split a spark dataframe into multiple columns in Spark 1.6 - pyspark

I have a spark dataframe of the below format:
+--------------------+
|value |
+--------------------+
|Id,date |
|000027,2017-11-14 |
|000045,2017-11-15 |
|000056,2018-09-09 |
|C000056,2018-07-01 |
+--------------------+
I need to loop through each row, split it by comma (,) and then place the values in different columns (Id and date as two separate columns).
I am new to spark, not sure whether it could be done through lambda function. Any suggestions would be appreciated.

import org.apache.spark.sql.Row
import org.apache.spark.sql.types._
import org.apache.spark.sql.SparkSession
val spark=SparkSession.builder().appName("Demo").getOrCreate()
var df=Seq("a,b,c,f","d,f,g,h").toDF("value")
df.show //show the dataFrame
+-------+
| value|
+-------+
|a,b,c,f|
|d,f,g,h|
+-------+
//splitting out the dataFrame with "," delimeter and creating rdd[Row]
var rdd=df.rdd.map(x=>Row(x.getString(0).split(","):_*))
var schema= StructType(Array("name","class","rank","grade").map(x=>StructField(x,StringType,true)))
spark.createDataFrame(rdd,schema).show
+----+-----+----+-----+
|name|class|rank|grade|
+----+-----+----+-----+
| a| b| c| f|
| d| f| g| h|
+----+-----+----+-----+

Related

How is the string column in DataFrame split into multiple columns when Spark Structed Streaming

This is the current code:
from pyspark.sql import SparkSession
park_session = SparkSession\
.builder\
.appName("test")\
.getOrCreate()
lines = spark_session\
.readStream\
.format("socket")\
.option("host", "127.0.0.1")\
.option("port", 9998)\
.load()
The 'lines' looks like this:
+-------------+
| value |
+-------------+
| a,b,c |
+-------------+
But I want to look like this:
+---+---+---+
| a | b | c |
+---+---+---+
I tried using the 'split()' method, but it didn't work. You could only split each string into a list in a column, not into multiple columns
What should I do?
Split the value column and by accessing array index (or) element_at(from spark-2.4) (or) getItem() functions to create new columns.
from pyspark.sql.functions import *
lines.withColumn("tmp",split(col("value"),',')).\
withColumn("col1",col("tmp")[0]).\
withColumn("col2",col("tmp").getItem(1)).\
withColumn("col3",element_at(col("tmp"),3))
drop("tmp","value").\
show()
#+----+----+----+
#|col1|col2|col3|
#+----+----+----+
#| a| b| c|
#+----+----+----+
from pyspark.sql.functions import *
import pyspark.sql.functions as f
from pyspark.sql import SparkSession
spark_session = SparkSession\
.builder\
.appName("test")\
.getOrCreate()
lines = spark_session\
.readStream\
.format("socket")\
.option("host", "127.0.0.1")\
.option("port", 9998)\
.load()
split_col = f.split(lines['value'], ",")
df = df.withColumn('col1', split_col.getItem(0))
df = df.withColumn('col2', split_col.getItem(1))
df = df.withColumn('col2', split_col.getItem(2))
df.show()
Incase you have different numbers of delimiters and not just 3 for each row , you can use the below:
Input:
+-------+
|value |
+-------+
|a,b,c |
|d,e,f,g|
+-------+
Solution
import pyspark.sql.functions as F
max_size = df.select(F.max(F.length(F.regexp_replace('value','[^,]','')))).first()[0]
out = df.select([F.split("value",',')[x].alias(f"Col{x+1}") for x in range(max_size+1)])
Output
out.show()
+----+----+----+----+
|Col1|Col2|Col3|Col4|
+----+----+----+----+
| a| b| c|null|
| d| e| f| g|
+----+----+----+----+

PySpark difference between pyspark.sql.functions.col and pyspark.sql.functions.lit

