Passing an entire row as an argument to spark udf through spark dataframe - throws AnalysisException - apache-spark

I am trying to pass an entire row to the spark udf along with few other arguments, I am not using spark sql rather I am using dataframe withColumn api, but I am getting the following exception:
Exception in thread "main" org.apache.spark.sql.AnalysisException: Resolved attribute(s) col3#9 missing from col1#7,col2#8,col3#13 in operator !Project [col1#7, col2#8, col3#13, UDF(col3#9, col2, named_struct(col1, col1#7, col2, col2#8, col3, col3#9)) AS contcatenated#17]. Attribute(s) with the same name appear in the operation: col3. Please check if the right attribute(s) are used.;;
The above exception can be replicated using the below code:
addRowUDF() // call invokes
def addRowUDF() {
import org.apache.spark.SparkConf
import org.apache.spark.sql.SparkSession
val spark = SparkSession.builder().config(new SparkConf().set("master", "local[*]")).appName(this.getClass.getSimpleName).getOrCreate()
import spark.implicits._
val df = Seq(
("a", "b", "c"),
("a1", "b1", "c1")).toDF("col1", "col2", "col3")
execute(df)
}
def execute(df: org.apache.spark.sql.DataFrame) {
import org.apache.spark.sql.Row
def concatFunc(x: Any, y: String, row: Row) = x.toString + ":" + y + ":" + row.mkString(", ")
import org.apache.spark.sql.functions.{ udf, struct }
val combineUdf = udf((x: Any, y: String, row: Row) => concatFunc(x, y, row))
def udf_execute(udf: String, args: org.apache.spark.sql.Column*) = (combineUdf)(args: _*)
val columns = df.columns.map(df(_))
val df2 = df.withColumn("col3", lit("xxxxxxxxxxx"))
val df3 = df2.withColumn("contcatenated", udf_execute("uudf", df2.col("col3"), lit("col2"), struct(columns: _*)))
df3.show(false)
}
output should be:
+----+----+-----------+----------------------------+
|col1|col2|col3 |contcatenated |
+----+----+-----------+----------------------------+
|a |b |xxxxxxxxxxx|xxxxxxxxxxx:col2:a, b, c |
|a1 |b1 |xxxxxxxxxxx|xxxxxxxxxxx:col2:a1, b1, c1 |
+----+----+-----------+----------------------------+

That happens because you refer to column that is no longer in the scope. When you call:
val df2 = df.withColumn("col3", lit("xxxxxxxxxxx"))
it shades the original col3 column, effectively making preceding columns with the same name accessible. Even if it wasn't the case, let's say after:
val df2 = df.select($"*", lit("xxxxxxxxxxx") as "col3")
the new col3 would be ambiguous, and indistinguishable by name from the one defined brought by *.
So to achieve the required output you'll have to use another name:
val df2 = df.withColumn("col3_", lit("xxxxxxxxxxx"))
and then adjust the rest of your code accordingly:
df2.withColumn(
"contcatenated",
udf_execute("uudf", df2.col("col3_") as "col3",
lit("col2"), struct(columns: _*))
).drop("_3")
If the logic is as simple as the one in the example, you can of course just inline things:
df.withColumn(
"contcatenated",
udf_execute("uudf", lit("xxxxxxxxxxx") as "col3",
lit("col2"), struct(columns: _*))
).drop("_3")

Related

Data type conversion in spark

I have an column id which had type int but later changed to bigint.
It has both types of values.
from pyspark.sql.functions import *
from pyspark.sql.types import *
df = spark.read.parquet('hdfs path')
df = df.select("id", "code")
df=df.withColumn("id1", df["id"].cast(LongType()))
res1=df.select("id1", "code")
res1.show(1, False)
It shows me the data frame but when i try to perform some operations on them
example:
res1.groupBy('code').agg(countDistinct("id1")).show(1, False)
I get Column: [id], Expected: int, Found: INT64
I tried mergeSchema did not work either.
from pyspark.sql.functions import *
from pyspark.sql.types import *
df1 = spark.read.parquet('hdfs path')
df2 = df1.select("id", "code")
df3 = df2.withColumn("id1", df2["id"].cast(LongType()))
res1=df3.select("id1", "code")
res1.show(1, False)
res1.groupBy("code").agg(countDistinct("id1")).show(1, False)
This should work. In spark Dataframes are immutable so you should not assign the value of transformation operation to a same df variable, you should use a different variable name. In scala it would give you compile time error but in python its allowed so you don't notice it.
if you want you could also chain all of your transformation and get a single df variable and perform groupby operation on it as below :
df = spark.read.parquet('hdfs path').select("id", "code").withColumn("id1", col("id").cast(LongType())).select("id1", "code")
df.groupBy("code").agg(countDistinct("id1")).show(1, False)

