I'm reading in data (as show below) into a list of lists, and I want to convert it into a dataframe with seven columns. The error I get is: requirement failed: number of columns doesn't match. Old column names (1): value, new column names (7): <list of columns>
What am I doing incorrectly and how can I fix it?
Data:
Column1, Column2, Column3, Column4, Column5, Column6, Column7
a,b,c,d,e,f,g
a2,b2,c2,d2,e2,f2,g2
Code:
val spark = SparkSession.builder.appName("er").master("local").getOrCreate()
import spark.implicits._
val erResponse = response.body.toString.split("\\\n")
val header = erResponse(0)
val body = erResponse.drop(1).map(x => x.split(",").toList).toList
val erDf = body.toDF()
erDf.show()
You get this number of columns doesn't match error because your erDf dataframe contains only one column, that contains an array:
+----------------------------+
|value |
+----------------------------+
|[a, b, c, d, e, f, g] |
|[a2, b2, c2, d2, e2, f2, g2]|
+----------------------------+
You can't match this unique column with the seven columns contained in your header.
The solution here is, given this erDf dataframe, to iterate over your header columns list to build columns one by one. Your complete code thus become:
val spark = SparkSession.builder.appName("er").master("local").getOrCreate()
import spark.implicits._
val erResponse = response.body.toString.split("\\\n")
val header = erResponse(0).split(", ") // build header columns list
val body = erResponse.drop(1).map(x => x.split(",").toList).toList
val erDf = header
.zipWithIndex
.foldLeft(body.toDF())((acc, elem) => acc.withColumn(elem._1, col("value")(elem._2)))
.drop("value")
That will give you the following erDf dataframe:
+-------+-------+-------+-------+-------+-------+-------+
|Column1|Column2|Column3|Column4|Column5|Column6|Column7|
+-------+-------+-------+-------+-------+-------+-------+
| a| b| c| d| e| f| g|
| a2| b2| c2| d2| e2| f2| g2|
+-------+-------+-------+-------+-------+-------+-------+
Related
I have a dataframe like this.
+---+---+---+---+
| M| c2| c3| d1|
+---+---+---+---+
| 1|2_1|4_3|1_2|
| 2|3_4|4_5|1_2|
+---+---+---+---+
I have to transform this df should look like below. Here, c_max = max(c2,c3) after splitting with _.ie, all the columns (c2 and c3) have to be splitted with _ and then getting the max.
In the actual scenario, I have 50 columns ie, c2,c3....c50 and need to take the max from this.
+---+---+---+---+------+
| M| c2| c3| d1|c_Max |
+---+---+---+---+------+
| 1|2_1|4_3|1_2| 4 |
| 2|3_4|4_5|1_2| 5 |
+---+---+---+---+------+
Here is one way using expr and build-in array functions for Spark >= 2.4.0:
import org.apache.spark.sql.functions.{expr, array_max, array}
val df = Seq(
(1, "2_1", "3_4", "1_2"),
(2, "3_4", "4_5", "1_2")
).toDF("M", "c2", "c3", "d1")
// get max c for each c column
val c_cols = df.columns.filter(_.startsWith("c")).map{ c =>
expr(s"array_max(cast(split(${c}, '_') as array<int>))")
}
df.withColumn("max_c", array_max(array(c_cols:_*))).show
Output:
+---+---+---+---+-----+
| M| c2| c3| d1|max_c|
+---+---+---+---+-----+
| 1|2_1|3_4|1_2| 4|
| 2|3_4|4_5|1_2| 5|
+---+---+---+---+-----+
For older versions use the next code:
val c_cols = df.columns.filter(_.startsWith("c")).map{ c =>
val c_ar = split(col(c), "_").cast("array<int>")
when(c_ar.getItem(0) > c_ar.getItem(1), c_ar.getItem(0)).otherwise(c_ar.getItem(1))
}
df.withColumn("max_c", greatest(c_cols:_*)).show
Use greatest function:
val df = Seq((1, "2_1", "3_4", "1_2"),(2, "3_4", "4_5", "1_2"),
).toDF("M", "c2", "c3", "d1")
// get all `c` columns and split by `_` to get the values after the underscore
val c_cols = df.columns.filter(_.startsWith("c"))
.flatMap{
c => Seq(split(col(c), "_").getItem(0).cast("int"),
split(col(c), "_").getItem(1).cast("int")
)
}
// apply greatest func
val c_max = greatest(c_cols: _*)
// add new column
df.withColumn("c_Max", c_max).show()
Gives:
+---+---+---+---+-----+
| M| c2| c3| d1|c_Max|
+---+---+---+---+-----+
| 1|2_1|3_4|1_2| 4|
| 2|3_4|4_5|1_2| 5|
+---+---+---+---+-----+
In spark >= 2.4.0, you can use the array_max function and get some code that would work even with columns containing more than 2 values. The idea is to start by concatenating all the columns (concat column). For that, I use concat_ws on an array of all the columns I want to concat, that I obtain with array(cols.map(col) :_*). Then I split the resulting string to get a big array of strings containing all the values of all the columns. I cast it to an array of ints and I call array_max on it.
