similar to: R for loop stops after 4 iterations

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 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,

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)

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

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

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

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 <-

I am trying to model the observed leaching of wood preservative chemicals from treated wood during an outdoor experiment where leaching is caused by rainfall events. For each rainfall event, the amount of rainfall was recorded as well as the amount of preservative chemical leached. A number of climatic variables were measured, but the most important is the amount of rainfall. I have tried a

Dear all,   I did ANOVA for many response variables (Var1, Var2, ....Var75000), and i got the result of p-value like below. Now, I want to select those predictors, which have pvalue less than or equal to 0.05 for each response variable. For example, X1, X2, X3, X4, X5 and X6 in case of Var1, and similarly, X1, X2.......X5 in case of Var2, only X1 in case of Var3 and none of the predictors in case

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)

hello, i'm pete ,how can i order rows of matrix by max to min value? I have a matrix of membership degrees, with 82 (i) rows and K coloumns, K are clusters. I need first and second largest elements of the i-th row. for example 1 0.66 0.04 0.01 0.30 2 0.02 0.89 0.09 0.00 3 0.06 0.92 0.01 0.01 4 0.07 0.71 0.21 0.01 5 0.10 0.85 0.04 0.01 6 0.91 0.04 0.02 0.02 7 0.00 0.01 0.98 0.00 8 0.02

aa <- (structure(list(X.0.85 = c(-1.02, -1.17, -1.29, -1.39, -1.46, -1.5, -1.52, -1.5, -1.46, -1.39, -1.3, -1.19, -1.07, -0.93, -0.79, -0.65, -0.5, -0.36, -0.22, -0.08, 0.05, 0.18, 0.3, 0.41, 0.52, 0.62, 0.72, 0.81, 0.89, 0.98, 1.05, 1.13, 1.19, 1.25, 1.29, 1.31, 1.31, 1.29, 1.24, 1.16, 1.06, 0.93, 0.77, 0.58, 0.38, 0.16, -0.07, -0.31, -0.89, -1.05, -1.19, -1.31, -1.41, -1.47, -1.51, -1.51,

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

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

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 >

Dear all,I am a new user of WinBUGS and need your help. After running the following code, I got parameters of beta0 through beta4 (stats, density), but I don't know how to get the prediction of the last value of h, the variable I set to NA and want to model it using the following code.Does anyone can given me a hint? Any advice would be greatly appreciated.Best

Dear list and Hadley, The new plyr package seems to provide a clean and consistent way to apply a function on several arguments. However, I don't understand why the following example does not work like the standard mapply, library(plyr) df <- data.frame(a=1:10 , b=1:10) foo1 <- function(a, b, cc=0, d=0){ a + b + cc + d } mdply(df, foo1, cc=1) # fine mdply(df, foo1, d=1) #

Hi there, I'm trying to make a scatterplot of removed versus duration for each type of bee. No matter what I try, I can't seem to get my code to work. Any help would be appreciated. Thanks! My r-code: dat$BEE <- with(dat, factor(BEE, c(1,2))) plot(REMOVED~DURATION,pch=BEE, col=BEE) REMOVED DURATION BEE 1 0.07 2 QUEEN 2 0.10 5 QUEEN 3 0.11 7 QUEEN 4 0.12 11 QUEEN 5 0.15 12 QUEEN 6 0.19

Hi all, I have a matrix called 'data', which looks like: > data[1:4,1:4] Probe_ID Gene_Symbol M1601 M1602 1 A_23_P105862 13CDNA73 -1.6 0.16 2 A_23_P76435 15E1.2 0.18 0.59 3 A_24_P402115 15E1.2 1.63 -0.62 4 A_32_P227764 15E1.2 -0.76 -0.42 > dim(data) [1]