Pyspark 中同一列的递归操作

recursive operation on the same column in Pyspark

我有一个这样的数据框:

数据框:

|SEQ_ID |TIME_STAMP             |_MS               |
+-------+-----------------------+------------------+
|3879826|2021-07-29 11:24:20.525|NaN               |
|3879826|2021-07-29 11:25:56.934|21.262409581399556|
|3879826|2021-07-29 11:27:43.264|27.247600203353613|
|3879826|2021-07-29 11:29:27.613|18.13528511851038 |
|3879826|2021-07-29 11:31:10.512|2.520896614376871 |
|3879826|2021-07-29 11:32:54.252|2.7081931585605541|
|3879826|2021-07-29 11:34:36.995|2.9832290627235505|
|3879826|2021-07-29 11:36:19.128|13.011968111650264|
|3879826|2021-07-29 11:38:10.919|17.762006254598797|
|3879826|2021-07-29 11:40:01.929|1.9661930950977457|

_MS >=3 且前一个 _MS 小于当前 _MS 我想将新列 drift_MS 增加 100。但是如果_MS <3 and previous _MS < current _MS 我想将 drift_MS 加 1。如果 none 的条件满足,我想设置值 0

预期输出:

|SEQ_ID |TIME_STAMP             |_MS               |drift_MS|
+-------+-----------------------+------------------+--------+
|3879826|2021-07-29 11:24:20.525|NaN               |0       |
|3879826|2021-07-29 11:25:56.934|21.262409581399556|0       |
|3879826|2021-07-29 11:27:43.264|27.247600203353613|100     |
|3879826|2021-07-29 11:29:27.613|18.13528511851038 |0       |
|3879826|2021-07-29 11:31:10.512|2.520896614376871 |0       |
|3879826|2021-07-29 11:32:54.252|2.7081931585605541|1       |
|3879826|2021-07-29 11:34:36.995|2.9832290627235505|2       |
|3879826|2021-07-29 11:36:19.128|13.011968111650264|102     |
|3879826|2021-07-29 11:38:10.919|17.762006254598797|202     |
|3879826|2021-07-29 11:40:01.929|1.9661930950977457|0       |

这个问题我有一个不同的版本,我只是想保持以前的值不变,一个非常有帮助的贡献者建议我像这样使用求和函数;

import pyspark.sql.functions as f

w1=Window.partitionBy('SEQ_ID').orderBy(col('TIME_STAMP').asc())
    
prev_MS = (f.lag(col('_MS'),1).over(w1))
df.withColumn('drift_MS', 
  f.sum(
    when((col('_MS') < 3) & (prev_MS < col('_MS')), 1)
    .when((col('_MS') >= 3) & (prev_MS < col('_MS')), 100)
    .otherwise(0)
 ).over(w1))

如果 none 的条件得到满足,我希望之前的 drift_MS 值保持不变,这非常有效。但是,如果不满足条件,我现在需要将其重置为零。 我试图弄清楚,但我一直在碰壁,我需要迭代循环回到前一行,这通常不会在 pyspark 或大数据中完成,因为它对列操作最有效

以下代码对我不起作用:

import pyspark.sql.functions as f

w1=Window.partitionBy('SEQ_ID').orderBy(col('TIME_STAMP').asc())
prev_drift_MS_temp = (f.lag(col('drift_MS_temp'),1).over(w1))
prev_drift_MS = (f.lag(col('drift_MS'),1).over(w1))
    
prev_MS = (f.lag(col('_MS'),1).over(w1))
df.withColumn('drift_MS_temp', 
  f.sum(
    when((col('_MS') < 3) & (prev_MS < col('_MS')), 1)
    .when((col('_MS') >= 3) & (prev_MS < col('_MS')), 100)
    .otherwise(0)
 ).over(w1))\
  .withColumn('drift_MS',when(prev_drift_MS_temp==col('drift_MS_temp'),0)
  .otherwise(col('drift_MS_temp') - prev_drift_MS_temp + prev_drift_MS))

有什么想法可以解决这个问题吗?