val cols = (2 to 50).map("c"+_)
val result = df
.withColumn("concat", concat_ws("_", array(cols.map(col) :_*)))
.withColumn("array_of_ints", split('concat, "_").cast(ArrayType(IntegerType)))
.withColumn("c_max", array_max('array_of_ints))
.drop("concat", "array_of_ints")
In spark < 2.4, you can define array_max yourself like this:
val array_max = udf((s : Seq[Int]) => s.max)
The previous code does not need to be modified. Note however that UDFs can be slower than predefined spark SQL functions.
i have record as string with 1000 fields with delimiter as comma in dataframe like
"a,b,c,d,e.......upto 1000" -1st record
"p,q,r,s,t ......upto 1000" - 2nd record
I am using below suggested solution from stackoverflow
Split 1 column into 3 columns in spark scala
df.withColumn("_tmp", split($"columnToSplit", "\\.")).select($"_tmp".getItem(0).as("col1"),$"_tmp".getItem(1).as("col2"),$"_tmp".getItem(2).as("col3")).drop("_tmp")
however in my case i am having 1000 columns which i have in JSON schema which i can retrive like
column_seq:Seq[Array]=Schema_func.map(_.name)
for(i <-o to column_seq.length-1){println(i+" " + column_seq(i))}
which returns like
0 col1
1 col2
2 col3
3 col4
Now I need to pass all this indexes and column names to below function of DataFrame
df.withColumn("_tmp", split($"columnToSplit", "\\.")).select($"_tmp".getItem(0).as("col1"),$"_tmp".getItem(1).as("col2"),$"_tmp".getItem(2).as("col3")).drop("_tmp")
in
$"_tmp".getItem(0).as("col1"),$"_tmp".getItem(1).as("col2"),
as i cant create the long statement with all 1000 columns , is there any effective way to pass all this arguments from above mentioned json schema to select function , so that i can split the columns , add the header and then covert the DF to parquet.
You can build a series of org.apache.spark.sql.Column, where each one is the result of selecting the right item and has the right name, and then select these columns:
val columns: Seq[Column] = Schema_func.map(_.name)
.zipWithIndex // attach index to names
.map { case (name, index) => $"_tmp".getItem(index) as name }
val result = df
.withColumn("_tmp", split($"columnToSplit", "\\."))
.select(columns: _*)
For example, for this input:
case class A(name: String)
val Schema_func = Seq(A("c1"), A("c2"), A("c3"), A("c4"), A("c5"))
val df = Seq("a.b.c.d.e").toDF("columnToSplit")
The result would be:
// +---+---+---+---+---+
// | c1| c2| c3| c4| c5|
// +---+---+---+---+---+
// | a| b| c| d| e|
// +---+---+---+---+---+
I'm trying to add a new column to a DataFrame. The value of this column is the value of another column whose name depends on other columns from the same DataFrame.
For instance, given this:
+---+---+----+----+
| A| B| A_1| B_2|
+---+---+----+----+
| A| 1| 0.1| 0.3|
| B| 2| 0.2| 0.4|
+---+---+----+----+
I'd like to obtain this:
+---+---+----+----+----+
| A| B| A_1| B_2| C|
+---+---+----+----+----+
| A| 1| 0.1| 0.3| 0.1|
| B| 2| 0.2| 0.4| 0.4|
+---+---+----+----+----+
That is, I added column C whose value came from either column A_1 or B_2. The name of the source column A_1 comes from concatenating the value of columns A and B.
I know that I can add a new column based on another and a constant like this:
df.withColumn("C", $"B" + 1)
I also know that the name of the column can come from a variable like this:
val name = "A_1"
df.withColumn("C", col(name) + 1)
However, what I'd like to do is something like this:
df.withColumn("C", col(s"${col("A")}_${col("B")}"))
Which doesn't work.
NOTE: I'm coding in Scala 2.11 and Spark 2.2.
You can achieve your requirement by writing a udf function. I am suggesting udf, as your requirement is to process dataframe row by row contradicting to inbuilt functions which functions column by column.
