search for: wyrs

Using the dataset below, I got close to what I'm after, but not quite all the way there. Any suggestions appreciated: Daily <- read.table(textConnection(" Date wyr Q 1911-04-01 1990 4.530695 1911-04-02 1990 4.700596 1911-04-03 1990 4.898814 1911-04-04 1990 5.097032 1911-04-05 1991 5.295250 1911-04-06 1991 6.569508 1911-04-07 1991 5.861587 1911-04-08 1991 5.153666

cumsum() seems to be what you need. This can probably be done more elegantly, but ... out <- aggregate(Q ~ wyr, data = Daily, which.max) tbl <- table(Daily$wyr) out$Q <- out$Q + cumsum(c(0,tbl[-length(tbl)])) out ## yields wyr Q 1 1990 4 2 1991 6 3 1992 9 4 1993 15 5 1994 18 I leave the matter of Julian dates to you or others. Cheers, Bert Bert Gunter "The trouble

...ich 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&...

...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$waterYear) x <- as.Date(sapply( wyrs, function(yr) { Df <- Daily[Daily$waterYear==yr,]; Df$Date[com(Df$Q)] } ), "1970-01-01") On Mon, Dec 18, 2017 at 4:47 PM, Morway, Eric <emorway at usgs.gov> wrote: > Eric B's response provided just the kind of...

Hello, I am trying to define a different interval for a "year". In hydrology, a "water year" is defined as the period between October 1st and September 30 of the following year. I was wondering how I might do this in R. Say I have a data.frame like the following and I want to extract a variable with the water year specs as defined above:

Hi all, I'm very new to R. I have installed R 1.9.0 on Linux (Fedora). Now I got an self-made package comprising R functions as well as C-Code which are used in several R functions. I installed the package without any error (see install log below). Then, I checked in /usr/lib/R/library if the package izbi exists and it exists. But whenever I try to load the library on the command line I

...(yr) for(n in 1:len){ WY[n]<-if(mo[n]>=10){ yr[n]+1 }else{ yr[n]} } partialpeakind<-which(round(q,2)>=round(pds_TD[i],2)) partialpeaksd<-dates[partialpeakind] partialpeaksq<-q[partialpeakind] partialpeaksWY<-WY[partialpeakind] for(q in 1:length(partialpeaksWY)){ wyrs<-seq(min(partialpeaksWY),max(partialpeaksWY),by=1) POTWY[[q]]<-sum(partialpeaksWY == wyrs[q]) POTWY_cut<-as.numeric(na.omit(POTWY)) paste(pds_gagehandles[[i]],"_pdswy",sep="")<-POTWY_cut} pds_dropdates<-data.frame(partialpeaksWY,partialpeaksd,partialpeaksq) w...

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 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 <-

Signed-off-by: Scott Seago <sseago at redhat.com> --- AUTHORS | 17 ++++++ README | 10 +++ conf/ovirt-agent | 12 ++++ conf/ovirt-db-omatic | 12 ++++ conf/ovirt-host-browser | 12 ++++