I've 2 dataframes and I want to find the records with all columns equal except 2 (surrogate_key,current)
And then I want to save those records with new surrogate_key value.
Following is my code :
val seq = csvDataFrame.columns.toSeq
var exceptDF = csvDataFrame.except(csvDataFrame.as('a).join(table.as('b),seq).drop("surrogate_key","current"))
exceptDF.show()
exceptDF = exceptDF.withColumn("surrogate_key", makeSurrogate(csvDataFrame("name"), lit("ecc")))
exceptDF = exceptDF.withColumn("current", lit("Y"))
exceptDF.show()
exceptDF.write.option("driver","org.postgresql.Driver").mode(SaveMode.Append).jdbc(postgreSQLProp.getProperty("url"), tableName, postgreSQLProp)
This code gives correct results, but get stuck while writing those results to postgre.
Not sure what's the issue. Also is there any better approach for this??
Regards,
Sorabh
By Default spark-sql creates 200 partitions, which means when you are trying to save the datafrmae it will be saved in 200 parquet files. you can reduce the number of partitions for Dataframe using below techniques.
At application level. Set the parameter "spark.sql.shuffle.partitions" as follows :
sqlContext.setConf("spark.sql.shuffle.partitions", "10")
Reduce the number of partition for a particular DataFrame as follows :
df.coalesce(10).write.save(...)
Using the var for dataframe are not suggested, You should always use val and create a new Dataframe after performing some transformation in dataframe.
Please remove all the var and replace with val.
Hope this helps!
Related
We are currrently facing an issue where we cannot insert more than 600K records in oracle db using AWS glue. We are getting connection reset error and DBA's are currently looking into it. As a temporary solution we thought of adding data in chunks by splitting a dataframe into multiple dataframe and looping this list of dataframe to add data. We are sure that splitting algorithm works fine and here is the code we use
def split_by_row_index(df, num_partitions=10):
# Let's assume you don't have a row_id column that has the row order
t = df.withColumn('_row_id', monotonically_increasing_id())
# Using ntile() because monotonically_increasing_id is discontinuous across partitions
t = t.withColumn('_partition', ntile(num_partitions).over(Window.orderBy(t._row_id)))
return [t.filter(t._partition == i + 1) for i in range(num_partitions)]
Here each DF have unique data but somehow when we convert this df in dynamic frame in loop it is we are getting common data in each dynamic frame. here is small snippet for this example
df_trns_details_list = split_by_row_index(df_trns_details, int(df_trns_details.count() / 100000))
trnsDetails1 = DynamicFrame.fromDF(df_trns_details_list[0], glueContext, "trnsDetails1")
trnsDetails2 = DynamicFrame.fromDF(df_trns_details_list[1], glueContext, "trnsDetails2")
print(df_trns_details_list[0].count())# counts are same
print(trnsDetails1.count())
print('-------------------------------')
print(df_trns_details_list[1].count()) # counts are same
print(trnsDetails2.count())
print('-------------------------------')
subDf1 = trnsDetails1.toDF().select(col("id"), col("details_id"))
subDf2 = trnsDetails2.toDF().select(col("id"), col("details_id"))
common = subDf1.intersect(subDf2)
# ------------------ common data exists----------------
print(common.count())
subDf3 = df_trns_details_list[0].select(col("id"), col("details_id"))
subDf4 = df_trns_details_list[1].select(col("id"), col("details_id"))
#------------------0 common data----------------
common1 = subDf3.intersect(subDf4)
print(common1.count())
here Id and details_id combination will be unique
We used this logic in multiple areas where it worked not sure why this is happening.
We are also quite new to Python and AWS Glue so any suggestion to improve it also welcomed. Thanks
I have many files containing millions of rows in format:
id, created_date, some_value_a, some_value_b, some_value_c
This way of repartitioning was super slow and created for me over million of small ~500b files:
rdd_df = rdd.toDF(["id", "created_time", "a", "b", "c"])
rdd_df.write.partitionBy("id").csv("output")
I would like to achieve output files, where each file contains like 10000 unique IDs and all their rows.
How could I achieve something like this?
You can repartition by adding a Random Salt key.
val totRows = rdd_df.count
val maxRowsForAnId = rdd_df.groupBy("id").count().agg(max("count"))
val numParts1 = totRows/maxRowsForAnId
val totalUniqueIds = rdd_df.select("id").distinct.count
val numParts2 = totRows/(10000*totalUniqueIds)
val numPart = numParts1.min(numParts2)
rdd_df
.repartition(numPart,col("id"),rand)
.csv("output")
The main concept is each partition will be written as 1 file. SO you would have bring your required rows in to 1 partition by repartition(numPart,col("id"),rand).
