Class RDD

x.__init__(...) initializes x; see help(type(x)) for signature

Overrides: object.__init__

(inherited documentation)

The SparkContext that this RDD was created on.

Decorators:

• @property

Mark this RDD for checkpointing. It will be saved to a file inside the checkpoint directory set with SparkContext.setCheckpointDir() and all references to its parent RDDs will be removed. This function must be called before any job has been executed on this RDD. It is strongly recommended that this RDD is persisted in memory, otherwise saving it on a file will require recomputation.

Return a new RDD by first applying a function to all elements of this RDD, and then flattening the results.

>>> rdd = sc.parallelize([2, 3, 4])
>>> sorted(rdd.flatMap(lambda x: range(1, x)).collect())
[1, 1, 1, 2, 2, 3]
>>> sorted(rdd.flatMap(lambda x: [(x, x), (x, x)]).collect())
[(2, 2), (2, 2), (3, 3), (3, 3), (4, 4), (4, 4)]

Return a new RDD by applying a function to each partition of this RDD.

>>> rdd = sc.parallelize([1, 2, 3, 4], 2)
>>> def f(iterator): yield sum(iterator)
>>> rdd.mapPartitions(f).collect()
[3, 7]

Return a new RDD by applying a function to each partition of this RDD, while tracking the index of the original partition.

>>> rdd = sc.parallelize([1, 2, 3, 4], 4)
>>> def f(splitIndex, iterator): yield splitIndex
>>> rdd.mapPartitionsWithSplit(f).sum()
6

Return a new RDD containing only the elements that satisfy a predicate.

>>> rdd = sc.parallelize([1, 2, 3, 4, 5])
>>> rdd.filter(lambda x: x % 2 == 0).collect()
[2, 4]

Return a new RDD containing the distinct elements in this RDD.

>>> sorted(sc.parallelize([1, 1, 2, 3]).distinct().collect())
[1, 2, 3]

Return the union of this RDD and another one.

>>> rdd = sc.parallelize([1, 1, 2, 3])
>>> rdd.union(rdd).collect()
[1, 1, 2, 3, 1, 1, 2, 3]

Return the union of this RDD and another one.

>>> rdd = sc.parallelize([1, 1, 2, 3])
>>> (rdd + rdd).collect()
[1, 1, 2, 3, 1, 1, 2, 3]

Return an RDD created by coalescing all elements within each partition into a list.

>>> rdd = sc.parallelize([1, 2, 3, 4], 2)
>>> sorted(rdd.glom().collect())
[[1, 2], [3, 4]]

Return the Cartesian product of this RDD and another one, that is, the RDD of all pairs of elements (a, b) where a is in self and b is in other.

>>> rdd = sc.parallelize([1, 2])
>>> sorted(rdd.cartesian(rdd).collect())
[(1, 1), (1, 2), (2, 1), (2, 2)]

Return an RDD of grouped items.

>>> rdd = sc.parallelize([1, 1, 2, 3, 5, 8])
>>> result = rdd.groupBy(lambda x: x % 2).collect()
>>> sorted([(x, sorted(y)) for (x, y) in result])
[(0, [2, 8]), (1, [1, 1, 3, 5])]

Return an RDD created by piping elements to a forked external process.

>>> sc.parallelize([1, 2, 3]).pipe('cat').collect()
['1', '2', '3']

Applies a function to all elements of this RDD.

>>> def f(x): print x
>>> sc.parallelize([1, 2, 3, 4, 5]).foreach(f)

Reduces the elements of this RDD using the specified commutative and associative binary operator.

>>> from operator import add
>>> sc.parallelize([1, 2, 3, 4, 5]).reduce(add)
15
>>> sc.parallelize((2 for _ in range(10))).map(lambda x: 1).cache().reduce(add)
10

Aggregate the elements of each partition, and then the results for all the partitions, using a given associative function and a neutral "zero value."

The function op(t1, t2) is allowed to modify t1 and return it as its result value to avoid object allocation; however, it should not modify t2.

>>> from operator import add
>>> sc.parallelize([1, 2, 3, 4, 5]).fold(0, add)
15

Add up the elements in this RDD.

>>> sc.parallelize([1.0, 2.0, 3.0]).sum()
6.0

Return the number of elements in this RDD.

>>> sc.parallelize([2, 3, 4]).count()
3

Return the count of each unique value in this RDD as a dictionary of (value, count) pairs.

>>> sorted(sc.parallelize([1, 2, 1, 2, 2], 2).countByValue().items())
[(1, 2), (2, 3)]

Take the first num elements of the RDD.

This currently scans the partitions *one by one*, so it will be slow if a lot of partitions are required. In that case, use collect to get the whole RDD instead.

>>> sc.parallelize([2, 3, 4, 5, 6]).cache().take(2)
[2, 3]
>>> sc.parallelize([2, 3, 4, 5, 6]).take(10)
[2, 3, 4, 5, 6]

Return the first element in this RDD.