HiveQL to PySpark - issue with aggregated column in SELECT statement

I have following HQL script which needs to be puti nto pyspark, spark 1.6
insert into table db.temp_avg
select
a,
avg(b) ,
c
from db.temp WHERE flag is not null GROUP BY a, c;
I created few versions of spark code, but I'm stuggling how to get this averaged column into select.
Also I found out that groupped data cannot be write this way:
df3 = df2.groupBy...
df3.write.mode('overwrite').saveAsTable('db.temp_avg')
part of pyspark code:
temp_table = sqlContext.table("db.temp")
df = temp_table.select('a', 'avg(b)', 'c', 'flag').toDF('a', 'avg(b)', 'c', 'flag')
df = df.where(['flag'] != 'null'))
# this ofc does not work along with the avg(b)
df2 = df.groupBy('a', 'c')
df3.write.mode('overwrite').saveAsTable('db.temp_avg')
Thx for your help.
Correct solution:
import pyspark.sql.functions as F
df = sqlContext.sql("SELECT * FROM db.temp_avg").alias("temp")
df = df.select('a', 'b', 'c')\
.filter(F.col("temp.flag").isNotNULL())\
.groupby('a', 'c')\
.agg(F.avg('b').alias("avg_b"))
import pyspark.sql.functions as F
df = sqlContext.sql("select * from db.temp_avg")
df = df.select('a',
b,
'c')\
.filter(F.col("flag").isNotNULL())\
.groupby('a', 'c')\
.agg(F.avg('b').alias("avg_b"))
Then you can save the table by
df.saveAsTable("tabe_name")

Error when converting from pyspark RDD to DataFrame: Cannot infer schema of type 'unicode' [duplicate]

Could someone help me solve this problem I have with Spark DataFrame?
When I do myFloatRDD.toDF() I get an error:
TypeError: Can not infer schema for type: type 'float'
I don't understand why...
Example:
myFloatRdd = sc.parallelize([1.0,2.0,3.0])
df = myFloatRdd.toDF()
Thanks
SparkSession.createDataFrame, which is used under the hood, requires an RDD / list of Row/tuple/list/dict* or pandas.DataFrame, unless schema with DataType is provided. Try to convert float to tuple like this:
myFloatRdd.map(lambda x: (x, )).toDF()
or even better:
from pyspark.sql import Row
row = Row("val") # Or some other column name
myFloatRdd.map(row).toDF()
To create a DataFrame from a list of scalars you'll have to use SparkSession.createDataFrame directly and provide a schema***:
from pyspark.sql.types import FloatType
df = spark.createDataFrame([1.0, 2.0, 3.0], FloatType())
df.show()
## +-----+
## |value|
## +-----+
## | 1.0|
## | 2.0|
## | 3.0|
## +-----+
but for a simple range it would be better to use SparkSession.range:
from pyspark.sql.functions import col
spark.range(1, 4).select(col("id").cast("double"))
* No longer supported.
** Spark SQL also provides a limited support for schema inference on Python objects exposing __dict__.
*** Supported only in Spark 2.0 or later.
from pyspark.sql.types import IntegerType, Row
mylist = [1, 2, 3, 4, None ]
l = map(lambda x : Row(x), mylist)
# notice the parens after the type name
df=spark.createDataFrame(l,["id"])
df.where(df.id.isNull() == False).show()
Basiclly, you need to init your int into Row(), then we can use the schema
Inferring the Schema Using Reflection
from pyspark.sql import Row
# spark - sparkSession
sc = spark.sparkContext
# Load a text file and convert each line to a Row.
orders = sc.textFile("/practicedata/orders")
#Split on delimiters
parts = orders.map(lambda l: l.split(","))
#Convert to Row
orders_struct = parts.map(lambda p: Row(order_id=int(p[0]), order_date=p[1], customer_id=p[2], order_status=p[3]))
for i in orders_struct.take(5): print(i)
#convert the RDD to DataFrame
orders_df = spark.createDataFrame(orders_struct)
Programmatically Specifying the Schema
from pyspark.sql import Row
# spark - sparkSession
sc = spark.sparkContext
# Load a text file and convert each line to a Row.
orders = sc.textFile("/practicedata/orders")
#Split on delimiters
parts = orders.map(lambda l: l.split(","))
#Convert to tuple
orders_struct = parts.map(lambda p: (p[0], p[1], p[2], p[3].strip()))
#convert the RDD to DataFrame
orders_df = spark.createDataFrame(orders_struct)
# The schema is encoded in a string.
schemaString = "order_id order_date customer_id status"
fields = [StructField(field_name, StringType(), True) for field_name in schemaString.split()]
schema = Struct
ordersDf = spark.createDataFrame(orders_struct, schema)
Type(fields)
from pyspark.sql import Row
myFloatRdd.map(lambda x: Row(x)).toDF()

