code: def countApprox(self, timeout, confidence=0.95):
          
          drdd = self.mapPartitions(lambda it: [float(sum(1 for i in it))])
          return int(drdd.sumApprox(timeout, confidence))
docs: .. note:: Experimental
  
          Approximate version of count() that returns a potentially incomplete
          result within a timeout, even if not all tasks have finished.
  
          >>> rdd = sc.parallelize(range(1000), 10)
          >>> rdd.countApprox(1000, 1.0)
          1000