I have json files of the following structure:
{"names":[{"name":"John","lastName":"Doe"},
{"name":"John","lastName":"Marcus"},
{"name":"David","lastName":"Luis"}
]}
I want to read several such json files and distinct them based on the "name" column inside names.
I tried
df.dropDuplicates(Array("names.name"))
but it didn't do the magic.
This seems to be a regression that was added in spark 2.0. If you bring the nested column to the highest level you can drop the duplicates. If we create a new column based on the columns you want to dedup on. Then we drop the columns and finally drop the column. The following function will work for composite keys as well.
val columns = Seq("names.name")
df.withColumn("DEDUP_KEY", concat_ws(",", columns:_*))
.dropDuplicates("DEDUP_KEY")
.drop("DEDUP_KEY")
just for future reference, the solution looks like
val uniqueNams = allNames.withColumn("DEDUP_NAME_KEY",
org.apache.spark.sql.functions.explode(new Column("names.name")))
.cache()
.dropDuplicates(Array("DEDUP_NAME_KEY"))
.drop("DEDUP_NAME_KEY")
Related
Oversimplified Scenario:
A process which generates monthly data in a s3 file. The number of fields could be different in each monthly run. Based on this data in s3,we load the data to a table and we manually (as number of fields could change in each run with addition or deletion of few columns) run a SQL for few metrics.There are more calculations/transforms on this data,but to have starter Im presenting the simpler version of the usecase.
Approach:
Considering the schema-less nature, as the number of fields in the s3 file could differ in each run with addition/deletion of few fields,which requires manual changes every-time in the SQL, Im planning to explore Spark/Scala, so that we can directly read from s3 and dynamically generate SQL based on the fields.
Query:
How I can achieve this in scala/spark-SQL/dataframe? s3 file contains only the required fields from each run.Hence there is no issue reading the dynamic fields from s3 as it is taken care by dataframe.The issue is how can we generate SQL dataframe-API/spark-SQL code to handle.
I can read s3 file via dataframe and register the dataframe as createOrReplaceTempView to write SQL, but I dont think it helps manually changing the spark-SQL, during addition of a new field in s3 during next run. what is the best way to dynamically generate the sql/any better ways to handle the issue?
Usecase-1:
First-run
dataframe: customer,1st_month_count (here dataframe directly points to s3, which has only required attributes)
--sample code
SELECT customer,sum(month_1_count)
FROM dataframe
GROUP BY customer
--Dataframe API/SparkSQL
dataframe.groupBy("customer").sum("month_1_count").show()
Second-Run - One additional column was added
dataframe: customer,month_1_count,month_2_count) (here dataframe directly points to s3, which has only required attributes)
--Sample SQL
SELECT customer,sum(month_1_count),sum(month_2_count)
FROM dataframe
GROUP BY customer
--Dataframe API/SparkSQL
dataframe.groupBy("customer").sum("month_1_count","month_2_count").show()
Im new to Spark/Scala, would be helpful if you can provide the direction so that I can explore further.
It sounds like you want to perform the same operation over and over again on new columns as they appear in the dataframe schema? This works:
from pyspark.sql import functions
#search for column names you want to sum, I put in "month"
column_search = lambda col_names: 'month' in col_names
#get column names of temp dataframe w/ only the columns you want to sum
relevant_columns = original_df.select(*filter(column_search, original_df.columns)).columns
#create dictionary with relevant column names to be passed to the agg function
columns = {col_names: "sum" for col_names in relevant_columns}
#apply agg function with your groupBy, passing in columns dictionary
grouped_df = original_df.groupBy("customer").agg(columns)
#show result
grouped_df.show()
Some important concepts can help you to learn:
DataFrames have data attributes stored in a list: dataframe.columns
Functions can be applied to lists to create new lists as in "column_search"
Agg function accepts multiple expressions in a dictionary as explained here which is what I pass into "columns"
Spark is lazy so it doesn't change data state or perform operations until you perform an action like show(). This means writing out temporary dataframes to use one element of the dataframe like column as I do is not costly even though it may seem inefficient if you're used to SQL.