how to select by max(date) with spark dataframe API

Given the following dataset
id v date
1 a1 1
1 a2 2
2 b1 3
2 b2 4
I want to select only the last value (regarding the date) for each id.
I've came up with this code :
scala> val df = sc.parallelize(List((41,"a1",1), (1, "a2", 2), (2, "b1", 3), (2, "b2", 4))).toDF("id", "v", "date")
df: org.apache.spark.sql.DataFrame = [id: int, v: string, date: int]
scala> val agg = df.groupBy("id").max("date")
agg: org.apache.spark.sql.DataFrame = [id: int, max(date): int]
scala> val res = df.join(agg, df("id") === agg("id") && df("date") === agg("max(date)"))
16/11/14 22:25:01 WARN sql.Column: Constructing trivially true equals predicate, 'id#3 = id#3'. Perhaps you need to use aliases.
res: org.apache.spark.sql.DataFrame = [id: int, v: string, date: int, id: int, max(date): int]
Is there a better way (more idiomatic, …) ?
Bonus : how to perform a max on a date column and avoid this error Aggregation function can only be applied on a numeric column. ?
You can try agg() with the max function :
import static org.apache.spark.sql.functions.*
df.groupBy("id").agg(max("date"))
For me it only worked taht way:
df = df.groupBy('CPF').agg({'DATA': 'max'})

