I have a dataframe which consists lists in columns similar to the following. The length of the lists in all columns is not same.
Name Age Subjects Grades
[Bob] [16] [Maths,Physics,Chemistry] [A,B,C]
I want to explode the dataframe in such a way that i get the following output-
Name Age Subjects Grades
Bob 16 Maths A
Bob 16 Physics B
Bob 16 Chemistry C
How can I achieve this?
PySpark has added an arrays_zip function in 2.4, which eliminates the need for a Python UDF to zip the arrays.
import pyspark.sql.functions as F
from pyspark.sql.types import *
df = sql.createDataFrame(
[(['Bob'], [16], ['Maths','Physics','Chemistry'], ['A','B','C'])],
['Name','Age','Subjects', 'Grades'])
df = df.withColumn("new", F.arrays_zip("Subjects", "Grades"))\
.withColumn("new", F.explode("new"))\
.select("Name", "Age", F.col("new.Subjects").alias("Subjects"), F.col("new.Grades").alias("Grades"))
df.show()
+-----+----+---------+------+
| Name| Age| Subjects|Grades|
+-----+----+---------+------+
|[Bob]|[16]| Maths| A|
|[Bob]|[16]| Physics| B|
|[Bob]|[16]|Chemistry| C|
+-----+----+---------+------+
This works,
import pyspark.sql.functions as F
from pyspark.sql.types import *
df = sql.createDataFrame(
[(['Bob'], [16], ['Maths','Physics','Chemistry'], ['A','B','C'])],
['Name','Age','Subjects', 'Grades'])
df.show()
+-----+----+--------------------+---------+
| Name| Age| Subjects| Grades|
+-----+----+--------------------+---------+
|[Bob]|[16]|[Maths, Physics, ...|[A, B, C]|
+-----+----+--------------------+---------+
Use udf with zip. Those columns needed to explode have to be merged before exploding.
combine = F.udf(lambda x, y: list(zip(x, y)),
ArrayType(StructType([StructField("subs", StringType()),
StructField("grades", StringType())])))
df = df.withColumn("new", combine("Subjects", "Grades"))\
.withColumn("new", F.explode("new"))\
.select("Name", "Age", F.col("new.subs").alias("Subjects"), F.col("new.grades").alias("Grades"))
df.show()
+-----+----+---------+------+
| Name| Age| Subjects|Grades|
+-----+----+---------+------+
|[Bob]|[16]| Maths| A|
|[Bob]|[16]| Physics| B|
|[Bob]|[16]|Chemistry| C|
+-----+----+---------+------+
Arriving late to the party :-)
The simplest way to go is by using inline that doesn't have python API but is supported by selectExpr.
df.selectExpr('Name[0] as Name','Age[0] as Age','inline(arrays_zip(Subjects,Grades))').show()
+----+---+---------+------+
|Name|Age| Subjects|Grades|
+----+---+---------+------+
| Bob| 16| Maths| A|
| Bob| 16| Physics| B|
| Bob| 16|Chemistry| C|
+----+---+---------+------+
Have you tried this
df.select(explode(split(col("Subjects"))).alias("Subjects")).show()
you can convert the data frame to an RDD.
For an RDD you can use a flatMap function to separate the Subjects.
Copy/paste function if you need to repeat this quickly and easily across a large number of columns in a dataset
cols = ["word", "stem", "pos", "ner"]
def explode_cols(self, data, cols):
data = data.withColumn('exp_combo', f.arrays_zip(*cols))
data = data.withColumn('exp_combo', f.explode('exp_combo'))
for col in cols:
data = data.withColumn(col, f.col('exp_combo.' + col))
return data.drop(f.col('exp_combo'))
result = explode_cols(data, cols)
Your welcome :)
When Exploding multiple columns, the above solution comes in handy only when the length of array is same, but if they are not.
It is better to explode them separately and take distinct values each time.
df = sql.createDataFrame(
[(['Bob'], [16], ['Maths','Physics','Chemistry'], ['A','B','C'])],
['Name','Age','Subjects', 'Grades'])
df = df.withColumn('Subjects',F.explode('Subjects')).select('Name','Age','Subjects', 'Grades').distinct()
df = df.withColumn('Grades',F.explode('Grades')).select('Name','Age','Subjects', 'Grades').distinct()
df.show()
+----+---+---------+------+
|Name|Age| Subjects|Grades|
+----+---+---------+------+
| Bob| 16| Maths| A|
| Bob| 16| Physics| B|
| Bob| 16|Chemistry| C|
+----+---+---------+------+
Thanks #nasty for saving the day.
