I'm trying to use Spark dataframes instead of RDDs since they appear to be more high-level than RDDs and tend to produce more readable code.
In a 14-nodes Google Dataproc cluster, I have about 6 millions names that are translated to ids by two different systems: sa and sb. Each Row contains name, id_sa and id_sb. My goal is to produce a mapping from id_sa to id_sb such that for each id_sa, the corresponding id_sb is the most frequent id among all names attached to id_sa.
Let's try to clarify with an example. If I have the following rows:
[Row(name='n1', id_sa='a1', id_sb='b1'),
Row(name='n2', id_sa='a1', id_sb='b2'),
Row(name='n3', id_sa='a1', id_sb='b2'),
Row(name='n4', id_sa='a2', id_sb='b2')]
My goal is to produce a mapping from a1 to b2. Indeed, the names associated to a1 are n1, n2 and n3, which map respectively to b1, b2 and b2, so b2 is the most frequent mapping in the names associated to a1. In the same way, a2 will be mapped to b2. It's OK to assume that there will always be a winner: no need to break ties.
I was hoping that I could use groupBy(df.id_sa) on my dataframe, but I don't know what to do next. I was hoping for an aggregation that could produce, in the end, the following rows:
[Row(id_sa=a1, max_id_sb=b2),
Row(id_sa=a2, max_id_sb=b2)]
But maybe I'm trying to use the wrong tool and I should just go back to using RDDs.
Using join (it will result in more than one row in group in case of ties):
import pyspark.sql.functions as F
from pyspark.sql.functions import count, col
cnts = df.groupBy("id_sa", "id_sb").agg(count("*").alias("cnt")).alias("cnts")
maxs = cnts.groupBy("id_sa").agg(F.max("cnt").alias("mx")).alias("maxs")
cnts.join(maxs,
(col("cnt") == col("mx")) & (col("cnts.id_sa") == col("maxs.id_sa"))
).select(col("cnts.id_sa"), col("cnts.id_sb"))
Using window functions (will drop ties):
from pyspark.sql.functions import row_number
from pyspark.sql.window import Window
w = Window().partitionBy("id_sa").orderBy(col("cnt").desc())
(cnts
.withColumn("rn", row_number().over(w))
.where(col("rn") == 1)
.select("id_sa", "id_sb"))
Using struct ordering:
from pyspark.sql.functions import struct
(cnts
.groupBy("id_sa")
.agg(F.max(struct(col("cnt"), col("id_sb"))).alias("max"))
.select(col("id_sa"), col("max.id_sb")))
See also How to select the first row of each group?
I think what you might be looking for are window functions:
http://spark.apache.org/docs/latest/api/python/pyspark.sql.html?highlight=window#pyspark.sql.Window
https://databricks.com/blog/2015/07/15/introducing-window-functions-in-spark-sql.html
Here is an example in Scala (I don't have a Spark Shell with Hive available right now, so I was not able to test the code, but I think it should work):
case class MyRow(name: String, id_sa: String, id_sb: String)
val myDF = sc.parallelize(Array(
MyRow("n1", "a1", "b1"),
MyRow("n2", "a1", "b2"),
MyRow("n3", "a1", "b2"),
MyRow("n1", "a2", "b2")
)).toDF("name", "id_sa", "id_sb")
import org.apache.spark.sql.expressions.Window
val windowSpec = Window.partitionBy(myDF("id_sa")).orderBy(myDF("id_sb").desc)
myDF.withColumn("max_id_b", first(myDF("id_sb")).over(windowSpec).as("max_id_sb")).filter("id_sb = max_id_sb")
There are probably more efficient ways to achieve the same results with Window functions, but I hope this points you in the right direction.
Related
I have a data frame like this
df = [id1, id2, name1, name2, address1, address2, DOB1, DOB2]
I would like get Jaro_winkler score (in a new column) for the column1 and column2 in the Pyspark DataFrame. I am trying to use jellyfish python package.
Thanks
This response stems mainly from viewing a similar question in StackOverflow here. In their example, they investigate how null values can be dealt with when running a jellyfish string comparison.