I find it hard to understand the difference between these two methods from pyspark.sql.functions as the documentation on PySpark official website is not very informative. For example the following code:
import pyspark.sql.functions as F
print(F.col('col_name'))
print(F.lit('col_name'))
The results are:
Column<b'col_name'>
Column<b'col_name'>
so what are the difference between the two and when should I use one and not the other?
The doc says:
col:
Returns a Column based on the given column name.
lit:
Creates a Column of literal value
Say if we have a data frame as below:
>>> import pyspark.sql.functions as F
>>> from pyspark.sql.types import *
>>> schema = StructType([StructField('A', StringType(), True)])
>>> df = spark.createDataFrame([("a",), ("b",), ("c",)], schema)
>>> df.show()
+---+
| A|
+---+
| a|
| b|
| c|
+---+
If using col to create a new column from A:
>>> df.withColumn("new", F.col("A")).show()
+---+---+
| A|new|
+---+---+
| a| a|
| b| b|
| c| c|
+---+---+
So col grabs an existing column with the given name, F.col("A") is equivalent to df.A or df["A"] here.
If using F.lit("A") to create the column:
>>> df.withColumn("new", F.lit("A")).show()
+---+---+
| A|new|
+---+---+
| a| A|
| b| A|
| c| A|
+---+---+
While lit will create a constant column with the given string as the values.
Both of them return a Column object but the content and meaning are different.
To explain in a very succinct manner, col is typically used to refer to an existing column in a DataFrame, as opposed to lit which is typically used to set the value of a column to a literal
To illustrate with an example:
Assume i have a DataFrame df containing two columns of IntegerType, col_a and col_b
If i wanted a column total which were the sum of the two columns:
df.withColumn('total', col('col_a') + col('col_b'))
Instead of i wanted a column fixed_val having the value "Hello" for all rows of the DataFrame df:
df.withColumn('fixed_val', lit('Hello'))

Appending a new column to existing dataframe Spark scala [duplicate]

I am using Apache Spark 2.0 Dataframe/Dataset API
I want to add a new column to my dataframe from List of values. My list has same number of values like given dataframe.
val list = List(4,5,10,7,2)
val df = List("a","b","c","d","e").toDF("row1")
I would like to do something like:
val appendedDF = df.withColumn("row2",somefunc(list))
df.show()
// +----+------+
// |row1 |row2 |
// +----+------+
// |a |4 |
// |b |5 |
// |c |10 |
// |d |7 |
// |e |2 |
// +----+------+
For any ideas I would be greatful, my dataframe in reality contains more columns.
You could do it like this:
import org.apache.spark.sql.Row
import org.apache.spark.sql.types._
// create rdd from the list
val rdd = sc.parallelize(List(4,5,10,7,2))
// rdd: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[31] at parallelize at <console>:28
// zip the data frame with rdd
val rdd_new = df.rdd.zip(rdd).map(r => Row.fromSeq(r._1.toSeq ++ Seq(r._2)))
// rdd_new: org.apache.spark.rdd.RDD[org.apache.spark.sql.Row] = MapPartitionsRDD[33] at map at <console>:32
// create a new data frame from the rdd_new with modified schema
spark.createDataFrame(rdd_new, df.schema.add("new_col", IntegerType)).show
+----+-------+
|row1|new_col|
+----+-------+
| a| 4|
| b| 5|
| c| 10|
| d| 7|
| e| 2|
+----+-------+
Adding for completeness: the fact that the input list (which exists in driver memory) has the same size as the DataFrame suggests that this is a small DataFrame to begin with - so you might consider collect()-ing it, zipping with list, and converting back into a DataFrame if needed:
df.collect()
.map(_.getAs[String]("row1"))
.zip(list).toList
.toDF("row1", "row2")
That won't be faster, but if the data is really small it might be negligible and the code is (arguably) clearer.