Just small tweaks to get the code working.
def explode_cols( df, cl):
df = df.withColumn('exp_combo', arrays_zip(*cl))
df = df.withColumn('exp_combo', explode('exp_combo'))
for colm in cl:
final_col = 'exp_combo.'+ colm
df = df.withColumn(final_col, col(final_col))
#print col
#print ('exp_combo.'+ colm)
return df.drop(col('exp_combo'))
Related
I am trying to partition spark dataframe and sum elements in each partition using pyspark. But I am unable to do this inside a called function "sumByHour". Basically, I am unable to access dataframe columns inside "sumByHour".
Basically, I am partitioning by "hour" column and trying to sum the elements based on "hour" partition. So expected output is: 6,15,24 for 0,1,2 hour respectively. Tried below with no luck.
from pyspark.sql.functions import *
from pyspark.sql.types import *
import pandas as pd
def sumByHour(ip):
print(ip)
pandasDF = pd.DataFrame({'hour': [0,0,0,1,1,1,2,2,2], 'numlist': [1,2,3,4,5,6,7,8,9]})
myschema = StructType(
[StructField('hour', IntegerType(), False),
StructField('numlist', IntegerType(), False)]
)
myDf = spark.createDataFrame(pandasDF, schema=myschema)
mydf = myDf.repartition(3, "hour")
myDf.foreachPartition(sumByHour)
I am able to solve this with "window.partitionBy". But I want to know if it can be solved by "foreachPartition".
Thanks in Advance,
Sri
Thanks for the code sample it made this easy. Here's a really simple example modifies you sumByHour code:
def sumByHour(ip):
mySum = 0
myPartition = ""
for x in ip:
mySum += x.numlist
myPartition = x.hour
myString = '{}_{}'.format(mySum, myPartition)
print(myString)
mydf = myDf.repartition(5,"hour") #wait 5 I wanted 3!!!
You get almost the expected result:
>>> mydf.foreachPartition(sumByHour)
0_
0_
24_2
6_0
15_1
>>>
You might ask why partition by '5' and not the '3'? Well turns out the hash formula used for 3 partitions has collision for (0,1) into the same partition and then has an empty partition.(Bad luck) So this will work but, you only want to use it on an array that will fit into memory.
You can use a Window to do that and add the sumByHour as a new column.
from pyspark.sql import functions, Window
w = Window.partitionBy("hour")
myDf = myDf.withColumn("sumByHour", functions.sum("numlist").over(w))
myDf.show()
+----+-------+---------+
|hour|numlist|sumByHour|
+----+-------+---------+
| 1| 4| 15|
| 1| 5| 15|
| 1| 6| 15|
| 2| 7| 24|
| 2| 8| 24|
| 2| 9| 24|
| 0| 1| 6|
| 0| 2| 6|
| 0| 3| 6|
+----+-------+---------+
I have a dataframe (df1) with m rows and n columns in Spark. I have another dataframe (df2) with 1 row and n columns. How can I efficiently compute the dot product of each row of df1 with the single row of df2?
We can use VectorAssembler to do dot product.
Create sample DataFrames:
from pyspark.ml.linalg import Vectors, DenseVector
from pyspark.sql import functions as F
from pyspark.ml.feature import VectorAssembler
from pyspark.sql.types import *
v = [('a', 1,2,3),
('b', 4,5,6),
('c', 9,8,7)]
df1 = spark.createDataFrame(v, ['id', 'v1', 'v2', 'v3'])
df2 = spark.createDataFrame([('d',3,2,1)], ['id', 'v1', 'v2', 'v3'])
df1.show()
df2.show()
They look like this:
+---+---+---+---+
| id| v1| v2| v3|
+---+---+---+---+
| a| 1| 2| 3|
| b| 4| 5| 6|
| c| 9| 8| 7|
+---+---+---+---+
+---+---+---+---+
| id| v1| v2| v3|
+---+---+---+---+
| d| 3| 2| 1|
+---+---+---+---+
Use VectorAssembler to convert the columns to Vector
vecAssembler = VectorAssembler(inputCols=["v1", "v2", "v3"], outputCol="values")
dfv1 = vecAssembler.transform(df1)
dfv2 = vecAssembler.transform(df2)
dfv1.show()
dfv2.show()
Now they look like this:
+---+---+---+---+-------------+
| id| v1| v2| v3| values|
+---+---+---+---+-------------+
| a| 1| 2| 3|[1.0,2.0,3.0]|
| b| 4| 5| 6|[4.0,5.0,6.0]|
| c| 9| 8| 7|[9.0,8.0,7.0]|
+---+---+---+---+-------------+
+---+---+---+---+-------------+
| id| v1| v2| v3| values|
+---+---+---+---+-------------+
| d| 3| 2| 1|[3.0,2.0,1.0]|
+---+---+---+---+-------------+
Define a udf to do the dot product
# Get the fixed vector from DataFrame dfv2
vm = Vectors.dense(dfv2.take(1)[0]['values'])
dot_prod_udf = F.udf(lambda v: float(v.dot(vm)), FloatType())
dfv1 = dfv1.withColumn('dot_prod', dot_prod_udf('values'))
dfv1.show()
The final result is:
+---+---+---+---+-------------+--------+
| id| v1| v2| v3| values|dot_prod|
+---+---+---+---+-------------+--------+
| a| 1| 2| 3|[1.0,2.0,3.0]| 10.0|
| b| 4| 5| 6|[4.0,5.0,6.0]| 28.0|
| c| 9| 8| 7|[9.0,8.0,7.0]| 50.0|
+---+---+---+---+-------------+--------+
You can simply form a matrix with the first data frame and another matrix with the second data frame and multiply them. Here is a code snippet to use (here I'm using block matrix since I assume your data frame can not be stored in your local machine)
v = [('a', [1,2,3]),
('b', [4,5,6]),
('c', [9,8,7])]
df1 = spark.createDataFrame(v, ['id', 'vec'])
df2 = spark.createDataFrame([('d',[3,2,1])], ['id', 'vec'])
m1 = matdf1.toBlockMatrix(100,100)
m2 = matdf2.toBlockMatrix(100,100)
m1.multiply(m2.transpose())
Since Python udfs are not very performant, its probably better to implement it in Scala. But if you want a pure Python implementation, you can try the following hack.
A dot product is equivalent to a linear prediction. So you can leverage LinearRegressionModel.transform if you create one with the desired coefficients. You can do that by "estimating" a regression with the features as an identity matrix and the labels as your desired coefficients. See the following implementation:
from pyspark.ml.feature import OneHotEncoder
from pyspark.ml.regression import LinearRegression
from pyspark.sql import Row, DataFrame
class DotProduct:
_regressors_col = 'regressors'
_dot_product_col = 'dot_product'
_coef_col = 'coef'
_index_col = 'index'
def __init__(
self,
coefficients: np.ndarray
):
self._coefficients = coefficients
def transform(self, dataframe: DataFrame) -> DataFrame:
coef_df = self._convert_coefficients_to_dataframe()
w_identity_df = self._add_identity_matrix_as_regressors(coef_df)
linear_reg = LinearRegression(
featuresCol=self._regressors_col,
labelCol=self._coef_col,
predictionCol=self._dot_product_col,
fitIntercept=False,
standardization=False
)
model = linear_reg.fit(w_identity_df)
return model.transform(dataset=dataframe)
def _convert_coefficients_to_dataframe(self) -> DataFrame:
rows = [
Row(**{self._index_col: index, self._coef_col: float(coef)})
for index, coef
in enumerate(self._coefficients)
]
return spark_session.createDataFrame(rows)
def _add_identity_matrix_as_regressors(
self,
coefficients_df: DataFrame
) -> DataFrame:
encoder = OneHotEncoder(
inputCols=[self._index_col],
outputCols=[self._regressors_col],
dropLast=False
)
model = encoder.fit(coefficients_df)
return model.transform(coefficients_df)
I have a dataframe which has one row, and several columns. Some of the columns are single values, and others are lists. All list columns are the same length. I want to split each list column into a separate row, while keeping any non-list column as is.
Sample DF:
from pyspark import Row
from pyspark.sql import SQLContext
from pyspark.sql.functions import explode
sqlc = SQLContext(sc)
df = sqlc.createDataFrame([Row(a=1, b=[1,2,3],c=[7,8,9], d='foo')])
# +---+---------+---------+---+
# | a| b| c| d|
# +---+---------+---------+---+
# | 1|[1, 2, 3]|[7, 8, 9]|foo|
# +---+---------+---------+---+
What I want:
+---+---+----+------+
| a| b| c | d |
+---+---+----+------+
| 1| 1| 7 | foo |
| 1| 2| 8 | foo |
| 1| 3| 9 | foo |
+---+---+----+------+
If I only had one list column, this would be easy by just doing an explode:
df_exploded = df.withColumn('b', explode('b'))
# >>> df_exploded.show()
# +---+---+---------+---+
# | a| b| c| d|
# +---+---+---------+---+
# | 1| 1|[7, 8, 9]|foo|
# | 1| 2|[7, 8, 9]|foo|
# | 1| 3|[7, 8, 9]|foo|
# +---+---+---------+---+
However, if I try to also explode the c column, I end up with a dataframe with a length the square of what I want:
df_exploded_again = df_exploded.withColumn('c', explode('c'))
# >>> df_exploded_again.show()
# +---+---+---+---+
# | a| b| c| d|
# +---+---+---+---+
# | 1| 1| 7|foo|
# | 1| 1| 8|foo|
# | 1| 1| 9|foo|
# | 1| 2| 7|foo|
# | 1| 2| 8|foo|
# | 1| 2| 9|foo|
# | 1| 3| 7|foo|
# | 1| 3| 8|foo|
# | 1| 3| 9|foo|
# +---+---+---+---+
What I want is - for each column, take the nth element of the array in that column and add that to a new row. I've tried mapping an explode accross all columns in the dataframe, but that doesn't seem to work either:
df_split = df.rdd.map(lambda col: df.withColumn(col, explode(col))).toDF()
Spark >= 2.4
You can replace zip_ udf with arrays_zip function
from pyspark.sql.functions import arrays_zip, col, explode
(df
.withColumn("tmp", arrays_zip("b", "c"))
.withColumn("tmp", explode("tmp"))
.select("a", col("tmp.b"), col("tmp.c"), "d"))
Spark < 2.4
With DataFrames and UDF:
from pyspark.sql.types import ArrayType, StructType, StructField, IntegerType
from pyspark.sql.functions import col, udf, explode
zip_ = udf(
lambda x, y: list(zip(x, y)),
ArrayType(StructType([
# Adjust types to reflect data types
StructField("first", IntegerType()),
StructField("second", IntegerType())
]))
)
(df
.withColumn("tmp", zip_("b", "c"))
# UDF output cannot be directly passed to explode
.withColumn("tmp", explode("tmp"))
.select("a", col("tmp.first").alias("b"), col("tmp.second").alias("c"), "d"))
With RDDs:
(df
.rdd
.flatMap(lambda row: [(row.a, b, c, row.d) for b, c in zip(row.b, row.c)])
.toDF(["a", "b", "c", "d"]))
Both solutions are inefficient due to Python communication overhead. If data size is fixed you can do something like this:
from functools import reduce
from pyspark.sql import DataFrame
# Length of array
n = 3
# For legacy Python you'll need a separate function
# in place of method accessor
reduce(
DataFrame.unionAll,
(df.select("a", col("b").getItem(i), col("c").getItem(i), "d")
for i in range(n))
).toDF("a", "b", "c", "d")
or even:
from pyspark.sql.functions import array, struct
# SQL level zip of arrays of known size
# followed by explode
tmp = explode(array(*[
struct(col("b").getItem(i).alias("b"), col("c").getItem(i).alias("c"))
for i in range(n)
]))
(df
.withColumn("tmp", tmp)
.select("a", col("tmp").getItem("b"), col("tmp").getItem("c"), "d"))
This should be significantly faster compared to UDF or RDD. Generalized to support an arbitrary number of columns:
# This uses keyword only arguments
# If you use legacy Python you'll have to change signature
# Body of the function can stay the same
def zip_and_explode(*colnames, n):
return explode(array(*[
struct(*[col(c).getItem(i).alias(c) for c in colnames])
for i in range(n)
]))
df.withColumn("tmp", zip_and_explode("b", "c", n=3))
You'd need to use flatMap, not map as you want to make multiple output rows out of each input row.
