Reputation: 3584
I come from Pandas background and new to Spark. I have a dataframe which has id
, dob
, age
as columns. I want to get the age of the user from his dob
(in some cases age
column is NULL).
+----+------+----------+
| id | age | dob |
+----+------+----------+
| 1 | 24 | NULL |
| 2 | 25 | NULL |
| 3 | NULL | 1/1/1973 |
| 4 | NULL | 6/6/1980 |
| 5 | 46 | |
| 6 | NULL | 1/1/1971 |
+----+------+----------+
I want a new column which will calculate age from dob
and current date.
I tried this, but not getting any results from it:
df.withColumn("diff",
datediff(to_date(lit("01-06-2020")),
to_date(unix_timestamp('dob', "dd-MM-yyyy").cast("timestamp")))).show()
Upvotes: 4
Views: 15743
Reputation: 24478
Using months_between
like in this answer, but in a different approach:
'age'
column yet;.cast('int')
.from pyspark.sql import functions as F
df = df.withColumn('age', (F.months_between(current_date(), F.col('dob')) / 12).cast('int'))
If system date is UTC and your locale is different, a separate date function may be needed:
from pyspark.sql import functions as F
def current_local_date():
return F.from_utc_timestamp(F.current_timestamp(), 'Europe/Riga').cast('date')
df = df.withColumn('age', (F.months_between(current_local_date(), F.col('dob')) / 12).cast('int'))
Upvotes: 3
Reputation: 6338
I believe it is more appropriate to use months_between
when it comes to year difference
. we should use datediff
only when if you need difference in days
val data =
"""
| id | age | dob
| 1 | 24 |
| 2 | 25 |
| 3 | | 1/1/1973
| 4 | | 6/6/1980
| 5 | 46 |
| 6 | | 1/1/1971
""".stripMargin
val stringDS = data.split(System.lineSeparator())
.map(_.split("\\|").map(_.replaceAll("""^[ \t]+|[ \t]+$""", "")).mkString(","))
.toSeq.toDS()
val df = spark.read
.option("sep", ",")
.option("inferSchema", "true")
.option("header", "true")
.csv(stringDS)
df.show(false)
df.printSchema()
/**
* +---+----+--------+
* |id |age |dob |
* +---+----+--------+
* |1 |24 |null |
* |2 |25 |null |
* |3 |null|1/1/1973|
* |4 |null|6/6/1980|
* |5 |46 |null |
* |6 |null|1/1/1971|
* +---+----+--------+
*
* root
* |-- id: integer (nullable = true)
* |-- age: integer (nullable = true)
* |-- dob: string (nullable = true)
*/
df.withColumn("diff",
coalesce(col("age"),
round(months_between(current_date(),to_date(col("dob"), "d/M/yyyy"),true).divide(12),2)
)
).show()
/**
* +---+----+--------+-----+
* | id| age| dob| diff|
* +---+----+--------+-----+
* | 1| 24| null| 24.0|
* | 2| 25| null| 25.0|
* | 3|null|1/1/1973|47.42|
* | 4|null|6/6/1980|39.99|
* | 5| 46| null| 46.0|
* | 6|null|1/1/1971|49.42|
* +---+----+--------+-----+
*/
round it to
0
if you want age in whole number
Upvotes: 4
Reputation: 45339
You need to compute the date difference and convert the result to years, something like this:
df.withColumn('diff',
when(col('age').isNull(),
floor(datediff(current_date(), to_date(col('dob'), 'M/d/yyyy'))/365.25))\
.otherwise(col('age'))).show()
Which produces:
+---+----+--------+----+
| id| age| dob|diff|
+---+----+--------+----+
| 1| 24| null| 24|
| 2| 25| null| 25|
| 3|null|1/1/1973| 47|
| 4|null|6/6/1980| 39|
| 5| 46| null| 46|
| 6|null|1/1/1971| 49|
+---+----+--------+----+
It preserves the age
column where not null and computes the difference (in days) between dob
and today where age
is null. The result is then converted to years (by dividing by 365.25; you may want to confirm this) then floor
ed.
Upvotes: 7