But before that you would need array of column names
val columns = df.columns
Then write a udf function as
import org.apache.spark.sql.functions._
def getValue = udf((A: String, B: String, array: mutable.WrappedArray[String]) => array(columns.indexOf(A+"_"+B)))
where
A is the first column value
B is the second column value
array is the Array of all the columns values
Now just call the udf function using withColumn api
df.withColumn("C", getValue($"A", $"B", array(columns.map(col): _*))).show(false)
You should get your desired output dataframe.
You can select from a map. Define map which translates name to column value:
import org.apache.spark.sql.functions.{col, concat_ws, lit, map}
val dataMap = map(
df.columns.diff(Seq("A", "B")).flatMap(c => lit(c) :: col(c) :: Nil): _*
)
df.select(dataMap).show(false)
+---------------------------+
|map(A_1, A_1, B_2, B_2) |
+---------------------------+
|Map(A_1 -> 0.1, B_2 -> 0.3)|
|Map(A_1 -> 0.2, B_2 -> 0.4)|
+---------------------------+
and select from it with apply:
df.withColumn("C", dataMap(concat_ws("_", $"A", $"B"))).show
+---+---+---+---+---+
| A| B|A_1|B_2| C|
+---+---+---+---+---+
| A| 1|0.1|0.3|0.1|
| B| 2|0.2|0.4|0.4|
+---+---+---+---+---+
You can also try mapping, but I suspect it won't perform well with very wide data:
import org.apache.spark.sql.catalyst.encoders.RowEncoder
import org.apache.spark.sql.types._
import org.apache.spark.sql.Row
val outputEncoder = RowEncoder(df.schema.add(StructField("C", DoubleType)))
df.map(row => {
val a = row.getAs[String]("A")
val b = row.getAs[String]("B")
val key = s"${a}_${b}"
Row.fromSeq(row.toSeq :+ row.getAs[Double](key))
})(outputEncoder).show
+---+---+---+---+---+
| A| B|A_1|B_2| C|
+---+---+---+---+---+
| A| 1|0.1|0.3|0.1|
| B| 2|0.2|0.4|0.4|
+---+---+---+---+---+
and in general I wouldn't recommend this approach.
If data comes from csv you might consider skipping default csv reader and use custom logic to push column selection directly into parsing process. With pseudocode:
spark.read.text(...).map { line => {
val a = ??? // parse A
val b = ??? // parse B
val c = ??? // find c, based on a and b
(a, b, c)
}}
I am very new to scala and spark.
I have read a text file into a dataframe, and successfully split the single column into columns (essentially the file is SPACE delimited csv)
val irisDF:DataFrame = spark.read.csv("src/test/resources/iris-in.txt")
irisDF.show()
val dfnew:DataFrame = irisDF.withColumn("_tmp", split($"_c0", " ")).select(
$"_tmp".getItem(0).as("col1"),
$"_tmp".getItem(1).as("col2"),
$"_tmp".getItem(2).as("col3"),
$"_tmp".getItem(3).as("col4")
).drop("_tmp")
This works.
BUT what if I do not know how many columns there are in the datafile? How do I dynamically generate the columns depending on the number of items generated by the split function?
You can create a sequence of select expressions, and then apply all of them to select method with :_* syntax:
Example Data:
val df = Seq("a b c d", "e f g").toDF("c0")
df.show
+-------+
| c0|
+-------+
|a b c d|
| e f g|
+-------+
If you want five columns from the c0 column, which you need to determine before doing this:
val selectExprs = 0 until 5 map (i => $"temp".getItem(i).as(s"col$i"))
df.withColumn("temp", split($"c0", " ")).select(selectExprs:_*).show
+----+----+----+----+----+
|col0|col1|col2|col3|col4|
+----+----+----+----+----+
| a| b| c| d|null|
| e| f| g|null|null|
+----+----+----+----+----+
I have a dataframe in Spark using scala that has a column that I need split.
scala> test.show
+-------------+
|columnToSplit|
+-------------+
| a.b.c|
| d.e.f|
+-------------+
I need this column split out to look like this:
+--------------+
|col1|col2|col3|
| a| b| c|
| d| e| f|
+--------------+
I'm using Spark 2.0.0
Thanks
Try:
import sparkObject.spark.implicits._
import org.apache.spark.sql.functions.split
df.withColumn("_tmp", split($"columnToSplit", "\\.")).select(
$"_tmp".getItem(0).as("col1"),
$"_tmp".getItem(1).as("col2"),
$"_tmp".getItem(2).as("col3")
)
The important point to note here is that the sparkObject is the SparkSession object you might have already initialized. So, the (1) import statement has to be compulsorily put inline within the code, not before the class definition.