更新: 因此,在为此苦思冥想之后,到目前为止我想到的最好的逻辑是从 drift_MS 创建一个不同的列,然后在差异列不是 0 时有一个条件累积和 所以像这样:

|SEQ_ID |TIME_STAMP             |_MS               |drift_MS|_diff   |drift   |
+-------+-----------------------+------------------+--------+--------+--------+
|3879826|2021-07-29 11:24:20.525|NaN               |0       |0       |0       |
|3879826|2021-07-29 11:25:56.934|21.262409581399556|0       |0       |0       |
|3879826|2021-07-29 11:27:43.264|27.247600203353613|100     |100     |100     |
|3879826|2021-07-29 11:29:27.613|18.13528511851038 |100     |0       |0       |
|3879826|2021-07-29 11:31:10.512|2.520896614376871 |100     |0       |0       |
|3879826|2021-07-29 11:32:54.252|2.7081931585605541|101     |1       |1       |
|3879826|2021-07-29 11:34:36.995|2.9832290627235505|102     |1       |1       |
|3879826|2021-07-29 11:36:19.128|13.011968111650264|202     |100     |102     |
|3879826|2021-07-29 11:38:10.919|17.762006254598797|302     |100     |202     |
|3879826|2021-07-29 11:40:01.929|1.9661930950977457|302     |0       |0       |

我设想的伪代码如下所示:

import pyspark.sql.functions as f

w1=Window.partitionBy('SEQ_ID').orderBy(col('TIME_STAMP').asc())
prev_drift_MS = (f.lag(col('drift_MS'),1).over(w1))
prev_diff= (f.lag(col('_diff'),1).over(w1))

prev_MS = (f.lag(col('_MS'),1).over(w1))
df.withColumn('drift_MS', 
  f.sum(
    when((col('_MS') < 3) & (prev_MS < col('_MS')), 1)
    .when((col('_MS') >= 3) & (prev_MS < col('_MS')), 100)
    .otherwise(0)
 ).over(w1))\
 .withColumn('_diff', prev_drift_MS - col('drift_MS'))\
 .withColumn('drift', when(prev_diff==0, 0).otherwise(f.sum(col('drift')).over(w1)))

以这种方式获取它的正确语法是什么?

我们可以使用的一个选项是在获得最终的 drift_MS 列之前创建一堆辅助列。让我们逐步尝试。

  1. 通过应用您定义的增量条件创建列 x
  2. 创建列 y 作为标志,其中列 x 中的值重置为零。
  3. 创建列 z 以将标志之间的行组合在一起。我们可以在当前行和无限后续行之间的行内使用累积和。
  4. 最终创建列 drift_MS 作为按 SEQ_ID 分组的行和按 TIME_STAMP.
  5. 排序的辅助列 z 的累积总和

将这些步骤放入代码中会像这样(在 SQL 表达式中更容易阅读)

import pyspark.sql.functions as F

expr_x = F.expr("""
    case 
    when _MS >= 3 AND lag(_MS) over (partition by SEQ_ID  order by TIME_STAMP) < _MS then 100
    when _MS < 3 AND lag(_MS) over (partition by SEQ_ID order by TIME_STAMP) < _MS then 1
    else 0 end  """)

expr_y = F.expr("""
    case 
    when x <> 0 and lead(x) over (partition by SEQ_ID order by TIME_STAMP) = 0 then 1
    else null end """)

expr_z = F.expr("""
    sum(y) over(partition by SEQ_ID 
                order by TIME_STAMP 
                rows between 0 preceding and unbounded following) """)

expr_drift = F.expr("""
    sum(x) over (partition by SEQ_ID, z 
                 order by TIME_STAMP 
                 rows between unbounded preceding and 0 following) """)

df = (df
      .withColumn('x', expr_x)
      .withColumn('y', expr_y)
      .withColumn('z', expr_z)
      .withColumn("drift_MS", expr_drift))
df.show()

# +-------+--------------------+------------------+---+----+----+--------+
# | SEQ_ID|          TIME_STAMP|               _MS|  x|   y|   z|drift_MS|
# +-------+--------------------+------------------+---+----+----+--------+
# |3879826|2021-07-29 11:24:...|               NaN|  0|null|   2|       0|
# |3879826|2021-07-29 11:25:...|21.262409581399556|  0|null|   2|       0|
# |3879826|2021-07-29 11:27:...|27.247600203353613|100|   1|   2|     100|
# |3879826|2021-07-29 11:29:...| 18.13528511851038|  0|null|   1|       0|
# |3879826|2021-07-29 11:31:...| 2.520896614376871|  0|null|   1|       0|
# |3879826|2021-07-29 11:32:...| 2.708193158560554|  1|null|   1|       1|
# |3879826|2021-07-29 11:34:...|2.9832290627235505|  1|null|   1|       2|
# |3879826|2021-07-29 11:36:...|13.011968111650264|100|null|   1|     102|
# |3879826|2021-07-29 11:38:...|  17.7620062545988|100|   1|   1|     202|
# |3879826|2021-07-29 11:40:...|1.9661930950977458|  0|null|null|       0|
# +-------+--------------------+------------------+---+----+----+--------+