如何在pyspark中以秒为单位获得datediff（）？

Question

我已尝试过（this_post）中的代码，无法在几秒钟内获得日期差异。我只是在下面的'Attributes_Timestamp_fix'和'lagged_date'列之间使用datediff（）。任何提示？在我的代码和输出下面。

eg = eg.withColumn("lagged_date", lag(eg.Attributes_Timestamp_fix, 1)
.over(Window.partitionBy("id")
.orderBy("Attributes_Timestamp_fix")))

eg = eg.withColumn("time_diff", 
datediff(eg.Attributes_Timestamp_fix, eg.lagged_date))

        id      Attributes_Timestamp_fix time_diff
0   3.531611e+14    2018-04-01 00:01:02 NaN
1   3.531611e+14    2018-04-01 00:01:02 0.0
2   3.531611e+14    2018-04-01 00:03:13 0.0
3   3.531611e+14    2018-04-01 00:03:13 0.0
4   3.531611e+14    2018-04-01 00:03:13 0.0
5   3.531611e+14    2018-04-01 00:03:13 0.0

Answer 1

在pyspark.sql.functions，有一个函数datediff，不幸的是只计算天数的差异。要解决此问题，您可以在unix时间戳（以秒为单位）中转换这两个日期并计算差异。

让我们创建一些样本数据，计算滞后，然后计算秒数的差异。

from pyspark.sql.functions import col, lag, unix_timestamp
from pyspark.sql.window import Window
import datetime

d = [{'id' : 1, 't' : datetime.datetime(2018,01,01)},\
 {'id' : 1, 't' : datetime.datetime(2018,01,02)},\
 {'id' : 1, 't' : datetime.datetime(2018,01,04)},\
 {'id' : 1, 't' : datetime.datetime(2018,01,07)}]

df = spark.createDataFrame(d)
df.show()
+---+-------------------+
| id|                  t|
+---+-------------------+
|  1|2018-01-01 00:00:00|
|  1|2018-01-02 00:00:00|
|  1|2018-01-04 00:00:00|
|  1|2018-01-07 00:00:00|
+---+-------------------+

w = Window.partitionBy('id').orderBy('t')
df.withColumn("previous_t", lag(df.t, 1).over(w))\
  .select(df.t, (unix_timestamp(df.t) - unix_timestamp(col('previous_t'))).alias('diff'))\
  .show()

+-------------------+------+
|                  t|  diff|
+-------------------+------+
|2018-01-01 00:00:00|  null|
|2018-01-02 00:00:00| 86400|
|2018-01-04 00:00:00|172800|
|2018-01-07 00:00:00|259200|
+-------------------+------+