Spark: Merge 2 dataframes by adding row index/number on both dataframes

Q: Is there is any way to merge two dataframes or copy a column of a dataframe to another in PySpark?
For example, I have two Dataframes:
DF1
C1 C2
23397414 20875.7353
5213970 20497.5582
41323308 20935.7956
123276113 18884.0477
76456078 18389.9269
the seconde dataframe
DF2
C3 C4
2008-02-04 262.00
2008-02-05 257.25
2008-02-06 262.75
2008-02-07 237.00
2008-02-08 231.00
Then i want to add C3 of DF2 to DF1 like this:
New DF
C1 C2 C3
23397414 20875.7353 2008-02-04
5213970 20497.5582 2008-02-05
41323308 20935.7956 2008-02-06
123276113 18884.0477 2008-02-07
76456078 18389.9269 2008-02-08
I hope this example was clear.
rownum + window function i.e solution 1 or zipWithIndex.map i.e solution 2 should help in this case.
Solution 1 : You can use window functions to get this kind of
Then I would suggest you to add rownumber as additional column name to Dataframe say df1.
DF1
C1 C2 columnindex
23397414 20875.7353 1
5213970 20497.5582 2
41323308 20935.7956 3
123276113 18884.0477 4
76456078 18389.9269 5
the second dataframe
DF2
C3 C4 columnindex
2008-02-04 262.00 1
2008-02-05 257.25 2
2008-02-06 262.75 3
2008-02-07 237.00 4
2008-02-08 231.00 5
Now .. do inner join of df1 and df2 that's all...
you will get below ouput
something like this
from pyspark.sql.window import Window
from pyspark.sql.functions import rowNumber
w = Window().orderBy()
df1 = .... // as showed above df1
df2 = .... // as shown above df2
df11 = df1.withColumn("columnindex", rowNumber().over(w))
df22 = df2.withColumn("columnindex", rowNumber().over(w))
newDF = df11.join(df22, df11.columnindex == df22.columnindex, 'inner').drop(df22.columnindex)
newDF.show()
New DF
C1 C2 C3
23397414 20875.7353 2008-02-04
5213970 20497.5582 2008-02-05
41323308 20935.7956 2008-02-06
123276113 18884.0477 2008-02-07
76456078 18389.9269 2008-02-08
Solution 2 : Another good way(probably this is best :)) in scala, which you can translate to pyspark :
/**
* Add Column Index to dataframe
*/
def addColumnIndex(df: DataFrame) = sqlContext.createDataFrame(
// Add Column index
df.rdd.zipWithIndex.map{case (row, columnindex) => Row.fromSeq(row.toSeq :+ columnindex)},
// Create schema
StructType(df.schema.fields :+ StructField("columnindex", LongType, false))
)
// Add index now...
val df1WithIndex = addColumnIndex(df1)
val df2WithIndex = addColumnIndex(df2)
// Now time to join ...
val newone = df1WithIndex
.join(df2WithIndex , Seq("columnindex"))
.drop("columnindex")
I thought I would share the python (pyspark) translation for answer #2 above from #Ram Ghadiyaram:
from pyspark.sql.functions import col
def addColumnIndex(df):
# Create new column names
oldColumns = df.schema.names
newColumns = oldColumns + ["columnindex"]
# Add Column index
df_indexed = df.rdd.zipWithIndex().map(lambda (row, columnindex): \
row + (columnindex,)).toDF()
#Rename all the columns
new_df = reduce(lambda data, idx: data.withColumnRenamed(oldColumns[idx],
newColumns[idx]), xrange(len(oldColumns)), df_indexed)
return new_df
# Add index now...
df1WithIndex = addColumnIndex(df1)
df2WithIndex = addColumnIndex(df2)
#Now time to join ...
newone = df1WithIndex.join(df2WithIndex, col("columnindex"),
'inner').drop("columnindex")
for python3 version,
from pyspark.sql.types import StructType, StructField, LongType
def with_column_index(sdf):
new_schema = StructType(sdf.schema.fields + [StructField("ColumnIndex", LongType(), False),])
return sdf.rdd.zipWithIndex().map(lambda row: row[0] + (row[1],)).toDF(schema=new_schema)
df1_ci = with_column_index(df1)
df2_ci = with_column_index(df2)
join_on_index = df1_ci.join(df2_ci, df1_ci.ColumnIndex == df2_ci.ColumnIndex, 'inner').drop("ColumnIndex")
I referred to his(#Jed) answer
from pyspark.sql.functions import col
def addColumnIndex(df):
# Get old columns names and add a column "columnindex"
oldColumns = df.columns
newColumns = oldColumns + ["columnindex"]
# Add Column index
df_indexed = df.rdd.zipWithIndex().map(lambda (row, columnindex): \
row + (columnindex,)).toDF()
#Rename all the columns
oldColumns = df_indexed.columns
new_df = reduce(lambda data, idx:data.withColumnRenamed(oldColumns[idx],
newColumns[idx]), xrange(len(oldColumns)), df_indexed)
return new_df
# Add index now...
df1WithIndex = addColumnIndex(df1)
