极端降水的百分位法阈值计算ncl或fortran程序

呜呼啊

求求大佬救救孩子叭

YQY优优

undef("percent_to_value")
function percent_to_value( \
  i_data           : numeric, \
  i_percentiles
: numeric  \
)
local None
begin

  retVal = new( dimsizes(i_percentiles), float )

  data1d = ndtooned( i_data )
  notMissing = data1d( ind(.not.ismissing(data1d) ) )

  qsort(notMissing)

  do p = 0, dimsizes(i_percentiles)-1
;   pInd = round( i_percentiles(p) * .01 * dimsizes(notMissing) + 0.5, 3 ) -1
;   pInd = where( pInd.ge.dimsizes(notMissing), dimsizes(notMissing)-1, pInd )
    floatInd = i_percentiles(p) * .01 * dimsizes(notMissing) - 0.5
    floorInd = toint( floor(floatInd) )
    floorInd = where( floorInd.lt.0, 0, floorInd )
    ceilInd = toint( ceil(floatInd) )
    ceilInd = where( ceilInd.ge.dimsizes(notMissing), \
        dimsizes(notMissing)-1, ceilInd )
;   print(pInd + " " + dimsizes(notMissing))
    if( ceilInd.eq.floorInd ) then
      retVal(p) = notMissing(floorInd)
    else
      retVal(p) = notMissing(floorInd) * ( ceilInd - floatInd ) \
          + notMissing(ceilInd) * ( floatInd - floorInd )
    end if
  end do

  return(retVal)
end ; percent_to_value
调用这个子程序就可以了

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wsx

YQY优优 发表于 2021-4-12 16:40
undef("percent_to_value")
function percent_to_value( \
  i_data           : numeric, \

请问能分享下原理吗？

duoci01

楼主解决了没？能否分享一些脚本