You'll want to set up a UDF call to utilize the parallel processing powers of pyspark. See code below:
from pyspark.sql.functions import udf
from pyspark.sql.functions import col
from pyspark.sql.types import DoubleType
import jellyfish
# initiate user defined function (UDF) call.
#udf(DoubleType())
def jaro_winkler(s1, s2):
return jellyfish.jaro_winkler(s1, s2)
# to create a new column
df = df.withColumn('new_column',jaro_winkler(col('column1'),col('column2')))
# to show top 20 results
df.select('new_column').show()
for a similar functionality with the option to deal with null values, I would suggest altering your function to incorporate the following change:
#udf(DoubleType())
def jaro_winkler(s1, s2):
if s1 is None or s2 is None:
out = 0
else:
out = jellyfish.jaro_winkler(s1, s2)
return out
This is probably far to simple of a question.
But I'm not getting very far on my own.
I'm trying to use PySpark in Databricks to do the SQL equivalent of a lookup:
select
a.*
, b.MASTER_ID as PLAYER_ID
from vGame a
join PLAYER_XREF b
on a.PLAYER_NAME = b.PLAYER
Note that the two attributes on both sides of the on are NOT named the same.
Can you show me the pyspark version of the same?
Seems to me the numerous tangential posts here for this are over the top complex compared to than this.
I found this and this is really close but the returned dataframe is all columns of ta & tb.
inner_join = ta.join(tb, ta.name == tb.name)
I can list out all the ta columns individually & alias the one tb column with:
from pyspark.sql.functions import *
inner_join = ta.join(tb, ta.PLAYER_NAME == tb.PLAYER).select('<taCol1>', '<taCol2>', ... col('MASTER_ID').alias('PLAYER_ID'))
display(inner_join)
logic:
1.) We first rename player_name in ta dataframe to player so that we can join
2.) Once the columnNames are same we can use a join using square brackets []
3.) also we dynamically select columns from data frame ta
code:
ta = ta.withColumn("player_name","player")
inner_join = ta.join(tb,["player"]).select(col(x) for x in ta.columns])
The question is pretty much in the title: Is there an efficient way to count the distinct values in every column in a DataFrame?
The describe method provides only the count but not the distinct count, and I wonder if there is a a way to get the distinct count for all (or some selected) columns.
In pySpark you could do something like this, using countDistinct():
from pyspark.sql.functions import col, countDistinct
df.agg(*(countDistinct(col(c)).alias(c) for c in df.columns))
Similarly in Scala :
import org.apache.spark.sql.functions.countDistinct
import org.apache.spark.sql.functions.col
df.select(df.columns.map(c => countDistinct(col(c)).alias(c)): _*)
If you want to speed things up at the potential loss of accuracy, you could also use approxCountDistinct().
Multiple aggregations would be quite expensive to compute. I suggest that you use approximation methods instead. In this case, approxating distinct count:
val df = Seq((1,3,4),(1,2,3),(2,3,4),(2,3,5)).toDF("col1","col2","col3")
val exprs = df.columns.map((_ -> "approx_count_distinct")).toMap
df.agg(exprs).show()
// +---------------------------+---------------------------+---------------------------+
// |approx_count_distinct(col1)|approx_count_distinct(col2)|approx_count_distinct(col3)|
// +---------------------------+---------------------------+---------------------------+
// | 2| 2| 3|
// +---------------------------+---------------------------+---------------------------+
The approx_count_distinct method relies on HyperLogLog under the hood.
The HyperLogLog algorithm and its variant HyperLogLog++ (implemented in Spark) relies on the following clever observation.
If the numbers are spread uniformly across a range, then the count of distinct elements can be approximated from the largest number of leading zeros in the binary representation of the numbers.
For example, if we observe a number whose digits in binary form are of the form 0…(k times)…01…1, then we can estimate that there are in the order of 2^k elements in the set. This is a very crude estimate but it can be refined to great precision with a sketching algorithm.
A thorough explanation of the mechanics behind this algorithm can be found in the original paper.
Note: Starting Spark 1.6, when Spark calls SELECT SOME_AGG(DISTINCT foo)), SOME_AGG(DISTINCT bar)) FROM df each clause should trigger separate aggregation for each clause. Whereas this is different than SELECT SOME_AGG(foo), SOME_AGG(bar) FROM df where we aggregate once. Thus the performance won't be comparable when using a count(distinct(_)) and approxCountDistinct (or approx_count_distinct).
It's one of the changes of behavior since Spark 1.6 :
With the improved query planner for queries having distinct aggregations (SPARK-9241), the plan of a query having a single distinct aggregation has been changed to a more robust version. To switch back to the plan generated by Spark 1.5’s planner, please set spark.sql.specializeSingleDistinctAggPlanning to true. (SPARK-12077)
Reference : Approximate Algorithms in Apache Spark: HyperLogLog and Quantiles.
if you just want to count for particular column then following could help. Although its late answer. it might help someone. (pyspark 2.2.0 tested)
from pyspark.sql.functions import col, countDistinct
df.agg(countDistinct(col("colName")).alias("count")).show()
Adding to desaiankitb's answer, this would provide you a more intuitive answer :
from pyspark.sql.functions import count
df.groupBy(colname).count().show()
You can use the count(column name) function of SQL
Alternatively if you are using data analysis and want a rough estimation and not exact count of each and every column you can use approx_count_distinct function
approx_count_distinct(expr[, relativeSD])
This is one way to create dataframe with every column counts :
> df = df.to_pandas_on_spark()
> collect_df = []
> for i in df.columns:
> collect_df.append({"field_name": i , "unique_count": df[i].nunique()})
> uniquedf = spark.createDataFrame(collect_df)
Output would like below. I used this with another dataframe to compare values if columns names are same.Other dataframe was also created way then joined.
df_prod_merged = uniquedf1.join(uniquedf2, on='field_name', how="left")
This is easy way to do it might be expensive on very huge data like 1 tb to process but still very efficient when used to_pandas_on_spark()
The question is pretty much in the title: Is there an efficient way to count the distinct values in every column in a DataFrame?
The describe method provides only the count but not the distinct count, and I wonder if there is a a way to get the distinct count for all (or some selected) columns.
In pySpark you could do something like this, using countDistinct():
from pyspark.sql.functions import col, countDistinct
df.agg(*(countDistinct(col(c)).alias(c) for c in df.columns))
Similarly in Scala :
import org.apache.spark.sql.functions.countDistinct
import org.apache.spark.sql.functions.col
df.select(df.columns.map(c => countDistinct(col(c)).alias(c)): _*)
If you want to speed things up at the potential loss of accuracy, you could also use approxCountDistinct().
Multiple aggregations would be quite expensive to compute. I suggest that you use approximation methods instead. In this case, approxating distinct count:
val df = Seq((1,3,4),(1,2,3),(2,3,4),(2,3,5)).toDF("col1","col2","col3")
val exprs = df.columns.map((_ -> "approx_count_distinct")).toMap
df.agg(exprs).show()
// +---------------------------+---------------------------+---------------------------+
// |approx_count_distinct(col1)|approx_count_distinct(col2)|approx_count_distinct(col3)|
// +---------------------------+---------------------------+---------------------------+
// | 2| 2| 3|
// +---------------------------+---------------------------+---------------------------+
The approx_count_distinct method relies on HyperLogLog under the hood.
The HyperLogLog algorithm and its variant HyperLogLog++ (implemented in Spark) relies on the following clever observation.
If the numbers are spread uniformly across a range, then the count of distinct elements can be approximated from the largest number of leading zeros in the binary representation of the numbers.
For example, if we observe a number whose digits in binary form are of the form 0…(k times)…01…1, then we can estimate that there are in the order of 2^k elements in the set. This is a very crude estimate but it can be refined to great precision with a sketching algorithm.
A thorough explanation of the mechanics behind this algorithm can be found in the original paper.
Note: Starting Spark 1.6, when Spark calls SELECT SOME_AGG(DISTINCT foo)), SOME_AGG(DISTINCT bar)) FROM df each clause should trigger separate aggregation for each clause. Whereas this is different than SELECT SOME_AGG(foo), SOME_AGG(bar) FROM df where we aggregate once. Thus the performance won't be comparable when using a count(distinct(_)) and approxCountDistinct (or approx_count_distinct).
