我有3个scala函数,运行良好,我在函数中只有错误
我做了另一个函数equals:通过调用上面的3个函数来计算2个日期之间的小时数:toEnd,ToStart,jourOuvree。 我在scala对象中编写了所有函数。但是当我运行函数equals时,它会给我一个错误,如下所示:
^
我应该使用该函数并将其应用于我的数据框,以计算两个日期之间的小时数。
我在java.time API中表现不佳。 有人可以帮我解决这些错误吗?
答案 0 :(得分:1)
我改写了你的解决方案。请参阅内联评论。
import java.time._
import java.time.temporal.ChronoUnit
import scala.annotation.tailrec
import scala.concurrent.duration._
object Demo extends App {
val start = LocalTime.of(9, 0)
val midEnd = LocalTime.of(13, 0)
val midStart = LocalTime.of(14, 0)
val end = LocalTime.of(18, 0)
val firstHalf = start.until(midEnd, ChronoUnit.MILLIS).millis
val secondHalf = midStart.until(end, ChronoUnit.MILLIS).millis
implicit class formatter(d: FiniteDuration) {
def withMinutes = {
val l = d.toMinutes
s"${l / 60}:${l % 60}"
}
def withSeconds = s"${d.toHours}:${d.toMinutes % 60}:${d.toSeconds % 60}"
}
// check if day is working. Could be easily adjusted to check not only weekend but holidays as well
def isWorkingDay(d: LocalDate) = !Set(DayOfWeek.SATURDAY, DayOfWeek.SUNDAY).contains(d.getDayOfWeek)
// check if current day is working day and if not switch to next start of day/previous end of day
// depending on ask parameter
@tailrec
def adjust(t: LocalDateTime, asc: Boolean = true): LocalDateTime = {
if (!isWorkingDay(t.toLocalDate) && asc) adjust(t.toLocalDate.plusDays(1).atTime(start), asc)
else if (!isWorkingDay(t.toLocalDate) && !asc) adjust(t.toLocalDate.minusDays(1).atTime(end), asc)
else t
}
def toEnd(t: LocalTime) = {
if (t.isBefore(start)) firstHalf + secondHalf
else if (t.isBefore(midEnd)) t.until(midEnd, ChronoUnit.MILLIS).millis + secondHalf
else if (t.isBefore(midStart)) secondHalf
else if (t.isBefore(end)) t.until(end, ChronoUnit.MILLIS).millis
else 0.hours
}
def toStart(t: LocalTime) = {
if (t.isBefore(start)) 0.hours
else if (t.isBefore(midEnd)) start.until(t, ChronoUnit.MILLIS).millis
else if (t.isBefore(midStart)) firstHalf
else if (t.isBefore(end)) firstHalf + midStart.until(t, ChronoUnit.MILLIS).millis
else firstHalf + secondHalf
}
// count amount of working days between two dates.
// if dates are the same - means 0 days
def jourOuvree(d1: LocalDate, d2: LocalDate): Int = {
@tailrec
def count(n: LocalDate, acc: Int): Int = {
if (n.isEqual(d2) || n.isAfter(d2)) acc
else if (!isWorkingDay(n)) count(n.plusDays(1), acc)
else count(n.plusDays(1), acc + 1)
}
count(d1, 0)
}
// evaluate duration of working time between to date/times
// take into account that start/end could be non-working and
// adjust
def toEquals(rd1: LocalDateTime, rd2: LocalDateTime) = {
val d1 = adjust(rd1)
val d2 = adjust(rd2, asc = false)
// if we are in the same day ==> I should compute just a difference between hours
if (d1.isAfter(d2)) 0.hours
else if (d1.toLocalDate.isEqual(d2.toLocalDate)) {
toEnd(d1.toLocalTime) - toEnd(d2.toLocalTime)
}
else {
toEnd(d1.toLocalTime) + jourOuvree(d1.toLocalDate.plusDays(1), d2.toLocalDate.minusDays(1)) * 8.hours + toStart(d2.toLocalTime)
}
}
val thursdayStart = LocalDate.of(2018, 3, 1).atTime(start)
val thursdayEnd = LocalDate.of(2018, 3, 1).atTime(end)
val fridayStart = LocalDate.of(2018, 3, 2).atTime(start)
val fridayEnd = LocalDate.of(2018, 3, 2).atTime(end)
val saturdayStart = LocalDate.of(2018, 3, 3).atTime(start)
val saturdayEnd = LocalDate.of(2018, 3, 3).atTime(end)
val sundayStart = LocalDate.of(2018, 3, 4).atTime(start)
val sundayEnd = LocalDate.of(2018, 3, 4).atTime(end)
val mondayStart = LocalDate.of(2018, 3, 5).atTime(start)
val mondayEnd = LocalDate.of(2018, 3, 6).atTime(end)
println(toEquals(thursdayStart, sundayStart).withSeconds)
println(toEquals(thursdayStart, saturdayStart).withSeconds)
println(toEquals(thursdayStart, mondayStart).withSeconds)
println(toEquals(thursdayStart, mondayEnd).withSeconds)
println(toEquals(thursdayStart, thursdayEnd).withSeconds)
// formatter to use to parse string input
import java.time.format.DateTimeFormatter
val DATE_TIME_FORMATTER = DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss")
println(toEquals(LocalDateTime.parse("2018-03-01 11:22:33", DATE_TIME_FORMATTER), LocalDateTime.parse("2018-03-02 11:22:33", DATE_TIME_FORMATTER)).withSeconds)
}