R help - Dec 2017 - Finding center of mass in a hydrologic time series

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]]

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 <emorway at 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]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/
> posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
	[[alternative HTML version deleted]]

Slightly faster: sum(cumsum(hyd) <= .5 * sum(hyd))

Best regards,

ir. Thierry Onkelinx
Statisticus / Statistician

Vlaamse Overheid / Government of Flanders
INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR NATURE
AND FOREST
Team Biometrie & Kwaliteitszorg / Team Biometrics & Quality Assurance
thierry.onkelinx at inbo.be
Kliniekstraat 25, B-1070 Brussel
www.inbo.be

///////////////////////////////////////////////////////////////////////////////////////////
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more than asking him to perform a post-mortem examination: he may be
able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data. ~ Roger Brinner
The combination of some data and an aching desire for an answer does
not ensure that a reasonable answer can be extracted from a given body
of data. ~ John Tukey
///////////////////////////////////////////////////////////////////////////////////////////


Van 14 tot en met 19 december 2017 verhuizen we uit onze vestiging in
Brussel naar het Herman Teirlinckgebouw op de site Thurn & Taxis.
Vanaf dan ben je welkom op het nieuwe adres: Havenlaan 88 bus 73, 1000 Brussel.

///////////////////////////////////////////////////////////////////////////////////////////



2017-12-16 14:32 GMT+01:00 Eric Berger <ericjberger at
gmail.com>:
> 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 <emorway at 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]]
>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/
>> posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>         [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

Eric B's response provided just the kind of quick & simple solution I
was
hoping for (appears as the function com below).  However, I once again
failed to take advantage of the power of R and have reverted back to using
a for loop for the next step of the processing.  The example below (which
requires the library EGRET for pulling an example dataset) works, but
probably can be replaced with some version of the apply functionality?  So
far, I've been unable to figure out how to enter the arguments to the apply
function.  The idea is this: for each unique water year (variable 'wyrs'
in
example below) in a 27 year continuous time series of daily values, find
the date of the 'center of mass', and build a vector of those dates.
Thanks, -Eric M

library(EGRET)

StartDate <- "1990-10-01"
EndDate <- "2017-09-30"
siteNumber <- "10310000"
QParameterCd <- "00060"

Daily <- readNWISDaily(siteNumber, QParameterCd, StartDate, EndDate)

# Define 'center of mass' function
com <- function(x) {
    match(TRUE, cumsum(x/sum(x)) > 0.5) - 1
}


wyrs <- unique(Daily$waterYear)
for(i in (1:length(wyrs))){
    OneYr <- Daily[Daily$waterYear==wyrs[i], ]
    mid <- com(OneYr$Q)
    if(i==1){
        midPts <- as.Date(OneYr$Date[mid])
    } else {
        midPts <- c(midPts, as.Date(OneYr$Date[mid]))
    }
}



Eric Morway
Research Hydrologist
Nevada Water Science Center
U.S. Geological Survey
2730 N. Deer Run Rd.
Carson City, NV 89701
(775) 887-7668
*orcid*:  0000-0002-8553-6140 <http://orcid.org/0000-0002-8553-6140>



On Sat, Dec 16, 2017 at 5:32 AM, Eric Berger <ericjberger at gmail.com>
wrote:
> 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 <emorway at 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]]
>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posti
>> ng-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
	[[alternative HTML version deleted]]