It's one of the changes of behavior since Spark 1.6 :
With the improved query planner for queries having distinct aggregations (SPARK-9241), the plan of a query having a single distinct aggregation has been changed to a more robust version. To switch back to the plan generated by Spark 1.5’s planner, please set spark.sql.specializeSingleDistinctAggPlanning to true. (SPARK-12077)
Reference : Approximate Algorithms in Apache Spark: HyperLogLog and Quantiles.
if you just want to count for particular column then following could help. Although its late answer. it might help someone. (pyspark 2.2.0 tested)
from pyspark.sql.functions import col, countDistinct
df.agg(countDistinct(col("colName")).alias("count")).show()
Adding to desaiankitb's answer, this would provide you a more intuitive answer :
from pyspark.sql.functions import count
df.groupBy(colname).count().show()
You can use the count(column name) function of SQL
Alternatively if you are using data analysis and want a rough estimation and not exact count of each and every column you can use approx_count_distinct function
approx_count_distinct(expr[, relativeSD])
This is one way to create dataframe with every column counts :
> df = df.to_pandas_on_spark()
> collect_df = []
> for i in df.columns:
> collect_df.append({"field_name": i , "unique_count": df[i].nunique()})
> uniquedf = spark.createDataFrame(collect_df)
Output would like below. I used this with another dataframe to compare values if columns names are same.Other dataframe was also created way then joined.
df_prod_merged = uniquedf1.join(uniquedf2, on='field_name', how="left")
This is easy way to do it might be expensive on very huge data like 1 tb to process but still very efficient when used to_pandas_on_spark()
I'm trying to use SQLContext.subtract() in Spark 1.6.1 to remove rows from a dataframe based on a column from another dataframe. Let's use an example:
from pyspark.sql import Row
df1 = sqlContext.createDataFrame([
Row(name='Alice', age=2),
Row(name='Bob', age=1),
]).alias('df1')
df2 = sqlContext.createDataFrame([
Row(name='Bob'),
])
df1_with_df2 = df1.join(df2, 'name').select('df1.*')
df1_without_df2 = df1.subtract(df1_with_df2)
Since I want all rows from df1 which don't include name='Bob' I expect Row(age=2, name='Alice'). But I also retrieve Bob:
print(df1_without_df2.collect())
# [Row(age='1', name='Bob'), Row(age='2', name='Alice')]
After various experiments to get down to this MCVE, I found out that the issue is with the age key. If I omit it:
df1_noage = sqlContext.createDataFrame([
Row(name='Alice'),
Row(name='Bob'),
]).alias('df1_noage')
df1_noage_with_df2 = df1_noage.join(df2, 'name').select('df1_noage.*')
df1_noage_without_df2 = df1_noage.subtract(df1_noage_with_df2)
print(df1_noage_without_df2.collect())
# [Row(name='Alice')]
Then I only get Alice as expected. The weirdest observation I made is that it's possible to add keys, as long as they're after (in the lexicographical order sense) the key I use in the join:
df1_zage = sqlContext.createDataFrame([
Row(zage=2, name='Alice'),
Row(zage=1, name='Bob'),
]).alias('df1_zage')
df1_zage_with_df2 = df1_zage.join(df2, 'name').select('df1_zage.*')
df1_zage_without_df2 = df1_zage.subtract(df1_zage_with_df2)
print(df1_zage_without_df2.collect())
# [Row(name='Alice', zage=2)]
I correctly get Alice (with her zage)! In my real examples, I'm interested in all columns, not only the ones that are after name.
Well there are some bugs here (the first issue looks like related to to the same problem as SPARK-6231) and JIRA looks like a good idea, but SUBTRACT / EXCEPT is no the right choice for partial matches.
Instead, as of Spark 2.0, you can use anti-join:
df1.join(df1_with_df2, ["name"], "leftanti").show()
In 1.6 you can do pretty much the same thing with standard outer join:
import pyspark.sql.functions as F
ref = df1_with_df2.select("name").alias("ref")
(df1
.join(ref, ref.name == df1.name, "leftouter")
.filter(F.isnull("ref.name"))
.drop(F.col("ref.name")))