I'm working on a spark mllib algorithm. The dataset I have is in this form
Company":"XXXX","CurrentTitle":"XYZ","Edu_Title":"ABC","Exp_mnth":.(there are more values similar to these)
Im trying to raw code String values to Numeric values. So, I tried using zipwithuniqueID for unique value for each of the string values.For some reason I'm not able to save the modified dataset to the disk. Can I do this in any way using spark SQL? or what would be the better approach for this?
Scala
import org.apache.spark.sql.functions.monotonically_increasing_id
val dataFrame1 = dataFrame0.withColumn("index",monotonically_increasing_id())
Java
Import org.apache.spark.sql.functions;
Dataset<Row> dataFrame1 = dataFrame0.withColumn("index",functions.monotonically_increasing_id());
Related
How could I union or concatenate a static dataframe with only one row to a stream dataframe with around 500 rows in spark. it is somehow put a stamp or mark (add a row to each table) for each streaming dataframe.
Adding more, my streaming data has no timestamp and I'm wondering if I can use foreach() or foreachBatch() or not?
I really appreciate it if you can help me.
In python or R, there are ways to slice DataFrame using index.
For example, in pandas:
df.iloc[5:10,:]
Is there a similar way in pyspark to slice data based on location of rows?
Short Answer
If you already have an index column (suppose it was called 'id') you can filter using pyspark.sql.Column.between:
from pyspark.sql.functions import col
df.where(col("id").between(5, 10))
If you don't already have an index column, you can add one yourself and then use the code above. You should have some ordering built in to your data based on some other columns (orderBy("someColumn")).
Full Explanation
No it is not easily possible to slice a Spark DataFrame by index, unless the index is already present as a column.
Spark DataFrames are inherently unordered and do not support random access. (There is no concept of a built-in index as there is in pandas). Each row is treated as an independent collection of structured data, and that is what allows for distributed parallel processing. Thus, any executor can take any chunk of the data and process it without regard for the order of the rows.
Now obviously it is possible to perform operations that do involve ordering (lead, lag, etc), but these will be slower because it requires spark to shuffle data between the executors. (The shuffling of data is typically one of the slowest components of a spark job.)
Related/Futher Reading
PySpark DataFrames - way to enumerate without converting to Pandas?
PySpark - get row number for each row in a group
how to add Row id in pySpark dataframes
You can convert your spark dataframe to koalas dataframe.
Koalas is a dataframe by Databricks to give an almost pandas like interface to spark dataframe. See here https://pypi.org/project/koalas/
import databricks.koalas as ks
kdf = ks.DataFrame(your_spark_df)
kdf[0:500] # your indexes here
I am a newbie in Spark.I want to write the dataframe data into hive table. Hive table is partitioned on mutliple column. Through, Hivemetastore client I am getting the partition column and passing that as a variable in partitionby clause in write method of dataframe.
var1="country","state" (Getting the partiton column names of hive table)
dataframe1.write.partitionBy(s"$var1").mode("overwrite").save(s"$hive_warehouse/$dbname.db/$temp_table/")
When I am executing the above code,it is giving me error partiton "country","state" does not exists.
I think it is taking "country","state" as a string.
Can you please help me out.
The partitionBy function takes a varargs not a list. You can use this as
dataframe1.write.partitionBy("country","state").mode("overwrite").save(s"$hive_warehouse/$dbname.db/$temp_table/")
Or in scala you can convert a list into a varargs like
val columns = Seq("country","state")
dataframe1.write.partitionBy(columns:_*).mode("overwrite").save(s"$hive_warehouse/$dbname.db/$temp_table/")
is there a way, how to convert an input string field into orc table with column specified as varchar(xx) in sparksql select query? Or I have to use some workaroud? I'm using Spark 1.6.
I found on Cloudera forum, Spark does not care about length, it saves the value as string with no size limit.
The table is inserted into Hive OK, but I'm little bit worried about data quality.
temp_table = sqlContext.table(ext)
df = temp_table.select(temp_dable.day.cast('string'))
I would like to see something like that :)))
df = temp_table.select(temp_dable.day.cast('varchar(100)'))
Edit:
df.write.partitionBy(part).mode('overwrite').insertInto(int)
Table I'm inserting into is saved as an ORC file (the line above probably should have .format('orc')).
I found here, that If I specify a column as a varchar(xx) type, than the input string will be cutoff to the xx length.
Thx
I have to read some data from a relational database to do some machine learning in Spark. However, the table I have to read has some sparse columns. Also, it has a column called "SpecialPurposeColumns" which contains non-zero data in XML format, like:
<Age>76</Age><ID>78</ID><Income>87000</Income> ... <ZIP>96733</ZIP>
what is the good way to create a dataframe in spark using this data?
Thanks in advance