[英]pyspark lag function with inconsistent time series
import pyspark.sql.functions as F
from pyspark.sql.window import Window
我想使用窗口函数从 4 个周期前的列中查找值。
假设我的数据 (df) 看起来像这样(实际上我有许多不同的 ID):
ID | value | period
a | 100 | 1
a | 200 | 2
a | 300 | 3
a | 400 | 5
a | 500 | 6
a | 600 | 7
如果时间序列是一致的(例如周期 1-6),我可以使用F.lag(df['value'], count=4).over(Window.partitionBy('id').orderBy('period'))
但是,由于时间序列具有不连续性,因此这些值将被置换。
我想要的输出是这样的:
ID | value | period | 4_lag_value
a | 100 | 1 | nan
a | 200 | 2 | nan
a | 300 | 3 | nan
a | 400 | 5 | 100
a | 500 | 6 | 200
a | 600 | 7 | 300
我怎样才能在 pyspark 中做到这一点?
这可能是你正在寻找的:
from pyspark.sql import Window, functions as F
def pyspark_timed_lag_values(df, lags, avg_diff, state_id='state_id', ds='ds', y='y'):
interval_expr = 'sequence(min_ds, max_ds, interval {0} day)'.format(avg_diff)
all_comb = (df.groupBy(F.col(state_id))
.agg(F.min(ds).alias('min_ds'), F.max(ds).alias('max_ds'))
.withColumn(ds, F.explode(F.expr(interval_expr)))
.select(*[state_id, ds]))
all_comb = all_comb.join(df.withColumn('exists', F.lit(True)), on=[state_id, ds], how='left')
window = Window.partitionBy(state_id).orderBy(F.col(ds).asc())
for lag in lags:
all_comb = all_comb.withColumn("{0}_{1}".format(y, lag), F.lag(y, lag).over(window))
all_comb = all_comb.filter(F.col('exists')).drop(*['exists'])
return all_comb
让我们将其应用于示例:
data = spark.sparkContext.parallelize([
(1,"2021-01-03",100),
(1,"2021-01-10",830),
(1,"2021-01-17",300),
(1,"2021-02-07",450),
(2,"2021-01-03",500),
(2,"2021-01-17",800),
(2,"2021-02-14",800)])
example = spark.createDataFrame(data, ['state_id','ds','y'])
example = example.withColumn('ds', F.to_date(F.col('ds')))
lags = list(range(1, n_periods + 1))
result = timed_lag_values(example, lags = lags, avg_diff = 7)
导致以下结果:
+--------+----------+---+----+----+----+----+----+----+----+
|state_id| ds| y| y_1| y_2| y_3| y_4| y_5| y_6| y_7|
+--------+----------+---+----+----+----+----+----+----+----+
| 1|2021-01-03|100|null|null|null|null|null|null|null|
| 1|2021-01-10|830| 100|null|null|null|null|null|null|
| 1|2021-01-17|300| 830| 100|null|null|null|null|null|
| 1|2021-02-07|450|null|null| 300| 830| 100|null|null|
| 2|2021-01-03|500|null|null|null|null|null|null|null|
| 2|2021-01-17|800|null| 500|null|null|null|null|null|
| 2|2021-02-14|800|null|null|null| 800|null| 500|null|
+--------+----------+---+----+----+----+----+----+----+----+
现在它已经为日期做好了准备,但经过一些小的调整,它应该适用于各种用例。 在这种情况下,缺点是需要使用 expand 来创建所有可能的日期组合并创建助手 DataFrame all_comb
。
这个解决方案的真正好处是它适用于大多数处理时间序列的用例,因为参数avg_diff
定义了时间段之间的预期距离。
只是提一下,可能有一个更干净的 Hive SQL 替代方案。
我想出了一个解决方案,但它似乎不必要地丑陋,欢迎任何更好的东西!
data = spark.sparkContext.parallelize([
('a',100,1),
('a',200,2),
('a',300,3),
('a',400,5),
('a',500,6),
('a',600,7)])
df = spark.createDataFrame(data, ['id','value','period'])
window = Window.partitionBy('id').orderBy('period')
# look 1, 2, 3 and 4 rows behind:
for diff in [1,2,3,4]:
df = df.withColumn('{}_diff'.format(diff),
df['period'] - F.lag(df['period'], count=diff).over(window))
# if any of these are 4, that's the lag we need
# if not, there is no 4 period lagged return, so return None
#initialise col
df = df.withColumn('4_lag_value', F.lit(None))
# loop:
for diff in [1,2,3,4]:
df = df.withColumn('4_lag_value',
F.when(df['{}_diff'.format(diff)] == 4,
F.lag(df['value'], count=diff).over(window))
.otherwise(df['4_lag_value']))
# drop working cols
df = df.drop(*['{}_diff'.format(diff) for diff in [1,2,3,4]])
这将返回所需的输出。
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