What you're getting back is an object which allows you to iterate over the results. You can turn the results of groupByKey into a list by calling list() on the values, e.g.
example = sc.parallelize([(0, u'D'), (0, u'D'), (1, u'E'), (2, u'F')])
example.groupByKey().collect()
# Gives [(0, <pyspark.resultiterable.ResultIterable object ......]
example.groupByKey().map(lambda x : (x[0], list(x[1]))).collect()
# Gives [(0, [u'D', u'D']), (1, [u'E']), (2, [u'F'])]
# OR:
example.groupByKey().mapValues(list)
Hey Ron,
It was pretty much exactly as Sean had depicted. I just needed to provide
count an anonymous function to tell it which elements to count. Since I
wanted to count them all, the function is simply "true".
val grouped = rdd.groupByKey().mapValues { mcs =>
val values = mcs.map(_.foo.toDouble)
val n = values.count(x => true)
val sum = values.sum
val sumSquares = values.map(x => x * x).sum
val stddev = math.sqrt(n * sumSquares - sum * sum) / n
print("stddev: " + stddev)
stddev
}
I hope that helps
Just don't. Use reduce by key:
lines.map(lambda x: (x[1][0:4], (x[0], float(x[3])))).map(lambda x: (x, x)) \
.reduceByKey(lambda x, y: (
min(x[0], y[0], key=lambda x: x[1]),
max(x[1], y[1], , key=lambda x: x[1])))
