similar to: mean of runoff for several years

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)

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

Dear R-Help, I have a generated file that looks like the following: ----- Begin file ----- # # Output File # float Version 2002.700000000000 int Numdays 31 int NumOFEs 1 # # Hillslope-specific variables # char HillVarNames[ 3 ] {Days In Simulation} {Hillslope: Precipitation (mm)}

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

Hello, I have a question regarding a way to control the appreance of output exported by R when I use capture.output( x, file = "Directory/file.txt") , I get a text file which when I paste to a word file looks like the first table 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

Hi, I'm attempting to add a "Julian 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",

The US water year extends from 01 October yyyy-1 through 30 September yyyy and is referenced by the year starting on the included 01 January yyyy. I'd like to be able to find the annual means for the water year. To do so I've taken the input date-time, which is in the usual format "1991-10-07 10:35:00" changed it by: w$d<-as.POSIXct(w$date.time) Now I can add an

I have x, y, z data. The x, y fields dont change but Z does. How do I add a very small number onto the end of each x, y data point. For example: Original (X) Original (Y) Original (Z) 15 20 30 15 20 40 New (X) New (Y)

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

Dear R community, 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

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 13.585175

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 <-

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

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

Dear all,   I have the following written example > coords <- "51°30'48.58\"N" > > as.integer(strsplit(coords, "°")[[1]][1]) [1] 51 > as.integer(strsplit(strsplit(coords, "°")[[1]][2], "'")[[1]][1]) [1] 30 > as.numeric(strsplit(strsplit(strsplit(coords, "°")[[1]][2], "'")[[1]][2],

Does anyone have information about using the National Practitioner Data Bank (NPDB) in R? It's public use data available from the U.S. Department of Health and Human Services that can be downloaded as a .dat or .por file. I can't even figure out how to open the file in R. [[alternative HTML version deleted]]

A data set I obtained has the hours running from 01 through 24 rather than the conventional 00 through 23. My favorite, strptime, balks at hour 24. I thought it would be easy to correct but it must be too late on Friday for my brain and caffeine isn't helping. TIA for a hint, Clint -- Clint Bowman INTERNET: clint at ecy.wa.gov Air Quality Modeler INTERNET: clint at math.utah.edu

Folks: I thought the following excerpt from Bruce McCullough's post would be a good candidate for the R fortunes package -- except that it's about Excel, not R! So I nominate it... but leave it to others to say whether it's really "qualified" to be nominated. ---- "The idea that the Excel solver "has a good reputation for being fast and accurate" does not

Hello there, would you please look into my codes? Here I have following: > set.seed(100) > samp <- sample(c(1,-1,0), 20, replace=T); samp [1] 1 1 -1 1 -1 -1 0 -1 -1 1 -1 0 1 -1 0 0 1 -1 -1 0 Here I want to calculate the length of each unique number for above vector. How can I do that? Thanks in advance [[alternative HTML version deleted]]

I'm trying to plot smooth trend using smoothTrend in OpenAir but I'm having problems. I used the following code. --------------------------------------------------------------------------------- #Set my working dir to the dir with my files setwd("c:/R") #Load the openair library library(openair) #Load the data mydata <- read.table("MCNP-pH.csv", header=TRUE,