from pyspark.sql import Row
def dualExplode(r):
rowDict = r.asDict()
bList = rowDict.pop('b')
cList = rowDict.pop('c')
for b,c in zip(bList, cList):
newDict = dict(rowDict)
newDict['b'] = b
newDict['c'] = c
yield Row(**newDict)
df_split = sqlContext.createDataFrame(df.rdd.flatMap(dualExplode))
One liner (for Spark>=2.4.0):
df.withColumn("bc", arrays_zip("b","c"))
.select("a", explode("bc").alias("tbc"))
.select("a", col"tbc.b", "tbc.c").show()
Import required:
from pyspark.sql.functions import arrays_zip
Steps -
Create a column bc which is an array_zip of columns b and c
Explode bc to get a struct tbc
Select the required columns a, b and c (all exploded as required).
Output:
> df.withColumn("bc", arrays_zip("b","c")).select("a", explode("bc").alias("tbc")).select("a", "tbc.b", col("tbc.c")).show()
+---+---+---+
| a| b| c|
+---+---+---+
| 1| 1| 7|
| 1| 2| 8|
| 1| 3| 9|
+---+---+---+
I have a dataframe which has one row, and several columns. Some of the columns are single values, and others are lists. All list columns are the same length. I want to split each list column into a separate row, while keeping any non-list column as is.
Sample DF:
from pyspark import Row
from pyspark.sql import SQLContext
from pyspark.sql.functions import explode
sqlc = SQLContext(sc)
df = sqlc.createDataFrame([Row(a=1, b=[1,2,3],c=[7,8,9], d='foo')])
# +---+---------+---------+---+
# | a| b| c| d|
# +---+---------+---------+---+
# | 1|[1, 2, 3]|[7, 8, 9]|foo|
# +---+---------+---------+---+
What I want:
+---+---+----+------+
| a| b| c | d |
+---+---+----+------+
| 1| 1| 7 | foo |
| 1| 2| 8 | foo |
| 1| 3| 9 | foo |
+---+---+----+------+
If I only had one list column, this would be easy by just doing an explode:
df_exploded = df.withColumn('b', explode('b'))
# >>> df_exploded.show()
# +---+---+---------+---+
# | a| b| c| d|
# +---+---+---------+---+
# | 1| 1|[7, 8, 9]|foo|
# | 1| 2|[7, 8, 9]|foo|
# | 1| 3|[7, 8, 9]|foo|
# +---+---+---------+---+
However, if I try to also explode the c column, I end up with a dataframe with a length the square of what I want:
df_exploded_again = df_exploded.withColumn('c', explode('c'))
# >>> df_exploded_again.show()
# +---+---+---+---+
# | a| b| c| d|
# +---+---+---+---+
# | 1| 1| 7|foo|
# | 1| 1| 8|foo|
# | 1| 1| 9|foo|
# | 1| 2| 7|foo|
# | 1| 2| 8|foo|
# | 1| 2| 9|foo|
# | 1| 3| 7|foo|
# | 1| 3| 8|foo|
# | 1| 3| 9|foo|
# +---+---+---+---+
What I want is - for each column, take the nth element of the array in that column and add that to a new row. I've tried mapping an explode accross all columns in the dataframe, but that doesn't seem to work either:
df_split = df.rdd.map(lambda col: df.withColumn(col, explode(col))).toDF()
Spark >= 2.4
You can replace zip_ udf with arrays_zip function
from pyspark.sql.functions import arrays_zip, col, explode
(df
.withColumn("tmp", arrays_zip("b", "c"))
.withColumn("tmp", explode("tmp"))
.select("a", col("tmp.b"), col("tmp.c"), "d"))
Spark < 2.4
With DataFrames and UDF:
from pyspark.sql.types import ArrayType, StructType, StructField, IntegerType
from pyspark.sql.functions import col, udf, explode
zip_ = udf(
lambda x, y: list(zip(x, y)),
ArrayType(StructType([
# Adjust types to reflect data types
StructField("first", IntegerType()),
StructField("second", IntegerType())
]))
)
(df
.withColumn("tmp", zip_("b", "c"))
# UDF output cannot be directly passed to explode
.withColumn("tmp", explode("tmp"))
.select("a", col("tmp.first").alias("b"), col("tmp.second").alias("c"), "d"))
With RDDs:
(df
.rdd
.flatMap(lambda row: [(row.a, b, c, row.d) for b, c in zip(row.b, row.c)])
.toDF(["a", "b", "c", "d"]))
Both solutions are inefficient due to Python communication overhead. If data size is fixed you can do something like this:
