Github user nrchandan commented on a diff in the pull request:
https://github.com/apache/spark/pull/1791#discussion_r15858417
--- Diff: python/pyspark/rdd.py ---
@@ -854,6 +884,97 @@ def redFunc(left_counter, right_counter):
return self.mapPartitions(lambda i:
[StatCounter(i)]).reduce(redFunc)
+ def histogram(self, buckets, even=False):
+ """
+ Compute a histogram using the provided buckets. The buckets
+ are all open to the right except for the last which is closed.
+ e.g. [1,10,20,50] means the buckets are [1,10) [10,20) [20,50],
+ which means 1<=x<10, 10<=x<20, 20<=x<=50. And on the input of 1
+ and 50 we would have a histogram of 1,0,1.
+
+ If your histogram is evenly spaced (e.g. [0, 10, 20, 30]),
+ this can be switched from an O(log n) inseration to O(1) per
+ element(where n = # buckets), if you set `even` to True.
+
+ Buckets must be sorted and not contain any duplicates, must be
+ at least two elements.
+
+ If `buckets` is a number, it will generates buckets which is
+ evenly spaced between the minimum and maximum of the RDD. For
+ example, if the min value is 0 and the max is 100, given buckets
+ as 2, the resulting buckets will be [0,50) [50,100]. buckets must
+ be at least 1 If the RDD contains infinity, NaN throws an exception
+ If the elements in RDD do not vary (max == min) always returns
+ a single bucket.
+
+ It will return an tuple of buckets and histogram.
+
+ >>> rdd = sc.parallelize(range(51))
+ >>> rdd.histogram(2)
+ ([0, 25, 50], [25, 26])
+ >>> rdd.histogram([0, 5, 25, 50])
+ ([0, 5, 25, 50], [5, 20, 26])
+ >>> rdd.histogram([0, 15, 30, 45, 60], True)
+ ([0, 15, 30, 45, 60], [15, 15, 15, 6])
+ """
+
+ if isinstance(buckets, (int, long)):
+ if buckets < 1:
+ raise ValueError("buckets should not less than 1")
+
+ # faster than stats()
+ def minmax(it):
+ minv, maxv = float("inf"), float("-inf")
+ for v in it:
+ minv = min(minv, v)
+ maxv = max(maxv, v)
+ return [(minv, maxv)]
+
+ def _merge(a, b):
+ return (min(a[0], b[0]), max(a[1], b[1]))
+
+ minv, maxv = self.mapPartitions(minmax).reduce(_merge)
+
+ if minv == maxv or buckets == 1:
+ return [minv, maxv], [self.count()]
+
+ inc = (maxv - minv) / buckets
+ # keep them as integer if possible
+ if inc * buckets != maxv - minv:
--- End diff --
This was smart!
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