Mixing
pyspark lag function with inconsistent time series
import pyspark.sql.functions as F
from pyspark.sql.window import Window
I would like to use a window function to find the value from a column 4 periods ago.
Suppose my data (df) looks like this (in reality i have many different IDs):
ID | value | period
a | 100 | 1
a | 200 | 2
a | 300 | 3
a | 400 | 5
a | 500 | 6
a | 600 | 7
If the time series was consistent (e.g. period 1-6) I could just use F.lag(df['value'], count=4).over(Window.partitionBy('id').orderBy('period'))
However, because the time series has a discontinuity, the values would be displaced.
My desired output would be this:
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
How can I do this in pyspark?
python pyspark pyspark-sql
asked Nov 20 '18 at 13:45
DanDan
858
add a comment |
import pyspark.sql.functions as F
from pyspark.sql.window import Window
I would like to use a window function to find the value from a column 4 periods ago.
Suppose my data (df) looks like this (in reality i have many different IDs):
ID | value | period
a | 100 | 1
a | 200 | 2
a | 300 | 3
a | 400 | 5
a | 500 | 6
a | 600 | 7
If the time series was consistent (e.g. period 1-6) I could just use F.lag(df['value'], count=4).over(Window.partitionBy('id').orderBy('period'))
However, because the time series has a discontinuity, the values would be displaced.
My desired output would be this:
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
How can I do this in pyspark?
python pyspark pyspark-sql
asked Nov 20 '18 at 13:45
DanDan
858
add a comment |
import pyspark.sql.functions as F
from pyspark.sql.window import Window
I would like to use a window function to find the value from a column 4 periods ago.
Suppose my data (df) looks like this (in reality i have many different IDs):
ID | value | period
a | 100 | 1
a | 200 | 2
a | 300 | 3
a | 400 | 5
a | 500 | 6
a | 600 | 7
If the time series was consistent (e.g. period 1-6) I could just use F.lag(df['value'], count=4).over(Window.partitionBy('id').orderBy('period'))
However, because the time series has a discontinuity, the values would be displaced.
My desired output would be this:
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
How can I do this in pyspark?
python pyspark pyspark-sql
asked Nov 20 '18 at 13:45
DanDan
858
import pyspark.sql.functions as F
from pyspark.sql.window import Window
I would like to use a window function to find the value from a column 4 periods ago.
Suppose my data (df) looks like this (in reality i have many different IDs):
ID | value | period
a | 100 | 1
a | 200 | 2
a | 300 | 3
a | 400 | 5
a | 500 | 6
a | 600 | 7
If the time series was consistent (e.g. period 1-6) I could just use F.lag(df['value'], count=4).over(Window.partitionBy('id').orderBy('period'))
However, because the time series has a discontinuity, the values would be displaced.
My desired output would be this:
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
How can I do this in pyspark?
python pyspark pyspark-sql
python pyspark pyspark-sql
asked Nov 20 '18 at 13:45
DanDan
858
asked Nov 20 '18 at 13:45
DanDan
858
edited Nov 20 '18 at 13:51
Dan
asked Nov 20 '18 at 13:45
DanDan
858
asked Nov 20 '18 at 13:45
DanDan
858
asked Nov 20 '18 at 13:45
DanDan
858
858
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I've come up with a solution, but it seems unnecessarily ugly, would welcome anything better!
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]])
This returns the desired output.
answered Nov 20 '18 at 15:51
DanDan
858
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I've come up with a solution, but it seems unnecessarily ugly, would welcome anything better!
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]])
This returns the desired output.
answered Nov 20 '18 at 15:51
DanDan
858
add a comment |
I've come up with a solution, but it seems unnecessarily ugly, would welcome anything better!
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]])
This returns the desired output.
answered Nov 20 '18 at 15:51
DanDan
858
add a comment |
I've come up with a solution, but it seems unnecessarily ugly, would welcome anything better!
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]])
This returns the desired output.
answered Nov 20 '18 at 15:51
DanDan
858
I've come up with a solution, but it seems unnecessarily ugly, would welcome anything better!
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]])
This returns the desired output.
answered Nov 20 '18 at 15:51
DanDan
858
answered Nov 20 '18 at 15:51
DanDan
858
answered Nov 20 '18 at 15:51
DanDan
858
answered Nov 20 '18 at 15:51
DanDan
858
858
add a comment |
add a comment |
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