To do this programmatically, you can create a sequence of expressions with (0 until 3).map(i => col("temp").getItem(i).as(s"col$i")) (assume you need 3 columns as result) and then apply it to select with : _* syntax:
df.withColumn("temp", split(col("columnToSplit"), "\\.")).select(
(0 until 3).map(i => col("temp").getItem(i).as(s"col$i")): _*
).show
+----+----+----+
|col0|col1|col2|
+----+----+----+
| a| b| c|
| d| e| f|
+----+----+----+
To keep all columns:
df.withColumn("temp", split(col("columnToSplit"), "\\.")).select(
col("*") +: (0 until 3).map(i => col("temp").getItem(i).as(s"col$i")): _*
).show
+-------------+---------+----+----+----+
|columnToSplit| temp|col0|col1|col2|
+-------------+---------+----+----+----+
| a.b.c|[a, b, c]| a| b| c|
| d.e.f|[d, e, f]| d| e| f|
+-------------+---------+----+----+----+
If you are using pyspark, use a list comprehension to replace the map in scala:
df = spark.createDataFrame([['a.b.c'], ['d.e.f']], ['columnToSplit'])
from pyspark.sql.functions import col, split
(df.withColumn('temp', split('columnToSplit', '\\.'))
.select(*(col('temp').getItem(i).alias(f'col{i}') for i in range(3))
).show()
+----+----+----+
|col0|col1|col2|
+----+----+----+
| a| b| c|
| d| e| f|
+----+----+----+
A solution which avoids the select part. This is helpful when you just want to append the new columns:
case class Message(others: String, text: String)
val r1 = Message("foo1", "a.b.c")
val r2 = Message("foo2", "d.e.f")
val records = Seq(r1, r2)
val df = spark.createDataFrame(records)
df.withColumn("col1", split(col("text"), "\\.").getItem(0))
.withColumn("col2", split(col("text"), "\\.").getItem(1))
.withColumn("col3", split(col("text"), "\\.").getItem(2))
.show(false)
+------+-----+----+----+----+
|others|text |col1|col2|col3|
+------+-----+----+----+----+
|foo1 |a.b.c|a |b |c |
|foo2 |d.e.f|d |e |f |
+------+-----+----+----+----+
Update: I highly recommend to use Psidom's implementation to avoid splitting three times.
This appends columns to the original DataFrame and doesn't use select, and only splits once using a temporary column:
import spark.implicits._
df.withColumn("_tmp", split($"columnToSplit", "\\."))
.withColumn("col1", $"_tmp".getItem(0))
.withColumn("col2", $"_tmp".getItem(1))
.withColumn("col3", $"_tmp".getItem(2))
.drop("_tmp")
This expands on Psidom's answer and shows how to do the split dynamically, without hardcoding the number of columns. This answer runs a query to calculate the number of columns.
val df = Seq(
"a.b.c",
"d.e.f"
).toDF("my_str")
.withColumn("letters", split(col("my_str"), "\\."))
val numCols = df
.withColumn("letters_size", size($"letters"))
.agg(max($"letters_size"))
.head()
.getInt(0)
df
.select(
(0 until numCols).map(i => $"letters".getItem(i).as(s"col$i")): _*
)
.show()
We can write using for with yield in Scala :-
If your number of columns exceeds just add it to desired column and play with it. :)
val aDF = Seq("Deepak.Singh.Delhi").toDF("name")
val desiredColumn = Seq("name","Lname","City")
val colsize = desiredColumn.size
val columList = for (i <- 0 until colsize) yield split(col("name"),".").getItem(i).alias(desiredColumn(i))
aDF.select(columList: _ *).show(false)
Output:-
+------+------+-----+--+
|name |Lname |city |
+-----+------+-----+---+
|Deepak|Singh |Delhi|
+---+------+-----+-----+
If you don't need name column then, drop the column and just use withColumn.
Example:
Without using the select statement.
Lets assume we have a dataframe having a set of columns and we want to split a column having column name as name
import spark.implicits._
val columns = Seq("name","age","address")
val data = Seq(("Amit.Mehta", 25, "1 Main st, Newark, NJ, 92537"),
("Rituraj.Mehta", 28,"3456 Walnut st, Newark, NJ, 94732"))
var dfFromData = spark.createDataFrame(data).toDF(columns:_*)
dfFromData.printSchema()
val newDF = dfFromData.map(f=>{
val nameSplit = f.getAs[String](0).split("\\.").map(_.trim)
(nameSplit(0),nameSplit(1),f.getAs[Int](1),f.getAs[String](2))
})
val finalDF = newDF.toDF("First Name","Last Name", "Age","Address")
finalDF.printSchema()
finalDF.show(false)
output: