spark 按照key 分组 然后统计每个key对应的最大、最小、平均值思路——使用groupby，或者reduceby

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])))