Question

我有一个类型为RDD [String]的rdd作为示例，如下所示：

1990,1990-07-08
1994,1994-06-18
1994,1994-06-18
1994,1994-06-22
1994,1994-06-22
1994,1994-06-26
1994,1994-06-26
1954,1954-06-20
2002,2002-06-26
1954,1954-06-23
2002,2002-06-29
1954,1954-06-16
2002,2002-06-30
...

结果：（1982,52）（2006,64）（1962,32）（1966,32）（1986,52）（2002,64）（1994,52）（1974,38）（1990,52）（2010,64）（1978，38）（1954,26）（2014,64）（1958,35）（1998,64）（1970,32）

I group it nicely, but my problem is this v.size part, I do not know to to calculate that length.

Just to put it in perspective, here are expected results:

It is not a mistake that there is two times for 2002. But ignore that.

Answer 1

定义日期格式：

val formatter = DateTimeFormatter.ofPattern("yyyy-MM-dd")

和顺序：

implicit val localDateOrdering: Ordering[LocalDate] = Ordering.by(_.toEpochDay)

创建一个接收“ v”并返回MAX（date_of_matching_year）-MIN（date_of_matching_year））=长度（以天为单位）的函数：

def f(v: Iterable[Array[String]]): Int = {
    val parsedDates = v.map(LocalDate.parse(_(1), formatter))
    parsedDates.max.getDayOfYear - parsedDates.min.getDayOfYear

然后将v.size替换为f（v）

如何获取每年的日期之间的最小值，最大值和长度？

1 个答案: