similar to: Is this a bug in `[`?

?s 15:51 de 04/08/2018, I?aki ?car escreveu: > El s?b., 4 ago. 2018 a las 15:32, Rui Barradas > (<ruipbarradas at sapo.pt>) escribi?: >> >> Hello, >> >> Maybe I am not understanding how negative indexing works but >> >> 1) This is right. >> >> (1:10)[-1] >> #[1] 2 3 4 5 6 7 8 9 10 >> >> 2) Are these right? They

Thanks. This is exactly the doubt I had. Rui Barradas ?s 05:26 de 05/08/2018, Kenny Bell escreveu: > This should more clearly illustrate the issue: > > c(1, 2, 3, 4)[-seq_len(4)] > #> numeric(0) > c(1, 2, 3, 4)[-seq_len(3)] > #> [1] 4 > c(1, 2, 3, 4)[-seq_len(2)] > #> [1] 3 4 > c(1, 2, 3, 4)[-seq_len(1)] > #> [1] 2 3 4 > c(1, 2, 3, 4)[-seq_len(0)]

El dom., 5 ago. 2018 a las 6:27, Kenny Bell (<kmbell56 at gmail.com>) escribi?: > > This should more clearly illustrate the issue: > > c(1, 2, 3, 4)[-seq_len(4)] > #> numeric(0) > c(1, 2, 3, 4)[-seq_len(3)] > #> [1] 4 > c(1, 2, 3, 4)[-seq_len(2)] > #> [1] 3 4 > c(1, 2, 3, 4)[-seq_len(1)] > #> [1] 2 3 4 > c(1, 2, 3, 4)[-seq_len(0)] > #>

Dear R users, I want to aggregate all *d *data from all combination of n *plots* taken by k. Thank very much! My data is like that: plot d 1 14 1 13 1 12 1 14 1 18 1 20 1 21 1 43 1 108 1 43 2 41 2 61 2 83 2 61 2 84 2 45 2 21 2 12 2 11 ... 100 10 100 12 -- --- Catalin-Constantin ROIBU Forestry engineer, PhD Forestry Faculty of Suceava

Short question: Is it possible to use statistical tests, like the Augmented Dickey-Fuller test, in functions with for-loops? If not, are there any alternative ways to scale measures? Detailed explanation: I am working with time-series, and I want to flag curves that are not stationary and which display pulses, trends, or level shifts. >df DATE ID VALUE2012-03-06 1

Hi, A column of my df looks like A 10/20/30 40/20 60/10/10/5 80/10 I want to split it such that the last column has the last composition and if there are not enough the middle columns get the 0s. That way my df would look like A1 A2 A3 A4 10 20 0 30 40 0 0 20 60 10 10 5 80 0 0 10 How can I do that ?? [[alternative HTML version deleted]]

General problem: I have 20 projects that can be invested in and I need to decide which combinations meet a certain set of standards. The total possible combinations comes out to 2^20. However I know for a fact that the number of projects must be greater than 5 and less than 13. So far the the code below is the best I can come up with for iteratively creating a set to check against my set of

Dear List, I want to compare the presence and absence of bird species based on the sites in a matrix. The matrix has 5 rows for Island A, B, C, D, and E. It has 100 columns for bird species D001-D100. In each cell of the matrix, the presence-absence of bird species will be recorded as 1 or 0. (For example, if species D001 is found on Island D, the matrix cell of species D001 and Island D

Hello, Another way would be n <- nrow(expand.grid(1:nrow(D1), 1:nrow(D2))) D5 <- data.frame(distance=integer(n),difference=integer(n)) D5[] <- do.call(rbind, lapply(seq_len(nrow(D1)), function(i) t(sapply(seq_len(nrow(D2)), function(j){ c(distance=sqrt(sum((D1[i,1:2]-D2[j,1:2])^2)),difference=(D1[i,3]-D2[j,3])^2) } )))) identical(D3, D5) In my first answer I forgot to say that

I am nowaware that I should not post this type of questions on this group. However, Iwould like to have some clarifications related to the response you've?alreadyprovided. The code you provided yields accurate results, however I still haveissues grasping the loop process in case 1 & 2. In case1,?the use of?"p+1" and "q+1" is still blurry tome? Likewise "0L"

This is Circle 8..1.13 of the R Inferno. On 05/08/2018 06:57, Rui Barradas wrote: > Thanks. > This is exactly the doubt I had. > > Rui Barradas > > ?s 05:26 de 05/08/2018, Kenny Bell escreveu: >> This should more clearly illustrate the issue: >> >> c(1, 2, 3, 4)[-seq_len(4)] >> #> numeric(0) >> c(1, 2, 3, 4)[-seq_len(3)] >> #> [1] 4

Hello, See if this is it: priceStore_Grps$StoreID <- paste("Store", seq_len(nrow(priceStore_Grps)), sep = "_") Hope this helps, Rui Barradas On 5/26/2018 2:03 PM, Jeff Reichman wrote: > ALCON > > > > I'm trying to figure out how to rename groups in a data frame after groups > by selected variabels. I am using the dplyr library to group my

Hello, Sorry, but I think my first answer is wrong. You probably want something along the lines of sp <- split(priceStore_Grps, priceStore_Grps$StorePC) res <- lapply(seq_along(sp), function(i){ sp[[i]]$StoreID <- paste("Store", i, sep = "_") sp[[i]] }) res <- do.call(rbind, res) row.names(res) <- NULL Hope this helps, Rui Barradas On 5/26/2018

I am lousy at simple regex and I have not found a solution to a simple problem. I have a vector with some character values that I want to split. Sample data dd1 <- c( "XXY (mat harry)","XXY (jim bob)", "CAMP (joe blow)", "ALP (max jack)") Desired result dd2 <- data.frame( xx = c("XXY", "XXY", "CAMP",

Dear @Rui Barradas, thank you for the solution. It works perfectly. On 13/10/2017, 23:35, "Rui Barradas" <ruipbarradas at sapo.pt> wrote: Hello, Try the following. myDF1$studyno <- as.character(myDF1$studyno) myDF2$studyno <- as.character(myDF2$studyno) i1 <- which(names(myDF1) == "pf_mcl") merge(myDF1[-i1], myDF2,

This should more clearly illustrate the issue: c(1, 2, 3, 4)[-seq_len(4)] #> numeric(0) c(1, 2, 3, 4)[-seq_len(3)] #> [1] 4 c(1, 2, 3, 4)[-seq_len(2)] #> [1] 3 4 c(1, 2, 3, 4)[-seq_len(1)] #> [1] 2 3 4 c(1, 2, 3, 4)[-seq_len(0)] #> numeric(0) Created on 2018-08-05 by the reprex package (v0.2.0.9000). On Sun, Aug 5, 2018 at 3:58 AM Rui Barradas <ruipbarradas at sapo.pt>

It works!!! Thank you so much for your help! Sent from my iPhone > On Feb 3, 2020, at 3:47 AM, Rui Barradas <ruipbarradas at sapo.pt> wrote: > > ?Hello, > > You can solve the problem in two different ways. > > 1. Redefine storage1 as a matrix and extract the aic *in* the loop. > > storage1 <- matrix(0, 4, 4) > for(p in 0:3){ > for(q in 0:3){ >

FYI, this behavior is documented in Section 3.4.1 'Indexing by vectors' of 'R Language Definition' (accessible for instance via help.start()): "*Integer* [...] A special case is the zero index, which has null effects: x[0] is an empty vector and otherwise including zeros among positive or negative indices has the same effect as if they were omitted." The rest of that

Hello I am trying to extract AICs from an ARIMA estimation with different combinations of p & q ( p =0,1,2,3 and q=0,1.2,3). I have tried using the following code unsucessfully. Can anyone help? code: storage1 <- numeric(16) for (p in 0:3){ ? ? for (q in 0:3){ ? ? ? storage1[p]? <- arima(x,order=c(p,0,q), method="ML")} } storage1$aic [[alternative HTML version deleted]]

Hi Kevin, I think there are issues with Rui's proposed solution. For example, if there are rows in myDF1 which have a studyno which does not match any row in myDF2, then you will lose those rows. In your original request you said that you wanted to keep those rows. To demonstrate my point I need to modify your sample data. Specifically, I changed some studyno settings in myDF1, and also the