>>> sc.parallelize([2, 3, 4]).first()
2

Save this RDD as a text file, using string representations of elements.

>>> tempFile = NamedTemporaryFile(delete=True)
>>> tempFile.close()
>>> sc.parallelize(range(10)).saveAsTextFile(tempFile.name)
>>> from fileinput import input
>>> from glob import glob
>>> ''.join(sorted(input(glob(tempFile.name + "/part-0000*"))))
'0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n'

Return the key-value pairs in this RDD to the master as a dictionary.

>>> m = sc.parallelize([(1, 2), (3, 4)]).collectAsMap()
>>> m[1]
2
>>> m[3]
4

Merge the values for each key using an associative reduce function.

This will also perform the merging locally on each mapper before sending results to a reducer, similarly to a "combiner" in MapReduce.

Output will be hash-partitioned with numPartitions partitions, or the default parallelism level if numPartitions is not specified.

>>> from operator import add
>>> rdd = sc.parallelize([("a", 1), ("b", 1), ("a", 1)])
>>> sorted(rdd.reduceByKey(add).collect())
[('a', 2), ('b', 1)]

Merge the values for each key using an associative reduce function, but return the results immediately to the master as a dictionary.

This will also perform the merging locally on each mapper before sending results to a reducer, similarly to a "combiner" in MapReduce.

>>> from operator import add
>>> rdd = sc.parallelize([("a", 1), ("b", 1), ("a", 1)])
>>> sorted(rdd.reduceByKeyLocally(add).items())
[('a', 2), ('b', 1)]

Count the number of elements for each key, and return the result to the master as a dictionary.

>>> rdd = sc.parallelize([("a", 1), ("b", 1), ("a", 1)])
>>> sorted(rdd.countByKey().items())
[('a', 2), ('b', 1)]

Return an RDD containing all pairs of elements with matching keys in self and other.

Each pair of elements will be returned as a (k, (v1, v2)) tuple, where (k, v1) is in self and (k, v2) is in other.

Performs a hash join across the cluster.

>>> x = sc.parallelize([("a", 1), ("b", 4)])
>>> y = sc.parallelize([("a", 2), ("a", 3)])
>>> sorted(x.join(y).collect())
[('a', (1, 2)), ('a', (1, 3))]

Perform a left outer join of self and other.

For each element (k, v) in self, the resulting RDD will either contain all pairs (k, (v, w)) for w in other, or the pair (k, (v, None)) if no elements in other have key k.

Hash-partitions the resulting RDD into the given number of partitions.

>>> x = sc.parallelize([("a", 1), ("b", 4)])
>>> y = sc.parallelize([("a", 2)])
>>> sorted(x.leftOuterJoin(y).collect())
[('a', (1, 2)), ('b', (4, None))]

Perform a right outer join of self and other.

For each element (k, w) in other, the resulting RDD will either contain all pairs (k, (v, w)) for v in this, or the pair (k, (None, w)) if no elements in self have key k.

Hash-partitions the resulting RDD into the given number of partitions.

>>> x = sc.parallelize([("a", 1), ("b", 4)])
>>> y = sc.parallelize([("a", 2)])
>>> sorted(y.rightOuterJoin(x).collect())
[('a', (2, 1)), ('b', (None, 4))]

Return a copy of the RDD partitioned using the specified partitioner.

>>> pairs = sc.parallelize([1, 2, 3, 4, 2, 4, 1]).map(lambda x: (x, x))
>>> sets = pairs.partitionBy(2).glom().collect()
>>> set(sets[0]).intersection(set(sets[1]))
set([])

Generic function to combine the elements for each key using a custom set of aggregation functions.

Turns an RDD[(K, V)] into a result of type RDD[(K, C)], for a "combined type" C. Note that V and C can be different -- for example, one might group an RDD of type (Int, Int) into an RDD of type (Int, List[Int]).

Users provide three functions:

• createCombiner, which turns a V into a C (e.g., creates a one-element list)
• mergeValue, to merge a V into a C (e.g., adds it to the end of a list)
• mergeCombiners, to combine two C's into a single one.

In addition, users can control the partitioning of the output RDD.

>>> x = sc.parallelize([("a", 1), ("b", 1), ("a", 1)])
>>> def f(x): return x
>>> def add(a, b): return a + str(b)
>>> sorted(x.combineByKey(str, add, add).collect())
[('a', '11'), ('b', '1')]

Group the values for each key in the RDD into a single sequence. Hash-partitions the resulting RDD with into numPartitions partitions.

>>> x = sc.parallelize([("a", 1), ("b", 1), ("a", 1)])
>>> sorted(x.groupByKey().collect())
[('a', [1, 1]), ('b', [1])]

For each key k in self or other, return a resulting RDD that contains a tuple with the list of values for that key in self as well as other.

>>> x = sc.parallelize([("a", 1), ("b", 4)])
>>> y = sc.parallelize([("a", 2)])
>>> sorted(x.cogroup(y).collect())
[('a', ([1], [2])), ('b', ([4], []))]