from functools import reduce
from pyspark.sql import DataFrame
# Length of array
n = 3
# For legacy Python you'll need a separate function
# in place of method accessor
reduce(
DataFrame.unionAll,
(df.select("a", col("b").getItem(i), col("c").getItem(i), "d")
for i in range(n))
).toDF("a", "b", "c", "d")
or even:
from pyspark.sql.functions import array, struct
# SQL level zip of arrays of known size
# followed by explode
tmp = explode(array(*[
struct(col("b").getItem(i).alias("b"), col("c").getItem(i).alias("c"))
for i in range(n)
]))
(df
.withColumn("tmp", tmp)
.select("a", col("tmp").getItem("b"), col("tmp").getItem("c"), "d"))
This should be significantly faster compared to UDF or RDD. Generalized to support an arbitrary number of columns:
# This uses keyword only arguments
# If you use legacy Python you'll have to change signature
# Body of the function can stay the same
def zip_and_explode(*colnames, n):
return explode(array(*[
struct(*[col(c).getItem(i).alias(c) for c in colnames])
for i in range(n)
]))
df.withColumn("tmp", zip_and_explode("b", "c", n=3))
You'd need to use flatMap, not map as you want to make multiple output rows out of each input row.
from pyspark.sql import Row
def dualExplode(r):
rowDict = r.asDict()
bList = rowDict.pop('b')
cList = rowDict.pop('c')
for b,c in zip(bList, cList):
newDict = dict(rowDict)
newDict['b'] = b
newDict['c'] = c
yield Row(**newDict)
df_split = sqlContext.createDataFrame(df.rdd.flatMap(dualExplode))
One liner (for Spark>=2.4.0):
df.withColumn("bc", arrays_zip("b","c"))
.select("a", explode("bc").alias("tbc"))
.select("a", col"tbc.b", "tbc.c").show()
Import required:
from pyspark.sql.functions import arrays_zip
Steps -
Create a column bc which is an array_zip of columns b and c
Explode bc to get a struct tbc
Select the required columns a, b and c (all exploded as required).
Output:
> df.withColumn("bc", arrays_zip("b","c")).select("a", explode("bc").alias("tbc")).select("a", "tbc.b", col("tbc.c")).show()
+---+---+---+
| a| b| c|
+---+---+---+
| 1| 1| 7|
| 1| 2| 8|
| 1| 3| 9|
+---+---+---+
Let's say I have a rather large dataset in the following form:
data = sc.parallelize([('Foo',41,'US',3),
('Foo',39,'UK',1),
('Bar',57,'CA',2),
('Bar',72,'CA',2),
('Baz',22,'US',6),
('Baz',36,'US',6)])
What I would like to do is remove duplicate rows based on the values of the first,third and fourth columns only.
Removing entirely duplicate rows is straightforward:
data = data.distinct()
and either row 5 or row 6 will be removed
But how do I only remove duplicate rows based on columns 1, 3 and 4 only? i.e. remove either one one of these:
('Baz',22,'US',6)
('Baz',36,'US',6)
In Python, this could be done by specifying columns with .drop_duplicates(). How can I achieve the same in Spark/Pyspark?
Pyspark does include a dropDuplicates() method, which was introduced in 1.4. https://spark.apache.org/docs/3.1.2/api/python/reference/api/pyspark.sql.DataFrame.dropDuplicates.html
>>> from pyspark.sql import Row
>>> df = sc.parallelize([ \
... Row(name='Alice', age=5, height=80), \
... Row(name='Alice', age=5, height=80), \
... Row(name='Alice', age=10, height=80)]).toDF()
>>> df.dropDuplicates().show()
+---+------+-----+
|age|height| name|
+---+------+-----+
| 5| 80|Alice|
| 10| 80|Alice|
+---+------+-----+
>>> df.dropDuplicates(['name', 'height']).show()
+---+------+-----+
|age|height| name|
+---+------+-----+
| 5| 80|Alice|
+---+------+-----+
From your question, it is unclear as-to which columns you want to use to determine duplicates. The general idea behind the solution is to create a key based on the values of the columns that identify duplicates. Then, you can use the reduceByKey or reduce operations to eliminate duplicates.
