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Doing the same for 2nd partition returns (7, 2).ДЄpply combOp to each local result to form the final, global result: (3,2) + (7,2) = (10, 4)Д®xample described in 'figure': (0, 0) <- zeroValue The local result is now (3, 2), which will be the final result from the 1st partition, since they are no other elements in the sublist of the 1st partition. local_result gets updated from (0, 0), to (1, 1).

The local result is (1, 1), which means the sum is 1 and the length 1 for the 1st partition after processing only the first element. For example, (0, 0) and list_element is the first element of the list: 0 + 1 = 1

Local_result gets initialized to the zeroValue parameter aggregate() was provided with. This applies the seqOp to each element of that list, which produces a local result - A pair of (sum, length) that will reflect the result locally, only in that first partition. Then aggregated: listRDD.aggregate( (0, 0), seqOp, combOp) Then define combOp: combOp = (lambda some_local_result, another_local_result: (some_local_result + another_local_result, some_local_result + another_local_result) ) Then define seqOp: seqOp = (lambda local_result, list_element: (local_result + list_element, local_result + 1) ) In a Spark shell, create a list with 4 elements, with 2 partitions: listRDD = sc.parallelize(, 2) Return the result in a pair of (sum, length). combOp : Defines how the resulted objects (one for every partition),Д¬ompute the sum of a list and the length of that list.seqOp : The operation you want to apply to RDD records.zeroValue : The initialization value, for your result, in the desired.Using spark-submit : spark-submit -versionДЄggregate() lets you take an RDD and generate a single value that is of a different type than what was stored in the original RDD. Generally in a program: SparkContext.version

Download spark and run wordcount example