The first 4-5 operations is just to calculate how many partitions we need to achieve almost 10000 ids per file.
Calculate assuming 10000 ids per partition
Corner case : if a single id has too many rows and doesn't fit in the above calculated partition size.
Hence we calculate no of paritition according to the largest count of ID present
Take min of the 2 noOfPartitons
rand is necessary so, that we can bring multiple IDs in a single partition
NOTE : Although this will give you larger files and each file will contain a set of unique ids for sure. But this involves shuffling , due to which your operation actually might be slower than the code you have mentioned in question.
You would need something like this:
rdd_df.repartition(*number of partitions you want*).write.csv("output", header = True)
or honestly - just let the job decide the number partitions instead of repartitioning. In theory, that should be faster:
rdd_df.write.csv("output", header = True)
I am trying to read a Hive table in Spark. Below is the Hive Table format:
# Storage Information
SerDe Library: org.apache.hadoop.hive.ql.io.orc.OrcSerde
InputFormat: org.apache.hadoop.hive.ql.io.orc.OrcInputFormat
OutputFormat: org.apache.hadoop.hive.ql.io.orc.OrcOutputFormat
Compressed: No
Num Buckets: -1
Bucket Columns: []
Sort Columns: []
Storage Desc Params:
field.delim \u0001
serialization.format \u0001
When I am trying to read it using the Spark SQL with the below command:
val c = hiveContext.sql("""select
a
from c_db.c cs
where dt >= '2016-05-12' """)
c. show
I am getting the below warning:-
18/07/02 18:02:02 WARN ReaderImpl: Cannot find field for: a in _col0,
_col1, _col2, _col3, _col4, _col5, _col6, _col7, _col8, _col9, _col10, _col11, _col12, _col13, _col14, _col15, _col16, _col17, _col18, _col19, _col20, _col21, _col22, _col23, _col24, _col25, _col26, _col27, _col28, _col29, _col30, _col31, _col32, _col33, _col34, _col35, _col36, _col37, _col38, _col39, _col40, _col41, _col42, _col43, _col44, _col45, _col46, _col47, _col48, _col49, _col50, _col51, _col52, _col53, _col54, _col55, _col56, _col57, _col58, _col59, _col60, _col61, _col62, _col63, _col64, _col65, _col66, _col67,
The read starts but it is very slow and getting network time out.
When i am trying to read the Hive table directory directly i am getting the below error.
val hiveContext = new org.apache.spark.sql.hive.HiveContext(sc)
hiveContext.setConf("spark.sql.orc.filterPushdown", "true")
val c = hiveContext.read.format("orc").load("/a/warehouse/c_db.db/c")
c.select("a").show()
org.apache.spark.sql.AnalysisException: cannot resolve 'a' given input
columns: [_col18, _col3, _col8, _col66, _col45, _col42, _col31,
_col17, _col52, _col58, _col50, _col26, _col63, _col12, _col27, _col23, _col6, _col28, _col54, _col48, _col33, _col56, _col22, _col35, _col44, _col67, _col15, _col32, _col9, _col11, _col41, _col20, _col2, _col25, _col24, _col64, _col40, _col34, _col61, _col49, _col14, _col13, _col19, _col43, _col65, _col29, _col10, _col7, _col21, _col39, _col46, _col4, _col5, _col62, _col0, _col30, _col47, trans_dt, _col57, _col16, _col36, _col38, _col59, _col1, _col37, _col55, _col51, _col60, _col53];
at org.apache.spark.sql.catalyst.analysis.package$AnalysisErrorAt.failAnalysis(package.scala:42)
I can convert the Hive table to TextInputFormat but that should be my last option as i would like to get the benefit of OrcInputFormat to compress the table size.
Really appreciate your suggestion.
I found workaround with reading table such way:
val schema = spark.table("db.name").schema
spark.read.schema(schema).orc("/path/to/table")
The issue occurs generally with large tables, as it fails to read to max field length. I added meta-store read as true (set spark.sql.hive.convertMetastoreOrc=true;) and it worked for me.
I think the table doesnt have named columns or if it has, Spark isnt able to read the names probably.
You can use the default column names that Spark has given as mentioned in the Error. Or also set column names in the Spark code.
Use printSchema and toDF method to rename the columns. But yes, you will need the mappings. This might require selecting and showing columns individually.
