search for: discharge

...quot;"),header=T,sep=",") This allows me to load the data into an array of length 3, with the res array containing my data from the csv files. I would like to parse the data in the res array so that I can access my specific data columns. For instance, I have field names called 'Discharge,' 'Volume,' and 'Id'. I would like to aggregate these fields for all the loaded files in the array so that I can perform some basic comparisons of distributions, etc. Does anyone have any example code that would allow me to parse the array so that I can extract the column names...

i have the following constructed and running very well,, thanks to Gabor Grothendieck for his help. >data.info <- c("station.id", "year", "date", "max.discharge") > > for(i in 1:num.files) { + station.id <- substring(data[i], 1,8) + DF <- read.table(data[i], sep=",", blank.lines.skip = TRUE) + z <- zoo(DF[,4], as.Date(as.character(DF[,3]), "%m/%d/%Y")) + f <- function(x) time(x) [which.max(x)] + ix <- tapply(z,...

...take advantage of #the agency column) L.USGS <- grep("^USGS", L, value = TRUE) DF <- read.table(textConnection(L.USGS), fill = TRUE) colnames(DF) <- c("agency", "gauge", "date", "time", "time_zone", "gauge_height", "discharge", "precipitation") pat <- "^# +USGS +([0-9]+) +(.*)" L.DD <- grep(pat, L, value = TRUE) library(gsubfn) DD <- strapply(L.DD, pat, c, simplify = rbind) DDdf <- data.frame(gauge = as.numeric(DD[,1]), gauge_name = DD[,2]) both <- merge(DF, DDdf, by = "gauge...

Look at the bottom of the message for my question #here is a little function that I wrote USGS <- function(input="discharge", days=7){ library(chron) library(gsubfn) #021973269 is the Waynesboro Gauge on the Savannah River Proper (SRS) #02102908 is the Flat Creek Gauge (ftbrfcms) #02133500 is the Drowning Creek (ftbrbmcm) #02341800 is the Upatoi Creek Near Columbus (ftbn) #02342500 is the Uchee Creek Near Fort Mitc...

...a.frame(date1=seq.Date(as.Date("1930.1.1",format="%Y.%m.%d"),as.Date("2010.12.31",format="%Y.%m.%d"),by="day")) elsplit1<- lapply(elsplit,function(x) data.frame(date1=as.Date(paste(x[,2],x[,3],x[,4],sep="-"),format="%Y-%m-%d"),discharge=x[,5])) ?elsplit2<-lapply(elsplit1,function(x) x[order(x[,1]),]) library(plyr) elsplit3<-lapply(elsplit2,function(x) join(datetrial,x,by="date1",type="full")) ?elsplit4<-lapply(elsplit3,function(x) {x[,2][is.na(x[,2])]<- "-9999.000";x}) elsplit5<-lapply(...

hi everyone! hope you can help me here. i am a new R user. what i am trying to do is to find the maximum annual discharge from a daily record. i have a data.frame which includes date and the discharge. somewhat like this.. 10/1/1989 2410 10/2/1989 2460 10/3/1989 2890 ... ... ... 12/31/2005 5730 i have been browsing through the archives and fount out about the aggregate function and the zoo pack...

...t;F:/R_code/exampledata.txt") #head(data) # shows header attach(data) #attaches data to current run datasort=data[order(Stage), ] #sorts data so predicted values follow in systematic order #datasort # shows sorted data starts = list(const =0.00001, alpha = 20) fm <- nls(Discharge ~ const*Stage ^alpha, start=starts) #runs a non linear least squares regression model #predict(fm) # fitted values at observed times #summary(fm) coef=coefficients(fm) const_val=round(coef["const"],4) #makes a variable for the equation in the plot and rounds it to...

Hello, In R I am trying to use Kalman filtering to find a solution for an hydrological problem. With Kalman Filtering I want to estimate the discharge comming from three storage bassins. I have programmed a function in R which can run KalmanSmooth. When I'm asking for the function and putting in values, R detects the following error: "Error in as.vector(data) : Argument "S1" is missing, with no default". I have try to find...

...lained loudly and frequently. I know upsd was running (on the connected server as well as all the clients), because if it was not, upsmon would have complained loudly and frequently. there were no indications in my logs that nut was paying any attention to the situation. why did my UPS decide to discharge its battery when there appears to be no power outage? (just looking for suggestions and guesses here from other MGE users...) why did nut not see that my UPS was discharging and the battery percentage running down and take appropriate action? "OL DISCHRG" status in the logs seems nonsen...

