Spark DataFrame: count distinct values of every column - apache-spark

Question

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.

Answer 1

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().

Answer 2

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.

Answer 3

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()

Answer 4

Adding to desaiankitb's answer, this would provide you a more intuitive answer :
from pyspark.sql.functions import count
df.groupBy(colname).count().show()

Answer 5

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])

Answer 6

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()