Setting (set spark.sql.hive.convertMetastoreOrc=true;) conf is working. But its trying to modify metadata of hive table. Can you please explain me, What is going to modify and does it effect the table.
Thanks
I am converting my legacy Python code to Spark using PySpark.
I would like to get a PySpark equivalent of:
usersofinterest = actdataall[actdataall['ORDValue'].isin(orddata['ORDER_ID'].unique())]['User ID']
Both, actdataall and orddata are Spark dataframes.
I don't want to use toPandas() function given the drawback associated with it.
If both dataframes are big, you should consider using an inner join which will work as a filter:
First let's create a dataframe containing the order IDs we want to keep:
orderid_df = orddata.select(orddata.ORDER_ID.alias("ORDValue")).distinct()
Now let's join it with our actdataall dataframe:
usersofinterest = actdataall.join(orderid_df, "ORDValue", "inner").select('User ID').distinct()
If your target list of order IDs is small then you can use the pyspark.sql isin function as mentioned in furianpandit's post, don't forget to broadcast your variable before using it (spark will copy the object to every node making their tasks a lot faster):
orderid_list = orddata.select('ORDER_ID').distinct().rdd.flatMap(lambda x:x).collect()[0]
sc.broadcast(orderid_list)
The most direct translation of your code would be:
from pyspark.sql import functions as F
# collect all the unique ORDER_IDs to the driver
order_ids = [x.ORDER_ID for x in orddata.select('ORDER_ID').distinct().collect()]
# filter ORDValue column by list of order_ids, then select only User ID column
usersofinterest = actdataall.filter(F.col('ORDValue').isin(order_ids)).select('User ID')
However, you should only filter like this only if number of 'ORDER_ID' is definitely small (perhaps <100,000 or so).
If the number of 'ORDER_ID's is large, you should use a broadcast variable which sends the list of order_ids to each executor so it can compare against the order_ids locally for faster processing. Note, this will work even if 'ORDER_ID' is small.
order_ids = [x.ORDER_ID for x in orddata.select('ORDER_ID').distinct().collect()]
order_ids_broadcast = sc.broadcast(order_ids) # send to broadcast variable
usersofinterest = actdataall.filter(F.col('ORDValue').isin(order_ids_broadcast.value)).select('User ID')
For more information on broadcast variables, check out: https://jaceklaskowski.gitbooks.io/mastering-apache-spark/spark-broadcast.html
So, you have two spark dataframe. One is actdataall and other is orddata, then use following command to get your desire result.
usersofinterest = actdataall.where(actdataall['ORDValue'].isin(orddata.select('ORDER_ID').distinct().rdd.flatMap(lambda x:x).collect()[0])).select('User ID')
I'm sure this is very simple, but despite by trying and research I can't find the solution. I'm working with flight info here.
I have an rdd with contents of :
[u'2007-09-22,9E,20363,TUL,OK,36.19,-95.88,MSP,MN,44.88,-93.22,1745,1737,-8,1953,1934,-19', u'2004-02-12,NW,19386,DEN,CO,39.86,-104.67,MSP,MN,44.88,-93.2
2,1050,1050,0,1341,1342,1', u'2007-05-07,F9,20436,DEN,CO,39.86,-104.67,MSP,MN,44.88,-93.22,1030,1040,10,1325,1347,22']
What transform do I need in order to make a new RDD with all the 2nd fields in it.
[u'9E',u'NW',u'F9']
I've tried filtering but can't make it work. This just gives me the entire line and I only want the 2nd field from each line.
new_rdd = current_rdd.filter(lambda x: x.split(',')[1])
Here is the solution :
data = [u'2007-09-22,9E,20363,TUL,OK,36.19,-95.88,MSP,MN,44.88,-93.22,1745,1737,-8,1953,1934,-19', u'2004-02-12,NW,19386,DEN,CO,39.86,-104.67,MSP,MN,44.88,-93.22,1050,1050,0,1341,1342,1', u'2007-05-07,F9,20436,DEN,CO,39.86,-104.67,MSP,MN,44.88,-93.22,1030,1040,10,1325,1347,22']
current_rdd = sc.parallelize(data)
rdd = current_rdd.map(lambda x : x.split(',')[1])
rdd.take(10)
# [u'9E', u'NW', u'F9']
You are using filter for the wrong purpose. So let's recall the definition of the filter function :
filter(f) - Return a new RDD containing only the elements that satisfy a predicate.
where as map returns a new RDD by applying a function to each element of this RDD, and that's what you need.
I advice to read the PythonRDD API documentation here to learn more about it.