I have a csv file; which i convert to DataFrame(df) in pyspark; after some transformation; I want to add a column in df; which should be simple row id (starting from 0 or 1 to N).
I converted df in rdd and use “zipwithindex”. I converted resulting rdd back to df. this approach works but it generated 250k tasks and takes a lot of time in execution. I was wondering if there is other way to do it which takes less runtime.
following is snippet of my code; the csv file I am processing is BIG; contains billions of rows.
debug_csv_rdd = (sc.textFile("debug.csv") .filter(lambda x: x.find('header') == -1) .map(lambda x : x.replace("NULL","0")).map(lambda p: p.split(',')) .map(lambda x:Row(c1=int(x),c2=int(x),c3=int(x),c4=int(x)))) debug_csv_df = sqlContext.createDataFrame(debug_csv_rdd) debug_csv_df.registerTempTable("debug_csv_table") sqlContext.cacheTable("debug_csv_table") r0 = sqlContext.sql("SELECT c2 FROM debug_csv_table WHERE c1 = 'str'") r0.registerTempTable("r0_table") r0_1 = (r0.flatMap(lambda x:x) .zipWithIndex() .map(lambda x: Row(c1=x,id=int(x)))) r0_df=sqlContext.createDataFrame(r0_2) r0_df.show(10)
You can use also use a function from sql package. It will generate a unique id, however it will not be sequential as it depends on the number of partitions. I believe it is available in Spark 1.5 +
from pyspark.sql.functions import monotonicallyIncreasingId # This will return a new DF with all the columns + id res = df.withColumn("id", monotonicallyIncreasingId())
As commented by @Sean
monotonically_increasing_id() instead from Spark 1.6 and on