search for: runoff

Hello everyone I have a dataframe with 4 colums (year, month, day, runoff) for 1993-2009. Now I like to calculate the average runoff for each day. Finally I like to plot the median runoff for all this years. I tried with some loops, but it didn't work. Do you have any Tips for my problem? Any help is greatly appreciated! Thank you very much. Dominique -- View thi...

Hi, For my master thesis I have 24 micro-plots on which I did measurements during 3 months. The measurements were: - Rainfall and runoff events throughout 3monts (runoff being dependant on the rainfall, a coefficient (%) has been made per rainfall event and per 3 months) - Soil texture (3 different textures were differentiated) - Slope (3 classes of slopes) - Stoniness (one time measurement) - Random roughness (throughout 3 months)...

Hi I think I have been struggling to use replace correctly, I usually work my way around this using a loop, but I think this is in fact inefficient. I have a dataset with runoff from three plots and associated rainfall. However either the datarecording was sloppy, or the rainfall very patchy. So I am trying to remove data from my dataset for which the runoff is larger than the rainfall on the same day. (Don't worry, the rainfall data is in fact the rainfall associated...

...{Hillslope: Precipitation (mm)} {Hillslope: Average detachment (kg/m**2)} # # OFE-specific variables # char OFEVarNames[ 3 ] {Irrigation depth (mm)} {Irrigation_volume_supplied/unit_area (mm)} {Runoff (mm)} # # Daily values: # 1 5.40000 0.00000 0.00000 0.00000 0.00000 2 0.00000 0.00000 0.00000 0.00000 0.00000 3 2.30000 0.00000 0.00000 0.00000 0.00000 4 0.00000 0.00000 0.00000 0.0...

Dear R Users, I have a data frame of the nature: > head(aggregate_1986) Latitude Mean Annual Simulated Runoff per 1? Latitudinal Band 1 -55 574.09287 2 -54 247.23078 3 -53 103.40756 4 -52 86.99991 5 -51...

...below. The following table has its clumns spaced closely so when I paste it to a word file it looks continuous. Is there any option in R to make the outputs like the second table? Any help will be highly appreciated. Thank you very much in advance. Model 1 4/70-3/80 Year 2 Source Area Runoff Nitrogen Phosphorus V1 LDR_imperv 203 52.6 1496.0 190.0 V2 LDR_perv 1147 5.7 489.6 163.6 V3 MDR_imperv 281 52.6 4141.7 654.7 V4 MDR_Perv 654 5.7 511.8 227.4 V5 IND_imperv 97 52.6 1604.4 226.0 V6 IND_perv 42 5.7 17.9...

...lian Day" column to a data frame. Here is my code and the resulting data frame: vic.data <- read.table("C:/VIC/data/vic.data.csv", header=F) names(vic.data) <- c("year", "month", "day", "precip", "evap", "runoff", "baseflow", "Tsup", "SM1", "SM2", "SM3", "SWE") vic.data$temp.date <- as.Date(paste(vic.data$year, "/", vic.data$month, "/", vic.data$day, sep="")) vic.data$julian.day <- julian...

...DF file, due to the data input requirements of a model I hope to use. I am using the ncdf package, version 1.6. The data are global-scale water values, on a monthly basis for 10 years (ie. 120 months of data in total; at present the data are separated by year, with 12 months of data in each file - mrunoff_1986 through to mrunoff_1995). For each month there are 720 longitude x 360 latitude values, each with a corresponding runoff value (although some of these may be NAs). Here is my code so far: # READ IN NetCDF FILES FROM DISK library(ncdf) year <- 1986:1995 file_list <- paste("mruno...

...DF file, due to the data input requirements of a model I hope to use. I am using the ncdf package, version 1.6. The data are global-scale water values, on a monthly basis for 10 years (ie. 120 months of data in total; at present the data are separated by year, with 12 months of data in each file - mrunoff_1986 through to mrunoff_1995). For each month there are 720 longitude x 360 latitude values, each with a corresponding runoff value (although some of these may be NAs). My problem is that I'm getting an error with the put.var.ncdf command, as shown below. Here is my code so far: # READ IN...

Looks like the c(x)[...] bit used to be as.matrix(x)[...]. Not sure why the change was made many years ago, but this was before names were handled explicitly. It would definitely be better to not force the duplicate, at least in the case where we are sure c() and [ would not dispatch. Best, luke On Mon, 4 May 2015, peter dalgaard wrote: > >> On 04 May 2015, at 19:59 , franknarf

...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]]

Dear R Users, I'm finding that when I execute the following bit of code, that the new line argument is actually displayed as text in the graphics device. How do I avoid this happening? mtext(side=2, line=5.5, expression(paste("Monthly Summed Runoff (mm/month)", "/n", "and Summed Monthly Precipitation (mm x ",10^2,"/month)"))) I suspect that I've done, or omitted, something fairly obvious, but as yet cannot see it! Thanks for your help, Steve

I have a matrix with 12 rows (one for each month), 2 columns (baseflow, runoff). I would like to make a barplot similar to Excel’s “clustered column chart”. Here is my matrix ‘x’ 8.258754 13.300710 10.180953 10.760465 11.012184 13.954887 10.910870 13.839839 9.023519 11.511129 7.189241 12.519830 5.925576 17.101491 5.211613...

...I'm hoping someone out there has perhaps done this and can share their code and/or expertise with me. I need to pull recession periods out of a hydrograph - can anyone help me with this? I want to create a subset from streamflow data that consists of just the recession curves - the decreasing runoff after the passage of a peak flow. would really appreciate any help on this! Janet -- Tague Team Lab Manager 1005 Bren Hall UCSB, Santa Barbara, CA. [[alternative HTML version deleted]]

Dear useRs and developeRs, In my diploma thesis I work with a daily time series of glacier runoff data. I did already aggregate them to monthly means etc. Now i want to use just the summer values (I am indecisive by now what that means, but let's make it easy and use months like June). Is there a way to extract the data off this zoo into another zoo with frequency=1 ? Do you have alterna...

...# 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 UNSU...

...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-pro...

...ready 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]] >>> >>> ___________________________________________...

...entries in the logo contest. We had a very good turnout with 11 high quality designs submitted. We will now have 3 days of discussion and unlimited modifications to these 11 designs after which we'll have a poll lasting 3 days. Since this brings us to Christmas Eve there will be no time for runoff polls before the holidays in the event one design fails to win a majority (which is pretty likely when we have 11 to choose from). There is some discussion here http://forum.go-compiz.org/viewtopic.php?t=193 regarding how we should handle additional polls and when we would have them. Please mak...

...- nrow else if (length(x) == 1L && nargs() == 1L) { n <- as.integer(x) x <- 1 } else n <- length(x) if (!missing(nrow)) n <- nrow if (missing(ncol)) ncol <- n } nc <- 10 set.seed(1) m <- matrix(sample(letters,nc^2,replace=TRUE), ncol = nc) runoff <- microbenchmark( diaga = diag(m), diagb = diag2(m) ) Regards, Steve Bronder Website: stevebronder.com Phone: 412-719-1282 Email: sbronder at stevebronder.com On Tue, May 5, 2015 at 9:46 AM, <luke-tierney at uiowa.edu> wrote: > Looks like the c(x)[...] bit used to be as.matrix...