similar to: estimating NA values against selected slots

Dear arun and all R users, I will first of all try to simply define my issue.. I have data in the following format Year Discharge dd/mm/yyyy x .. … … … There are some NA values in the discharge which I would like to predict by using “predict command”. I cant figure out the way to write the coding for that. Could you please help me on that??? I have also ,written

All, What would be the most efficient way to load the data at the following address into a dataframe? http://ratings.fide.com/top.phtml?list=men Thanks, David -- View this message in context: http://r.789695.n4.nabble.com/Loading-Chess-Data-tp4642006.html Sent from the R help mailing list archive at Nabble.com.

Dear R users, [in case the format of email is changed or you dont finf it easy to understand, i have attached a text file of my question] i have the data in the following format and i want to convert it in the format given at the end. Ta ans Sa are the names of certain cities. there are 69 cities in my data. column 1 is representing station name (i am writing the data of only two cities for

Dear R Users,i am very new to R. I want your help on an issue regarding distance matrix and cluster analysis i had discharge data of 4 rivers(a,b,c,d) in 4 vectors each having 364 values > dput(qmu)structure(list(a = c(0.26, 0.25, 0.25, 0.25, 0.24, 0.23, 0.22, 0.21, 0.21, 0.21, 0.2, 0.19, 0.19, 0.19, 0.19, 0.18, 0.18, 0.18, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17,

Dear r-users, I would like to construct 3-day moving average for block maxima series. I tried this: bmthree <- lapply(split(dt, dt$Year), function(x) max(sapply(1:(nrow(x)-2), function(i) with(x, mean(Amount[i:(i+2)],na.rm=TRUE))))) bmthree and got the following output. $`1971` [1] 70.81667 $`1972` [1] 68.94553 $`1973` [1] 102.7236 $`1974` [1] 73.6625 $`1975` [1]

I have a problem with R. I try to compute the confidence interval for my df. When I want to create the plot I have this problem: Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differ. I try this code: library(dplR) df.rwi <- detrend(rwl = df, method = "Spline",nyrs=NULL) write.table(df.rwi,file="rwi.txt",quote=FALSE,row.names=TRUE)

Hi, I have a 3d array as below, I want to make this array to a matrix of p=50(rows) and n=20(columns) with the coverage values . The code before the array is: library(binom) Loading required package: lattice pi.seq<-seq(from = 0.01, to = 0.5, by = 0.01) no.seq<-seq(from = 5, to = 100, by = 5) cp.all = binom.coverage( p = pi.seq, n = no.seq , conf.level = 0.95, method = "exact")

Dear all, I have weekly data by city (variable citycode). I would like to take the average of the previous two, three, four weeks (without the current week) of the variable called value. This is what I have tried to compute the average of the two previous weeks; df = df %>% mutate(value.lag1 = lag(value, n = 1)) %>% mutate(value .2.previous = rollapply(data = value.lag1,

On Mon, Dec 16, 2013 at 04:16:29PM -0800, Michael Dalton wrote: > Commit 2613af0ed18a ("virtio_net: migrate mergeable rx buffers to page frag > allocators") changed the mergeable receive buffer size from PAGE_SIZE to > MTU-size, introducing a single-stream regression for benchmarks with large > average packet size. There is no single optimal buffer size for all >

On Mon, Dec 16, 2013 at 04:16:29PM -0800, Michael Dalton wrote: > Commit 2613af0ed18a ("virtio_net: migrate mergeable rx buffers to page frag > allocators") changed the mergeable receive buffer size from PAGE_SIZE to > MTU-size, introducing a single-stream regression for benchmarks with large > average packet size. There is no single optimal buffer size for all >

I am sure that this sort of thing has been asked and answered before, so in case my suggestions don't work for you, just search the archives a bit more. I am also sure that it can be handled directly by numerous functions in numerous packages, e.g. via time series methods or by calculating running means of suitably shifted series. However, as it seems to be a straightforward task, I'll

Dear Bert, Thank you very much.This works. I was wondering if the fact that I want to create new variables (sorry for not stating that fact) makes any difference? Thank you again. Sincerely, Milu On Sun, Mar 25, 2018 at 10:05 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > I am sure that this sort of thing has been asked and answered before, > so in case my suggestions

On 2016-05-25 19:13, Kelly Lesperance wrote: > Hdparm didn?t get far: > > [root at r1k1 ~] # hdparm -tT /dev/sda > > /dev/sda: > Timing cached reads: Alarm clock > [root at r1k1 ~] # Hi Kelly, Try running 'iostat -xdmc 1'. Look for a single drive that has substantially greater await than ~10msec. If all the drives except one are taking 6-8msec, but one is very

Hello~   Did any one have used the package 'orcutt' ?   I find that it can not work smoothly in a single variable regression. I use the example following, it function very well.   But when I regress "cons" on "price" (use the "reg1<-lm(cons~price+income+temp)") , then  use "reg11<-cochrane.orcutt(reg1) ". There is an error message “Error in

Hello everyone,           I have a basic question regarding logical operators. > x<-seq(-1,1,by=0.02) > x   [1] -1.00 -0.98 -0.96 -0.94 -0.92 -0.90 -0.88 -0.86 -0.84 -0.82 -0.80 -0.78  [13] -0.76 -0.74 -0.72 -0.70 -0.68 -0.66 -0.64 -0.62 -0.60 -0.58 -0.56 -0.54  [25] -0.52 -0.50 -0.48 -0.46 -0.44 -0.42 -0.40 -0.38 -0.36 -0.34 -0.32 -0.30  [37] -0.28 -0.26 -0.24 -0.22 -0.20 -0.18 -0.16

Hi all, I have 2 questions: 1)How do I calculate the mean on an imported txt file? I've imported the file below and that's what it looks like imported. How do I then calcuate the mean, median, or mode on the column LeafArea using the desktop R package? Any help would be greatly appreciated!! Thanks, Nat LeafType Leaflets LeafArea ShapeRatio LeafWeight LeafThickness 1 1

Hey, I have a set of income data which I'd like to fit to a gamma distribution. How can I estimate the two parameters of the gamma distribution for a vector, e.g. c(2039L, 2088L, 5966L, 2353L, 1966L, 2312L, 3305L, 2013L, 3376L, 3363L, 3567L, 4798L, 2032L, 1699L, 3001L, 2329L, 3944L, 2568L, 1699L, 4545L) I sense this will be a very easy one-liner, but my searching didn't come up with

Greetings.? My wife is teaching an introductory stat class at UC Davis.? The class emphasizes the use of simulations, rather than mathematics, to get insight into statistics, and R is the mandated tool.?? A student in the class recently inquired about different approaches to sampling from a binomial distribution.? I've appended some code that exhibits the idea, the gist of which is that using

Dear list, I have observed a weird behaviour from R --- apologies if I am missing something obvious! df3f826f28 df3f826f28 Say you type in R: >c.preec <- 10074 >c.gd <- 2200 >p1 <- .2 >c.neo <- p1*9451 + (1-p1)*3883 >n.preec <- 3710 >n.gd <- 2650 >n.neo <- 2120 >n.pcos <- 53000 >unit.met <- 94 >cost.met <- 94*n.pcos >effect <-

Hi there, Here is a minimum working example: ---------------------------------------------------------------- lower = 0 upper = 1 n_bins = 50 interval = (upper - lower) / n_bins bins = vector(mode="numeric", length=n_bins) breaks = seq(from=lower + interval, to=upper, by=interval) for(idx in breaks) { bins[idx / interval] = idx } print(bins)