Here is my current code:
pipe_exec_df_final_grouped = pipe_exec_df_final.groupBy("application_id").agg(collect_list("table_name").alias("tables"))
However, in my collected list, I would like multiple column values, so the aggregated column would be an array of arrays. Currently the result look like this:
1|[a,b,c,d]
2|[e,f,g,h]
However, I would also like to keep another column attached to the aggragation (lets call it 'status' column name). So my new output would be:
1|[[a,pass],[b,fail],[c,fail],[d,pass]]
...
I tried collect_list("table_name, status") however collect_list only takes one column name. How can I accomplish what I am trying to do?
Use array to collect columns into an array column first, then apply collect_list:
df.groupBy(...).agg(collect_list(array("table_name", "status")))
I have a basic DataFrame containing all the data and several derivative DataFrames that I've been subsequently creating from the basic DF making grouping, joins etc.
Every time I want to append a column to the last DataFrame containing the most relevant data I have to do something like this:
val theMostRelevantFinalDf = olderDF.withColumn("new_date_", to_utc_timestamp(unix_timestamp(col("new_date"))
.cast(TimestampType), "UTC").cast(StringType)).drop($"new_date")
As you may see I have to change the original column name to new_date_
But I want the column name to remain the same.
However if I don't change the name the column gets dropped. So renaming is just a not too pretty workaround.
How can I preserve the original column name when appending the column?
As far as I know you can not create two columns with the same name in a DataFrame transformation. I rename the new column to the older´s name like
val theMostRelevantFinalDf = olderDF.withColumn("new_date_", to_utc_timestamp(unix_timestamp(col("new_date"))
.cast(TimestampType), "UTC").cast(StringType)).drop($"new_date").withColumnRenamed("new_date_", "new_date")
Problem statement
I have a table called employee from which i am creating a data-frame .There are some columns which don't have any record.I want to remove that columns from data frame.i also don't know how many columns of the data frame has no record in it.
You cannot remove a column from the dataFrame AFAIK !
What you can do is make another dataframe from the old dataFrame and extract the column names that you actually want to !
Example:
oldDFSchema like this(id,name,badColumn,email)
then
val newDf=oldDF.select("id","name","email")
Or there is one more thing that you can use is :
the .drop() function on dataframe that takes the column names and drops them and returns you a new dataframe !
You can find about it here : https://spark.apache.org/docs/2.0.0/api/scala/index.html#org.apache.spark.sql.Dataset#drop(col:org.apache.spark.sql.Column):org.apache.spark.sql.DataFrame
I hope this might solve your use case !
In order to run a few ML algorithms, I need to create extra columns of data. Each of these columns involves some fairly intense calculations that involves keeping moving averages and recording information as you go through each row (and updating it meanwhile). I've done a mock through with a simple Python script and it works, and I am currently looking to translate it to a Scala Spark script that could be run on a larger data set.
The issue is it seems that for these to be highly efficient, using Spark SQL, it is preferred to use the built in syntax and operations (which are SQL-like). Encoding the logic in a SQL expression seems to be a very thought-intensive process, so I'm wondering what the downsides will be if I just manually create the new column values by iterating through each row, keeping track of variables and inserting the column value at the end.
You can convert an rdd into dataframe. Then use map on the data frame and process each row as you wish. If you need to add new column, then you can use, withColumn. However this will only allow one column to be added and it happens for the entire dataframe. If you want more columns to be added, then inside map method,
a. you can gather new values based on the calculations
b. Add these new column values to main rdd as below
val newColumns: Seq[Any] = Seq(newcol1,newcol2)
Row.fromSeq(row.toSeq.init ++ newColumns)
Here row, is the reference of row in map method
c. Create new schema as below
val newColumnsStructType = StructType{Seq(new StructField("newcolName1",IntegerType),new StructField("newColName2", IntegerType))
d. Add to the old schema
val newSchema = StructType(mainDataFrame.schema.init ++ newColumnsStructType)
e. Create new dataframe with new columns
val newDataFrame = sqlContext.createDataFrame(newRDD, newSchema)