my question is how to split a column to multiple columns.
I don't know why df.toPandas() does not work.
For example, I would like to change 'df_test' to 'df_test2'. I saw many examples using the pandas module. Is there another way? Thank you in advance.
df_test = sqlContext.createDataFrame([
(1, '14-Jul-15'),
(2, '14-Jun-15'),
(3, '11-Oct-15'),
], ('id', 'date'))
df_test2
id day month year
1 14 Jul 15
2 14 Jun 15
1 11 Oct 15
Spark >= 2.2
You can skip unix_timestamp and cast and use to_date or to_timestamp:
from pyspark.sql.functions import to_date, to_timestamp
df_test.withColumn("date", to_date("date", "dd-MMM-yy")).show()
## +---+----------+
## | id| date|
## +---+----------+
## | 1|2015-07-14|
## | 2|2015-06-14|
## | 3|2015-10-11|
## +---+----------+
df_test.withColumn("date", to_timestamp("date", "dd-MMM-yy")).show()
## +---+-------------------+
## | id| date|
## +---+-------------------+
## | 1|2015-07-14 00:00:00|
## | 2|2015-06-14 00:00:00|
## | 3|2015-10-11 00:00:00|
## +---+-------------------+
and then apply other datetime functions shown below.
Spark < 2.2
It is not possible to derive multiple top level columns in a single access. You can use structs or collection types with an UDF like this:
from pyspark.sql.types import StringType, StructType, StructField
from pyspark.sql import Row
from pyspark.sql.functions import udf, col
schema = StructType([
StructField("day", StringType(), True),
StructField("month", StringType(), True),
StructField("year", StringType(), True)
])
def split_date_(s):
try:
d, m, y = s.split("-")
return d, m, y
except:
return None
split_date = udf(split_date_, schema)
transformed = df_test.withColumn("date", split_date(col("date")))
transformed.printSchema()
## root
## |-- id: long (nullable = true)
## |-- date: struct (nullable = true)
## | |-- day: string (nullable = true)
## | |-- month: string (nullable = true)
## | |-- year: string (nullable = true)
but it is not only quite verbose in PySpark, but also expensive.
For date based transformations you can simply use built-in functions:
from pyspark.sql.functions import unix_timestamp, dayofmonth, year, date_format
transformed = (df_test
.withColumn("ts",
unix_timestamp(col("date"), "dd-MMM-yy").cast("timestamp"))
.withColumn("day", dayofmonth(col("ts")).cast("string"))
.withColumn("month", date_format(col("ts"), "MMM"))
.withColumn("year", year(col("ts")).cast("string"))
.drop("ts"))
Similarly you could use regexp_extract to split date string.
See also Derive multiple columns from a single column in a Spark DataFrame
Note:
If you use version not patched against SPARK-11724 this will require correction after unix_timestamp(...) and before cast("timestamp").
The Solution here is to use pyspark.sql.functions.split() function.
df = sqlContext.createDataFrame([
(1, '14-Jul-15'),
(2, '14-Jun-15'),
(3, '11-Oct-15'),
], ('id', 'date'))
split_col = pyspark.sql.functions.split(df['date'], '-')
df = df.withColumn('day', split_col.getItem(0))
df = df.withColumn('month', split_col.getItem(1))
df = df.withColumn('year', split_col.getItem(2))
df = df.drop("date")