Here is some code to get you started:
def get_key(x):
return "{0}{1}{2}".format(x[0],x[2],x[3])
m = data.map(lambda x: (get_key(x),x))
Now, you have a key-value RDD that is keyed by columns 1,3 and 4.
The next step would be either a reduceByKey or groupByKey and filter.
This would eliminate duplicates.
r = m.reduceByKey(lambda x,y: (x))
I know you already accepted the other answer, but if you want to do this as a
DataFrame, just use groupBy and agg. Assuming you had a DF already created (with columns named "col1", "col2", etc) you could do:
myDF.groupBy($"col1", $"col3", $"col4").agg($"col1", max($"col2"), $"col3", $"col4")
Note that in this case, I chose the Max of col2, but you could do avg, min, etc.
Agree with David. To add on, it may not be the case that we want to groupBy all columns other than the column(s) in aggregate function i.e, if we want to remove duplicates purely based on a subset of columns and retain all columns in the original dataframe. So the better way to do this could be using dropDuplicates Dataframe api available in Spark 1.4.0
For reference, see: https://spark.apache.org/docs/1.4.0/api/scala/index.html#org.apache.spark.sql.DataFrame
I used inbuilt function dropDuplicates(). Scala code given below
val data = sc.parallelize(List(("Foo",41,"US",3),
("Foo",39,"UK",1),
("Bar",57,"CA",2),
("Bar",72,"CA",2),
("Baz",22,"US",6),
("Baz",36,"US",6))).toDF("x","y","z","count")
data.dropDuplicates(Array("x","count")).show()
Output :
+---+---+---+-----+
| x| y| z|count|
+---+---+---+-----+
|Baz| 22| US| 6|
|Foo| 39| UK| 1|
|Foo| 41| US| 3|
|Bar| 57| CA| 2|
+---+---+---+-----+
The below programme will help you drop duplicates on whole , or if you want to drop duplicates based on certain columns , you can even do that:
import org.apache.spark.sql.SparkSession
object DropDuplicates {
def main(args: Array[String]) {
val spark =
SparkSession.builder()
.appName("DataFrame-DropDuplicates")
.master("local[4]")
.getOrCreate()
import spark.implicits._
// create an RDD of tuples with some data
val custs = Seq(
(1, "Widget Co", 120000.00, 0.00, "AZ"),
(2, "Acme Widgets", 410500.00, 500.00, "CA"),
(3, "Widgetry", 410500.00, 200.00, "CA"),
(4, "Widgets R Us", 410500.00, 0.0, "CA"),
(3, "Widgetry", 410500.00, 200.00, "CA"),
(5, "Ye Olde Widgete", 500.00, 0.0, "MA"),
(6, "Widget Co", 12000.00, 10.00, "AZ")
)
val customerRows = spark.sparkContext.parallelize(custs, 4)
// convert RDD of tuples to DataFrame by supplying column names
val customerDF = customerRows.toDF("id", "name", "sales", "discount", "state")
println("*** Here's the whole DataFrame with duplicates")
customerDF.printSchema()
customerDF.show()
// drop fully identical rows
val withoutDuplicates = customerDF.dropDuplicates()
println("*** Now without duplicates")
withoutDuplicates.show()
val withoutPartials = customerDF.dropDuplicates(Seq("name", "state"))
println("*** Now without partial duplicates too")
withoutPartials.show()
}
}
This is my Df contain 4 is repeated twice so here will remove repeated values.
scala> df.show
+-----+
|value|
+-----+
| 1|
| 4|
| 3|
| 5|
| 4|
| 18|
+-----+
scala> val newdf=df.dropDuplicates
scala> newdf.show
+-----+
|value|
+-----+
| 1|
| 3|
| 5|
| 4|
| 18|
+-----+