df2WithIndex = addColumnIndex(df2)
#Now time to join ...
newone = df1WithIndex.join(df2WithIndex, col("columnindex"),
'inner').drop("columnindex")
This answer solved it for me:
import pyspark.sql.functions as sparkf
# This will return a new DF with all the columns + id
res = df.withColumn('id', sparkf.monotonically_increasing_id())
Credit to Arkadi T
Here is an simple example that can help you even if you have already solve the issue.
//create First Dataframe
val df1 = spark.sparkContext.parallelize(Seq(1,2,1)).toDF("lavel1")
//create second Dataframe
val df2 = spark.sparkContext.parallelize(Seq((1.0, 12.1), (12.1, 1.3), (1.1, 0.3))). toDF("f1", "f2")
//Combine both dataframe
val combinedRow = df1.rdd.zip(df2.rdd). map({
//convert both dataframe to Seq and join them and return as a row
case (df1Data, df2Data) => Row.fromSeq(df1Data.toSeq ++ df2Data.toSeq)
})
// create new Schema from both the dataframe's schema
val combinedschema = StructType(df1.schema.fields ++ df2.schema.fields)
// Create a new dataframe from new row and new schema
val finalDF = spark.sqlContext.createDataFrame(combinedRow, combinedschema)
finalDF.show
Expanding on Jed's answer, in response to Ajinkya's comment:
To get the same old column names, you need to replace "old_cols" with a column list of the newly named indexed columns. See my modified version of the function below
def add_column_index(df):
new_cols = df.schema.names + ['ix']
ix_df = df.rdd.zipWithIndex().map(lambda (row, ix): row + (ix,)).toDF()
tmp_cols = ix_df.schema.names
return reduce(lambda data, idx: data.withColumnRenamed(tmp_cols[idx], new_cols[idx]), xrange(len(tmp_cols)), ix_df)
Not the better way performance wise.
df3=df1.crossJoin(df2).show(3)
To merge columns from two different dataframe you have first to create a column index and then join the two dataframes. Indeed, two dataframes are similar to two SQL tables. To make a connection you have to join them.
If you don't care about the final order of the rows you can generate the index column with monotonically_increasing_id().
Using the following code you can check that monotonically_increasing_id generates the same index column in both dataframes (at least up to a billion of rows), so you won't have any error in the merged dataframe.
from pyspark.sql import SparkSession
import pyspark.sql.functions as F
sample_size = 1E9
sdf1 = spark.range(1, sample_size).select(F.col("id").alias("id1"))
sdf2 = spark.range(1, sample_size).select(F.col("id").alias("id2"))
sdf1 = sdf1.withColumn("idx", sf.monotonically_increasing_id())
sdf2 = sdf2.withColumn("idx", sf.monotonically_increasing_id())
sdf3 = sdf1.join(sdf2, 'idx', 'inner')
sdf3 = sdf3.withColumn("diff", F.col("id1")-F.col("id2")).select("diff")
sdf3.filter(F.col("diff") != 0 ).show()
You can use a combination of monotonically_increasing_id (guaranteed to always be increasing) and row_number (guaranteed to always give the same sequence). You cannot use row_number alone because it needs to be ordered by something. So here we order by monotonically_increasing_id. I am using Spark 2.3.1 and Python 2.7.13.
from pandas import DataFrame
from pyspark.sql.functions import (
monotonically_increasing_id,
row_number)
from pyspark.sql import Window
DF1 = spark.createDataFrame(DataFrame({
'C1': [23397414, 5213970, 41323308, 123276113, 76456078],
'C2': [20875.7353, 20497.5582, 20935.7956, 18884.0477, 18389.9269]}))
DF2 = spark.createDataFrame(DataFrame({
'C3':['2008-02-04', '2008-02-05', '2008-02-06', '2008-02-07', '2008-02-08']}))
DF1_idx = (
DF1
.withColumn('id', monotonically_increasing_id())
.withColumn('columnindex', row_number().over(Window.orderBy('id')))
.select('columnindex', 'C1', 'C2'))
DF2_idx = (
DF2
.withColumn('id', monotonically_increasing_id())
.withColumn('columnindex', row_number().over(Window.orderBy('id')))
.select('columnindex', 'C3'))
DF_complete = (
DF1_idx
.join(
other=DF2_idx,
on=['columnindex'],
how='inner')
.select('C1', 'C2', 'C3'))
DF_complete.show()
+---------+----------+----------+
| C1| C2| C3|
+---------+----------+----------+
| 23397414|20875.7353|2008-02-04|
| 5213970|20497.5582|2008-02-05|
| 41323308|20935.7956|2008-02-06|
|123276113|18884.0477|2008-02-07|
| 76456078|18389.9269|2008-02-08|
+---------+----------+----------+

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