I have a data set of mean velocity, discharge, and mean depth. I need to find out which model best fits them out of log linear, linear, some other kind of model... Using excel I have found that linear is not that bad and log10(discharge) vs. the other two variables (I am trying to predict velocity and depth from discharge) is not that bad ei...

OK. I'll try it, but it really puzzles me how the battery could have gone bad for just 15 months. The online power is pretty stable, so the battery hasn't gone through many charge/discharge cycles... Regards, -- Georgi On 11.1.2020 at 14:24, Manuel Wolfshant wrote: > > indicates that the UPS thought that battery was close to depleted and decided to shutdown. I suggest to fully retest your battery ( doing a full charge/discharge cycle with a dummy load ), I suspect it is not a...

...uot;physician", "orders", "treatment", "times", "care", "plan", "strategies", "concerns", "meetings", "equipment", "treatment", "options", "delivery", "care", "discharge", "welfare", "violations", "HIPPS", "professionalism", "lack", "boundaries crossing", "transportation", "benefits", "assistance", "beneficiary", "complaint", "grievance", &qu...

Citeren "David C. Rankin" <drankinatty at suddenlinkmail.com>: > Discharged battery.voltage: > > 17.4 Volts - (indicated on shutdown, discharging battery.charge 77%) > 16.9 Volts - (indicated on shutdown, discharging battery.charge 51%%) > 16.6 Volts - (indicated on shutdown, discharging battery.charge 33%) > > -- Shutdown Occurred at < 3% -- B...

Dear all, I'm using the curve() function to plot discharge Q against water depth a. However, I would like to have a graph of water depth a plotted against discharge Q. How can this be done? Minimal working example: S0 = 0.004 n = 0.04 tanalpha = 1.4/1.5 par(mar = c(5,5,1,1)) # b, l, t, r curve((sqrt(S0)/n)*(0.035+(0.7+(x-0.1)/tanalpha)*(x-0.1))*((0.035+(0....

Hi, I am trying to create a loess smooth from hydrologic data. My goal is to create a smooth line that describes discharge at a certain point in time. I have done this using the 'lowess' function and had no problem, but I'm having some difficulty with loess. I am inputting the date ('date') and discharge ('q') values using the 'scan' function, then inputting the following: > da...

...ended by sealed lead acid battery manufacturers should keep it under 42V. Overcharged or worn batteries might have a highere clamping voltage, but more than 42V will move it into the regions of 48V power supplies. - Battery manufacturers recommend a minimum of 10.8V for a 12V battery as the cutoff discharge voltage (further discharge will cause battery damage) but the UPS is likely to shut off before that, so I guesstimated 11V for a 12V cell to be a reasonable "empty battery" voltage. - The float charge (clamping) voltage for a fully charged 36V battery if 40.9V but I've subtracted 10...

From the moment when the UPS rapports "Low Battery" to the end of the shutdown, too mutch time is passing. So the computer does not have enough time shut down properly. Can I shut down the computer on the "On Batery" status? Or can I shut down when the UPS rapports 50% of the battery? Thank you.

On Dec 18, 2014, at 9:16 AM, Frantz de Germain <frantz at info.univ-angers.fr> wrote: > What is the difference between "test.battery.start.deep" and > "test.battery.start.quick" command (upscmd) ? The intent for the quick test is to only discharge the battery a little bit, whereas the deep test would discharge until the low battery threshold is reached (or the test is stopped). In practice, the difference is determined by the UPS firmware. -- Charles Lepple clepple at gmail

Dear R-users! I am using R 1.7.0, under Windows XP. Having some hospital discharge data (admission date and discharge date for each patient), I want to get the number of patients in the hospital on a given date. My data look like (simple example): > x <- data.frame(patid=c("pat1", "pat2"), adm.date = c("15.03.2002", "16.03.2002"),...

...39;:??? 506953 obs. of? 5 variables: # $ st?????? : chr? "AGOMO" "AGOMO" "AGOMO" "AGOMO" ... # $ year???? : int? 2004 2004 2004 2004 2004 2004 2004 2004 2004 2004 ... # $ month??? : int? 1 1 1 1 1 1 1 1 1 1 ... # $ day????? : int? 1 2 3 4 5 6 7 8 9 10 ... # $ discharge: num? 6.75 7.6 5.58 5.02 4.8 ... library(reshape2) dat2<- do.call(rbind,lapply(split(dat1,list(dat1$st,dat1$year)),function(x) {x$day<-seq_along(x$day);x})) row.names(dat2)<-1:nrow(dat2) ?str(dat2) #'data.frame':??? 506953 obs. of? 5 variables: # $ st?????? : chr? "TANNU&quot...