I am loading some data into Spark with a wrapper function:
def load_data( filename ):
df = sqlContext.read.format("com.databricks.spark.csv")\
.option("delimiter", "\t")\
.option("header", "false")\
.option("mode", "DROPMALFORMED")\
.load(filename)
# add the filename base as hostname
( hostname, _ ) = os.path.splitext( os.path.basename(filename) )
( hostname, _ ) = os.path.splitext( hostname )
df = df.withColumn('hostname', lit(hostname))
return df
specifically, I am using a glob to load a bunch of files at once:
df = load_data( '/scratch/*.txt.gz' )
the files are:
/scratch/host1.txt.gz
/scratch/host2.txt.gz
...
I would like the column 'hostname' to actually contain the real name of the file being loaded rather than the glob (ie host1
, host2
etc, rather than *
).
How can I do this?
You can use input_file_name
which:
Creates a string column for the file name of the current Spark task.
from pyspark.sql.functions import input_file_name
df.withColumn("filename", input_file_name())
Same thing in Scala:
import org.apache.spark.sql.functions.input_file_name
df.withColumn("filename", input_file_name)
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