timestamp column changes of format in a csv file spark

[#1]

Hi guys i am trying to save a dataframe to a csv file , that contains a timestamp. The problem that this column changes of format one written in the csv file .when showing via df.show i got a correct format

when i check the csv file i got this format

i also tried some think like this ,and still got the same problem

finalresult.coalesce(1).write.option("header",true).option("inferSchema","true").option("dateFormat","yyyy-MM-dd HH:mm:ss").csv("C:/mydata.csv")

val spark =SparkSession.builder.master("local").appName("my-spark-app").getOrCreate()val df = spark.read.option("header",true).option("inferSchema","true").csv("C:/Users/mhattabi/Desktop/dataTest2.csv")//val df = spark.read.option("header",true).option("inferSchema", "true").csv("C:\dataSet.csv\datasetTest.csv")//convert all column to numeric value in order to apply aggregation function 
    df.columns.map { c  =>df.withColumn(c, col(c).cast("int"))}//add a new column inluding the new timestamp columnval result2=df.withColumn("new_time",((unix_timestamp(col("time"))/300).cast("long")*300).cast("timestamp")).drop("time")val finalresult=result2.groupBy("new_time").agg(result2.drop("new_time").columns.map((_ ->"mean")).toMap).sort("new_time")//agg(avg(all columns..)   finalresult.coalesce(1).write.option("header",true).option("inferSchema","true").csv("C:/mydata.csv")

[#2]

A quick hack would be to use scala "substring"

http://alvinalexander.com/scala/scala-string-examples-collection-cheat-sheet

So what you can do is write a UDF and run the "new_time" column through it and grab upto time stamp you want. For example if you want just "yyyy-MM-dd HH:MM" as seen when you run the "df.show", your sub string code will be

new_time.substring(0,15)

which will yield "2015-12-06 12:40"

pseudo code

def getDateTimeSplit = udf((new_time:String) => {
    val s = new_time.substring(0,15)
    return s
})