PySpark: implementing subtractByKey(), subtract() and keyBy()

python/pyspark/rdd.py

@@ -754,6 +754,43 @@ class RDD(object):
         """
         return python_cogroup(self, other, numPartitions)
 
+    def subtractByKey(self, other, numPartitions=None):
+        """
+        Return each (key, value) pair in C{self} that has no pair with matching key
+        in C{other}.
+
+        >>> x = sc.parallelize([("a", 1), ("b", 4), ("b", 5), ("a", 2)])
+        >>> y = sc.parallelize([("a", 3), ("c", None)])
+        >>> sorted(x.subtractByKey(y).collect())
+        [('b', 4), ('b', 5)]
+        """ 
+        filter_func = lambda tpl: len(tpl[1][0]) > 0 and len(tpl[1][1]) == 0
+        map_func = lambda tpl: [(tpl[0], val) for val in tpl[1][0]]
+        return self.cogroup(other, numPartitions).filter(filter_func).flatMap(map_func)
+
+    def subtract(self, other, numPartitions=None):
+        """
+        Return each value in C{self} that is not contained in C{other}.
+
+        >>> x = sc.parallelize([("a", 1), ("b", 4), ("b", 5), ("a", 3)])
+        >>> y = sc.parallelize([("a", 3), ("c", None)])
+        >>> sorted(x.subtract(y).collect())
+        [('a', 1), ('b', 4), ('b', 5)]
+        """
+        rdd = other.map(lambda x: (x, True)) # note: here 'True' is just a placeholder
+        return self.map(lambda x: (x, True)).subtractByKey(rdd).map(lambda tpl: tpl[0]) # note: here 'True' is just a placeholder
+
+    def keyBy(self, f):
+        """
+        Creates tuples of the elements in this RDD by applying C{f}.
+
+        >>> x = sc.parallelize(range(0,3)).keyBy(lambda x: x*x)
+        >>> y = sc.parallelize(zip(range(0,5), range(0,5)))
+        >>> sorted(x.cogroup(y).collect())
+        [(0, ([0], [0])), (1, ([1], [1])), (2, ([], [2])), (3, ([], [3])), (4, ([2], [4]))]
+        """
+        return self.map(lambda x: (f(x), x))
+
     # TODO: `lookup` is disabled because we can't make direct comparisons based
     # on the key; we need to compare the hash of the key to the hash of the
     # keys in the pairs.  This could be an expensive operation, since those