Hi Eric,
How about
match( TRUE, cumsum(hyd/sum(hyd)) > .5 ) - 1
HTH,
Eric
On Sat, Dec 16, 2017 at 3:18 PM, Morway, Eric <emor...@usgs.gov> wrote:
> The small bit of script below is an example of what I'm attempting to do -
> find the day on which the 'center of mass' occurs. In case that is the
> wrong term, I'd like to know the day that essentially cuts the area under
> the curve in to two equal parts:
>
> set.seed(4004)
> Date <- seq(as.Date('2000-09-01'), as.Date('2000-09-30'), by='day')
> hyd <- ((100*(sin(seq(0.5,4.5,length.out=30))+10) +
> seq(45,1,length.out=30)) + rnorm(30)*8) - 800
>
> # View the example curve
> plot(Date, hyd, las=1)
>
> # By trial-and-error, the day on which the center of mass occurs is the
> 11th day:
> # Add up the area under the curve for the first 11 days and compare
> # with the last 19 days:
>
> sum(hyd[1:11])
> # 3546.364
> sum(hyd[12:30])
> # 3947.553
>
> # Add up the area under the curve for the first 12 days and compare
> # with the last 18 days:
>
> sum(hyd[1:12])
> # 3875.753
> sum(hyd[13:30])
> # 3618.164
>
> By day 12, the halfway point has already been passed, so the answer that
> would be returned would be:
>
> Date[11]
> # "2000-09-11"
>
> For the larger problem, it'd be handy if the proposed function could
> process a multi-year time series (a runoff hydrograph) and return the day
> of the center of mass for each year in the time series.
>
> I appreciate any pointers...Eric
>
> [[alternative HTML version deleted]]
>
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