Github user mattf commented on a diff in the pull request:
https://github.com/apache/spark/pull/2091#discussion_r16630390
--- Diff: python/pyspark/rdd.py ---
@@ -856,6 +856,104 @@ def redFunc(left_counter, right_counter):
return self.mapPartitions(lambda i:
[StatCounter(i)]).reduce(redFunc)
+ def histogram(self, buckets, evenBuckets=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")
+
+ # filter out non-comparable elements
+ self = self.filter(lambda x: x is not None and not isnan(x))
+
+ # faster than stats()
+ def minmax(a, b):
+ return min(a[0], b[0]), max(a[1], b[1])
+ try:
+ minv, maxv = self.map(lambda x: (x, x)).reduce(minmax)
+ except TypeError as e:
+ if e.message == "reduce() of empty sequence with no
initial value":
--- End diff --
the goal of propagating messages that do not expose implementation details
is good imho
are you confident that the "empty sequence" is the only exception that
could arise?
i was thinking about a mixed types in the rdd, but maybe that's not a
problem
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