Pass Array[seq[String]] to UDF in spark scala

I am new to UDF in spark. I have also read the answer here
Problem statement: I'm trying to find pattern matching from a dataframe col.
Ex: Dataframe
val df = Seq((1, Some("z")), (2, Some("abs,abc,dfg")),
(3,Some("a,b,c,d,e,f,abs,abc,dfg"))).toDF("id", "text")
df.show()
+---+--------------------+
| id| text|
+---+--------------------+
| 1| z|
| 2| abs,abc,dfg|
| 3|a,b,c,d,e,f,abs,a...|
+---+--------------------+
df.filter($"text".contains("abs,abc,dfg")).count()
//returns 2 as abs exits in 2nd row and 3rd row
Now I want to do this pattern matching for every row in column $text and add new column called count.
Result:
+---+--------------------+-----+
| id| text|count|
+---+--------------------+-----+
| 1| z| 1|
| 2| abs,abc,dfg| 2|
| 3|a,b,c,d,e,f,abs,a...| 1|
+---+--------------------+-----+
I tried to define a udf passing $text column as Array[Seq[String]. But I am not able to get what I intended.
What I tried so far:
val txt = df.select("text").collect.map(_.toSeq.map(_.toString)) //convert column to Array[Seq[String]
val valsum = udf((txt:Array[Seq[String],pattern:String)=> {txt.count(_ == pattern) } )
df.withColumn("newCol", valsum( lit(txt) ,df(text)) )).show()
Any help would be appreciated
You will have to know all the elements of text column which can be done using collect_list by grouping all the rows of your dataframe as one. Then just check if element in text column in the collected array and count them as in the following code.
import sqlContext.implicits._
import org.apache.spark.sql.functions._
import org.apache.spark.sql.expressions._
val df = Seq((1, Some("z")), (2, Some("abs,abc,dfg")),(3,Some("a,b,c,d,e,f,abs,abc,dfg"))).toDF("id", "text")
val valsum = udf((txt: String, array : mutable.WrappedArray[String])=> array.filter(element => element.contains(txt)).size)
df.withColumn("grouping", lit("g"))
.withColumn("array", collect_list("text").over(Window.partitionBy("grouping")))
.withColumn("count", valsum($"text", $"array"))
.drop("grouping", "array")
.show(false)
You should have following output
+---+-----------------------+-----+
|id |text |count|
+---+-----------------------+-----+
|1 |z |1 |
|2 |abs,abc,dfg |2 |
|3 |a,b,c,d,e,f,abs,abc,dfg|1 |
+---+-----------------------+-----+
I hope this is helpful.

Apache Spark how to append new column from list/array to Spark dataframe

I am using Apache Spark 2.0 Dataframe/Dataset API
I want to add a new column to my dataframe from List of values. My list has same number of values like given dataframe.
val list = List(4,5,10,7,2)
val df = List("a","b","c","d","e").toDF("row1")
I would like to do something like:
val appendedDF = df.withColumn("row2",somefunc(list))
df.show()
// +----+------+
// |row1 |row2 |
// +----+------+
// |a |4 |
// |b |5 |
// |c |10 |
// |d |7 |
// |e |2 |
// +----+------+
For any ideas I would be greatful, my dataframe in reality contains more columns.
You could do it like this:
import org.apache.spark.sql.Row
import org.apache.spark.sql.types._
// create rdd from the list
val rdd = sc.parallelize(List(4,5,10,7,2))
// rdd: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[31] at parallelize at <console>:28
// zip the data frame with rdd
val rdd_new = df.rdd.zip(rdd).map(r => Row.fromSeq(r._1.toSeq ++ Seq(r._2)))
// rdd_new: org.apache.spark.rdd.RDD[org.apache.spark.sql.Row] = MapPartitionsRDD[33] at map at <console>:32
// create a new data frame from the rdd_new with modified schema
spark.createDataFrame(rdd_new, df.schema.add("new_col", IntegerType)).show
+----+-------+
|row1|new_col|
+----+-------+
| a| 4|
| b| 5|
| c| 10|
| d| 7|
| e| 2|
+----+-------+
Adding for completeness: the fact that the input list (which exists in driver memory) has the same size as the DataFrame suggests that this is a small DataFrame to begin with - so you might consider collect()-ing it, zipping with list, and converting back into a DataFrame if needed:
df.collect()
.map(_.getAs[String]("row1"))
.zip(list).toList
.toDF("row1", "row2")
That won't be faster, but if the data is really small it might be negligible and the code is (arguably) clearer.