       | value
--- | ---------- | ------
1   | 2016-01-07 | 13.90
1   | 2016-01-16 | 14.50
2   | 2016-01-09 | 10.50
2   | 2016-01-28 | 5.50
3   | 2016-01-05 | 1.50

我正在尝试保留每个ID的最新值，如下所示：

id  | date       | value
--- | ---------- | ------
1   | 2016-01-16 | 14.50
2   | 2016-01-28 | 5.50
3   | 2016-01-05 | 1.50

我尝试按日期desc和drop duplicates排序：

new_df = df.orderBy(df.date.desc()).dropDuplicates(['id'])

我的问题是，dropDuplicates()会保留它找到的第一个重复值吗？有没有更好的方法来完成我想做的事情？顺便说一句，我使用的是python。

谢谢。

Answer 1

建议的窗口运算符可以很好地解决这个问题：

from datetime import date

rdd = sc.parallelize([
    [1, date(2016, 1, 7), 13.90],
    [1, date(2016, 1, 16), 14.50],
    [2, date(2016, 1, 9), 10.50],
    [2, date(2016, 1, 28), 5.50],
    [3, date(2016, 1, 5), 1.50]
])

df = rdd.toDF(['id','date','price'])
df.show()

+---+----------+-----+
| id|      date|price|
+---+----------+-----+
|  1|2016-01-07| 13.9|
|  1|2016-01-16| 14.5|
|  2|2016-01-09| 10.5|
|  2|2016-01-28|  5.5|
|  3|2016-01-05|  1.5|
+---+----------+-----+

df.registerTempTable("entries") // Replaced by createOrReplaceTempView in Spark 2.0

output = sqlContext.sql('''
    SELECT 
        *
    FROM (
        SELECT 
            *,
            dense_rank() OVER (PARTITION BY id ORDER BY date DESC) AS rank
        FROM entries
    ) vo WHERE rank = 1
''');

output.show();

+---+----------+-----+----+
| id|      date|price|rank|
+---+----------+-----+----+
|  1|2016-01-16| 14.5|   1|
|  2|2016-01-28|  5.5|   1|
|  3|2016-01-05|  1.5|   1|
+---+----------+-----+----+

Answer 2

如果您有相同日期的物品，那么您将获得与density_rank相同的副本。您应该使用row_number：

from pyspark.sql.window import Window
from datetime import date
import pyspark.sql.functions as F 

rdd = spark.sparkContext.parallelize([
    [1, date(2016, 1, 7), 13.90],
    [1, date(2016, 1, 7), 10.0 ], # I added this row to show the effect of duplicate
    [1, date(2016, 1, 16), 14.50],
    [2, date(2016, 1, 9), 10.50],
    [2, date(2016, 1, 28), 5.50],
    [3, date(2016, 1, 5), 1.50]]
)

df = rdd.toDF(['id','date','price'])
df.show(10)

+---+----------+-----+
| id|      date|price|
+---+----------+-----+
|  1|2016-01-07| 13.9|
|  1|2016-01-07| 10.0|
|  1|2016-01-16| 14.5|
|  2|2016-01-09| 10.5|
|  2|2016-01-28|  5.5|
|  3|2016-01-05|  1.5|
+---+----------+-----+


# row_number
df.withColumn("row_number",F.row_number().over(Window.partitionBy(df.id).orderBy(df.date))).filter(F.col("row_number")==1).show()

+---+----------+-----+----------+
| id|      date|price|row_number|
+---+----------+-----+----------+
|  3|2016-01-05|  1.5|         1|
|  1|2016-01-07| 13.9|         1|
|  2|2016-01-09| 10.5|         1|
+---+----------+-----+----------+

# dense_rank
df.withColumn("dense_rank",F.dense_rank().over(Window.partitionBy(df.id).orderBy(df.date))).filter(F.col("dense_rank")==1).show()

+---+----------+-----+----------+
| id|      date|price|dense_rank|
+---+----------+-----+----------+
|  3|2016-01-05|  1.5|         1|
|  1|2016-01-07| 13.9|         1|
|  1|2016-01-07| 10.0|         1|
|  2|2016-01-09| 10.5|         1|
+---+----------+-----+----------+

Answer 3

您可以使用row_number来获取具有最新日期的记录：

import pyspark.sql.functions as F
from pyspark.sql.window import Window

new_df = df.withColumn("row_number",F.row_number().over(Window.partitionBy(df.id).orderBy(df.date.desc()))).filter(F.col("row_number")==1).drop("row_number")

火花数据帧保持最近的记录

3 个答案: