Count the number of non-null values in a Spark DataFrame

Question

I have a data frame with some columns, and before doing analysis, I'd like to understand how complete the data frame is. So I want to filter the data frame and count for each column the number of non-null values, possibly returning a dataframe back.

Basically, I am trying to achieve the same result as expressed in this question but using Scala instead of Python.

Say you have:

val row = Row("x", "y", "z")
val df = sc.parallelize(Seq(row(0, 4, 3), row(None, 3, 4), row(None, None, 5))).toDF()

How can you summarize the number of non-null values for each column and return a dataframe with the same number of columns and just a single row with the answer?

Answer 1

One straight forward option is to use .describe() function to get a summary of your data frame, where the count row includes a count of non-null values:

df.describe().filter($"summary" === "count").show
+-------+---+---+---+
|summary|  x|  y|  z|
+-------+---+---+---+
|  count|  1|  2|  3|
+-------+---+---+---+

Answer 2

Although I like Psidoms answer, often I'm more interested in the fraction of null-values, because just the number of non-null values doesn't tell much...

You can do something like:

import org.apache.spark.sql.functions.{sum,when, count}

df.agg(
   (sum(when($"x".isNotNull,0).otherwise(1))/count("*")).as("x : fraction null"),
   (sum(when($"y".isNotNull,0).otherwise(1))/count("*")).as("y : fraction null"),
   (sum(when($"z".isNotNull,0).otherwise(1))/count("*")).as("z : fraction null")
 ).show()

EDIT: sum(when($"x".isNotNull,0).otherwise(1)) can also just be replaced by count($"x") which only counts non-null values. As I find this not obvious, I tend to use the sum notation which is more clear

Answer 3

Here's how I did it in Scala 2.11, Spark 2.3.1:

import org.apache.spark.sql.functions._
import org.apache.spark.sql.types._

df.agg(
    count("x").divide(count(lit(1)))
        .as("x: percent non-null")
    // ...copy paste that for columns y and z
).head()

count(*) counts non-null rows, count(1) runs on every row.

If you instead want to count percent null in population, find the complement of our count-based equation:

lit(1).minus(
    count("x").divide(count(lit(1)))
    )
    .as("x: percent null")

---

It's also worth knowing that you can cast nullness to an integer, then sum it.
But it's probably less performant:

// cast null-ness to an integer
sum(col("x").isNull.cast(IntegerType))
    .divide(count(lit(1)))
    .as("x: percent null")