Is posible to use the function partitionBy or other without returning the path "ColumnName=Value"?
I´m using a python notebook in azure databricks to send a csv file to Azure Data Lake Store. The Cmd used is the following:
%scala
val filepath= "dbfs:/mnt/Test"
Sample
.coalesce(1)
.write
.mode("overwrite")
.partitionBy("Year","Month","Day")
.option("header", "true")
.option("delimiter",";")
.csv(filepath)
Expecting to have this path:
/Test/2018/12/11
Instead of:
/Test/Year=2018/Month=12/Day=11
This is expected behavior.
Spark uses directory path for partition with columns names.
If you need specific directory you should use downstream process to rename directory or you can filter your df and save one by one in the specific directory.
Related
I have a column with s3 file paths, I want to read all those paths, concatenate it later in PySpark
You can get the paths as a list using map and collect. Iterate over that list to read the paths and append the resulting spark dataframes into another list. Use the second list (which is a list of spark dataframes) to union all the dataframes.
# get all paths in a list
list_of_paths = data_sdf.rdd.map(lambda r: r.links).collect()
# read all paths and store the df in a list as element
list_of_sdf = []
for path in list_of_paths:
list_of_sdf.append(spark.read.parquet(path))
# check using list_of_sdf[0].show() or list_of_sdf[1].printSchema()
# run union on all of the stored dataframes
import pyspark
final_sdf = reduce(pyspark.sql.dataframe.DataFrame.unionByName, list_of_sdf)
Use the final_sdf dataframe to write to a new parquet file.
You can supply multiple paths to the Spark parquet read function. So, assuming these are paths to parquet files that you want to read into one DataFrame, you can do something like:
list_of_paths = [r.links for links_df.select("links").collect()]
aggregate_df = spark.read.parquet(*list_of_paths)
I have created a pyspark dataframe and trying to write the file in s3 bucket in csv format. here the file is writing in csv but the issue is it's writing the file twice(i.e., with actual data and another is with empty data). I have checked the data frame by printing fine only. please suggest any way to prevent that empty wouldn't create.
code snippet:
df = spark.createDataFrame(data=dt1, schema = op_df.columns)
df.write.option("header","true").csv("s3://"+ src_bucket_name+"/src/output/"+row.brand +'/'+fileN)
One possible solution to make sure that the output will include only one file is to do repartition(1) or coalesce(1) before writing.
So something like this:
df.repartition(1).write.option("header","true").csv("s3://"+ src_bucket_name+"/src/output/"+row.brand +'/'+fileN)
Note that having one partition doesn't not necessarily mean that it will result in one file as this can depend on the spark.sql.files.maxRecordsPerFile configuration as well. Assuming this config is set to 0 (the default) you should get only 1 file in the output.
I have a pyspark job which writes my resultant dataframe in local filesystem. Currently it is running in local mode and so I am doing coalesce(1) to get a single file as below
file_format = 'avro' # will be dynamic and so it will be like avro, json, csv, etc
df.coalesce.write.format(file_format).save('file:///pyspark_data/output')
But I see a lot of memory issues (OOM) and takes longer time as well. So I want to run this job with master as yarn and mode as client. And so to write the result df into a single file in localsystem, I need to use toLocalIterator which yields Rows. How can I stream these Rows into a file of required format (json/avro/csv/parquet and so on)?
file_format = 'avro'
for row in df.toLocalIterator():
# write the data into a single file
pass
You get OOM error because you try to retrieve all the data into a single partition with: coalesce(1)
I dont recommend to use toLocalIterator because you will re-rewrite a custom writer for every format and you wont have parallele writing.
You first solution is a good one :
df.write.format(file_format).save('file:///pyspark_data/output')
if you use hadoop you can retrieve all the data into one on filesysteme this way : (it work for csv, you can try for other) :
hadoop fs -getmerge <HDFS src> <FS destination>
I am trying to generate a file from a Dataframe that I have created in AWS-Glue, I am trying to give it a specific name, I see most answers on stack overflow actually uses Filesystem modules, but here this particular csv file is generated in S3, also I want to give the file a name while generating it, and not rename it after it is generated, is there any way to do that?
I have tried using df.save(s3:://PATH/filename.csv) which actually generates a new directory in S3 named filename.csv and then generates part-*.csv inside that directory
df.repartition(1).write.mode('append').format('csv').save('s3://PATH').option("header", "true")
I have some code like this
wordCounts
.map{ case (word, count) =>
Seq(
word,
count
).mkString("\t")
}
.coalesce(1,true)
.saveAsTextFile("s3n://mybucket/data/myfilename.csv")
However myfilename.csv was created as a directory in my S3 bucket and the file name is always something like myfilename.csv/part-00000? Is there a way I can change the name of the file I am writing to? Thanks!
I strongly suggest that you use the spark-csv package from Databrick to read and write csv files in Spark. One of the (many) benefits from using this package is that it allows you to specify the name of the output csv-file :)