[R] Finding center of mass in a hydrologic time series
Morway, Eric
emorway at usgs.gov
Sat Dec 16 14:18:35 CET 2017
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
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