Sort in descending order in PySpark

In PySpark 1.3 sort method doesn't take ascending parameter. You can use desc method instead:

from pyspark.sql.functions import col

(group_by_dataframe
    .count()
    .filter("`count` >= 10")
    .sort(col("count").desc()))

or desc function:

from pyspark.sql.functions import desc

(group_by_dataframe
    .count()
    .filter("`count` >= 10")
    .sort(desc("count"))

Both methods can be used with with Spark >= 1.3 (including Spark 2.x).

Use orderBy:

df.orderBy('column_name', ascending=False)

Complete answer:

group_by_dataframe.count().filter("`count` >= 10").orderBy('count', ascending=False)

http://spark.apache.org/docs/2.0.0/api/python/pyspark.sql.html

By far the most convenient way is using this:

df.orderBy(df.column_name.desc())

Doesn't require special imports.

you can use groupBy and orderBy as follows also

dataFrameWay = df.groupBy("firstName").count().withColumnRenamed("count","distinct_name").sort(desc("count"))

In pyspark 2.4.4

1) group_by_dataframe.count().filter("`count` >= 10").orderBy('count', ascending=False)

2) from pyspark.sql.functions import desc
   group_by_dataframe.count().filter("`count` >= 10").orderBy('count').sort(desc('count'))

No need to import in 1) and 1) is short & easy to read, So I prefer 1) over 2)

RDD.sortBy(keyfunc, ascending=True, numPartitions=None)

an example:

words =  rdd2.flatMap(lambda line: line.split(" "))
counter = words.map(lambda word: (word,1)).reduceByKey(lambda a,b: a+b)

print(counter.sortBy(lambda a: a[1],ascending=False).take(10))